The particle in a box problem

Question

I'm studying undergrad level chemistry with no strong background in physics. So the problem is a little confusing to me. A few questions for clarification:

Using the electron on a 2-dimensional box model,

1) I understand that the walls have infinite potential energy, and the particle inside the box has none. I understand that this means the particle can't leave the box. But I don't understand the physics of why? Why can't something with no potential energy go though a wall of infinite potential energy? What even is potential energy in this context?

2) There was some discussion of quantum numbers. I understand you have to use two quantum numbers when working in two dimensions. What is a quantum number? I understand the quantum numbers represented by n, l, and m(l), and what they mean, but in this example the quantum numbers were represented by n and m. It was also present in the formula:

$$\psi _{n,m}\left(x,y\right)\:=\:\frac{2}{L}\sin \left(\frac{n\pi x}{L}\right)\sin \left(\frac{m\pi y}{L}\right)$$

What determines the values of n and m? Why are they there?

Answer 1

1) Imagine the infinite potential box as a small room with infinitely tall, infinitely hard and perfectly vertical walls. It doesn't matter what you put in that room, it can never escape. In this model, the electron is in such a room, trapped by walls so strong that it could never ever escape. That's what infinite potential means.

2) Those quantum numbers are integers (1, 2, 3, etc) that tells you how many nodes the wave functions has. The electron inside a box is not localized. It is represented by a wave function $\psi$ so that the probability density of it's position is $|\psi|^2$. This wave function has a constraint: as the electron cannot be in the same place as the walls, the probability of finding an electron must be zero so $|\psi|^2 = 0$ and $\psi = 0$ right in the walls. By solving the Schrödinger equation we see that the wave function is a two dimensional sine function. But, remember, this sine function must be zero when evaluated on the walls. So the sine functions must have a node in the walls. $\psi = \frac{2}{L}\sin(\frac{\pi x}{l})\sin(\frac{\pi y}{l})$ would do, but also $\psi = \frac{2}{L}\sin(\frac{2 \pi x}{l})\sin(\frac{\pi y}{l})$ or $\psi = \frac{2}{L}\sin(\frac{3 \pi x}{l})\sin( \frac{7 \pi y}{l})$, or in general, any function $\psi = \frac{2}{L}\sin(\frac{n \pi x}{l})\sin(\frac{m \pi y}{l})$. For every pair (n,m) there will be a corresponding valid wave function.

Answer 2

1. Consider for the moment a 1D box with a FINITE potential $V_0>E$ outside the box. Solving Schrödinger equation $$-{\hbar^2\over 2m}{\partial^2\psi\over\partial x^2}+V_0\psi=E\psi$$ leads to solutions of the form $$\psi(x)=Ae^{qx}+Be^{-qx}$$ with the dispersion relation $$-{\hbar^2q^2\over 2m}+V_0=E\ \Leftrightarrow\ q={\sqrt{2m(V_0-E)}\over\hbar}$$ To get a wavefunction that can be normalized, one must set $A=0$ at the right of the box and $B=0$ at the left. In both cases, the wavefunction decays exponentially fast as $e^{-q|x|}$ when penetrating into the region of potential $V_0$. The wavefunction is roughly equal to zero after a distance $\simeq 1/q$. Now, in the case you are interested in, i.e. $V_0\rightarrow +\infty$ and therefore $q\rightarrow +\infty$, one gets a wavefunction that goes immediately to zero. This is the reason why one sets $\psi=0$ in any region where the potential is infinite.

2. I am not sure how to understand your second question. You can view the quantum numbers as a convenient way to label the different stationary states. For a particle in a 1D trap, there is a single quantum number $n=1,2,3,\ldots$. Think to the string of a guitar. $n=1$ is the fundamental frequency and $n>2$ the different harmonics. Their amplitude depends on how the string was touched. For a particle in a 2D trap, it is more convenient to label the stationary states with two quantum numbers because the Schrödinger equation can be separated into two one-dimensional Schrödinger equations. Each one will give a family of solutions labeled by a different quantum number. For the hydrogen atom, there are three quantum numbers $n,l,m$. Basically, the conditions $l<n$ and $-l\le m\le l$ come from the fact that $l$ and $m$ are related to equations for the spherical angles $\theta$ and $\phi$. It turns out that $l$ and $m$ are also the quantum numbers of the stationary states of $L^2$ and $L_z$ where $\vec L$ is the kinetic momentum. One can therefore give a physical interpretation to them.

Answer 3

1) Potential energy is the part of the energy of the system that gives rise to what we normally think of as force. The walls of a box might be more intuitive if thought as a thing that exerts a force as soon as you touch it. You might know that force is given by the negative of the gradient of the potential

$$ \mathbf{F}=-\nabla V$$

let's stick to $1$ dimension and suppose your potential is given by something like this

$$V_\epsilon(x)=\frac{\epsilon}{x^2+\epsilon^2} $$

you can play with it on a grapher tool and see that it's some sort of bell curve, and it resembles more and more an infinite wall as $\epsilon \rightarrow 0$. What's the force generated by this potential?

$$F=\frac{2\epsilon x}{(x^2+\epsilon^2)^2} $$

if you plot this, you'll notice that as $\epsilon \rightarrow 0$, it's very negative, crosses $0$ at $x=0$, and quickly becomes very positive as $x$ becomes positive. In other words, it repels anything that tries to touch it, and this effect is more localized and pronounced as the potential becomes more similar to an infinite spike. This is what we would imagine a wall to behave like.

If you imagine the potential as "a hill" that a ball must climb to roll on the other side, the picture is even clearer.

2) A complete answer would be the first half of an undergrad quantum mechanics book, but long story short, wave functions one usually considers in chemistry are energy eigenstates, i.e. solutions to the Schrödinger equation. You might now that physical quantities in QM are represented by Hermitian operators, and you might know that solving the time independent Schrödinger equation is equivalent to finding the eigenvectors and eigvenvalues of the Hamiltonian operator $H$. $$H|\psi_n\rangle =\lambda_n |\psi_n\rangle $$

Imagine now that you find another Hermitian operator, $A$, that commutes with $H$, i.e. $AH=HA$. It's a known fact of linear algebra that commuting diagonalizable operators are simultaneously diagonalizable, so there must be some basis of the Hilbert space

$$|\psi_{n,m}\rangle $$

such that

$$H|\psi_{n,m}\rangle=\lambda_n|\psi_{n,m}\rangle\qquad A|\psi_{n,m}\rangle=\mu_m|\psi_{n,m}\rangle$$

more generally if there are $k$ operators $A_1,\dots, A_k$ that commute with $H$ and each other, there is a basis akin to the one above with a set of eigenvalues for each operator.

This is what the quantum numbers label: given all the quantum numbers (and the eigenvalues they label) you can reconstruct the vector as the unique eigenvectors of all the commuting operators simultaneously.

What do these quantum numbers label? This is clearer if we shift to the Heisenberg picture: for an observable $O$ the Heisenberg equation (excluding explicit time dependence) is

$$\frac{d}{dt} O(t)=i[H,O(t)] $$

so it's clear that operators that commute with the Hamiltonian correspond to conserved quantities! In other words:

• Conserved quantities commute with the Hamiltonian, and in this way they can be perfectly defined together with the energy (the uncertainty principle does not affect them!).
• As a consequence, they can label eigenstates of the Hamiltonian (wave functions), and a particular state can be reconstructed entirely from these values.

Answer 4

Classically, a particle cannot have less energy than the potential energy it encounters.

$$p^2/2m=E-V$$

Where p is the momentum, E is the max energy and V is the potential energy.

If $E<V$, the squared magnitude of the momentum has to be negative, which is not possible for classical particles. Regions where $E<V$ is the energetically forbidden Zone.

In quantum mechanics we have some flexibility in part because of the uncertainty principle.

We can rearrange the Schrodinger Equation to get $$\nabla^2\Psi+\frac{2m}{\hbar^2}(E-V)\Psi=0$$

Solutions of the equation must not only solve it, they must also be normalizable. These restrictions do allow solutions in the case when $E<V$, but it can be shown there are still energetically forbidden regions, e.g. when $E<V_{min}$, the total energy is smaller than the minimum of the potential energy.

Quantum Tunneling allows passing through classically forbidden regions.

In those regions where the potential is infinite, all total energy values are smaller than the minimum, and so are forbidden. The solutions to the schrodinger equation can't be normalized.