Subsections


3.5 A Particle Confined Inside a Pipe

This section demonstrates the general procedure for analyzing quantum systems using a very elementary example. The system to be studied is that of a particle, say an electron, confined to the inside of a narrow pipe with sealed ends. This example will be studied in some detail, since if you understand it thoroughly, it becomes much easier not to get lost in the more advanced examples of quantum mechanics discussed later. And as the final subsection 3.5.9 shows, as well as much of chapter 6, the particle in a pipe is really quite interesting despite its simplicity.


3.5.1 The physical system

The system to be analyzed is shown in figure 3.4 as it would appear in classical nonquantum physics.

Figure 3.4: Classical picture of a particle in a closed pipe.
\begin{figure}
\centering
{}%
\epsffile{pipecl.eps}
\end{figure}

A particle is bouncing around between the two ends of a pipe. It is assumed that there is no friction, so the particle will keep bouncing back and forward forever. (Friction is a macroscopic effect that has no place in the sort of quantum-scale systems analyzed here.) Typically, classical physics draws the particles that it describes as little spheres, so that is what figure 3.4 shows.

The actual quantum system to be analyzed is shown in figure 3.5.

Figure 3.5: Quantum mechanics picture of a particle in a closed pipe.
\begin{figure}
\centering
{}%
\epsffile{pipeq.eps}
\end{figure}

A particle like an electron has no (known) specific shape or size, but it does have a wave function blob. So in quantum mechanics the equivalent of a particle bouncing around is a wave function blob bouncing around between the ends of the pipe.

Please do not ask what this impenetrable pipe is made off. It is obviously a crude idealization. You could imagine that the electron is a valence electron in a very tiny bar of copper. In that case the pipe walls would correspond to the surface of the copper bar, and it is assumed that the electron cannot get off the bar.

But of course, a copper bar would have nuclei, and other electrons, and the analysis here does not consider those. So maybe it is better to think of the particle as being a lone helium atom stuck inside a carbon nanotube.


Key Points
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An idealized problem of a particle bouncing about in a pipe will be considered.


3.5.2 Mathematical notations

The first step in the solution process is to describe the problem mathematically. To do so, an $x$-​coordinate that measures longitudinal position inside the pipe will be used, as shown in figure 3.6. Also,the length of the pipe will be called $\ell_x$.

Figure 3.6: Definitions for one-dimensional motion in a pipe.
\begin{figure}
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\beg...
...}}
\put(121,15){\makebox(0,0)[t]{$x=\ell_x$}}
\end{picture}
\end{figure}

To make the problem as easy to solve as possible, it will be assumed that the only position coordinate that exists is the longitudinal position $x$ along the pipe. For now, the existence of any coordinates $y$ and $z$ that measure the location in cross section will be completely ignored.


Key Points
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The only position coordinate to be considered for now is $x$.

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The notations have been defined.


3.5.3 The Hamiltonian

To analyze a quantum system you must find the Hamiltonian. The Hamiltonian is the total energy operator, equal to the sum of kinetic plus potential energy.

The potential energy $V$ is the easiest to find: since it is assumed that the particle does not experience forces inside the pipe, (until it hits the ends of the pipe, that is), the potential energy must be constant inside the pipe:

\begin{displaymath}
V = \mbox{constant}
\end{displaymath}

(The force is the derivative of the potential energy, so a constant potential energy produces zero force.) Further, since the value of the constant does not make any difference physically, you may as well assume that it is zero and save some writing:

\begin{displaymath}
V = 0
\end{displaymath}

Next, the kinetic energy operator ${\widehat T}$ is needed. You can just look up its precise form in section 3.3 and find it is:

\begin{displaymath}
{\widehat T}= - \frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2}
\end{displaymath}

Note that only the $x$ term is used here; the existence of the other two coordinates $y$ and $z$ is completely ignored. The constant $m$ is the mass of the particle, and $\hbar$ is Planck's constant.

Since the potential energy is zero, the Hamiltonian $H$ is just this kinetic energy:

\begin{displaymath}
H = - \frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2} %
\end{displaymath} (3.13)


Key Points
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The one-di­men­sion­al Hamiltonian (3.13) has been written down.


3.5.4 The Hamiltonian eigenvalue problem

With the Hamiltonian $H$ found, the next step is to formulate the Hamiltonian eigenvalue problem, (or “time-independent Schrö­din­ger equation.”). This problem is always of the form

\begin{displaymath}
H \psi = E \psi
\end{displaymath}

Any nonzero solution $\psi$ of this equation is called an energy eigenfunction and the corresponding constant $E$ is called the energy eigenvalue.

Substituting the Hamiltonian for the pipe as found in the previous subsection, the eigenvalue problem is:

\begin{displaymath}
- \frac{\hbar^2}{2m} \frac{\partial^2\psi}{\partial x^2}
= E \psi %
\end{displaymath} (3.14)

The problem is not complete yet. These problems also need so called boundary conditions, conditions that say what happens at the ends of the $x$ range. In this case, the ends of the $x$ range are the ends of the pipe. Now recall that the square magnitude of the wave function gives the probability of finding the particle. So the wave function must be zero wherever there is no possibility of finding the particle. That is outside the pipe: it is assumed that the particle is confined to the pipe. So the wave function is zero outside the pipe. And since the outside of the pipe starts at the ends of the pipe, that means that the wave function must be zero at the ends {N.5}:

\begin{displaymath}
\psi = 0 \mbox{ at $x=0$}
\qquad\mbox{and}\qquad
\psi = 0 \mbox{ at $x=\ell_x$} %
\end{displaymath} (3.15)


Key Points
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The Hamiltonian eigenvalue problem (3.14)has been found.

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It also includes the boundary conditions (3.15).


3.5.5 All solutions of the eigenvalue problem

The previous section found the Hamiltonian eigenvalue problem to be:

\begin{displaymath}
- \frac{\hbar^2}{2m} \frac{\partial^2\psi}{\partial x^2}
= E \psi
\end{displaymath}

Now you need to solve this equation. Mathematicians call an equation of this type an ordinary differential equation; differential because it has a derivative in it, and ordinary since there are no derivatives with respect to variables other than $x$.

If you do not know how to solve ordinary differential equations, it is no big deal. The best way is usually to look them up anyway. The equation above can be found in most mathematical table books, e.g. [40, item 19.7]. According to what it says there, (with changes in notation), if you assume that the energy $E$ is negative, the solution is

\begin{displaymath}
\psi = C_1 e^{\kappa x} + C_2 e^{-\kappa x}
\qquad \kappa = \frac{\sqrt{-2mE}}{\hbar}
\end{displaymath}

This solution may easily by checked by simply substituting it into the ordinary differential equation.

As far as the ordinary differential equation is concerned, the constants $C_1$ and $C_2$ could be any two numbers. But you also need to satisfy the two boundary conditions given in the previous subsection. The boundary condition that $\psi$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 when $x$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 produces, if $\psi$ is as above,

\begin{displaymath}
C_1 e^{0} + C_2 e^{0} = 0
\end{displaymath}

and since $e^0$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1, this can be used to find an expression for $C_2$:

\begin{displaymath}
C_2=-C_1
\end{displaymath}

The second boundary condition, that $\psi$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 at $x$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\ell_x$, produces

\begin{displaymath}
C_1 e^{\kappa \ell_x} + C_2 e^{-\kappa \ell_x} = 0
\end{displaymath}

or, since you just found out that $C_2$ $\vphantom0\raisebox{1.5pt}{$=$}$ $-C_1$,

\begin{displaymath}
C_1 \left(e^{\kappa \ell_x} - e^{-\kappa \ell_x}\right) = 0
\end{displaymath}

This equation spells trouble because the term between parentheses cannot be zero; the exponentials are not equal. Instead $C_1$ will have to be zero; that is bad news since it implies that $C_2$ $\vphantom0\raisebox{1.5pt}{$=$}$ $-C_1$ is zero too, and then so is the wave function $\psi$:

\begin{displaymath}
\psi = C_1 e^{\kappa x} + C_2 e^{-\kappa x} = 0
\end{displaymath}

A zero wave function is not acceptable, since there would be no possibility to find the particle anywhere!

Everything was done right. So the problem must be the initial assumption that the energy is negative. Apparently, the energy cannot be negative. This can be understood from the fact that for this particle, the energy is all kinetic energy. Classical physics would say that the kinetic energy cannot be negative because it is proportional to the square of the velocity. You now see that quantum mechanics agrees that the kinetic energy cannot be negative, but says it is because of the boundary conditions on the wave function.

Try again, but now assume that the energy $E$ is zero instead of negative. In that case the solution of the ordinary differential equation is according to [40, item 19.7]

\begin{displaymath}
\psi = C_1 + C_2 x
\end{displaymath}

The boundary condition that $\psi$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 at $x$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 now produces:

\begin{displaymath}
C_1 + C_2 0 = C_1 = 0
\end{displaymath}

so $C_1$ must be zero. The boundary condition that $\psi$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 at $x$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\ell_x$ gives:

\begin{displaymath}
0 + C_2 \ell_x = 0
\end{displaymath}

so $C_2$ must be zero too. Once again there is no nonzero solution, so the assumption that the energy $E$ can be zero must be wrong too.

Note that classically, it is perfectly OK for the energy to be zero: it would simply mean that the particle is sitting in the pipe at rest. But in quantum mechanics, zero kinetic energy is not acceptable, and it all has to do with Heisenberg's uncertainty principle. Since the particle is restricted to the inside of the pipe, its position is constrained, and so the uncertainty principle requires that the linear momentum must be uncertain. Uncertain momentum cannot be zero momentum; measurements will show a range of values for the momentum of the particle, implying that it is in motion and therefore has kinetic energy.

Try, try again. The only possibility left is that the energy $E$ is positive. In that case, the solution of the ordinary differential equation is according to [40, item 19.7]:

\begin{displaymath}
\psi = C_1 \cos(k x) + C_2 \sin(k x)
\qquad k = \frac{\sqrt{2mE}}{\hbar}
\end{displaymath}

Here the constant $k$ is called the “wave number.”

The boundary condition that $\psi$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 at $x$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 is:

\begin{displaymath}
C_1 1 + C_2 0 = C_1 = 0
\end{displaymath}

so $C_1$ must be zero. The boundary condition $\psi$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 at $x$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\ell_x$ is then:

\begin{displaymath}
0 + C_2 \sin(k\ell_x) = 0
\end{displaymath}

There finally is possibility to get a nonzero coefficient $C_2$: this equation can be satisfied if $\sin(k\ell_x)$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 instead of $C_2$. In fact, there is not just one possibility for this to happen: a sine is zero when its argument equals $\pi$, $2\pi$, $3\pi$, .... So there is a nonzero solution for each of the following values of the positive constant $k$:

\begin{displaymath}
k =\frac{\pi}{\ell_x},\;
k =\frac{2\pi}{\ell_x},\;
k =\frac{3\pi}{\ell_x},\;
\ldots
\end{displaymath}

Each of these possibilities gives one solution $\psi$. Different solutions $\psi$ will be distinguished by giving them a numeric subscript:

\begin{displaymath}
\psi_1 = C_2 \sin\left(\frac{\pi}{\ell_x} x\right),\;
\p...
...si_3 = C_2 \sin\left(\frac{3\pi}{\ell_x} x\right),\;
\ldots
\end{displaymath}

The generic solution can be written more concisely using a counter $n$ as:

\begin{displaymath}
\psi_n = C_2 \sin\left(\frac{n\pi}{\ell_x} x\right)
\quad\mbox{for } n = 1,2,3,\ldots
\end{displaymath}

Let’s check the solutions. Clearly each is zero when $x$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 and when $x$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\ell_x$. Also, substitution of each of the solutions into the ordinary differential equation

\begin{displaymath}
- \frac{\hbar^2}{2m} \frac{\partial^2\psi}{\partial x^2}
= E \psi
\end{displaymath}

shows that they all satisfy it, provided that their energy values are, respectively:

\begin{displaymath}
E_1 = \frac{\hbar^2\pi^2}{2m\ell_x^2},\;
E_2 = \frac{2^2...
...^2},\;
E_3 = \frac{3^2\hbar^2\pi^2}{2m\ell_x^2},\;
\ldots
\end{displaymath}

or generically:

\begin{displaymath}
E_n = \frac{n^2\hbar^2\pi^2}{2m\ell_x^2}
\quad\mbox{for } n = 1,2,3,\ldots
\end{displaymath}

There is one more condition that must be satisfied: each solution must be normalized so that the total probability of finding the particle integrated over all possible positions is 1 (certainty). That requires:

\begin{displaymath}
1 = \langle\psi_n\vert\psi_n\rangle =
\int_{x=0}^{\ell_x...
..._2\vert^2 \sin^2\left(\frac{n\pi}{\ell_x} x\right) {\,\rm d}x
\end{displaymath}

which after integration fixes $C_2$ (assuming you choose it to be a positive real number):

\begin{displaymath}
C_2 = \sqrt{\frac{2}{\ell_x}}
\end{displaymath}

Summarizing the results of this subsection, there is not just one energy eigenfunction and corresponding eigenvalue, but an infinite set of them:

\begin{displaymath}
\renewedcommand{arraystretch}{2.9}
\begin{array}{ll}
\...
...l_x^2}} \\
\qquad \!\vdots & \qquad \vdots
\end{array} %
\end{displaymath} (3.16)

or in generic form:
\begin{displaymath}
\psi_n = \sqrt{\frac{2}{\ell_x}} \sin\left(\frac{n\pi}{\el...
...2\pi^2}{2m\ell_x^2}
\quad\mbox{for } n = 1,2,3,4,5,\ldots %
\end{displaymath} (3.17)

The next thing will be to take a better look at these results.


Key Points
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After a lot of grinding mathematics armed with table books, the energy eigenfunctions and eigenvalues have finally been found

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There are infinitely many of them.

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They are as listed in (3.17) above. The first few are also written out explicitly in (3.16).

3.5.5 Review Questions
1.

Write down eigenfunction number 6.

Solution piped-a

2.

Write down eigenvalue number 6.

Solution piped-b


3.5.6 Discussion of the energy values

This subsection discusses the energy that the particle in the pipe can have. It was already discovered in the previous subsection that the particle cannot have negative energy, nor zero energy. In fact, according to the orthodox interpretation, the only values that the total energy of the particle can take are the energy eigenvalues

\begin{displaymath}
E_1 = \frac{\hbar^2\pi^2}{2m\ell_x^2},\;
E_2 = \frac{2^2...
...^2},\;
E_3 = \frac{3^2\hbar^2\pi^2}{2m\ell_x^2},\;
\ldots
\end{displaymath}

derived in the previous subsection.

Energy values are typically shown graphically in the form of an energy spectrum, as in figure 3.7.

Figure 3.7: One-dimensional energy spectrum for a particle in a pipe.
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...\line(1,0){50}}
\put(110,250){\line(1,0){50}}
\end{picture}
\end{figure}

Energy is plotted upwards, so the vertical height of each energy level indicates the amount of energy it has. To the right of each energy level, the solution counter, or quantum number, $n$ is listed.

Classically, the total energy of the particle can have any nonnegative value. But according to quantum mechanics, that is not true: the total energy must be one of the levels shown in the energy spectrum figure 3.7. It should be noted that for a macroscopic particle, you would not know the difference; the spacing between the energy levels is macroscopically very fine, since Planck's constant $\hbar$ is so small. However, for a quantum-scale system, the discreteness of the energy values can make a major difference.

Another point: at absolute zero temperature, the particle will be stuck in the lowest possible energy level, $E_1$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\hbar^2\pi^2$$\raisebox{.5pt}{$/$}$$2m\ell_x^2$, in the spectrum figure 3.7. This lowest possible energy level is called the ground state. Classically you would expect that at absolute zero the particle has no kinetic energy, so zero total energy. But quantum mechanics does not allow it. Heisenberg's principle requires some momentum, hence kinetic energy to remain for a confined particle even at zero temperature.


Key Points
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Energy values can be shown as an energy spectrum.

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The possible energy levels are discrete.

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But for a macroscopic particle, they are extremely close together.

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The ground state of lowest energy has nonzero kinetic energy.

3.5.6 Review Questions
1.

Plug the mass of an electron, $m$ $\vphantom0\raisebox{1.5pt}{$=$}$ 9.109,38 10$\POW9,{-31}$ kg, and the rough size of an hydrogen atom, call it $\ell_x$ $\vphantom0\raisebox{1.5pt}{$=$}$ 2 10$\POW9,{-10}$ m, into the expression for the ground state kinetic energy and see how big it is. Note that $\hbar$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1.054,57 10$\POW9,{-34}$ J s. Express in units of eV, where one eV equals 1.602,18 10$\POW9,{-19}$ J.

Solution pipee-a

2.

Just for fun, plug macroscopic values, $m$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 kg and $\ell_x$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 m, into the expression for the ground state energy and see how big it is. Note that $\hbar$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1.054,57 10$\POW9,{-34}$ J s.

Solution pipee-b

3.

What is the eigenfunction number, or quantum number, $n$ that produces a macroscopic amount of energy, 1 J, for macroscopic values $m$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 kg and $\ell_x$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 m? With that many energy levels involved, would you see the difference between successive ones?

Solution pipee-c


3.5.7 Discussion of the eigenfunctions

This subsection discusses the one-di­men­sion­al energy eigenfunctions of the particle in the pipe. The solution of subsection 3.5.5 found them to be:

\begin{displaymath}
\psi_1 = \sqrt{\frac{2}{\ell_x}} \sin\left(\frac{\pi}{\ell...
...}{\ell_x}} \sin\left(\frac{3\pi}{\ell_x} x\right),\;
\ldots
\end{displaymath}

The first one to look at is the ground state eigenfunction

\begin{displaymath}
\psi_1 = \sqrt{\frac{2}{\ell_x}} \sin\left(\frac{\pi}{\ell_x} x\right).
\end{displaymath}

It is plotted at the top of figure 3.8. As noted in section 3.1, it is the square magnitude of a wave function that gives the probability of finding the particle. So, the second graph in figure 3.8 shows the square of the ground state wave function, and the higher values of this function then give the locations where the particle is more likely to be found. This book shows regions where the particle is more likely to be found as darker regions, and in those terms the probability of finding the particle is as shown in the bottom graphic of figure 3.8.

Figure 3.8: One-dimensional ground state of a particle in a pipe.
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\begin{pic...
...t(0,0){\makebox(0,0)[b]{\epsffile{pipeo1.eps}}}
\end{picture}
\end{figure}

It is seen that in the ground state, the particle is much more likely to be found somewhere in the middle of the pipe than close to the ends.

Figure 3.9: Second and third lowest one-dimensional energy states.
\begin{figure}
\centering
{}%
\setlength{\unitlength}{1pt}
\begin{pic...
...t(0,0){\makebox(0,0)[b]{\epsffile{pipeo3.eps}}}
\end{picture}
\end{figure}

Figure 3.9 shows the two next lowest energy states

\begin{displaymath}
\psi_2 = \sqrt{\frac{2}{\ell_x}} \sin\left(\frac{2\pi}{\el...
...sqrt{\frac{2}{\ell_x}} \sin\left(\frac{3\pi}{\ell_x} x\right)
\end{displaymath}

as grey tones. Regions where the particle is relatively likely to be found alternate with ones where it is less likely to be found. And the higher the energy, the more such regions there are. Also note that in sharp contrast to the ground state, for eigenfunction $\psi_2$ there is almost no likelihood of finding the particle close to the center.

Needless to say, none of those energy states looks at all like the wave function blob bouncing around in figure 3.5. Moreover, it turns out that energy eigenstates are stationary states: the probabilities shown in figures 3.8 and 3.9 do not change with time.

In order to describe a localized wave function blob bouncing around, states of different energy must be combined. It will take until chapter 7.11.4 before the analytical tools to do so have been described. For now, the discussion must remain restricted to just finding the energy levels. And these are important enough by themselves anyway, sufficient for many practical applications of quantum mechanics.


Key Points
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In the energy eigenfunctions, the particle is not localized to within any particular small region of the pipe.

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\end{picture}$
In general there are regions where the particle may be found separated by regions in which there is little chance to find the particle.

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\end{picture}$
The higher the energy level, the more such regions there are.

3.5.7 Review Questions
1.

So how does, say, the one-di­men­sion­al eigenstate $\psi_6$ look?

Solution pipef-a

2.

Generalizing the results above, for eigenfunction $\psi_n$, any $n$, how many distinct regions are there where the particle may be found?

Solution pipef-b

3.

If you are up to a trick question, consider the following. There are no forces inside the pipe, so the particle has to keep moving until it hits an end of the pipe, then reflect backward until it hits the other side and so on. So, it has to cross the center of the pipe regularly. But in the energy eigenstate $\psi_2$, the particle has zero chance of ever being found at the center of the pipe. What gives?

Solution pipef-c


3.5.8 Three-dimensional solution

The solution for the particle stuck in a pipe that was obtained in the previous subsections cheated. It pretended that there was only one spatial coordinate $x$. Real life is three-di­men­sion­al. And yes, as a result, the solution as obtained is simply wrong.

Fortunately, it turns out that you can fix up the problem pretty easily if you assume that the pipe has a square cross section. There is a way of combining one-di­men­sion­al solutions for all three coordinates into full three-di­men­sion­al solutions. This is called the separation of variables idea: Solve each of the three variables $x$, $y$, and $z$ separately, then combine the results.

The full coordinate system for the problem is shown in figure 3.10: in addition to the $x$-​coordinate along the length of the pipe, there is also a $y$-​coordinate giving the vertical position in cross section, and similarly a $z$-​coordinate giving the position in cross section towards you.

Figure 3.10: Definition of all variables for motion in a pipe.
\begin{figure}
\centering
{}%
\setlength{\unitlength}{0.8225 pt}
\beg...
...}
\put(-195,36){\makebox(0,0)[l]{$z,\ell_z$}}
\end{picture}
\end{figure}

Now recall the one-di­men­sion­al solutions that were obtained assuming there is just an $x$-​coordinate, but add subscripts $x$ to keep them apart from any solutions for $y$ and $z$:

\begin{displaymath}
\renewedcommand{arraystretch}{2.9}
\begin{array}{ll}
\...
...hbar^2\pi^2}{2m\ell_x^2}} \\
\vdots & \vdots
\end{array}
\end{displaymath} (3.18)

or in generic form:
\begin{displaymath}
\psi_{xn_x} = \sqrt{\frac{2}{\ell_x}}\sin\left(\frac{n_x\p...
...bar^2\pi^2}{2m\ell_x^2}
\quad\mbox{for } n_x = 1,2,3,\ldots
\end{displaymath} (3.19)

Since it is assumed that the cross section of the pipe is square or rectangular of dimensions $\ell_y$ $\times$ $\ell_z$, the $y$ and $z$ directions have one-di­men­sion­al solutions completely equivalent to the $x$ direction:

\begin{displaymath}
\psi_{yn_y} = \sqrt{\frac{2}{\ell_y}}\sin\left(\frac{n_y\p...
...bar^2\pi^2}{2m\ell_y^2}
\quad\mbox{for } n_y = 1,2,3,\ldots
\end{displaymath} (3.20)

and
\begin{displaymath}
\psi_{zn_z} = \sqrt{\frac{2}{\ell_z}}\sin\left(\frac{n_z\p...
...bar^2\pi^2}{2m\ell_z^2}
\quad\mbox{for } n_z = 1,2,3,\ldots
\end{displaymath} (3.21)

After all, there is no fundamental difference between the three coordinate directions; each is along an edge of a rectangular box.

Now it turns out, {D.11}, that the full three-di­men­sion­al problem has eigenfunctions $\psi_{n_xn_yn_z}$ that are simply products of the one-di­men­sion­al ones:

\begin{displaymath}
\fbox{$\displaystyle
\psi_{n_xn_yn_z} = \sqrt{\frac{8}{\...
...y} y\right)
\sin\left(\frac{n_z\pi}{\ell_z} z\right)
$} %
\end{displaymath} (3.22)

There is one such three-di­men­sion­al eigenfunction for each set of three numbers $(n_x,n_y,n_z)$. These numbers are the three quantum numbers of the eigenfunction.

Further, the energy eigenvalues $E_{n_xn_yn_z}$ of the three-di­men­sion­al problem are the sum of those of the one-di­men­sion­al problems:

\begin{displaymath}
\fbox{$\displaystyle
E_{n_xn_yn_z} =
\frac{n_x^2\hbar^...
...}{2m\ell_y^2} +
\frac{n_z^2\hbar^2\pi^2}{2m\ell_z^2}
$} %
\end{displaymath} (3.23)

For example, the ground state of lowest energy occurs when all three quantum numbers are lowest, $n_x$ $\vphantom0\raisebox{1.5pt}{$=$}$ $n_y$ $\vphantom0\raisebox{1.5pt}{$=$}$ $n_z$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1. The three-di­men­sion­al ground state wave function is therefore:

\begin{displaymath}
\psi_{111} = \sqrt{\frac{8}{\ell_x\ell_y\ell_z}}
\sin\le...
...\pi}{\ell_y} y\right)
\sin\left(\frac{\pi}{\ell_z} z\right)
\end{displaymath} (3.24)

This ground state is shown in figure 3.11. The $y$ and $z$ factors ensure that the wave function is now zero at all the surfaces of the pipe.

Figure 3.11: True ground state of a particle in a pipe.
\begin{figure}
\centering
{}%
\setlength{\unitlength}{1pt}
\begin{pic...
...{dark}}
\put(-130,4){\makebox(0,0)[l]{light}}
\end{picture}
\end{figure}

The ground state energy is:

\begin{displaymath}
E_{111} =
\frac{\hbar^2\pi^2}{2m\ell_x^2} +
\frac{\hbar^2\pi^2}{2m\ell_y^2} +
\frac{\hbar^2\pi^2}{2m\ell_z^2}
\end{displaymath} (3.25)

Since the cross section dimensions $\ell_y$ and $\ell_z$ are small compared to the length of the pipe, the last two terms are large compared to the first one. They make the true ground state energy much larger than the one-di­men­sion­al value, which was just the first term.

Figure 3.12: True second and third lowest energy states.
\begin{figure}
\centering
{}%
\epsffile{pipet2.eps} \\ [10pt]
\epsffile{pipet3.eps}
\end{figure}

The next two lowest energy levels occur for $n_x$ $\vphantom0\raisebox{1.5pt}{$=$}$ 2, $n_y$ $\vphantom0\raisebox{1.5pt}{$=$}$ $n_z$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 respectively $n_x$ $\vphantom0\raisebox{1.5pt}{$=$}$ 3, $n_y$ $\vphantom0\raisebox{1.5pt}{$=$}$ $n_z$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1. (The latter assumes that the cross section dimensions are small enough that the alternative possibilities $n_y$ $\vphantom0\raisebox{1.5pt}{$=$}$ 2, $n_x$ $\vphantom0\raisebox{1.5pt}{$=$}$ $n_z$ $\vphantom0\raisebox{1.5pt}{$=$}$1 and $n_z$ $\vphantom0\raisebox{1.5pt}{$=$}$2, $n_x$ $\vphantom0\raisebox{1.5pt}{$=$}$ $n_y$ $\vphantom0\raisebox{1.5pt}{$=$}$1 have more energy.) The energy eigenfunctions

\begin{displaymath}
\psi_{211} = \sqrt{\frac{8}{\ell_x\ell_y\ell_z}}
\sin\le...
...\pi}{\ell_y} y\right)
\sin\left(\frac{\pi}{\ell_z} z\right)
\end{displaymath} (3.26)


\begin{displaymath}
\psi_{311} = \sqrt{\frac{8}{\ell_x\ell_y\ell_z}}
\sin\le...
...\pi}{\ell_y} y\right)
\sin\left(\frac{\pi}{\ell_z} z\right)
\end{displaymath} (3.27)

are shown in figure 3.12. They have energy levels:
\begin{displaymath}
E_{211} =
\frac{4\hbar^2\pi^2}{2m\ell_x^2} +
\frac{\hb...
...\hbar^2\pi^2}{2m\ell_y^2} +
\frac{\hbar^2\pi^2}{2m\ell_z^2}
\end{displaymath} (3.28)


Key Points
$\begin{picture}(15,5.5)(0,-3)
\put(2,0){\makebox(0,0){\scriptsize\bf0}}
\put(12...
...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}}
\end{picture}$
Three-di­men­sion­al energy eigenfunctions can be found as products of one-di­men­sion­al ones.

$\begin{picture}(15,5.5)(0,-3)
\put(2,0){\makebox(0,0){\scriptsize\bf0}}
\put(12...
...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}}
\end{picture}$
Three-di­men­sion­al energies can be found as sums of one-di­men­sion­al ones.

$\begin{picture}(15,5.5)(0,-3)
\put(2,0){\makebox(0,0){\scriptsize\bf0}}
\put(12...
...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}}
\end{picture}$
Example three-di­men­sion­al eigenstates have been shown.

3.5.8 Review Questions
1.

If the cross section dimensions $\ell_y$ and $\ell_z$ are one tenth the size of the pipe length, how much bigger are the energies $E_{y1}$ and $E_{z1}$ compared to $E_{x1}$? So, by what percentage is the one-di­men­sion­al ground state energy $E_{x1}$ as an approximation to the three-di­men­sion­al one, $E_{111}$, then in error?

Solution pipeg-a

2.

At what ratio of $\ell_y$$\raisebox{.5pt}{$/$}$$\ell_x$ does the energy $E_{121}$ become higher than the energy $E_{311}$?

Solution pipeg-b

3.

Shade the regions where the particle is likely to be found in the $\psi_{322}$ energy eigenstate.

Solution pipeg-c


3.5.9 Quantum confinement

Normally, motion in physics occurs in three dimensions. Even in a narrow pipe, in classical physics a point particle of zero size would be able to move in all three directions. But in quantum mechanics, if the pipe gets very narrow, the motion becomes truly one-di­men­sion­al.

To understand why, the first problem that must be addressed is what motion means in the first place, because normally motion is defined as change in position, and in quantum mechanics particles do not have a well-defined position.

Consider the particle in the ground state of lowest energy, shown in figure 3.11. This is one boring state; the picture never changes. You might be surprised by that; after all, it was found that the ground state has energy, and it is all kinetic energy. If the particle has kinetic energy, should not the positions where the particle is likely to be found change with time?

The answer is no; kinetic energy is not directly related to changes in likely positions of a particle; that is only an approximation valid for macroscopic systems. It is not necessarily true for quantum-scale systems, certainly not if they are in the ground state. Like it or not, in quantum mechanics kinetic energy is second-order derivatives of the wave function, and nothing else.

Next, as already pointed out, all the other energy eigenstates, like those in figure 3.12, have the same boring property of not changing with time.

Things only become somewhat interesting when you combine states of different energy. As the simplest possible example, consider the possibility that the particle has the wave function:

\begin{displaymath}
\Psi=\sqrt{{\textstyle\frac{4}{5}}} \psi_{111} + \sqrt{{\textstyle\frac{1}{5}}} \psi_{211}
\end{displaymath}

at some starting time, which will be taken as $t$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0. According to the orthodox interpretation, in an energy measurement this particle would have a ${\textstyle\frac{4}{5}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ 80% chance of being found at the ground state energy $E_{111}$ and a 20% chance of being found at the elevated energy level $E_{211}$. So there is now uncertainty in energy; that is critical.

Figure 3.13: A combination of $\psi_{111}$ and $\psi_{211}$ seen at some typical times.
 
a

b

c

d

Move your mouse over any figure to see the animation. Javascript must be enabled on your browser. Give it a few seconds for the animation to load, especially on a phone line.

In chapter 7.1 it will be found that for nonzero times, the wave function of this particle is given by

\begin{displaymath}
\Psi=\sqrt{{\textstyle\frac{4}{5}}} e^{-{\rm i}E_{111}t/\h...
...textstyle\frac{1}{5}}} e^{-{\rm i}E_{211}t/\hbar} \psi_{211}.
\end{displaymath}

Using this expression, the probability of finding the particle, $\vert\Psi\vert^2$, can be plotted for various times. That is done in figure 3.13 for four typical times. It shows that with uncertainty in energy, the wave function blob does move. It performs a periodic oscillation: after figure 3.13(d), the wave function returns to state 3.13(a), and the cycle repeats.

You would not yet want to call the particle localized, but at least the locations where the particle can be found are now bouncing back and forwards between the ends of the pipe. And if you add additional wave functions $\psi_{311}$, $\psi_{411}$, ..., you can get closer and closer to a localized wave function blob bouncing around.

But if you look closer at figure 3.13, you will note that the wave function blob does not move at all in the $y$-​direction; it remains at all times centered around the horizontal pipe centerline. It may seem that this is no big deal; just add one or more wave functions with an $n_y$ value greater than one, like $\psi_{121}$, and bingo, there will be interesting motion in the $y$-​direction too.

But there is a catch, and it has to do with the required energy. According to the previous section, the kinetic energy in the $y$-​direction takes the values

\begin{displaymath}
E_{y1}=\frac{\hbar^2\pi^2}{2m\ell_y^2} ,\;
E_{y2}=\frac{...
...2} ,\;
E_{y3}=\frac{9\hbar^2\pi^2}{2m\ell_y^2} ,\;
\ldots
\end{displaymath}

That will be very large energies for a narrow pipe in which $\ell_y$ is small. The particle will certainly have the large energy $E_{y1}$ in the $y$-​direction; if it is in the pipe at all it has at least that amount of energy. But if the pipe is really narrow, it will simply not have enough additional, say thermal, energy to get anywhere close to the next level $E_{y2}$. The kinetic energy in the $y$-​direction will therefore be stuck at the lowest possible level $E_{y1}$.

The result is that absolutely nothing interesting goes on in the $y$-​direction. As far as a particle in a narrow pipe is concerned, the $y$-​direction might just as well not exist. It is ironic that while the kinetic energy in the $y$-​direction, $E_{y1}$, is very large, nothing actually happens in that direction.

If the pipe is also narrow in the $z$-​direction, the only interesting motion is in the $x$-​direction, making the nontrivial physics truly one-di­men­sion­al. It becomes a “quantum wire”. However, if the pipe size in the $z$-​direction is relatively wide, the particle will have lots of different energy states in the $z$-​direction available too and the motion will be two-di­men­sion­al, a “quantum well”. Conversely, if the pipe is narrow in all three directions, you get a zero-di­men­sion­al “quantum dot” in which the particle does nothing unless it gets a sizable chunk of energy.

An isolated atom can be regarded as an example of a quantum dot; the electrons are confined to a small region around the nucleus and will be at a single energy level unless they are given a considerable amount of energy. But note that when people talk about quantum confinement, they are normally talking about semi-conductors, for which similar effects occur at significantly larger scales, maybe tens of times as large, making them much easier to manufacture. An actual quantum dot is often referred to as an “artificial atom”, and has similar properties as a real atom.

It may give you a rough idea of all the interesting things you can do in nanotechnology when you restrict the motion of particles, in particular of electrons, in various directions. You truly change the dimensionality of the normal three-di­men­sion­al world into a lower dimensional one. Only quantum mechanics can explain why, by making the energy levels discrete instead of continuously varying. And the lower dimensional worlds can have your choice of topology (a ring, a letter 8, a sphere, a cylinder, a Möbius strip?, ...,) to make things really exciting.


Key Points
$\begin{picture}(15,5.5)(0,-3)
\put(2,0){\makebox(0,0){\scriptsize\bf0}}
\put(12...
...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}}
\end{picture}$
Quantum mechanics allows you to create lower-di­men­sion­al worlds for particles.