Particle in a well

Part 5 in the quest for the hydrogen molecule

Last time we investigated the energy states of a free particle. That is of course a bit of a boring example: the particle does not really have any behavior except that it flies in one direction or another. This time we will derive our first results on how quantum mechanics leads to special bound states, which is a very important step towards understanding the hydrogen molecule.

Also we will have to roll up our sleeves, because we will have to start doing some tough mathematics do derive some real results. Do not be get discouraged, because mathematics is the language of the universe, and we need it to get to some tangible results now that we have established our first principles.

We will study the case of a single particle that is under the influence of a position-dependent potential \(V(x)\). The total energy of the particle is given by

$$E=\frac{p^2}{2m} + V(x).$$

We will have to choose a simple potential \(V(x)\) to work with. This is a very important choice, and for this post I would like to choose the “finite square well”,

Finite square well potential

Using the notation of Iverson brackets, the potential energy of the particle at position \(x\) is given by:

$$V(x) =-V_0[|x|<L].$$

As we did in the previous post, we will solve the energy equation to determine the energy states \(|\psi\rangle\):

$$\hat E |\psi\rangle = E |\psi\rangle.$$

We choose to express everything in terms of momentum states \(|p\rangle\). To make that more clear, we project both the left side and the right side on these momentum states, as was explained in part 3:

$$\langle p|\hat E |\psi\rangle = \langle p|E |\psi\rangle=E\phi(p),$$

where we have written the projection on the momentum states as a function which we call \(\phi(p)\equiv\langle p|\psi\rangle\).

A few mathematical tools

Before we continue, let’s develop a few formulas that will be useful. First, we should also remember from part 3 that we can always project a state on the full set of either the position or the momentum states. Let me repeat those formulas here:

$$| \psi \rangle = \int_{-\infty}^{+\infty} \langle x| \psi \rangle |x \rangle dx,\\
| \psi \rangle = \int_{-\infty}^{+\infty} \langle p| \psi \rangle |p \rangle dp.$$

Note that there is an interesting case when \(| \psi \rangle = | p’ \rangle\):

$$| p’ \rangle = \int_{-\infty}^{+\infty} \langle p| p’ \rangle |p \rangle dp,$$

and since the left hand side and right hand side have to be equal, we conclude that \(\langle p| p’ \rangle =0\) whenever \(p \neq p’\). In fact, \(\langle p| p’ \rangle \) satisfies

$$ \int_{-\infty}^{+\infty} f(p)\langle p| p’ \rangle dp=f(p’),$$

for any function \(f(p)\), which will come in handy.

Solving the energy equation

Great, now we are ready to calculate. Get ready for some serious mathematics! What I would like to calculate is for a state \(|\psi\rangle\) that satisfies the energy equation:

  • what energy \(E\) it might have, and
  • what is its projection on the momentum states \(\phi(p)\)

First I split the kinetic and potential energy terms,

$$\langle p|\hat E |\psi\rangle = \langle p|\hat E_k |\psi\rangle+\langle p|\hat E_p|\psi\rangle.$$

Kinetic energy

The first term, the kinetic energy, is relatively simple. To calculate it, I will insert a projection on momentum states \(|p’\rangle\).

\langle p|\hat E_k |\psi\rangle=\langle p|\frac{\hat p^2}{2m} |\psi\rangle=\\
\int_{-\infty}^{+\infty}\! dp’\langle p|\frac{\hat p^2}{2m} |p’ \rangle \langle p’| \psi \rangle=\\
\int_{-\infty}^{+\infty}\!dp’ \langle p|\frac{p’^2}{2m} |p’ \rangle \langle p’| \psi \rangle =\\
\int_{-\infty}^{+\infty}\!dp’ \frac{p’^2}{2m} \langle p|p’ \rangle \langle p’| \psi \rangle=\\
\frac{p^2}{2m} \langle p| \psi \rangle=\frac{p^2}{2m} \phi(p).$$

Or, to summarize, the \(\hat p\) operator is also allowed to operate on the projected-to states \(\langle p|\) to the left of it.

Potential energy

Good. Now I will work on the potential term. I will use a projection on position states \(|x\rangle\):

$$\langle p|\hat E_p|\psi\rangle=\langle p|V(\hat x)|\psi\rangle=\\
\int_{-\infty}^{+\infty} \!dx\langle p|V(\hat x)\langle x| \psi \rangle |x \rangle=\\
\int_{-\infty}^{+\infty} \!dx\langle p|V(x)\langle x| \psi \rangle |x \rangle=\\
-V_0\int_{-L}^{+L} dx\langle p|x \rangle\langle x| \psi \rangle .$$

Whoa! Now it becomes clear why the square well potential is such a convenient potential! Unfortunately, this term still contains an expression \(\langle x| \psi \rangle \), while the kinetic term was expressed in terms of \(\phi(p)\). To bring the terms together again, we add another projection here, this time on momentum states \(|p’\rangle\):

-V_0\int_{-L}^{+L} \!dx\langle p|x \rangle\langle x| \psi \rangle =\\
-V_0\int_{-\infty}^{+\infty}\!dp’\int_{-L}^{+L} \!dx\langle p|x \rangle\langle x|p’ \rangle\langle p’| \psi \rangle .$$

Now we remember from part 3 our postulate:

$$\langle x|p \rangle = e^{i x p / \hbar},$$

which actually has a twin brother:

$$\langle p|x \rangle = e^{-i x p / \hbar}.$$

Using also by definition \(\langle p’| \psi \rangle=\phi(p’)\), we manage to arrive at a quite normal mathematical equation:

\langle p|\hat E_p|\psi\rangle=-V_0\int_{-\infty}^{+\infty}\!dp’\int_{-L}^{+L} \!dx\,e^{-i x p / \hbar}e^{i x p’ / \hbar}\phi(p’) =\\
-V_0\int_{-\infty}^{+\infty}\!dp’\int_{-L}^{+L} \!dx\,e^{i x (p’-p) / \hbar}\phi(p’) .$$

The integration over \(x\) is something we can actually do!

\int_{-L}^{+L} \!dx\,e^{i x (p’-p) / \hbar} =\\
\frac{1}{i(p’-p)}e^{i x (p’-p) / \hbar} \Biggr|_{-L}^{+L}=\\
\frac{1}{i(p’-p)}\bigg[e^{i L (p’-p) / \hbar}-e^{i L (p’-p) / \hbar}\bigg]=\\
\frac{2\sin[L (p’-p) / \hbar]}{p’-p},$$

arriving at a final expression for the potential energy:

$$\langle p|\hat E_p|\psi\rangle=-V_0\int_{-\infty}^{+\infty}\!dp’\frac{2\sin[L (p’-p) / \hbar]}{p’-p}\phi(p’).$$

Final equation

In the end, the full energy equation thus becomes

$$\frac{p^2}{2m}\phi(p)-2V_0\int_{-\infty}^{+\infty}\!dp’\,\frac{\sin[L (p’-p) / \hbar]}{p’-p}\phi(p’) =E\phi(p),$$

which we have to solve for the energy value \(E\) and the function \(\phi\). How to proceed?

Numerical calculation

Unfortunately there is no analytical solution. But we can solve this equation numerically. The first thing to do is to apply some substitutions, to remove irrelevant parameters. We use.

\tilde{p} \equiv \frac{L}{\hbar}p,\\
\tilde{V} \equiv \frac{2mL^2}{\hbar^2}V_0,\\
\tilde{E} \equiv \frac{2mL^2}{\hbar^2} E,\\
\tilde{\phi}(\tilde{p})\equiv \phi(p).$$

Resulting in

$$\tilde{p}^2\tilde{\phi}(\tilde{p})-2\tilde{V}\int_{-\infty}^{+\infty}\!d\tilde{p}’\,\frac{\sin(\tilde{p}’-\tilde{p})}{\tilde{p}’-\tilde{p}}\tilde{\phi}(\tilde{p}’) =\tilde{E}\tilde{\phi}(\tilde{p}) ,$$

As we can see, the solutions to the energy value \(\tilde{E}\) only depend on a single input parameter \(\tilde{V}\). Next step to solve this numerically is to replace the integral by a summation using \(\tilde{p} = n\Delta \tilde{p}\) and \(\tilde{\phi}_n=\tilde{\phi}(n\Delta \tilde{p})\):

$$(\Delta \tilde{p})^2n^2\tilde{\phi}_n-2\tilde{V}\sum_{n’=-\infty}^{+\infty}\frac{\sin[\Delta \tilde{p}(n’-n)]}{n’-n}\tilde{\phi}_{n’} =\tilde{E}\tilde{\phi}_n,$$

which makes it look like a linear algebra problem,

$$\sum_{n’}K(n, n’)\phi_{n’}=E\phi_n,$$


$$K(n, n’)=((\Delta \tilde{p})^2n^2-2\tilde{V} \Delta \tilde{p})[n=n’]-2\tilde{V}\frac{\sin[\Delta \tilde{p}(n’-n)]}{n’-n}[n \neq n’],$$

which can be solved by Octave.


Free particle

First we look at the case \(\tilde{V}=0\). We expect to retrieve the energy spectrum of a free particle, since the potential well is not really there in that case. Indeed, we find the following spectrum:

All energy states have a positive energy. The reason that the energy spectrum is discrete rather than continuous is that we have replaced the integral by a summation (the distance between the states depends on the parameter \(\Delta \tilde{p}\)), but this does show nicely that the states get farther apart for higher energies.

First bound state

Moving on to a nonzero potential for the well, let’s pick \(\tilde{V}=0.5\). Here is the energy spectrum:

Remarkable! A single state was added with a negative energy. This is the energy of the bound state, and the most remarkable thing is that there is only one bound state. This does not align with our normal experience of the world. We think there are always many ways to put a marble in a shallow but wide cup: we can put it on the left or on the right, or in the middle, rolling with a little speed or not. What we learn is that this is in fact not true for very shallow cups: there is only one state, and the marble in the cup cannot carry any information.

This bound state has energy -2.1754. How does its projection on momentum states \(\phi(p)=\langle p| \psi \rangle\) and its projection on position states \(\phi(x)=\langle x| \phi \rangle\) look like?

As we can see, the bound state can be described as a mix of momentum states, or as a mix of position states. For the latter there is the unexpected discovery that the particle in the bound state does not actually stay within the limits of the well at \(x = \pm 1\). It is a strange world we live in!

Other bound states

Increasing the potential \(\tilde{V}\) lowers the energy of the first bound state, and at some point a second bound state appears. Here is a nice animation to illustrate this:

At \(\tilde{V}=3\) there are three bound states. For illustration, here are the projections on momentum states and on position states:


I have shown that a particle in a well has a number of discrete bound states, in contrast to the continuum of unbound states. That is an important conclusion, it is essentially where “quantum mechanics” derives its name from!

I can calculate the energy of these bound states, and also how these bound states can be written as a superposition of either momentum or position states. With respect to the momentum states, it can be noted that we have a superposition of many momentum states, and not just two of them, as one might expect. With respect to the position states, it is surprising that the particle position is not limited to within the well, although the particle in the bound state cannot escape the well.

But to me, the most remarkable thing is that we have derived all these experimentally verifiable conclusions from very few propositions. In fact we have skipped large parts of what is usually considered to be essential for understanding quantum mechanics: measurements, uncertainties and probabilities, normalization, time-evolution, the Schrödinger equation. It has been very interesting, to study this, but we are certainly not finished. In the next part, we will derive the Schrödinger equation.

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