Subsections


4.1 The Harmonic Oscillator

This section provides an in-depth discussion of a basic quantum system. The case to be analyzed is a particle that is constrained by some kind of forces to remain at approximately the same position. This can describe systems such as an atom in a solid or in a molecule. If the forces pushing the particle back to its nominal position are proportional to the distance that the particle moves away from it, you have what is called an harmonic oscillator. Even if the forces vary nonlinearly with position, they can often still be approximated to vary linearly as long as the distances from the nominal position remain small.

The particle’s displacement from the nominal position will be indicated by $(x,y,z)$. The forces keeping the particle constrained can be modeled as springs, as sketched in figure 4.1.

Figure 4.1: Classical picture of an harmonic oscillator.
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The stiffness of the springs is characterized by the so called spring constant $c$, giving the ratio between force and displacement. Note that it will be assumed that the three spring stiffnesses are equal.

For a quantum picture of a harmonic oscillator, imagine a light atom like a carbon atom surrounded by much heavier atoms. When the carbon atom tries to move away from its nominal position, the heavy atoms push it back. The harmonic oscillator is also the basic relativistic model for the quantum electromagnetic field.

According to classical Newtonian physics, the particle vibrates back and forth around its nominal position with a frequency

\begin{displaymath}
\omega=\sqrt{\frac{c}{m}}
\end{displaymath} (4.1)

in radians per second. In quantum mechanics, a particle does not have a precise position. But the natural frequency above remains a convenient computational quantity in the quantum solution.


Key Points
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The system to be described is that of a particle held in place by forces that increase proportional to the distance that the particle moves away from its equilibrium position.

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The relation between distance and force is assumed to be the same in all three coordinate directions.

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Number $c$ is a measure of the strength of the forces and $\omega$ is the frequency of vibration according to classical physics.


4.1.1 The Hamiltonian

In order to find the energy levels that the oscillating particle can have, you must first write down the total energy Hamiltonian.

As far as the potential energy is concerned, the spring in the $x$-​direction holds an amount of potential energy equal to $\frac12cx^2$, and similarly the ones in the $y$ and $z$ directions.

To this total potential energy, you need to add the kinetic energy operator ${\widehat T}$ from section 3.3 to get the Hamiltonian:

\begin{displaymath}
H =
- \frac{\hbar^2}{2m}
\left(
\frac{\partial^2}{\p...
...ight)
+ {\textstyle\frac{1}{2}}c \left(x^2+y^2+z^2\right) %
\end{displaymath} (4.2)


Key Points
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The Hamiltonian (4.2) has been found.


4.1.2 Solution using separation of variables

This section finds the energy eigenfunctions and eigenvalues of the harmonic oscillator using the Hamiltonian as found in the previous subsection. Every energy eigenfunction $\psi$ and its eigenvalue $E$ must satisfy the Hamiltonian eigenvalue problem, (or time-independent Schrö­din­ger equation):

\begin{displaymath}
\left[- \frac{\hbar^2}{2m}
\left(
\frac{\partial^2}{\p...
...frac{1}{2}}c \left(x^2+y^2+z^2\right)
\right] \psi = E \psi
\end{displaymath} (4.3)

The boundary condition is that $\psi$ becomes zero at large distance from the nominal position. After all, the magnitude of $\psi$ tells you the relative probability of finding the particle at that position, and because of the rapidly increasing potential energy, the chances of finding the particle very far from the nominal position should be vanishingly small.

Like for the particle in the pipe of the previous section, it will be assumed that each eigenfunction is a product of one-di­men­sion­al eigenfunctions, one in each direction:

\begin{displaymath}
\psi = \psi_x(x) \psi_y(y) \psi_z(z)
\end{displaymath} (4.4)

Finding the eigenfunctions and eigenvalues by making such an assumption is known in mathematics as the “method of separation of variables”.

Substituting the assumption in the eigenvalue problem above, and dividing everything by $\psi_x(x)\psi_y(y)\psi_z(z)$ reveals that E consists of three parts that will be called $E_x$, $E_y$, and $E_z$:

\begin{displaymath}
\begin{array}{l}
\displaystyle \rule[-1ex]{0pt}{2ex}
E...
...'(z)}{\psi_z(z)} + {\textstyle\frac{1}{2}}c z^2
\end{array}
\end{displaymath} (4.5)

where the primes indicate derivatives. The three parts represent the $x$, $y$, and $z$ dependent terms.

By the definition above, the quantity $E_x$ can only depend on $x$; variables $y$ and $z$ do not appear in its definition. But actually, $E_x$ cannot depend on $x$ either, since $E_x$ $\vphantom0\raisebox{1.5pt}{$=$}$ $E-E_y-E_z$, and none of those quantities depends on $x$. The inescapable conclusion is that $E_x$ must be a constant, independent of all three variables $(x,y,z)$. The same way $E_y$ and $E_z$ must be constants.

If now in the definition of $E_x$ above, both sides are multiplied by $\psi_x(x)$, a one-di­men­sion­al eigenvalue problem results:

\begin{displaymath}
\left[
- \frac{\hbar^2}{2m} \frac{\partial^2}{\partial x...
...{\textstyle\frac{1}{2}} c x^2
\right]
\psi_x = E_x \psi_x
\end{displaymath} (4.6)

The operator within the square brackets here, call it $H_x$, involves only the $x$-​related terms in the full Hamiltonian. Similar problems can be written down for $E_y$ and $E_z$. Separate problems in each of the three variables $x$, $y$, and $z$ have been obtained, explaining why this mathematical method is called separation of variables.

Solving the one-di­men­sion­al problem for $\psi_x$ can be done by fairly elementary but elaborate means. If you are interested, you can find how it is done in derivation {D.12}, but that is mathematics and it will not teach you much about quantum mechanics. It turns out that, like for the particle in the pipe of the previous section, there is again an infinite number of different solutions for $E_x$ and $\psi_x$:

\begin{displaymath}
\begin{array}{ll}
\strut^{\strut} E_{x0} = \frac12 \hbar...
...& \psi_{x2}(x)=h_2(x) \\
\;\vdots & \;\vdots
\end{array}
\end{displaymath} (4.7)

Unlike for the particle in the pipe, here by convention the solutions are numbered starting from 0, rather than from 1. So the first eigenvalue is $E_{x0}$ and the first eigenfunction $\psi_{x0}$. That is just how people choose to do it.

Also, the eigenfunctions are not sines like for the particle in the pipe; instead, as table 4.1 shows, they take the form of some polynomial times an exponential. But you will probably really not care much about what kind of functions they are anyway unless you end up writing a textbook on quantum mechanics and have to plot them. In that case, you can find a general expression, (D.4), in derivation {D.12}.


Table 4.1: First few one-dimensional eigenfunctions of the harmonic oscillator.
\begin{table}\begin{displaymath}
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\begin{a...
...ray}
\\ [1 in]\hline\hline
\end{array}
\end{displaymath}
\end{table}


But the eigenvalues are what you want to remember from this solution. According to the orthodox interpretation, these are the measurable values of the total energy in the $x$-​direction (potential energy in the $x$-​direction spring plus kinetic energy of the motion in the $x$-​direction.) Instead of writing them all out as was done above, they can be described using the generic expression:

\begin{displaymath}
E_{xn_x} = \frac{2n_x+1}2 \hbar \omega
\quad \mbox{for } n_x = 0, 1, 2, 3, \ldots
\end{displaymath} (4.8)

The eigenvalue problem has now been solved, because the equations for $Y$ and $Z$ are mathematically the same and must therefore have corresponding solutions:

\begin{displaymath}
E_{yn_y} = \frac{2n_y+1}2 \hbar \omega
\quad \mbox{for } n_y = 0, 1, 2, 3, \ldots
\end{displaymath} (4.9)


\begin{displaymath}
E_{zn_z} = \frac{2n_z+1}2 \hbar \omega
\quad \mbox{for } n_z = 0, 1, 2, 3, \ldots
\end{displaymath} (4.10)

The total energy $E$ of the complete system is the sum of $E_x$, $E_y$, and $E_z$. Any nonnegative choice for number $n_x$, combined with any nonnegative choice for number $n_y$, and for $n_z$, produces one combined total energy value $E_{xn_x}+E_{yn_y}+E_{zn_z}$, which will be indicated by $E_{n_xn_yn_z}$. Putting in the expressions for the three partial energies above, these total energy eigenvalues become:

\begin{displaymath}
E_{n_xn_yn_z} = \frac{2n_x+2n_y+2n_z+3}2\; \hbar \omega %
\end{displaymath} (4.11)

where the quantum numbers $n_x$, $n_y$, and $n_z$ may each have any value in the range 0, 1, 2, 3, ...

The corresponding eigenfunction of the complete system is:

\begin{displaymath}
\psi_{n_xn_yn_z}=h_{n_x}(x) h_{n_y}(y) h_{n_z}(z) %
\end{displaymath} (4.12)

where the functions $h_0$, $h_1$, ...are in table 4.1 or in (D.4) if you need them.

Note that the $n_x,n_y,n_z$ numbering system for the solutions arose naturally from the solution process; it was not imposed a priori.


Key Points
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The eigenvalues and eigenfunctions have been found, skipping a lot of tedious math that you can check when the weather is bad during spring break.

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Generic expressions for the eigenvalues are above in (4.11) and for the eigenfunctions in (4.12).

4.1.2 Review Questions
1.

Write out the ground state energy.

Solution harmb-a

2.

Write out the ground state wave function fully.

Solution harmb-b

3.

Write out the energy $E_{100}$.

Solution harmb-c

4.

Write out the eigenstate $\psi_{100}$ fully.

Solution harmb-d


4.1.3 Discussion of the eigenvalues

As the previous subsection showed, for every set of three nonnegative whole numbers $n_x,n_y,n_z$, there is one unique energy eigenfunction, or eigenstate, (4.12) and a corresponding energy eigenvalue (4.11). The “quantum numbers” $n_x$, $n_y$, and $n_z$ correspond to the numbering system of the one-di­men­sion­al solutions that make up the full solution.

This section will examine the energy eigenvalues. These are of great physical importance, because according to the orthodox interpretation, they are the only measurable values of the total energy, the only energy levels that the oscillator can ever be found at.

The energy levels can be plotted in the form of a so-called “energy spectrum”, as in figure 4.2. The energy values are listed along the vertical axis, and the sets of quantum numbers $n_x,n_y,n_z$ for which they occur are shown to the right of the plot.

Figure 4.2: The energy spectrum of the harmonic oscillator.
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The first point of interest illustrated by the energy spectrum is that the energy of the oscillating particle cannot take on any arbitrary value, but only certain discrete values. Of course, that is just like for the particle in the pipe of the previous section, but for the harmonic oscillator, the energy levels are evenly spaced. In particular the energy value is always an odd multiple of $\frac12\hbar\omega$. It contradicts the Newtonian notion that a harmonic oscillator can have any energy level. But since $\hbar$ is so small, about 10$\POW9,{-34}$ kg m$\POW9,{2}$/s, macroscopically the different energy levels are extremely close together. Though the old Newtonian theory is strictly speaking incorrect, it remains an excellent approximation for macroscopic oscillators.

Also note that the energy levels have no largest value; however high the energy of the particle in a true harmonic oscillator may be, it will never escape. The further it tries to go, the larger the forces that pull it back. It can’t win.

Another striking feature of the energy spectrum is that the lowest possible energy is again nonzero. The lowest energy occurs for $n_x$ $\vphantom0\raisebox{1.5pt}{$=$}$ $n_y$ $\vphantom0\raisebox{1.5pt}{$=$}$ $n_z$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 and has a value:

\begin{displaymath}
E_{000} = {\textstyle\frac{3}{2}} \hbar \omega
\end{displaymath} (4.13)

So, even at absolute zero temperature, the particle is not completely at rest at its nominal position; it still has $\frac32\hbar\omega$ worth of kinetic and potential energy left that it can never get rid of. This lowest energy state is the ground state.

The reason that the energy cannot be zero can be understood from the uncertainty principle. To get the potential energy to be zero, the particle would have to be at its nominal position for certain. But the uncertainty principle does not allow a precise position. Also, to get the kinetic energy to be zero, the linear momentum would have to be zero for certain, and the uncertainty principle does not allow that either.

The actual ground state is a compromise between uncertainties in momentum and position that make the total energy as small as Heisenberg's relationship allows. There is enough uncertainty in momentum to keep the particle near the nominal position, minimizing potential energy, but there is still enough uncertainty in position to keep the momentum low, minimizing kinetic energy. In fact, the compromise results in potential and kinetic energies that are exactly equal, {D.13}.

For energy levels above the ground state, figure 4.2 shows that there is a rapidly increasing number of different sets of quantum numbers $n_x$, $n_y$, and $n_z$ that all produce that energy. Since each set represents one eigenstate, it means that multiple states produce the same energy.


Key Points
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Energy values can be graphically represented as an energy spectrum.

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The energy values of the harmonic oscillator are equally spaced, with a constant energy difference of $\hbar\omega$ between successive levels.

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The ground state of lowest energy has nonzero kinetic and potential energy.

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For any energy level above the ground state, there is more than one eigenstate that produces that energy.

4.1.3 Review Questions
1.

Verify that the sets of quantum numbers shown in the spectrum figure 4.2 do indeed produce the indicated energy levels.

Solution harmc-a

2.

Verify that there are no sets of quantum numbers missing in the spectrum figure 4.2; the listed ones are the only ones that produce those energy levels.

Solution harmc-b


4.1.4 Discussion of the eigenfunctions

This section takes a look at the energy eigenfunctions of the harmonic oscillator to see what can be said about the position of the particle at various energy levels.

At absolute zero temperature, the particle will be in the ground state of lowest energy. The eigenfunction describing this state has the lowest possible numbering $n_x$ $\vphantom0\raisebox{1.5pt}{$=$}$ $n_y$ $\vphantom0\raisebox{1.5pt}{$=$}$ $n_z$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0, and is according to (4.12) of subsection 4.1.2 equal to

\begin{displaymath}
\psi_{000}=h_0(x) h_0(y) h_0(z)
\end{displaymath} (4.14)

where function $h_0$ is in table 4.1. The wave function in the ground state must be equal to the eigenfunction to within a constant:
\begin{displaymath}
\Psi_{\mbox{gs}}=c_{000} h_0(x) h_0(y) h_0(z)
\end{displaymath} (4.15)

where the magnitude of the constant $c_{000}$ must be one. Using the expression for function $h_0$ from table 4.1, the properties of the ground state can be explored.

As noted earlier in section 3.1, it is useful to plot the square magnitude of $\Psi$ as grey tones, because the darker regions will be the ones where the particle is more likely to be found. Such a plot for the ground state is shown in figure 4.3. It shows that in the ground state, the particle is most likely to be found near the nominal position, and that the probability of finding the particle falls off quickly to zero beyond a certain distance from the nominal position.

Figure 4.3: Ground state of the harmonic oscillator
\begin{figure}
\centering
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\epsffile{harm000.eps}
\end{figure}

The region in which the particle is likely to be found extends, roughly speaking, about a distance $\ell$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\sqrt{\hbar/m\omega}$ from the nominal position. For a macroscopic oscillator, this will be a very small distance because of the smallness of $\hbar$. That is somewhat comforting, because macroscopically, you would expect an oscillator to be able to be at rest at the nominal position. While quantum mechanics does not allow it, at least the distance $\ell$ from the nominal position, and the energy $\frac32\hbar\omega$ are extremely small.

But obviously, the bad news is that the ground state probability density of figure 4.3 does not at all resemble the classical Newtonian picture of a localized particle oscillating back and forwards. In fact, the probability density does not even depend on time: the chances of finding the particle in any given location are the same for all times. The probability density is also spherically symmetric; it only depends on the distance from the nominal position, and is the same at all angular orientations. To get something that can start to resemble a Newtonian spring-mass oscillator, one requirement is that the energy is well above the ground level.

Turning now to the second lowest energy level, this energy level is achieved by three different energy eigenfunctions, $\psi_{100}$, $\psi_{010}$, and $\psi_{001}$. The probability distribution of each of the three takes the form of two separate blobs; figure 4.4 shows $\psi_{100}$ and $\psi_{010}$ when seen along the $z$-​direction. In case of $\psi_{001}$, one blob hides the other, so this eigenfunction was not shown.

Figure 4.4: Wave functions $\psi_{100}$ and $\psi_{010}$.
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Obviously, these states too do not resemble a Newtonian oscillator at all. The probability distributions once again stay the same at all times. (This is a consequence of energy conservation, as discussed later in chapter 7.1.) Also, while in each case there are two blobs occupied by a single particle, the particle will never be be caught on the symmetry plane in between the blobs, which naively could be taken as a sign of the particle moving from one blob to the other.

The eigenfunctions for still higher energy levels show similar lack of resemblance to the classical motion. As an arbitrary example, figure 4.5 shows eigenfunction $\psi_{213}$ when looking along the $z$-​axis. To resemble a classical oscillator, the particle would need to be restricted to, maybe not an exact moving point, but at most a very small moving region. Instead, all energy eigenfunctions have steady probability distributions and the locations where the particle may be found extend over large regions. It turns out that there is an uncertainty principle involved here: in order to get some localization of the position of the particle, you need to allow some uncertainty in its energy. This will have to wait until much later, in chapter 7.11.4.

Figure 4.5: Energy eigenfunction $\psi_{213}$.
\begin{figure}
\centering
{}%
\epsffile{harm213.eps}
\end{figure}

The basic reason that quantum mechanics is so slow is simple. To analyze, say the $x$ motion, classical physics says: “the value of the total energy $E_x$ is

\begin{displaymath}
E_x = {\textstyle\frac{1}{2}} m \dot x^2 + {\textstyle\frac{1}{2}} c x^2,
\end{displaymath}

now go analyze the motion!”. Quantum mechanics says: “the total energy operator $H_x$ is

\begin{displaymath}
H_x = {\textstyle\frac{1}{2}} m
\left(\frac{\hbar}{{\rm ...
...ial x}\right)^2
+ {\textstyle\frac{1}{2}} c {\widehat x}^2,
\end{displaymath}

now first figure out the possible energy values $E_{x0},E_{x1},\ldots$ before you can even start thinking about analyzing the motion.”


Key Points
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The ground state wave function is spherically symmetric: it looks the same seen from any angle.

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In energy eigenstates the particle position is uncertain.

4.1.4 Review Questions
1.

Write out the ground state wave function and show that it is indeed spherically symmetric.

Solution harmd-a

2.

Show that the ground state wave function is maximal at the origin and, like all the other energy eigenfunctions, becomes zero at large distances from the origin.

Solution harmd-b

3.

Write down the explicit expression for the eigenstate $\psi_{213}$ using table 4.1, then verify that it looks like figure 4.5 when looking along the $z$-​axis, with the $x$-​axis horizontal and the $y$-​axis vertical.

Solution harmd-c


4.1.5 Degeneracy

As the energy spectrum figure 4.2 illustrated, the only energy level for which there is only a single energy eigenfunction is the ground state. All higher energy levels are what is called degenerate; there is more than one eigenfunction that produces that energy. (In other words, more than one set of three quantum numbers $n_x$, $n_y$, and $n_z$.)

It turns out that degeneracy always results in nonuniqueness of the eigenfunctions. That is important for a variety of reasons. For example, in the quantum mechanics of molecules, chemical bonds often select among nonunique theoretical solutions those that best fit the given conditions. Also, to find specific mathematical or numerical solutions for the eigenfunctions of a quantum system, the nonuniquenesses will somehow have to be resolved.

Nonuniqueness also poses problems for advanced analysis. For example, suppose you try to analyze the effect of various small perturbations that a harmonic oscillator might experience in real life. Analyzing the effect of small perturbations is typically a relatively easy mathematical problem: the perturbation will slightly change an eigenfunction, but it can still be approximated by the unperturbed one. So, if you know the unperturbed eigenfunction you are in business; unfortunately, if the unperturbed eigenfunction is not unique, you may not know which is the right one to use in the analysis.

The nonuniqueness arises from the fact that:

Linear combinations of eigenfunctions at the same energy level produce alternative eigenfunctions that still have that same energy level.

For example, the eigenfunctions $\psi_{100}$, and $\psi_{010}$ of the harmonic oscillator have the same energy $E_{100}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $E_{010}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\frac52\hbar\omega$ (as does $\psi_{001}$, but this example will be restricted to two eigenfunctions.) Any linear combination of the two has that energy too, so you could replace eigenfunctions $\psi_{100}$ and $\psi_{010}$ by two alternative ones such as:

\begin{displaymath}
\frac{\psi_{100}+\psi_{010}}{\sqrt2}\quad \mbox {and}\quad
\frac{\psi_{010}-\psi_{100}}{\sqrt2}
\end{displaymath}

It is readily verified these linear combinations are indeed still eigenfunctions with eigenvalue $E_{100}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $E_{010}$: applying the Hamiltonian $H$ to either one will multiply each term by $E_{100}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $E_{010}$, hence the entire combination by that amount. How do these alternative eigenfunctions look? Exactly like $\psi_{100}$ and $\psi_{010}$ in figure 4.4, except that they are rotated over 45 degrees. Clearly then, they are just as good as the originals, just seen under a different angle.

Which raises the question, how come the analysis ended up with the ones that it did in the first place? The answer is in the method of separation of variables that was used in subsection 4.1.2. It produced eigenfunctions of the form $h_{n_x}(x)h_{n_y}(y)h_{n_z}(z)$ that were not just eigenfunctions of the full Hamiltonian $H$, but also of the partial Hamiltonians $H_x$, $H_y$, and $H_z$, being the $x$, $y$, and $z$ parts of it.

For example, $\psi_{100}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $h_1(x)h_0(y)h_0(z)$ is an eigenfunction of $H_x$ with eigenvalue $E_{x1}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\frac32\hbar\omega$, of $H_y$ with eigenvalue $E_{y0}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\frac12\hbar\omega$, and of $H_z$ with eigenvalue $E_{z0}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\frac12\hbar\omega$, as well as of $H$ with eigenvalue $E_{100}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\frac52\hbar\omega$.

The alternative eigenfunctions are still eigenfunctions of $H$, but no longer of the partial Hamiltonians. For example,

\begin{displaymath}
\frac{\psi_{100}+\psi_{010}}{\sqrt2}
= \frac{h_1(x)h_0(y)h_0(z)+h_0(x)h_1(y)h_0(z)}{\sqrt2}
\end{displaymath}

is not an eigenfunction of $H_x$: taking $H_x$ times this eigenfunction would multiply the first term by $E_{x1}$ but the second term by $E_{x0}$.

So, the obtained eigenfunctions were really made determinate by ensuring that they are simultaneously eigenfunctions of $H$, $H_x$, $H_y$, and $H_z$. The nice thing about them is that they can answer questions not just about the total energy of the oscillator, but also about how much of that energy is in each of the three directions.


Key Points
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Degeneracy occurs when different eigenfunctions produce the same energy.

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It causes nonuniqueness: alternative eigenfunctions will exist.

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That can make various analysis a lot more complex.

4.1.5 Review Questions
1.

Just to check that this book is not lying, (you cannot be too careful), write down the analytical expression for $\psi_{100}$ and $\psi_{010}$ using table 4.1. Next write down $\left(\psi_{100}+\psi_{010}\right)$$\raisebox{.5pt}{$/$}$$\sqrt 2$ and $\left(\psi_{010}-\psi_{100}\right)$$\raisebox{.5pt}{$/$}$$\sqrt 2$. Verify that the latter two are the functions $\psi_{100}$ and $\psi_{010}$ in a coordinate system $(\bar{x},\bar{y},z)$ that is rotated 45 degrees counter-clockwise around the $z$-​axis compared to the original $(x,y,z)$ coordinate system.

Solution harme-a


4.1.6 Noneigenstates

It should not be thought that the harmonic oscillator only exists in energy eigenstates. The opposite is more like it. Anything that somewhat localizes the particle will produce an uncertainty in energy. This section explores the procedures to deal with states that are not energy eigenstates.

First, even if the wave function is not an energy eigenfunction, it can still always be written as a combination of the eigenfunctions:

\begin{displaymath}
\Psi(x,y,z,t) = \sum_{n_x=0}^\infty\sum_{n_y=0}^\infty\sum_{n_z=0}^\infty
c_{n_xn_yn_z} \psi_{n_xn_yn_z}
\end{displaymath} (4.16)

That this is always possible is a consequence of the completeness of the eigenfunctions of Hermitian operators such as the Hamiltonian. An arbitrary example of such a combination state is shown in figure 4.6.

Figure 4.6: Arbitrary wave function (not an energy eigenfunction).
\begin{figure}
\centering
{}%
\epsffile{harmsum.eps}
\end{figure}

The coefficients $c_{n_xn_yn_z}$ in the combination are important: according to the orthodox statistical interpretation, their square magnitude gives the probability to find the energy to be the corresponding eigenvalue $E_{n_xn_yn_z}$. For example, $\vert c_{000}\vert^2$ gives the probability of finding that the oscillator is in the ground state of lowest energy.

If the wave function $\Psi$ is in a known state, (maybe because the position of the particle was fairly accurately measured), then each coefficient $c_{n_xn_yn_z}$ can be found by computing an inner product:

\begin{displaymath}
c_{n_xn_yn_z} = \langle \psi_{n_xn_yn_z} \vert \Psi \rangle
\end{displaymath} (4.17)

The reason this works is orthonormality of the eigenfunctions. As an example, consider the case of coefficient $c_{100}$:

\begin{displaymath}
c_{100} =
\langle \psi_{100} \vert \Psi \rangle =
\lan...
... c_{001} \psi_{001} +
c_{200} \psi_{200} + \ldots
\rangle
\end{displaymath}

Now proper eigenfunctions of Hermitian operators are orthonormal; the inner product between different eigenfunctions is zero, and between identical eigenfunctions is one:

\begin{displaymath}
\langle \psi_{100}\vert\psi_{000}\rangle = 0 \quad
\lang...
...
\langle \psi_{100}\vert\psi_{001}\rangle = 0 \quad
\ldots
\end{displaymath}

So, the inner product above must indeed produce $c_{100}$.

Chapter 7.1 will discuss another reason why the coefficients are important: they determine the time evolution of the wave function. It may be recalled that the Hamiltonian, and hence the eigenfunctions derived from it, did not involve time. However, the coefficients do.

Even if the wave function is initially in a state involving many eigenfunctions, such as the one in figure 4.6, the orthodox interpretation says that energy measurement will collapse it into a single eigenfunction. For example, assume that the energies in all three coordinate directions are measured and that they return the values:

\begin{displaymath}
E_{x2}={\textstyle\frac{5}{2}}\hbar\omega \quad
E_{y1}={...
...\hbar\omega \quad
E_{z3}={\textstyle\frac{7}{2}}\hbar\omega
\end{displaymath}

for a total energy $E$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\frac{15}2\hbar\omega$. Quantum mechanics could not exactly predict that this was going to happen, but it did predict that the energies had to be odd multiples of $\frac12\hbar\omega$. Also, quantum mechanics gave the probability of measuring the given values to be whatever $\vert c_{213}\vert^2$ was. Or in other words, what $\vert\langle\psi_{213}\vert\Psi\rangle\vert^2$ was.

After the example measurement, the predictions become much more specific, because the wave function is now collapsed into the measured one:

\begin{displaymath}
\Psi^{\mbox{\scriptsize new}} = c^{\mbox{\scriptsize new}}_{213} \psi_{213}
\end{displaymath}

This eigenfunction was shown earlier in figure 4.5.

If another measurement of the energies is now done, the only values that can come out are $E_{x2}$, $E_{y1}$, and $E_{z3}$, the same as in the first measurement. There is now certainty of getting those values; the probability $\vert c^{\mbox{\scriptsize {new}}}_{213}\vert^2$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1. This will continue to be true for energy measurements until the system is disturbed, maybe by a position measurement.


Key Points
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The basic ideas of quantum mechanics were illustrated using an example.

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The energy eigenfunctions are not the only game in town. Their seemingly lowly coefficients are important too.

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When the wave function is known, the coefficient of any eigenfunction can be found by taking an inner product of the wave function with that eigenfunction.