Subsections


A.15 Quantum Field Theory in a Nanoshell

The classical quantum theory discussed in this book runs into major difficulties with truly relativistic effects. In particular, relativity allows particles to be created or destroyed. For example, a very energetic photon near a heavy nucleus might create an electron and a positron. Einstein’s $E$ $\vphantom0\raisebox{1.5pt}{$=$}$ $mc^2$ implies that that is possible because mass is equivalent to energy. The photon energy is converted into the electron and positron masses. Similarly, an electron and positron can annihilate each other, releasing their energy as photons. The quantum formalism in this book cannot deal with particles that appear out of nothing or disappear. A modified formulation called quantum field theory is needed.

And quantum field theory is not just for esoteric conditions like electron-positron pair creation. The photons of light are routinely created and destroyed under normal conditions. Still more basic to an engineer, so are their equivalents in solids, the phonons of crystal vibrations. Then there is the band theory of semiconductors: electrons are created within the conduction band, if they pick up enough energy, or annihilated when they lose it. And the same happens for the real-life equivalent of positrons, holes in the valence band.

Such phenomena are routinely described within the framework of quantum field theory. Almost unavoidably you will run into it in literature, [18,28]. Electron-phonon interactions are particularly important for engineering applications, leading to electrical resistance (along with crystal defects and impurities), and to the combination of electrons into Cooper pairs that act as bosons and so give rise to superconductivity.

This addendum explains some of the basic ideas of quantum field theory. It should allow you to recognize it when you see it. Addendum {A.23} uses the ideas to explain the quantization of the electromagnetic field. That then allows the quantum description of spontaneous emission of radiation by excited atoms or nuclei in {A.24}. Here a photon is created.

Unfortunately a full discussion of quantum field theory is far outside the scope of this book. Especially the fully relativistic theory is very involved. To explain quantum field theory in a nutshell takes Zee 500 pages, [52]. Tong [[17]] writes: “This is charming book, where emphasis is placed on physical understanding and the author isn’t afraid to hide the ugly truth when necessary. It contains many gems.” But you first need to learn linear algebra, at the minimum read all of chapter 1 on relativity, chapter 1.2.5 and {A.4} on index notation, chapter 12.12 and {A.35} on the Dirac equation, addendum {A.14} on the Klein-Gordon equation, {A.1} on Lagrangian mechanics, {A.12} on the Heisenberg interpretation, and pick up enough group theory. Learning something about the path integral approach to quantum mechanics, like from [22], cannot hurt either. In the absence of 1,000 pages and a willing author, the following discussion will truly be quantum field theory in a nanoshell.

If you want to get a start on a more advanced treatment of quantum field theory of elementary particles at a relatively low level of mathematics, Griffiths [24] is recommended.

And if you are just interested in relativistic quantum mechanics from an intellectual point of view, there is good news. Feynman gave a set of lectures on “quantum electrodynamics” for a general audience around 1983, and the text is readily available at low cost. Without doubt, this is the best exposition of the fundamentals of quantum mechanics that has ever been written, or ever will. The subject is reduced to its bare abstract axioms, and no more can be said. If the human race is still around a millennium or so from now, artificial intelligence may take care of the needed details of quantum mechanics. But those who need or want to understand what it means will still reach for Feynman. The 2006 edition, [19], has a foreword by Zee that gives a few hints how to relate the basic concepts in the discussion to more conventional mathematics like the complex numbers found in this book. It will not be much help applying quantum field theory to engineering problems, however.


A.15.1 Occupation numbers

The first concept that must be understood in quantum field theory is occupation numbers. They will be the new way to represent quantum wave functions.

Recall first the form of wave functions in classical quantum mechanics, as normally covered in this book. Assume a system of independent, or maybe weakly interacting particles. The energy eigenfunctions of such a system can be written in terms of whatever are the single-particle energy eigenfunctions

\begin{displaymath}
\pp1/{\skew0\vec r}//z/,\pp2/{\skew0\vec r}//z/,\pp3/{\skew0\vec r}//z/,\ldots
\end{displaymath}

For each single-particle eigenfunction, ${\skew0\vec r}$ indicates the position of the particle and $S_z$ its spin angular momentum in the chosen $z$-​direction.

Now consider a system of, say, 36 particles. A completely arbitrary example of an energy eigenfunction for such a system would be:

\begin{displaymath}
\begin{array}{l}
\psi^{\rm S}({\skew0\vec r}_1,S_{z1},{\...
...5//z5/ \ldots \pp54/{\skew0\vec r}_{36}//z36/
\end{array} %
\end{displaymath} (A.46)

This system eigenfunction has particle 1 in the single-particle state $\pp24////$, particle 2 in $\pp4////$, etcetera. The system energy is the sum of the separate energies of the 36 single-particle states involved:

\begin{displaymath}
{\vphantom' E}^{\rm S}= {\vphantom' E}^{\rm p}_{\psi_{24}}...
...\rm p}_{\psi_6} + \ldots + {\vphantom' E}^{\rm p}_{\psi_{54}}
\end{displaymath}

Figure A.2: Graphical depiction of an arbitrary system energy eigenfunction for 36 distinguishable particles.
\begin{figure}
\centering
\setlength{\unitlength}{1pt}
\begin{picture}(4...
...0,196,t'$\pp72////$'
\PB378,196,t'$\pp73////$'
\end{picture}
\end{figure}

Instead of writing out the example eigenfunction mathematically as done in (A.46) above, it can be graphically depicted as in figure A.2. In the figure the single-particle states are shown as boxes, and the particles that are in those particular single-particle states are shown inside the boxes. In the example, particle 1 is inside the $\pp24////$ box, particle 2 is inside the $\pp4////$ one, etcetera. It is just the reverse from the mathematical expression (A.46): the mathematical expression shows for each particle in turn what the single-particle eigenstate of that particle is. The figure shows for each single-particle eigenstate in turn what particles are in that eigenstate.

Figure A.3: Graphical depiction of an arbitrary system energy eigenfunction for 36 identical bosons.
\begin{figure}
\centering
\setlength{\unitlength}{1pt}
\begin{picture}(4...
...0,196,t'$\pp72////$'
\PB378,196,t'$\pp73////$'
\end{picture}
\end{figure}

However, if the 36 particles are identical bosons, (like photons or phonons), the example mathematical eigenfunction (A.46) and corresponding depiction figure A.2 is unacceptable. As chapter 5.7 explained, wave functions for bosons must be unchanged if two particles are swapped. But if, for example, particles 2 and 5 in eigenfunction (A.46) above are exchanged, it puts 2 in state 6 and 5 in state 4:

\begin{displaymath}
\begin{array}{l}
\psi^{\rm S}_{2\leftrightarrow5}({\skew...
...}_5//z5/ \ldots \pp54/{\skew0\vec r}_{36}//z36/
\end{array}
\end{displaymath}

That is simply a different energy eigenfunction. So neither (A.46) nor this swapped form are acceptable by themselves. To fix up the problem, eigenfunctions must be combined. To get a valid energy eigenfunction for bosons out of (A.46), all the different eigenfunctions that can be formed by swapping the 36 particles must be summed together. The normalized sum gives the correct eigenfunction for bosons. But note that there is a humongous number of different eigenfunctions that can be obtained by swapping the particles. Over 10$\POW9,{37}$ if you care to count them. As a result, there is no way that the gigantic expression for the resulting 36-boson energy eigenfunction could ever be written out here.

It is much easier in terms of the graphical depiction figure A.2: graphically all these countless system eigenfunctions differ only with respect to the numbers in the particles. And since in the final eigenfunction, all particles are present in exactly the same way, then so are their numbers within the particles. Every number appears equally in every particle. So the numbers do no longer add distinguishing information and can be left out. That makes the graphical depiction of the example eigenfunction for a system of identical bosons as in figure A.3. It illustrates why identical particles are commonly called “indistinguishable.”

Figure A.4: Graphical depiction of an arbitrary system energy eigenfunction for 33 identical fermions.
\begin{figure}
\centering
\setlength{\unitlength}{1pt}
\begin{picture}(4...
...0,196,t'$\pp72////$'
\PB378,196,t'$\pp73////$'
\end{picture}
\end{figure}

For a system of identical fermions, (like electrons or quarks), the eigenfunctions must change sign if two particles are swapped. As chapter 5.7 showed, that is very restrictive. It means that you cannot create an eigenfunction for a system of 36 fermions from the example eigenfunction (A.46) and the swapped versions of it. Various single-particle eigenfunctions appear multiple times in (A.46), like $\pp4////$, which is occupied by particles 2, 31, and 33. That cannot happen for fermions. A system eigenfunction for 36 identical fermions requires 36 different single-particle eigenfunctions.

It is the same graphically. The example figure A.3 for bosons is impossible for a system of identical fermions; there cannot be more than one fermion in a single-particle state. A depiction of an arbitrary energy eigenfunction that is acceptable for a system of 33 identical fermions is in figure A.4.

As explained in chapter 5.7, a neat way of writing down the system energy eigenfunction of the pictured example is to form a Slater determinant from the occupied states

\begin{displaymath}
\pp1////,
\pp2////,
\pp3////,
\ldots,
\pp{43}////,
\pp{45}////,
\pp{56}////.
\end{displaymath}

It is good to meet old friends again, isn’t it?

Now consider what happens in relativistic quantum mechanics. For example, suppose that an electron and positron annihilate each other. What are you going to do, leave holes in the parameter list of your wave function, where the electron and positron used to be? Like

\begin{displaymath}
\Psi({\skew0\vec r}_1,S_{z1},\mbox{[gone]},{\skew0\vec r}_...
...{\skew0\vec r}_5,S_{z5},\ldots,{\skew0\vec r}_{36},S_{z36};t)
\end{displaymath}

say? Or worse, what if a photon with very high energy hits an heavy nucleus and creates an electron-positron pair in the collision from scratch? Are you going to scribble in a set of additional parameters for the new particles into your parameter list? Scribble in another row and column in the Slater determinant for your electrons? That is voodoo mathematics. The classical way of writing wave functions fails.

And if positrons are too weird for you, consider photons, the particles of electromagnetic radiation, like ordinary light. As chapters 6.8 and 7.8 showed, the electrons in hot surfaces create and destroy photons readily when the thermal equilibrium shifts. Moving at the speed of light, with zero rest mass, photons are as relativistic as they come. Good luck scribbling in trillions of new states for the photons into your wave function when your black box heats up. Then there are solids; as chapter 11.14.6 shows, the phonons of crystal vibrational waves are the equivalent of the photons of electromagnetic waves.

One of the key insights of quantum field theory is to do away with classical mathematical forms of the wave function such as (A.46) and the Slater determinants. Instead, the graphical depictions, such as the examples in figures A.3 and A.4, are captured in terms of mathematics. How do you do that? By listing how many particles are in each type of single-particle state. In other words, you do it by listing the single-state “occupation numbers.”

Consider the example bosonic eigenfunction of figure A.3. The occupation numbers for that state would be

\begin{displaymath}
\left\vert 3,4, 1,3,2,2,2, 1,1,0,0,2,1,2, 0,1,1,1,1,1,0,0,0,
1,0,0,0,0,0,0,1,\ldots\right\rangle
\end{displaymath}

indicating that there are 3 bosons in single-particle state $\pp1////$, 4 in $\pp2////$, 1 in $\pp3////$, etcetera. Knowing those numbers is completely equivalent to knowing the classical system energy eigenfunction; it could be reconstructed from them. Similarly, the occupation numbers for the example fermionic eigenfunction of figure A.4 would be

\begin{displaymath}
\left\vert 1,1, 1,1,1,1,1, 1,1,1,1,1,1,1, 0,1,0,1,1,1,1,1,1,
1,0,1,0,0,1,1,1, \ldots\right\rangle
\end{displaymath}

Such sets of occupation numbers are called “Fock basis states.” Each describes one system energy eigenfunction.

General wave functions can be described by taking linear combinations of these basis states. The most general Fock wave function for a classical set of exactly $I$ particles is a linear combination of all the basis states whose occupation numbers add up to $I$. But Fock states make it also possible to describe systems like photons in a box with varying numbers of particles. Then the most general wave function is a linear combination of all the Fock basis states, regardless of the total number of particles. The set of all possible wave functions that can be formed from linear combinations of the Fock basis states is called the Fock space.

How about the case of distinguishable particles as in figure A.2? In that case, the numbers inside the particles also make a difference, so where do they go?? The answer of quantum field theory is to deny the existence of generic particles that take numbers. There are no generic particles in quantum field theory. There is a field of electrons, there is a field of protons, (or quarks, actually), there is a field of photons, etcetera, and each of these fields is granted its own set of occupation numbers. There is no way to describe a generic particle using a number. For example, if there is an electron in a single-particle state, in quantum field theory it means that the electron field has a particle in that energy state. The particle has no number.

Some physicist feel that this is a strong point in favor of believing that quantum field theory is the way nature really works. In the classical formulation of quantum mechanics, the (anti) symmetrization requirements under particle exchange are an additional ingredient, added to explain the data. In quantum field theory, it comes naturally: particles that are distinguishable simply cannot be described by the formalism. Still, our convenience in describing it is an uncertain motivator for nature.

The successful analysis of the blackbody spectrum in chapter 6.8 already testified to the usefulness of the Fock space. If you check the derivations in chapter 11 leading to it, they were all conducted based on occupation numbers. A classical wave function for the system of photons was never written down; that simply cannot be done.

Figure A.5: Example wave functions for a system with just one type of single particle state. Left: identical bosons; right: identical fermions.
\begin{figure}
\centering
\setlength{\unitlength}{1pt}
\begin{picture}(...
...t'$\pp////$'}
\PC8,5.5,\PC20,5.5,
\PD164,5.5,
\end{picture}
\end{figure}

There is a lot more involved in quantum field theory than just the blackbody spectrum, of course. To explain some of the basic ideas, simple examples can be helpful. The simplest example that can be studied involves just one single-particle state, say just a single-particle ground state. The graphical depiction of an arbitrary example wave function is then as in figure A.5. There is just one single-particle box. In nonrelativistic quantum mechanics, this would be a completely trivial quantum system. In the case of $I$ identical bosons, shown to the left in the figure, all of them would have to go into the only state there is. In the case of identical fermions, shown to the right, there can only be one fermion, and it has to go into the only state there is.

But when particles can be created or destroyed, things get more interesting. When there is no given number of particles, there can be any number of identical bosons within that single particle state. That allows $\vert\rangle$ (no particles,) $\vert 1\rangle$ (1 particle), $\vert 2\rangle$ (2 particles), etcetera. And the general wave function can be a linear combination of those possibilities. It is the same for identical fermions, except that there are now only the states $\vert\rangle$ (no particles) and $\vert 1\rangle$ (1 particle). The wave function can still be a combination of these two possibilities.

A relativistic system with just one type of single-particle state does seem very artificial. It raises the question how esoteric such an example is. But there are in fact two very well established classical systems that behave just like this:

1.
The one-di­men­sion­al harmonic oscillator of chapter 4.1 has energy levels that happen to be exactly equally spaced. It can pick up an energy above the ground state that is any whole multiple of $\hbar\omega$, where $\omega$ is its natural frequency. If you are willing to accept the particles to be quanta of energy of size $\hbar\omega$, then it provides a model of a bosonic system with just one single-particle state. The ground state, $h_0$ in the notations of chapter 4.1, is the state $\big\vert\big\rangle $. The first excited state $h_1$ is $\big\vert 1\big\rangle $; it has one additional energy quantum $\hbar\omega$. The second excited state $h_2$ is $\big\vert 2\big\rangle $, with two quanta more than the ground state, etcetera.

Recall from chapter 4.1 that there is an additional ground state energy of half a $\hbar\omega$ quantum. In a quantum field theory, this additional energy that exists even when there are no particles is called the vacuum energy.

The general wave function of a harmonic oscillator is a linear combination of the energy states. In terms of chapter 4.1, that expresses an uncertainty in energy. In the present context, it expresses an uncertainty in the number of these energy particles!

2.
A single electron has exactly two spin states. It can pick up exactly one unit $\hbar$ of $z$-​momentum above the spin-down state. If you accept the particles to be single quanta of $z$-​momentum of size $\hbar$, then it provides an example of a fermionic system with just one single-particle state. There can be either 0 or 1 quantum $\hbar$ of angular momentum in that single-particle state. The general wave function is a linear combination of the state with one angular momentum particle and the state with no angular momentum particle.

This example is less intuitive, since normally when you talk about a particle, you talk about an amount of energy, like in Einstein’s mass-energy relation. If it bothers you, think of the electron as being confined inside a magnetic field; then the spin-up state is associated with a corresponding increase in energy.

While the above two examples of relativistic systems with only one single-particle state are obviously made up, they do provide a very valuable sanity check on any relativistic analysis.

Not only that, the two examples are also very useful to understand the difference between a zero wave function and the so-called “vacuum state”

\begin{displaymath}
\fbox{$\displaystyle
\vert\vec 0\,\rangle \equiv \vert,0,0,\ldots\rangle
$} %
\end{displaymath} (A.47)

in which all occupation numbers are zero. The vacuum state is a normalized, nonzero, wave function just like the other possible sets of occupation numbers. It describes that there are no particles with certainty. You can see it from the two examples above. For the harmonic oscillator, the state $\vert\rangle$ is the ground state $h_0$ of the oscillator. For the electron-spin example, it is the spin-down state of the electron. These are completely normal eigenstates that the system can be in. They are not zero wave functions, which would be unable to describe a system.

Fock basis kets are taken to be orthonormal; an inner product between kets is zero unless all occupation numbers are equal. If they are all equal, the inner product is 1. In short:

\begin{displaymath}
\fbox{$\displaystyle
\big\langle\ldots,{\underline i}_3,...
...strut_1^1} 0 \mbox{ otherwise}
\end{array}
\right.
$} %
\end{displaymath} (A.48)

If the two kets have the same total number of particles, this orthonormality is required because the corresponding classical wave functions are orthonormal. Inner products between classical eigenfunctions that have even a single particle in a different state are zero. That is easily verified if the wave functions are simple products of single-particle ones. But then it also holds for sums of such eigenfunctions, as you have for bosons and fermions.

If the two kets have different total numbers of particles, the inner product between the classical wave functions does not exist. But basis kets are still orthonormal. To see that, take the two simple examples given above. For the harmonic oscillator example, different occupation numbers for the particles correspond to different energy eigenfunctions of the actual harmonic oscillator. These are orthonormal. It is similar for the spin example. The state of 0 particles is the spin-down state of the electron. The state of 1 particle is the spin-up state. These spin states are orthonormal states of the actual electron.


A.15.2 Creation and annihilation operators

The key to relativistic quantum mechanics is that particles can be created and annihilated. So it may not be surprising that it is very helpful to define operators that create and annihilate particles .

To keep the notations relatively simple, it will initially be assumed that there is just one type of single-particle state. Graphically that means that there is just one single-particle state box, like in figure A.5. However, there can be an arbitrary number of particles in that box.

The desired actions of the creation and annihilation operators are sketched in figure A.6. An annihilation operator $\widehat a$ turns a state $\big\vert i\big\rangle $ with $i$ particles into a state $\big\vert i{-}1\big\rangle $ with $i-1$ particles. A creation operator $\widehat a^\dagger $ turns a state $\big\vert i\big\rangle $ with $i$ particles into a state $\big\vert i{+}1\big\rangle $ with $i+1$ particles.

Figure A.6: Creation and annihilation operators for a system with just one type of single particle state. Left: identical bosons; right: identical fermions.
\begin{figure}
\centering
\setlength{\unitlength}{1pt}
\begin{picture}(...
...C5,131.5,\PC14,131.5,\PC23,131.5,
\PD164,47.5,
\end{picture}
\end{figure}

The operators are therefore defined by the relations

\begin{displaymath}
\widehat a\big\vert i\big\rangle = \alpha _i \big\vert i{-...
...t a^\dagger \big\vert 1\big\rangle = 0 \mbox{ for fermions} %
\end{displaymath} (A.49)

Here the $\alpha _i$ and $\alpha^\dagger _i$ are numerical constants still to be chosen.

Note that the above relations only specify what the operators $\widehat a$ and $\widehat a^\dagger $ do to basis kets. But that is enough information to define them. To figure out what these operators do to linear combinations of basis kets, just apply them to each term in the combination separately.

Mathematically you can always define whatever operators you want. But you must hope that they will turn out to be operators that are physically helpful. To help achieve that, you want to chose the numerical constants $\alpha _i$ and $\alpha^\dagger _i$ appropriately. Consider what happens if the operators are applied in sequence:

\begin{displaymath}
\widehat a^\dagger \widehat a\big\vert i\big\rangle = \wid...
...ngle = \alpha^\dagger _{i-1}\alpha _i \big\vert i\big\rangle
\end{displaymath}

Reading from right to left, the order in which the operators act on the state, first $\widehat a$ destroys a particle, then $\widehat a^\dagger $ restores it again. It gives the same state back, except for the numerical factor $\alpha^\dagger _{i-1}\alpha _i$. That makes every state $\big\vert i\big\rangle $ an eigenvector of the operator $\widehat a^\dagger \widehat a$ with eigenvalue $\alpha^\dagger _{i-1}\alpha _i$.

If the constants $\alpha^\dagger _{i-1}$ and $\alpha _i$ are chosen to make the eigenvalue a real number, then the operator $\widehat a^\dagger \widehat a$ will be Hermitian. More specifically, if they are chosen to make the eigenvalue equal to $i$, then $\widehat a^\dagger \widehat a$ will be the “particle number operator” whose eigenvalues are the number of particles in the single-particle state. The most logical choice for the constants to achieve that is clearly

\begin{displaymath}
\alpha _i=\sqrt{i}
\qquad
\alpha^\dagger _{i-1}=\sqrt{i}
\quad\Longrightarrow\quad
\alpha^\dagger _i=\sqrt{i+1}
\end{displaymath}

The full definition of the annihilation and creation operators can now be written in a nice symmetric way as

\begin{displaymath}
\fbox{$\displaystyle
\widehat a\big\vert i\big\rangle = ...
...\dagger \big\vert 1\big\rangle = 0\mbox{ for fermions}
$} %
\end{displaymath} (A.50)

In words, the annihilation operator $\widehat a$ kills off one particle and adds a factor $\sqrt{i}$. The operator $\widehat a^\dagger $ puts the particle back in and adds another factor $\sqrt{i}$.

These operators are particularly convenient since they are Hermitian conjugates. That means that if you take them to the other side in an inner product, they turn into each other. In particular, for inner products between basis kets,

\begin{displaymath}
\Big\langle \big\vert{\underline i}\big\rangle \Big\vert \...
...ngle \Big\vert \big\vert{\underline i}\big\rangle \Big\rangle
\end{displaymath}

Note that if such relations apply for basis kets, they also apply for all linear combinations of basis kets.

To verify that the above relations apply, recall from the previous subsection that kets are orthonormal. In the equalities above, the inner products are only nonzero if ${\underline i}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $i-1$: after lowering the particle number with $\widehat a$, or raising it with $\widehat a^\dagger $, the particle numbers must be the same at both sides of the inner product. And when ${\underline i}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $i-1$, according to the definitions (A.50) of $\widehat a$ and $\widehat a^\dagger $ all inner products above equal $\sqrt{i}$, so the equalities still apply.

It remains true for fermions that $\widehat a$ and $\widehat a^\dagger $ are Hermitian conjugates, despite the fact that $\widehat a^\dagger \big\vert 1\big\rangle $ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 instead of $\sqrt{2}\,\big\vert 2\big\rangle $. The reason is that it would only make a difference if there was a $\big\vert 2\big\rangle $ state in the other side of the inner product, and such a state does not exist.

The inner products are usually written in the more esthetic form

\begin{displaymath}
\big\langle{\underline i}\big\vert\widehat a\big\vert i\bi...
...rt\widehat a^\dagger \Big)\big\vert{\underline i}\big\rangle
\end{displaymath}

Here it is to be understood that, say, $\big\langle{\underline i}\big\vert\widehat a$ stands for $\widehat a^\dagger \big\vert{\underline i}\big\rangle $ pushed into the left hand side of an inner product, chapter 2.7.1.

You may well wonder why $\widehat a^\dagger \widehat a$ is the particle count operator; why not $\widehat a\widehat a^\dagger $? The reason is that $\widehat a\widehat a^\dagger $ would not work for the state $\big\vert\big\rangle $ unless you took $\widehat a^\dagger \big\vert\big\rangle $ to be zero or $\widehat a\big\vert 1\big\rangle $ to be zero, and then they could no longer create or annihilate $\big\vert 1\big\rangle $.

Still, it is interesting to see what the effect of $\widehat a\widehat a^\dagger $ is. It turns out that this depends on the type of particle. For bosons, using (A.50),

\begin{displaymath}
\widehat a_{\rm {b}}\widehat a^\dagger _{\rm {b}} \big\ver...
...i+1}\, \big\vert i\big\rangle = (i+1) \big\vert i\big\rangle
\end{displaymath}

So the operator $\widehat a_{\rm {b}}\widehat a^\dagger _{\rm {b}}$ has eigenvalues one greater than the number of particles. That means that if you subtract $\widehat a_{\rm {b}}\widehat a^\dagger _{\rm {b}}$ and $\widehat a^\dagger _{\rm {b}}\widehat a_{\rm {b}}$, you get the unit operator that leaves all states unchanged. And the difference between $\widehat a_{\rm {b}}\widehat a^\dagger _{\rm {b}}$ and $\widehat a^\dagger _{\rm {b}}\widehat a_{\rm {b}}$ is by definition the commutator of $\widehat a_{\rm {b}}$ and $\widehat a^\dagger _{\rm {b}}$, indicated by square brackets:
\begin{displaymath}
\fbox{$\displaystyle
[\widehat a_{\rm{b}},\widehat a^\da...
...- \widehat a^\dagger _{\rm{b}} \widehat a_{\rm{b}} = 1
$} %
\end{displaymath} (A.51)

Isn’t that cute! Of course, $[\widehat a_{\rm {b}},\widehat a_{\rm {b}}]$ and $[\widehat a^\dagger _{\rm {b}},\widehat a^\dagger _{\rm {b}}]$ are zero since everything commutes with itself. It turns out that you can learn a lot from these commutators, as seen in later subsections.

The same commutator does not apply to fermions, because if you apply $\widehat a_{\rm {f}}\widehat a^\dagger _{\rm {f}}$ to $\big\vert 1\big\rangle $, you get zero instead of $2\big\vert 1\big\rangle $. But for fermions, the only state for which $\widehat a_{\rm {f}}\widehat a^\dagger _{\rm {f}}$ produces something nonzero is $\big\vert\big\rangle $ and then it leaves the state unchanged. Similarly, the only state for which $\widehat a^\dagger _{\rm {f}}\widehat a_{\rm {f}}$ produces something nonzero is $\big\vert 1\big\rangle $ and then it leaves that state unchanged. That means that if you add $\widehat a_{\rm {f}}\widehat a^\dagger _{\rm {f}}$ and $\widehat a^\dagger _{\rm {f}}\widehat a_{\rm {f}}$ together, instead of subtract them, it reproduces the same state state whether it is $\big\vert\big\rangle $ or $\big\vert 1\big\rangle $ (or any combination of them). The sum of $\widehat a_{\rm {f}}\widehat a^\dagger _{\rm {f}}$ and $\widehat a^\dagger _{\rm {f}}\widehat a_{\rm {f}}$ is called the “anticommutator” of $\widehat a_{\rm {f}}$ and $\widehat a^\dagger _{\rm {f}}$; it is indicated by curly brackets:

\begin{displaymath}
\fbox{$\displaystyle
\{\widehat a_{\rm{f}},\widehat a^\d...
...+ \widehat a^\dagger _{\rm{f}} \widehat a_{\rm{f}} = 1
$} %
\end{displaymath} (A.52)

Isn’t that neat? Note also that $\{\widehat a_{\rm {f}},\widehat a_{\rm {f}}\}$ and $\{\widehat a_{\rm {f}},\widehat a_{\rm {f}}\}$ are zero since applying either operator twice ends up in a nonexisting state.

How about the Hamiltonian for the energy of the system of particles? Well, for noninteracting particles the energy of $i$ particles is $i$ times the single particle energy ${\vphantom' E}^{\rm p}$. And since the operator that gives the number of particles is $\widehat a^\dagger \widehat a$, that is ${\vphantom' E}^{\rm p}\widehat a^\dagger \widehat a$. The total Hamiltonian for noninteracting particles becomes therefore:

\begin{displaymath}
\fbox{$\displaystyle
H = {\vphantom' E}^{\rm p}\widehat a^\dagger \widehat a+ E_{\rm{ve}}
$} %
\end{displaymath} (A.53)

Here $E_{\rm {ve}}$ stands for any additional “vacuum energy” that exists even if there are no particles. That is the ground state energy of the system. The above Hamiltonian allows the Schrö­din­ger equation to be written in terms of occupation numbers and creation and annihilation operators.


A.15.3 The caHermitians

It is important to note that the creation and annihilation operators $\widehat a^\dagger $ and $\widehat a$ are not Hermitian. They cannot be taken unchanged to the other side of an inner product. And their eigenvalues are not real. Therefore they cannot correspond to physically observable quantities. But since they are Hermitian conjugates, it is easy to form operators from them that are Hermitian. For example, their products $\widehat a^\dagger \widehat a$ and $\widehat a\widehat a^\dagger $ are Hermitian. The Hamiltonian for noninteracting particles (A.53) given in the previous subsection illustrates that.

Hermitian operators can also be formed from linear combinations of the creation and annihilation operators. Two combinations that are often physically relevant are

\begin{displaymath}
\widehat P \equiv {\textstyle\frac{1}{2}}(\widehat a+\wide...
...extstyle\frac{1}{2}} {\rm i}(\widehat a- \widehat a^\dagger )
\end{displaymath}

In lack of a better name that the author knows of, this book will call $\widehat{P}$ and $\widehat{Q}$ the caHermitians.

Conversely, the annihilation and creation operators can be written in terms of the caHermitians as

\begin{displaymath}
\fbox{$\displaystyle
\widehat a= \widehat P - {\rm i}\wi...
...ad \widehat a^\dagger = \widehat P + {\rm i}\widehat Q
$} %
\end{displaymath} (A.54)

This can be verified by substituting in the definitions of $\widehat{P}$ and $\widehat{Q}$.

The Hamiltonian (A.53) for noninteracting particles can be written in terms of $\widehat{P}$ and $\widehat{Q}$ as

\begin{displaymath}
H = {\vphantom' E}^{\rm p}\left(\widehat P^2 + \widehat Q^2
- {\rm i}[\widehat P,\widehat Q]\right) + E_{\rm {ve}}
\end{displaymath}

Here ${\vphantom' E}^{\rm p}$ is again the single-particle energy and $E_{\rm {ve}}$ the vacuum energy. The square brackets indicate again the commutator of the enclosed operators.

What this Hamiltonian means depends on whether the particles being described are bosons or fermions. They have different commutators $[\widehat{P},\widehat{Q}]$.

Consider first the case that the particles are bosons. The previous subsection showed that the commutator $[\widehat a_{\rm {b}},\widehat a^\dagger _{\rm {b}}]$ is 1. From that the commutator of $P_{\rm {b}}$ and $Q_{\rm {b}}$ is readily found using the rules of chapter 4.5.4. It is:

\begin{displaymath}
\fbox{$\displaystyle
[\widehat P_{\rm{b}},\widehat Q_{\rm{b}}] = - {\textstyle\frac{1}{2}}{\rm i}
$} %
\end{displaymath} (A.55)

So the commutator is an imaginary constant. That is very much like Heisenberg’s canonical commutator between position and linear momentum in classical quantum mechanics. It implies a similar uncertainty principle, chapter 4.5.2 (4.46). In particular, $P_{\rm {b}}$ and $Q_{\rm {b}}$ cannot have definite values at the same time. Their values have uncertainties $\sigma_{P_{\rm {b}}}$ and $\sigma_{Q_{\rm {b}}}$ that are at least so big that

\begin{displaymath}
\sigma_{P_{\rm {b}}} \sigma_{Q_{\rm {b}}} \mathrel{\raisebox{-1pt}{$\geqslant$}}{\textstyle\frac{1}{4}}
\end{displaymath}

The Hamiltonian for bosons becomes, using the commutator above,

\begin{displaymath}
H_{\rm {b}}
= {\vphantom' E}^{\rm p}\left(\widehat P_{\r...
...{\rm {ve}} - {\textstyle\frac{1}{2}} {\vphantom' E}^{\rm p} %
\end{displaymath} (A.56)

Often, the Hamiltonian is simply the first term in the right hand side. In that case, the vacuum energy is half a particle.

For fermions, the following useful relations follow from the anticommutators for the creation and annihilation operators given in the previous subsection:

\begin{displaymath}
\widehat P_{\rm {f}}^2 = {\textstyle\frac{1}{4}} \qquad
\widehat Q_{\rm {f}}^2 = {\textstyle\frac{1}{4}} %
\end{displaymath} (A.57)

The Hamiltonian then becomes
\begin{displaymath}
H = {\vphantom' E}^{\rm p}\left({\textstyle\frac{1}{2}}-{\...
...t P_{\rm {f}},\widehat Q_{\rm {f}}]\right)
+ E_{\rm {ve}} %
\end{displaymath} (A.58)


A.15.4 Recasting a Hamiltonian as a quantum field one

The arguments of the previous subsection can be reversed. Given a suitable Hamiltonian, it can be recast in terms of annihilation and creation operators. This is often useful. It provides a way to quantize systems such as a harmonic oscillator or electromagnetic radiation.

Assume that some system has a Hamiltonian with the following properties:

\begin{displaymath}
\fbox{$\displaystyle
H = {\vphantom' E}^{\rm p}\left(\wi...
... P,\widehat Q\right] = - {\textstyle\frac{1}{2}} {\rm i}
$}
\end{displaymath} (A.59)

Here $\widehat{P}$ and $\widehat{Q}$ must be Hermitian operators and ${\vphantom' E}^{\rm p}$ and $E_{\rm {ref}}$ must be constants with units of energy.

It may be noted that typically $E_{\rm {ref}}$ is zero. It may also be noted that it suffices that the commutator is an imaginary constant. A different magnitude of the constant can be accommodated by rescaling $\widehat{P}$ and $\widehat{Q}$, and absorbing the scaling factor in ${\vphantom' E}^{\rm p}$. A sign change can be accommodated by swapping $\widehat{P}$ and $\widehat{Q}$.

From the given apparently limited amount of information, all of the following conclusions follow:

1.
The observable quantities $P$ and $Q$ corresponding to the Hermitian operators are always uncertain. As explained in chapter 4.4, if you measure an uncertain quantity, say $P$, for a lot of identical systems, you do get some average value. That average value is called the expectation value $\big\langle P\big\rangle $. However, the individual measured values will deviate from that expectation value. The average deviation is called the standard deviation or uncertainty $\sigma_P$. For the system above, the uncertainties in $P$ and $Q$ must satisfy the relation

\begin{displaymath}
\sigma_P \sigma_Q \mathrel{\raisebox{-1pt}{$\geqslant$}}{\textstyle\frac{1}{4}}
\end{displaymath}

Neither uncertainty can be zero, because that would make the other uncertainty infinite.
2.
The expectation values of the observables $P$ and $Q$ satisfy the equations

\begin{displaymath}
\frac{{\rm d}\big\langle P\big\rangle }{{\rm d}t} = - \ome...
...x{where } \omega \equiv \frac{{\vphantom' E}^{\rm p}}{\hbar}
\end{displaymath}

That means that the expectation values vary harmonically with time,

\begin{displaymath}
\big\langle P\big\rangle = A \cos(\omega t + \alpha)
\qquad
\big\langle Q\big\rangle = A \sin(\omega t + \alpha)
\end{displaymath}

Here the amplitude $A$ and the “phase angle” $\alpha$ are arbitrary constants.
3.
In energy eigenstates, the expectation values $\big\langle P\big\rangle $ and $\big\langle Q\big\rangle $ are always zero.
4.
The ground state energy of the system is

\begin{displaymath}
E_0 = {\textstyle\frac{1}{2}}{\vphantom' E}^{\rm p}+E_{\rm {ref}}
\end{displaymath}

For now it will be assumed that the ground state is unique. It will be indicated as $\big\vert\big\rangle $. It is often called the vacuum state.
5.
The higher energy states will be indicated by $\big\vert 1\big\rangle $, $\big\vert 2\big\rangle $, ...in order of increasing energy $E_1$, $E_2$, .... The states are unique and their energy is

\begin{displaymath}
\mbox{wave function: } \big\vert i\big\rangle
\qquad
...
...y: } E_i = (i + {\textstyle\frac{1}{2}}) E_p + E_{\rm {ref}}
\end{displaymath}

So a state $\big\vert i\big\rangle $ has $i$ additional quanta of energy ${\vphantom' E}^{\rm p}$ more than the vacuum state. In particular that means that the energy levels are equally spaced. There is no maximum energy.
6.
In energy eigenstates,

\begin{displaymath}
\big\langle{\vphantom' E}^{\rm p}P^2\big\rangle = \big\lan...
...e = {\textstyle\frac{1}{2}}(i + {\textstyle\frac{1}{2}}) E_p
\end{displaymath}

So the expectation values of these two terms in the Hamiltonian are equal. Each contributes half to the energy of the quanta.
7.
In the ground state, the two expectation energies above are the absolute minimum allowed by the uncertainty relation. Each expectation energy is then ${\textstyle\frac{1}{4}}{\vphantom' E}^{\rm p}$.
8.
Annihilation and creation operators can be defined as

\begin{displaymath}
\widehat a\equiv \widehat P - {\rm i}\widehat Q
\qquad
\widehat a^\dagger \equiv \widehat P + {\rm i}\widehat Q
\end{displaymath}

These have the following effects on the energy states:

\begin{displaymath}
\widehat a\big\vert i\big\rangle = \sqrt{i} \big\vert i{-}...
...big\vert i{-}1\big\rangle = \sqrt{i} \big\vert i\big\rangle
\end{displaymath}

(This does assume that the normalization factors in the energy eigenstates are chosen consistently. Otherwise there might be additional factors of magnitude 1.) The commutator $[\widehat a,\widehat a^\dagger ]$ is 1.
9.
The Hamiltonian can be rewritten as

\begin{displaymath}
H = {\vphantom' E}^{\rm p}\widehat a^\dagger \widehat a+ {\textstyle\frac{1}{2}} E_p + E_{\rm {ref}}
\end{displaymath}

Here the operator $\widehat a^\dagger \widehat a$ gives the number of energy quanta of the state it acts on.
10.
If the ground state is not unique, each independent ground state gives rise to its own set of energy eigenfunctions, with the above properties. Consider the example that the system describes an electron, and that the energy does not depend on the spin. In that case, there will be a spin-up and a spin-down version of the ground state, $\big\vert\big\rangle {\uparrow}$ and $\big\vert\big\rangle {\downarrow}$. These will give rise to two families of energy states $\big\vert i\big\rangle {\uparrow}$ respectively $\big\vert i\big\rangle {\downarrow}$. Each family will have the properties described above.

The derivation of the above properties is really quite simple and elegant. It can be found in {D.33}.

Note that various properties above are exactly the same as found in the analysis of bosons starting with the annihilation and creation operators. The difference in this subsection is that the starting point was a Hamiltonian in terms of two square Hermitian operators; and those merely needed to have a purely imaginary commutator.


A.15.5 The harmonic oscillator as a boson system

This subsection will illustrate the power of the introduced quantum field ideas by example. The objective is to use these ideas to rederive the one-di­men­sion­al harmonic oscillator from scratch. The derivation will be much cleaner than the elaborate algebraic derivation of chapter 4.1, and in particular {D.12}.

The Hamiltonian of a harmonic oscillator in classical quantum mechanics is, chapter 4.1,

\begin{displaymath}
H = \frac{1}{2m}{\widehat p}_x^2 + \frac{m}{2}\omega^2 x^2
\end{displaymath}

Here the first term is the kinetic energy and the second the potential energy.

According to the previous subsection, a system like this can be solved immediately if the commutator of ${\widehat p}_x$ and $x$ is an imaginary constant. It is, that is the famous “canonical commutator” of Heisenberg:

\begin{displaymath}[x,{\widehat p}_x]= {\rm i}\hbar
\end{displaymath}

To use the results of the previous subsection, first the Hamiltonian must be rewritten in the form

\begin{displaymath}
H = {\vphantom' E}^{\rm p}\left(\widehat P^2 + \widehat Q^2\right)
\end{displaymath}

where $\widehat{P}$ and $\widehat{Q}$ satisfy the commutation relationship for bosonic caHermitians:

\begin{displaymath}
\left[\widehat P,\widehat Q\right] = - {\textstyle\frac{1}{2}} {\rm i}
\end{displaymath}

That requires that you define

\begin{displaymath}
{\vphantom' E}^{\rm p}= \hbar\omega \qquad
\widehat P = ...
...t p}_x \qquad
\widehat Q = \sqrt{\frac{m\omega}{2\hbar}}\,x
\end{displaymath}

According to the previous subsection, the energy eigenvalues are

\begin{displaymath}
E_i = (i + {\textstyle\frac{1}{2}}) \hbar\omega
\end{displaymath}

So the spectrum has already been found.

And various other interesting properties of the solution may also be found in the previous subsection. Like the fact that there is half a quantum of energy left in the ground state. True, the zero level of energy is not important for the dynamics. But this half quantum does have a physical meaning. Assume that you have a lot of identical harmonic oscillators in the ground state, and that you do a measurement of the kinetic energy for each. You will not get zero kinetic energy. In fact, the average kinetic energy measured will be a quarter quantum, half of the total energy. The other quarter quantum is what you get on average if you do potential energy measurements.

Another observation of the previous subsection is that the expectation position of the particle will vary harmonically with time. It is a harmonic oscillator, after all.

The energy eigenfunctions will be indicated by $h_i$, rather than $\big\vert i\big\rangle $. What has not yet been found are specific expressions for these eigenfunctions. However, as figure A.6 shows, if you apply the annihilation operator $\widehat a$ on the ground state $h_0$, you get zero:

\begin{displaymath}
\widehat ah_0 = 0
\end{displaymath}

And also according to the previous subsection

\begin{displaymath}
\widehat a= \widehat P - {\rm i}\widehat Q
\end{displaymath}

Putting in the expressions for $\widehat{P}$ and $\widehat{Q}$ above, with ${\widehat p}_x$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\hbar\partial$$\raisebox{.5pt}{$/$}$${\rm i}\partial{x}$, and rearranging gives

\begin{displaymath}
\frac{1}{h_0} \frac{\partial h_0}{\partial x}
= - \frac{m\omega}{\hbar} x
\end{displaymath}

This can be simplified by defining a scaled $x$ coordinate:

\begin{displaymath}
\frac{1}{h_0} \frac{\partial h_0}{\partial \xi}
= - \xi
...
...ac{x}{\ell}
\qquad \ell \equiv \sqrt{\frac{\hbar}{m\omega}}
\end{displaymath}

Integrating both sides with respect to $\xi$ and cleaning up by taking an exponential gives the ground state as

\begin{displaymath}
h_0 = C e^{-\xi^2/2}
\end{displaymath}

The integration constant $C$ can be found from normalizing the wave function. The needed integral can be found under ! in the notations section. That gives the final ground state as

\begin{displaymath}
h_0 = \frac{1}{(\pi\ell^2)^{1/4}} e^{-\xi^2/2}
\end{displaymath}

To get the other eigenfunctions $h_i$ for $i$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1, 2, ..., apply the creation operator $\widehat a^\dagger $ repeatedly:

\begin{displaymath}
h_i = \frac{1}{\sqrt{i}} \widehat a^\dagger h_{i-1}
\end{displaymath}

According to the previous subsection, the creation operator is

\begin{displaymath}
\widehat a^\dagger = \widehat P + {\rm i}\widehat Q =
\...
...}{\sqrt{2}} \left(\xi - \frac{\partial}{\partial \xi}\right)
\end{displaymath}

So the entire process involves little more than a single differentiation for each energy eigenfunction found. In particular, unlike in {D.12}, no table books are needed. Note that factors ${\rm i}$ do not make a difference in eigenfunctions. So the ${\rm i}$ in the final expression for $\widehat a^\dagger $ may be left out to get real eigenfunctions. That gives table 4.1.

That was easy, wasn’t it?


A.15.6 Canonical (second) quantization

Canonical quantization” is a procedure to turn a classical system into the proper quantum one. If it is applied to a field, like the electromagnetic field, it is often called “second quantization.”

Recall the quantum analysis of the harmonic oscillator in the previous subsection. The key to the correct solution was the canonical commutator between position and momentum. Apparently, if you get the commutators right in quantum mechanics, you get the quantum mechanics right. That is the idea behind canonical quantization.

The basic idea can easily be illustrated for the harmonic oscillator. The standard harmonic oscillator in classical physics is a simple spring-mass system. The classical governing equations are:

\begin{displaymath}
\frac{{\rm d}x}{{\rm d}t} = v_x
\qquad
m \frac{{\rm d}v_x}{{\rm d}t} = - k x
\end{displaymath}

Here $x$ is the position of the oscillating mass $m$ and $k$ is the spring constant. The first of these equations is merely the definition of velocity. The second is Newton’s second law.

As you can readily check by substitution, the most general solution is

\begin{displaymath}
x = A \sin(\omega t +\alpha) \qquad v_x = A\omega \cos(\omega t +\alpha)
\qquad \omega \equiv \sqrt{\frac{k}{m}}
\end{displaymath}

Here the amplitude $A$ and the “phase angle” $\alpha$ are arbitrary constants. The frequency $\omega$ is given in terms of the known spring constant and mass.

This system is now to be quantized using canonical quantization. The process is somewhat round-about. First a “canonical momentum,” or “conjugate momentum,” or “generalized momentum,” $p_x$ is defined by taking the derivative of the kinetic energy, $\frac12mv_x^2$, (or more generally, of the Lagrangian {A.1}), with respect to the time derivative of $x$. Since the time derivative of $x$ is $v_x$, the momentum is $mv_x$. That is the usual linear momentum.

Next a classical Hamiltonian is defined. It is the total energy of the system expressed in terms of position and momentum:

\begin{displaymath}
H_{\rm cl} = \frac{p_x^2}{2 m} + \frac{m}{2} \omega^2 x^2
\end{displaymath}

Here the first term is the kinetic energy, with $v_x$ rewritten in terms of the momentum. The second term is the potential energy in the spring. The spring constant in it was rewritten as $m\omega^2$ because $m$ and $\omega$ are physically more important variables, and the symbol $k$ is already greatly overworked in quantum mechanics as it is. See {A.1} for more on classical Hamiltonians.

To quantize the system, the momentum and position in the Hamiltonian must be turned into operators. Actual values of momentum and position are then the eigenvalues of these operators. Basically, you just put a hat on the momentum and position in the Hamiltonian:

\begin{displaymath}
H = \frac{{\widehat p}_x^{\,2}}{2 m} + \frac{m}{2} \omega^2 {\widehat x}^{\,2}
\end{displaymath}

Note that the hat on $x$ is usually omitted. However, it is still an operator in the sense that it is supposed to multiply wave functions now. Now all you need is the right commutator between ${\widehat p}_x$ and ${\widehat x}$.

In general, you identify commutators in quantum mechanics with so-called Poisson brackets in classical mechanics. Assume that $A$ and $B$ are any two quantities that depend on $x$ and $p_x$. Then their Poisson bracket is defined as, {A.12},

\begin{displaymath}
\{A,B\}_{\rm P} \equiv
\frac{\partial A}{\partial x} \fr...
...\frac{\partial B}{\partial x} \frac{\partial A}{\partial p_x}
\end{displaymath}

From that it is immediately seen that

\begin{displaymath}
\{x,p_x\}_{\rm P} = 1 \qquad \{x,x\}_{\rm P} = 0 \qquad \{p_x,p_x\}_{\rm P} = 0
\end{displaymath}

Correspondingly, in quantum mechanics you take

\begin{displaymath}[x,{\widehat p}_x]= {\rm i}\hbar \qquad [x,x] = 0 \qquad [{\widehat p}_x,{\widehat p}_x] = 0
\end{displaymath}

In this way the nonzero Poisson brackets bring in Planck’s constant that defines quantum mechanics. (In case of fermions, anticommutators take the place of commutators.)

Because of reasons discussed for the Heisenberg picture of quantum mechanics, {A.12}, the procedure ensures that the quantum mechanics is consistent with the classical mechanics. And indeed, the results of the previous subsection confirmed that. You can check that the expectation position and momentum had the correct classical harmonic dependence on time.

Fundamentally, quantization of a classical system is just an educated guess. Classical mechanics is a special case of quantum mechanics, but quantum mechanics is not a special case of classical mechanics. For the material covered in this book, there are simpler ways to make an educated guess than canonical quantization. Being less mathematical, they are more understandable and intuitive. That might make them maybe more convincing too.


A.15.7 Spin as a fermion system

There is, of course, not much analysis that can be done with a fermion system with only one single-particle state. There are only two independent system states; no fermion or one fermion.

However, there is at least one physical example of such a simple system. As noted in subsection A.15.1, a particle with spin $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$ like an electron can be considered to be a model for it. The vacuum state $\big\vert\big\rangle $ is the spin-down state of the electron. The state $\big\vert 1\big\rangle $ is the spin-up state. This state has one unit $\hbar$ more angular momentum in the $z$-​direction. If the electron is in a magnetic field, that additional momentum corresponds to a quantum of energy.

One reasonable question that can now be asked is whether the annihilation and creation operators, and the caHermitians, have some physical meaning for this system. They do.

Recall that for fermions, the Hamiltonian was given in terms of the caHermitians $\widehat{P}_{\rm {f}}$ and $\widehat{Q}_{\rm {f}}$ as

\begin{displaymath}
H = {\vphantom' E}^{\rm p}\left({\textstyle\frac{1}{2}}-{\...
...hat P_{\rm {f}},\widehat Q_{\rm {f}}]\right)
+ E_{\rm {ve}}
\end{displaymath}

The expression between parentheses is the particle count operator, equal to zero for the spin-down state and 1 for the spin up state. So the second term within parentheses in the Hamiltonian must be the spin in the $z$-​direction, nondimensionalized by $\hbar$. (Recall that the spin in the $z$-​direction has the values $\pm\frac12\hbar$.) So apparently

\begin{displaymath}[\widehat P_{\rm {f}},\widehat Q_{\rm {f}}]= {\rm i}\frac{{\widehat S}_z}{\hbar}
\end{displaymath}

Reasonably speaking then, the caHermitians themselves should be the nondi­men­sion­al components of spin in the $x$ and $y$ directions,

\begin{displaymath}
\widehat P_{\rm {f}} = \frac{{\widehat S}_x}{\hbar}
\qquad
\widehat Q_{\rm {f}} = \frac{{\widehat S}_y}{\hbar}
\end{displaymath}

What other variables are there in this problem? And so it is. The commutator above, with the caHermitians equal to the nondi­men­sion­al spin components, is known as the “fundamental commutation relation.” Quantum field analysis is one way to understand that this relation applies.

Recall another property of the caHermitians for fermions:

\begin{displaymath}
\widehat P_{\rm {f}}^2 = {\textstyle\frac{1}{4}} \qquad \widehat Q_{\rm {f}}^2 = {\textstyle\frac{1}{4}}
\end{displaymath}

Apparently then, the square spin components are just constants with no uncertainty. Of course, that is no surprise since the only spin values in any direction are $\pm\frac12\hbar$.

Finally consider the annihilation and creation operators, multiplied by $\hbar$:

\begin{displaymath}
\hbar\widehat a= {\widehat S}_x - {\rm i}{\widehat S}_y \q...
...ar\widehat a^\dagger = {\widehat S}_x + {\rm i}{\widehat S}_y
\end{displaymath}

Apparently these operators can remove, respectively add a unit $\hbar$ of angular momentum in the $z$-​direction. That is often important in relativistic applications where a fermion emits or absorbs angular momentum in the $z$-​direction. This changes the spin of the fermion and that can be expressed by the operators above. So you will usually find $x$ and $y$ spin operators in the analysis of such processes.

Obviously, you can learn a lot by taking a quantum field type approach. To be sure, the current analysis applies only to particles with spin $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$. But advanced analysis of angular momentum in general is very similar to quantum field analysis, chapter 12. It resembles some mixture of the boson and fermion cases.


A.15.8 More single particle states

The previous subsections discussed quantum field theory when there is just one type of single-particle state for the particles. This subsection considers the case that there is more than one. An index $n$ will be used to number the states.

Graphically, the case of multiple single-particle states was illustrated in figures A.3 and A.4. There is now more than one box that particles can be in. Each box corresponds to one type of single-particle state $\pp{n}////$.

Each such single-particle state has an occupation number $i_n$ that gives the number of particles in that state. A complete set of such occupation numbers form a Fock space basis ket

\begin{displaymath}
\vert i_1,i_2,i_3,i_4,\ldots\rangle
\end{displaymath}

An annihilation operator $\widehat a_n$ and a creation operator $\widehat a^\dagger _n$ must be defined for every occupation number. The mathematical definition of these operators for bosons is

\begin{displaymath}
\fbox{$\displaystyle
\begin{array}{l}
\displaystyle\st...
...\ldots,i_{n-1},i_n,i_{n+1},\ldots\rangle
\end{array}
$} %
\end{displaymath} (A.60)

The commutator relations are

\begin{displaymath}
\fbox{$\displaystyle
\left[\widehat a_{{\rm{b}},n},\wide...
...{b}},{\underline n}}\right] = \delta_{n{\underline n}}
$} %
\end{displaymath} (A.61)

Here $\delta_{n{\underline n}}$ is the Kronecker delta, equal to one if $n$ $\vphantom0\raisebox{1.5pt}{$=$}$ ${\underline n}$, and zero in all other cases. These commutator relations apply for $n$ $\raisebox{.2pt}{$\ne$}$ ${\underline n}$ because then the operators do unrelated things to different single-particle states; in that case it does not make a difference in what order you apply them. That makes the commutator zero. For $n$ $\vphantom0\raisebox{1.5pt}{$=$}$ ${\underline n}$, the commutator relations are unchanged from the case of just one single-particle state.

For fermions it is a bit more complex. The graphical representation of the example fermionic energy eigenfunction figure A.4 cheats a bit, because it suggests that there is only one classical wave function for a given set of occupation numbers. Actually, there are two variations, based on how the particles are ordered. The two are the same except that they have the opposite sign. Suppose that you create a particle in a state $n$; classically you would want to call that particle 1, and then create a particle in a state ${\underline n}$, classically you would want to call it particle 2. Do the particle creation in the opposite order, and it is particle 1 that ends up in state ${\underline n}$ and particle 2 that ends up in state $n$. That means that the classical wave function will have changed sign. However, the Fock space ket will not unless you do something.

What you can do is define the annihilation and creation operators for fermions as follows:

\begin{displaymath}
\fbox{$\displaystyle
\begin{array}{l}
\displaystyle\st...
...dots,i_{n-1},1,i_{n+1},\ldots\rangle = 0
\end{array}
$} %
\end{displaymath} (A.62)

The only difference from the annihilation and creation operators for just one type of single-particle state is the potential sign changes due to the $(-1)^{\ldots}$. It adds a minus sign whenever you swap the order of annihilating/creating two particles in different states. For the annihilation and creation operators of the same state, it may change both their signs, but that does nothing much: it leaves the important products such as $\widehat a^\dagger _n\widehat a_n$ and the anticommutators unchanged.

Of course, you can define the annihilation and creation operators with whatever sign you want, but putting in the sign pattern above may produce easier mathematics. In fact, there is an immediate benefit already for the anticommutator relations; they take the same form as for bosons, except with anticommutators instead of commutators:

\begin{displaymath}
\fbox{$\displaystyle
\left\{\widehat a_{{\rm{f}},n},\wid...
...f}},{\underline n}}\right\} = \delta_{n{\underline n}}
$} %
\end{displaymath} (A.63)

These relationships apply for $n$ $\raisebox{.2pt}{$\ne$}$ ${\underline n}$ exactly because of the sign change caused by swapping the order of the operators. For $n$ $\vphantom0\raisebox{1.5pt}{$=$}$ ${\underline n}$, they are unchanged from the case of just one single-particle state.

The Hamiltonian for a system of noninteracting particles is like the one for just one single-particle state, except that you must now sum over all single-particle states:

\begin{displaymath}
\fbox{$\displaystyle
H = \sum_n {\vphantom' E}^{\rm p}_n \widehat a^\dagger _n \widehat a_n + E_{{\rm ve},n}
$} %
\end{displaymath} (A.64)


A.15.9 Field operators

As noted at the start of this section, quantum field theory is particularly suited for relativistic applications because the number of particles can vary. However, in relativistic applications, it is often necessary to work in terms of position coordinates instead of single-particle energy eigenfunctions. To be sure, practical quantum field computations are usually worked out in terms of relativistic energy-momentum states. But to understand them requires consideration of position and time. Relativistic applications must make sure that coordinate systems moving at different speeds are physically equivalent and related through the Lorentz transformation. There is also the “causality problem,” that an event at one location and time may not affect an event at another location and time that is not reachable with the speed of light. These conditions are posed in terms of position and time.

To handle such problems, the annihilation and creation operators can be converted into so-called field operators $\widehat a({\underline{\skew0\vec r}})$ and $\widehat a^\dagger ({\underline{\skew0\vec r}})$ that annihilate respectively create particles at a given position ${\underline{\skew0\vec r}}$ in space. At least, roughly speaking that is what they do.

Now in classical quantum mechanics, a particle at a given position ${\underline{\skew0\vec r}}$ corresponds to a wave function that is nonzero at only that single point. And if the wave function is concentrated at the single point ${\underline{\skew0\vec r}}$, it must then be infinitely large at that point. Relaxing the normalization condition a bit, the appropriate infinitely concentrated mathematical function is called the “delta function,” $\Psi$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\delta^3({\skew0\vec r}-{\underline{\skew0\vec r}})$, chapter 7.9. Here ${\underline{\skew0\vec r}}$ is the position of the particle and ${\skew0\vec r}$ the position at which the delta function is evaluated. If ${\skew0\vec r}$ is not equal to ${\underline{\skew0\vec r}}$, the delta function is zero; but at ${\skew0\vec r}$ $\vphantom0\raisebox{1.5pt}{$=$}$ ${\underline{\skew0\vec r}}$ it is infinite. A delta function by itself integrates to 1; its square magnitude would integrate to infinity. So it is definitely not normalized.

Like any function, a delta function can be written in terms of the single-particle energy eigenfunctions $\psi_n$ as

\begin{displaymath}
\delta^3({\skew0\vec r}-{\underline{\skew0\vec r}}) = \sum_{{\rm all\ }n} c_n \psi_n({\skew0\vec r})
\end{displaymath}

Here the coefficients $c_n$ can be found by taking inner products of both sides with an arbitrary eigenfunction $\psi_{\underline n}$. That gives, noting that orthonormality of the eigenfunctions only leaves $c_{\underline n}$ in the right-hand side,

\begin{displaymath}
c_{\underline n}= \int \psi_{\underline n}^*({\skew0\vec r...
...\vec r}-{\underline{\skew0\vec r}}) {\,\rm d}^3{\skew0\vec r}
\end{displaymath}

The integral is over all space. The index ${\underline n}$ can be renotated as $n$ since the above expression applies for all possible values of ${\underline n}$. Also, an inner product with a delta function can easily be evaluated. The inner product above simply picks out the value of $\psi_n^*$ at ${\underline{\skew0\vec r}}$. So

\begin{displaymath}
c_n = \psi_n^*({\underline{\skew0\vec r}})
\end{displaymath}

After all, ${\underline{\skew0\vec r}}$ is the only position where the delta function is nonzero. So finally

\begin{displaymath}
\delta^3({\skew0\vec r}-{\underline{\skew0\vec r}}) = \sum...
...} \psi^*_n({\underline{\skew0\vec r}}) \psi_n({\skew0\vec r})
\end{displaymath}

Since $\psi^*_n({\underline{\skew0\vec r}})$ is the amount of eigenfunction $\psi_n$ that must be created to create the delta function at ${\underline{\skew0\vec r}}$, the annihilation and creation field operators should presumably be

\begin{displaymath}
\widehat a({\underline{\skew0\vec r}}) = \sum_n \psi_n({\u...
...m_n \psi^*_n({\underline{\skew0\vec r}})\widehat a^\dagger _n
\end{displaymath} (A.65)

The annihilation operator is again the Hermitian conjugate of the creation operator.

In the case of noninteracting particles in free space, the energy eigenfunctions are the momentum eigenfunctions $e^{{\rm i}{\skew0\vec p}\cdot{\skew0\vec r}/\hbar}$. The combination ${\vec k}$ $\vphantom0\raisebox{1.5pt}{$=$}$ ${\skew0\vec p}$$\raisebox{.5pt}{$/$}$$\hbar$ is commonly referred to as the “wave number vector.” Note that in infinite free space, the sums become integrals called Fourier transforms; see chapter 7.9 and 7.10.1 for more details.

To check the appropriateness of the creation field operator as defined above, consider its consistency with classical quantum mechanics. A classical wave function $\Psi$ can always be written as a combination of the energy eigenfunctions:

\begin{displaymath}
\Psi({\skew0\vec r}) = \sum_n c_n \psi_n({\skew0\vec r})
...
...\skew0\vec r}) \Psi({\skew0\vec r}) {\,\rm d}^3{\skew0\vec r}
\end{displaymath}

That is the same as for the delta function case above. However, any normal function also always satisfies

\begin{displaymath}
\Psi({\skew0\vec r}) = \int \Psi({\underline{\skew0\vec r}...
...derline{\skew0\vec r}}) {\,\rm d}^3{\underline{\skew0\vec r}}
\end{displaymath}

That is because the delta function picks out the value of $\Psi({\underline{\skew0\vec r}})$ at ${\underline{\skew0\vec r}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ ${\skew0\vec r}$ as also already noted above. You can look at the expression above as follows: $\Psi({\skew0\vec r})$ is a combination of position states $\delta({\skew0\vec r}-{\underline{\skew0\vec r}}){\rm d}^3{\underline{\skew0\vec r}}$ with coefficients $\Psi({\underline{\skew0\vec r}})$. So here the classical wave function is written as a combination of position states instead of energy states.

Now this needs to be converted to quantum field form. The classical wave function then becomes a combination $\big\vert\Psi\big\rangle $ of Fock space kets. But by definition, the creation field operator $\widehat a^\dagger ({\underline{\skew0\vec r}})$ applied on the vacuum state $\big\vert\big\rangle $ should produce the Fock space equivalent of a delta function at ${\underline{\skew0\vec r}}$. So the above classical wave function should convert to a Fock space wave function as

\begin{displaymath}
\big\vert\Psi\big\rangle = \int \Psi({\underline{\skew0\ve...
...}})\big\vert\big\rangle {\,\rm d}^3{\underline{\skew0\vec r}}
\end{displaymath}

To check that, substitute in the definition of the creation field operator:

\begin{displaymath}
\big\vert\Psi\big\rangle = \sum_n \int \psi_n^*({\underlin...
...\skew0\vec r}}\;\; \widehat a^\dagger _n\big\vert\big\rangle
\end{displaymath}

But $\widehat a^\dagger _n\big\vert\big\rangle $ is the Fock space equivalent of the classical energy eigenfunction $\psi_n$. The reason is that $\widehat a^\dagger _n$ puts exactly one particle in the state $\psi_n$. And the integral is the same coefficient $c_n$ of this energy eigenstate as in the classical case. So the creation field operator as defined does produce the correct combination of energy states.

As a check on the appropriateness of the annihilation field operator, consider the Hamiltonian. The Hamiltonian of noninteracting particles satisfies

\begin{displaymath}
H \big\vert\Psi\big\rangle = \sum_n \widehat a^\dagger _n {\vphantom' E}^{\rm p}_n \widehat a_n \big\vert\Psi\big\rangle
\end{displaymath}

Here ${\vphantom' E}^{\rm p}_n$ is the single-particle energy and $\big\vert\Psi\big\rangle $ stands for a state described by Fock space kets. The ground state energy was taken zero for simplicity. Note the critical role of the trailing $\widehat a_n$. States with no particles should not produce energy. The trailing $\widehat a_n$ ensures that they do not; it produces 0 when state $n$ has no particles.

In terms of annihilation and creation field operators, you would like the Hamiltonian to be defined similarly:

\begin{displaymath}
H \big\vert\Psi\big\rangle = \int \widehat a^\dagger ({\sk...
...w0\vec r}) \big\vert\Psi\big\rangle {\,\rm d}^3{\skew0\vec r}
\end{displaymath}

Note that the sum has become an integral, as ${\skew0\vec r}$ is a continuous variable. Also, the single-particle energy ${\vphantom' E}^{\rm p}$ has become the single-particle Hamiltonian; that is necessary because position states are not energy eigenstates with definite energy. The trailing $\widehat a({\skew0\vec r})$ ensures that positions with no particles do not contribute to the Hamiltonian.

Now, if the definitions of the field operators are right, this Hamiltonian should still produce the same answer as before. Substituting in the definitions of the field operators gives

\begin{displaymath}
H \big\vert\Psi\big\rangle = \int
\sum_{\underline n}\ps...
...idehat a_n \big\vert\Psi\big\rangle {\,\rm d}^3{\skew0\vec r}
\end{displaymath}

The single-particle Hamiltonian $H^{\rm p}$ applied on $\psi_n$ gives a factor ${\vphantom' E}^{\rm p}_n$. And orthonormality of the eigenfunctions implies that the integral is zero unless ${\underline n}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $n$. And in that case, the square energy eigenfunction magnitude integrates to 1. That then implies that the Hamiltonian is indeed the same as before.

The above argument roughly follows [42, pp. 22-29], but note that this source puts a tilde on $\widehat a^\dagger _n$ and $\widehat a_n$ as defined here. See also [34, pp. 19-24] for a somewhat different approach, with a somewhat different definition of the annihilation and creation field operators.

One final question that is much more messy is in what sense these operators really create or annihilate a particle localized at ${\underline{\skew0\vec r}}$. An answer can be given using arguments like those used for the electromagnetic magnetic field in {A.23.4}. In particular, you want to leave some uncertainty in the number of particles created at position ${\underline{\skew0\vec r}}$. Then the expectation values for the observable field do become strongly localized near position ${\underline{\skew0\vec r}}$. The details will be skipped. But qualitatively, the fact that in quantum field theory there is uncertainty in the number of particles does of course add to the uncertainty in the measured quantities.

A big advantage of the way the annihilation and creation operators were defined now shows up: the annihilation and creation field operators satisfy essentially the same (anti)commutation relations. In particular

\begin{displaymath}
\fbox{$\displaystyle
\Big[\widehat a_{\rm{b}}({\skew0\ve...
... = \delta^3({\skew0\vec r}-{\underline{\skew0\vec r}})
$} %
\end{displaymath} (A.66)


\begin{displaymath}
\fbox{$\displaystyle
\Big\{\widehat a_{\rm{f}}({\skew0\v...
... = \delta^3({\skew0\vec r}-{\underline{\skew0\vec r}})
$} %
\end{displaymath} (A.67)

In other references you might see an additional constant multiplying the three-di­men­sion­al delta function, depending on how the position and momentum eigenfunctions were normalized.

To check these commutators, plug in the definitions of the field operators. Then the zero commutators above follow immediately from the ones for $a_n$ and $\widehat a^\dagger _n$, (A.61) and (A.63). For the nonzero commutator, multiply by a completely arbitrary function $f({\skew0\vec r})$ and integrate over ${\skew0\vec r}$. That gives $f({\underline{\skew0\vec r}})$, which is the same result as obtained from integrating against $\delta^3({\skew0\vec r}-{\underline{\skew0\vec r}})$. That can only be true for every function $f$ if the commutator is the delta function. (In fact, producing $f({\underline{\skew0\vec r}})$ for any $f({\skew0\vec r})$ is exactly the way how a delta function would be defined by a conscientious mathematician.)

Field operators help solve a vexing problem for relativistic quantum mechanics: how to put space and time on equal footing, [42, p. 7ff]. Relativity unavoidably mixes up position and time. But classical quantum mechanics, as covered in this book, needs to keep them rigidly apart.

Right at the beginning, this book told you that observable quantities are the eigenvalues of Hermitian operators. That was not completely true, there is an exception. Spatial coordinates are indeed the eigenvalues of Hermitian position operators, chapter 7.9. But time is not an eigenvalue of an operator. When this book wrote a wave function as, say, $\Psi({\skew0\vec r},S_z;t)$ the time $t$ was just a label. It indicated that at any given time, you have some wave function. Then you can apply purely spatial operators like $x$, ${\widehat p}_x$, $H$, etcetera to find out things about the measurable position, momentum, energy, etcetera at that time. At a different time you have a different wave function, for which you can do the same things. Time itself is left out in the cold.

Correspondingly, the classical Schrö­din­ger equation ${\rm i}\hbar\partial\Psi$$\raisebox{.5pt}{$/$}$$\partial{t}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $H\Psi$ treats space and time quite different. The spatial derivatives, in $H$, are second order, but the time derivative is first order. The first-order time derivative describes the change from one spatial wave function to the next one, a time $\partial{t}$ later. Of course, you cannot think of the spatial derivatives in the same way. Even if there was only one spatial coordinate instead of three, the second order spatial derivatives would not represent a change of wave function from one position to the next.

The different treatment of time and space causes problems in generalizing the Schrö­din­ger equation to the relativistic case.

For spinless particles, the simplest generalization of the Schrö­din­ger equation is the Klein-Gordon equation, {A.14}. However, this equation brings in states with negative energies, including negative rest mass energies. That is a problem. For example, what prevents a particle from transitioning to states of more and more negative energy, releasing infinite amounts of energy in the process? There is no clean way to deal with such problems within the bare context of the Klein-Gordon equation.

There is also the matter of what to make of the Klein-Gordon wave function. It appears as if a wave function for a single particle is being written down, like it would be for the Schrö­din­ger equation. But for the Schrö­din­ger equation the integrated square magnitude of the wave function is 1 and stays 1. That is taken to mean that the probability of finding the particle is 1 if you look everywhere. But the Klein-Gordon equation does not preserve the integrated square magnitude of the wave function in time. That is not surprising, since in relativity particles can be created out of energy or annihilated. But if that is so, in what sense could the Klein-Gordon equation possibly describe a wave function for a single, (i.e. exactly 1), particle?

(Of course, this is not a problem for single-particle energy eigenstates. Energy eigenstates are stationary, chapter 7.1.4. It is also not a problem if there are only particle states, or only antiparticle states, {D.32}. The real problems start when you try to add perturbations to the equation.)

For fermions with spin $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$, the appropriate generalization of the Schrö­din­ger equation is the Dirac equation, chapter 12.12. However, there are still those negative-energy solutions. Dirac postulated that all, infinitely many, negative energy states in the universe are already filled with electrons. That is obviously a rather ugly assumption. Worse, it would not work for bosons. Any number of bosons can go into a single state, they cannot fill them.

Quantum field theory can put space and time on a more equal footing, especially in the Heisenberg formulation, {A.12}. This formulation pushes time from the wave function onto the operator. To see how this works, consider some arbitrary inner product involving a Schrö­din­ger operator $\widehat{A}$:

\begin{displaymath}
\big\langle\Phi\big\vert\widehat A \Psi\big\rangle
\end{displaymath}

(Why look at inner products? Simply put, if you get all inner products right, you get the quantum mechanics right. Anything in quantum mechanics can be found by taking the right inner product.) Now recall that if a wave function $\Psi$ has definite energy $E$, it varies in time as $e^{-{{\rm i}}Et/\hbar}\Psi_0$ where $\Psi_0$ is independent of time, chapter 7.1.2. If $\Psi$ does not have definite energy, you can replace $E$ in the exponential by the Hamiltonian $H$. (Exponentials of operators are defined by their Taylor series.) So the inner product becomes

\begin{displaymath}
\big\langle\Phi_0\big\vert e^{{\rm i}H t/\hbar} \widehat A e^{-{\rm i}H t/\hbar} \Psi_0\big\rangle
\end{displaymath}

(Recall that ${\rm i}$ changes sign when taken to the other side of an inner product.) The Heisenberg $\widetilde{A}$ operator absorbs the exponentials:

\begin{displaymath}
\widetilde A \equiv e^{{\rm i}H t/\hbar} \widehat A e^{-{\rm i}H t/\hbar}
\end{displaymath}

Now note that if $\widehat{A}$ is a field operator, the position coordinates in it are not Hamiltonian operators. They are labels just like time. They label what position the particle is annihilated or created at. So space and time are now treated much more equally.

Here is where the term field in “quantum field theory” comes from. In classical physics, a field is a numerical function of position. For example, a pressure field in a moving fluid has a value, the pressure, at each position. An electric field has three values, the components of the electric field, at each position. However, in quantum field theory, a field does not consist of values, but of operators. Each position has one or more operator associated with it. Each particle type is associated with a field. This field will involve both creation and annihilation operators of that particle, or the associated antiparticle, at each position.

Within the quantum field framework, equations like the Klein-Gordon and Dirac ones can be given a clear meaning. The eigenfunctions of these equations give states that particles can be in. Since energy eigenfunctions are stationary, conservation of probability is not an issue.

It may be mentioned that there is an alternate way to put space and time on an equal footing, [42, p. 10]. Instead of turning spatial coordinates into labels, time can be turned into an operator. However, clearly wave functions do evolve with time, even if different observers may disagree about the details. So what to make of the time parameter in the Schrö­din­ger equation? Relativity offers an answer. The time in the Schrö­din­ger equation can be associated with the proper time of the considered particle. That is the time measured by an observer moving along with the particle, chapter 1.2.2. The time measured by an observer in an inertial coordinate system is then promoted to an operator. All this can be done. In fact, it is the starting point of the so-called “string theory.” In string theory, a second parameter is added to proper time. You might think of the second parameter as the arc length along a string that wiggles around in time. However, approaches along these lines are extremely complicated. Quantum field theory remains the workhorse of relativistic quantum mechanics.


A.15.10 Nonrelativistic quantum field theory

This example exercise from Srednicki [42, p. 11] uses quantum field theory to describe nonrelativistic quantum mechanics. It illustrates some of the mathematics that you will encounter in quantum field theories.

The objective is to convert the classical nonrelativistic Schrö­din­ger equation for $I$ particles,

\begin{displaymath}
{\rm i}\hbar \frac{\partial \Psi}{\partial t} = H_{\rm cl} \Psi %
\end{displaymath} (A.68)

into quantum field form. The classical wave function has the positions of the numbered particles and time as arguments:
\begin{displaymath}
\mbox{classical quantum mechanics:}\quad
\Psi=\Psi({\ske...
...\skew0\vec r}_2,{\skew0\vec r}_3,\ldots,{\skew0\vec r}_I;t) %
\end{displaymath} (A.69)

where ${\skew0\vec r}_1$ is the position of particle 1, ${\skew0\vec r}_2$ is the position of particle 2, etcetera. (You could include particle spin within the vectors ${\skew0\vec r}$ if you want. But particle spin is in fact relativistic, chapter 12.12.) The classical Hamiltonian is
\begin{displaymath}
H_{\rm cl}
= \sum_{i=1}^I\left(\frac{\hbar^2}{2m}\nabla^...
...e i}}^I
V({\skew0\vec r}_i-{\skew0\vec r}_{\underline i}) %
\end{displaymath} (A.70)

The $\nabla_i^2$ term represents the kinetic energy of particle number $i$. The potential $V_{\rm {ext}}$ represents forces on the particles by external sources, while the potential $V$ represents forces between particles.

In quantum field theory, the wave function for exactly $I$ particles takes the form

\begin{displaymath}
\big\vert\Psi\big\rangle =
\int_{{\rm all\ }{\skew0\vec ...
... {\,\rm d}^3{\skew0\vec r}_1\ldots{\rm d}^3{\skew0\vec r}_I %
\end{displaymath} (A.71)

Here the ket $\big\vert\Psi\big\rangle $ in the left hand side is the wave function expressed as a Fock space ket. The ket $\big\vert\vec0\big\rangle $ to the far right is the vacuum state where there are no particles. However, the preceding creation operators then put in the particles at positions ${\skew0\vec r}_1$, ${\skew0\vec r}_2$, .... That produces a ket state with the particles at these positions.

The quantum amplitude of that ket state is the preceding $\Psi$, a function, not a ket. This is the classical nonrelativistic wave function, the one found in the nonrelativistic Schrö­din­ger equation. After all, the classical wave function is supposed to give the quantum amplitude for the particles to be at given positions. In particular, its square magnitude gives the probability for them to be at given positions.

So far, all this gives just the ket for one particular set of particle positions. But then it is integrated over all possible particle positions.

The Fock space Schrö­din­ger equation for $\big\vert\Psi\big\rangle $ takes the form

\begin{displaymath}
{\rm i}\hbar\frac{{\rm d}\big\vert\Psi\big\rangle }{{\rm d}t} = H \big\vert\Psi\big\rangle %
\end{displaymath} (A.72)

That looks just like the classical case. However, the Fock space Hamiltonian $H$ is defined by
 $\displaystyle H \big\vert\Psi\big\rangle$ $\textstyle =$ $\displaystyle \int_{{\rm all\ }{\skew0\vec r}}
\widehat a^\dagger ({\skew0\vec ...
...]
\widehat a({\skew0\vec r}) \big\vert\Psi\big\rangle
{\,\rm d}^3{\skew0\vec r}$   
     $\displaystyle +
{\textstyle\frac{1}{2}} \int_{{\rm all\ }{\skew0\vec r}}\int_{{...
...\Psi\big\rangle
{\,\rm d}^3{\skew0\vec r}{\rm d}^3{\underline{\skew0\vec r}}%
$  (A.73)

In order for this to make some sense, note that the Fock space ket $\big\vert\Psi\big\rangle $ is an object that allows you to annihilate or create a particle at any arbitrary location ${\skew0\vec r}$. That is because it is a linear combination of basis kets that allow the same thing.

The goal is now to show that the Schrö­din­ger equation (A.72) for the Fock space ket $\big\vert\Psi\big\rangle $ produces the classical Schrö­din­ger equation (A.68) for classical wave function $\Psi(\ldots)$. This needs to be shown whether it is a system of identical bosons or a system of identical fermions.

Before trying to tackle this problem, it is probably a good idea to review representations of functions using delta functions. As the simplest example, a wave function $\Psi(x)$ of just one spatial coordinate can be written as

\begin{displaymath}
\Psi(x) =
\int_{{\rm all\ }{\underline x}}
\;
\under...
...nderline x}){\rm d}{\underline x}$}}
_{{\rm basis\ states}}
\end{displaymath}

The way to think about the above integral expression for $\Psi(x)$ is just like you would think about a vector in three dimensions being written as $\vec{v}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $v_1{\hat\imath}+v_2{\hat\jmath}+v_3{\hat k}$ or a vector in 30 dimensions as $\vec{v}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\sum_{i=1}^{30}v_i{\hat\imath}_i$. The $\Psi({\underline x})$ are the coefficients, corresponding to the $v_i$-​components of the vectors. The $\delta(x-{\underline x}){\rm d}{\underline x}$ are the basis states, just like the unit vectors ${\hat\imath}_i$. If you want a graphical illustration, each $\delta(x-{\underline x}){\rm d}{\underline x}$ would correspond to one spike of unit height at a position ${\underline x}$ in figure 2.3, and you need to sum (integrate) over them all, with their coefficients, to get the total vector.

Now assume that $H_1$ is the one-di­men­sion­al classical Hamiltonian. Then $H_1\Psi(x)$ is just another function of $x$, so it can be written similarly:

\begin{eqnarray*}
H_1 \Psi(x)
& = &
\int_{{\rm all\ }{\underline x}}
H_1...
... \right]
\delta(x - {\underline x})
{\,\rm d}{\underline x}
\end{eqnarray*}

Note that the Hamiltonian acts on the coefficients, not on the basis states.

You may be surprised by this, because if you straightforwardly apply the Hamiltonian $H_1$, in terms of $x$, on the integral expression for $\Psi(x)$, you get:

\begin{displaymath}
H_1 \Psi(x) =
\int_{{\rm all\ }{\underline x}}
\Psi({\...
...delta(x - {\underline x})
\right]
{\,\rm d}{\underline x}
\end{displaymath}

Here the Hamiltonian acts on the basis states, not the coefficients.

However, the two expressions are indeed the same. Whether there is an $x$ or ${\underline x}$ in the potential does not make a difference, because the multiplying delta function is only nonzero when $x$ $\vphantom0\raisebox{1.5pt}{$=$}$ ${\underline x}$. And you can use a couple of integrations by parts to get the derivatives off the delta function and on $\Psi({\underline x})$. Note here that differentiation of the delta function with respect to $x$ or ${\underline x}$ is the same save for a sign change.

The bottom line is that you do not want to use the expression in which the Hamiltonian is applied to the basis states, because derivatives of delta functions are highly singular objects that you should not touch with a ten foot pole. (And if you have mathematical integrity, you would not really want to use delta functions either. At least not the way that they do it in physics. But in that case, you better forget about quantum field theory.)

It may here be noted that if you do have to differentiate an integral for a function $\Psi(x)$ in terms of delta functions, there is a much better way to do it. If you first make a change of integration variable to $u$ $\vphantom0\raisebox{1.5pt}{$=$}$ ${\underline x}-x$, the differentiation is no longer on the nasty delta functions.

Still, there is an important observation here: you might either know what an operator does to the coefficients, leaving the basis states untouched, or what it does to the basis states, leaving the coefficients untouched. Either one will tell you the final effect of the operator, but the mathematics is different.

Now that the general terms of engagement have been discussed, it is time to start solving Srednicki’s problem. The Fock space wave function ket can be thought of the same way as the example:

\begin{displaymath}
\big\vert\Psi\big\rangle =
\int_{{\rm all\ }{\skew0\vec ...
...3{\skew0\vec r}_I}
_{{\rm Fock\ space\ basis\ state\ kets}}
\end{displaymath}

The basis states are Fock space kets in which a particle called 1 is in a delta function at a position ${\skew0\vec r}_1$, a particle called 2 in a delta function at position ${\skew0\vec r}_2$, etcetera. The classical wave function $\Psi(\ldots)$ gives the quantum amplitude of each such ket. The integration gives $\big\vert\Psi\big\rangle $ as a combined ket.

Note that Fock states do not know about particle numbers. A Fock basis state is the same regardless what the classical wave function calls the particles. It means that the same Fock basis state ket reappears in the integration above at all swapped positions of the particles. (For fermions read: the same except possibly a sign change, since swapping the order of application of any two $\widehat a^\dagger $ creation operators flips the sign, compare subsection A.15.2.) This will become important at the end of the derivation.

The left hand side of the Fock space Schrö­din­ger equation (A.72) is evaluated by pushing the time derivative inside the above integral for $\big\vert\Psi\big\rangle $:

\begin{displaymath}
{\rm i}\hbar\frac{{\rm d}\big\vert\Psi\big\rangle }{{\rm d...
...
{\,\rm d}^3{\skew0\vec r}_1\ldots{\rm d}^3{\skew0\vec r}_I
\end{displaymath}

so the time derivative drops down on the classical wave function in the normal way.

Applying the Fock-space Hamiltonian (A.73) on the wave function is quite a different story, however. It is best to start with just a single particle:

\begin{displaymath}
H \big\vert\Psi\big\rangle =
\int_{{\rm all\ }{\skew0\ve...
...0\rangle
{\,\rm d}^3{\skew0\vec r}_1{\rm d}^3{\skew0\vec r}
\end{displaymath}

The field operator $\widehat a({\skew0\vec r})$ may be pushed past the classical wave function $\Psi(\ldots)$; $\widehat a({\skew0\vec r})$ is defined by what it does to the Fock basis states while leaving their coefficients, here $\Psi(\ldots)$, unchanged. That gives:

\begin{displaymath}
H \big\vert\Psi\big\rangle =
\int_{{\rm all\ }{\skew0\ve...
...0\rangle
{\,\rm d}^3{\skew0\vec r}_1{\rm d}^3{\skew0\vec r}
\end{displaymath}

It is now that the (anti)commutator relations become useful. The fact that for bosons $[\widehat a({\skew0\vec r})\widehat a^\dagger ({\skew0\vec r}_1)]$ or for fermions $\{\widehat a({\skew0\vec r})\widehat a^\dagger ({\skew0\vec r}_1)\}$ equals $\delta^3({\skew0\vec r}-{\skew0\vec r}_1)$ means that you can swap the order of these operators as long as you add a delta function term:

\begin{eqnarray*}
& \widehat a_{\rm {b}}({\skew0\vec r})\widehat a^\dagger _{\...
...}({\skew0\vec r})
+ \delta^3({\skew0\vec r}-{\skew0\vec r}_1)
\end{eqnarray*}

But when you swap the order of these operators, you get a factor $\widehat a({\skew0\vec r})\vert\vec0\rangle$. That is zero, because applying an annihilation operator on the vacuum state produces zero, figure A.6. So the delta function term is all that remains:

\begin{displaymath}
H \big\vert\Psi\big\rangle =
\int_{{\rm all\ }{\skew0\ve...
...0\rangle
{\,\rm d}^3{\skew0\vec r}_1{\rm d}^3{\skew0\vec r}
\end{displaymath}

Integration over ${\skew0\vec r}_1$ now picks out the value $\Psi({\skew0\vec r},t)$ from function $\Psi({\skew0\vec r}_1,t)$, as delta functions do, so

\begin{displaymath}
H \big\vert\Psi\big\rangle =
\int_{{\rm all\ }{\skew0\ve...
...w0\vec r};t)
\vert\vec 0\rangle
{\,\rm d}^3{\skew0\vec r}
\end{displaymath}

Note that the term in square brackets is the classical Hamiltonian $H_{\rm {cl}}$ for a single particle. The creation operator $\widehat a^\dagger ({\skew0\vec r})$ can be pushed over the coefficient $H_{\rm {cl}}\Psi({\skew0\vec r};t)$ of the vacuum state ket for the same reason that $\widehat a({\skew0\vec r})$ could be pushed over $\Psi({\skew0\vec r}_1;t)$; these operators do not affect the coefficients of the Fock states, just the states themselves.

Then, renotating ${\skew0\vec r}$ to ${\skew0\vec r}_1$, the grand total Fock state Schrö­din­ger equation for a system of one particle becomes

\begin{eqnarray*}
\lefteqn{\int_{{\rm all\ }{\skew0\vec r}_1}
{\rm i}\hbar\f...
...skew0\vec r}_1)\vert\vec 0\rangle
{\,\rm d}^3{\skew0\vec r}_1
\end{eqnarray*}

It is now seen that if the classical wave function $\Psi({\skew0\vec r}_1;t)$ satisfies the classical Schrö­din­ger equation, the Fock-space Schrö­din­ger equation above is also satisfied. And so is the converse: if the Fock-space equation above is satisfied, the classical wave function must satisfy the classical Schrö­din­ger equation. The reason is that Fock states can only be equal if the coefficients of all the basis states are equal, just like vectors can only be equal if all their components are equal. Here that means that the coefficient of $\widehat a^\dagger ({\skew0\vec r}_1)\vert\vec0\rangle$ must be the same at both sides, for every single value of ${\skew0\vec r}_1$.

If there is more than one particle, however, the equivalent latter conclusion is not justified. Remember that the same Fock space kets reappear in the integration at swapped positions of the particles. It now makes a difference. The following example from basic vectors illustrates the problem: yes, $a{\hat\imath}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $a'{\hat\imath}$ implies that $a$ $\vphantom0\raisebox{1.5pt}{$=$}$ $a'$, but no, $(a+b){\hat\imath}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $(a'+b'){\hat\imath}$ does not imply that $a$ $\vphantom0\raisebox{1.5pt}{$=$}$ $a'$ and $b$ $\vphantom0\raisebox{1.5pt}{$=$}$ $b'$; it merely implies that $a+b$ $\vphantom0\raisebox{1.5pt}{$=$}$ $a'+b'$. However, if additionally it is postulated that the classical wave function has the symmetry properties appropriate for bosons or fermions, then the Fock-space Schrö­din­ger equation does imply the classical one. In terms of the example from vectors, $(a+a){\hat\imath}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $(a'+a'){\hat\imath}$ does imply that $a$ $\vphantom0\raisebox{1.5pt}{$=$}$ $a'$.

In any case, the problem has been solved for a system with one particle. Doing it for $I$ particles will be left as an exercise for your mathematical skills.