Energy degeneracy in a cubical box

[docnet]

Homework Statement

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Relevant Equations

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For one-dimensional binding potential, a unique energy corresponds to a unique quantum state of the bound particle. In contrast, a particle of unique energy bound in a three-dimensional potential may be in one of several different quantum states. For example, suppose that the three-dimensional box has edges of equal lengths $a=b=c$, so that it is a cube. Then the energy states are given by
$$E=\frac{h^2}{8ma^2}(n_x^2+n_y^2+n_z^2), n_x,n_y,n_z=1,2,3,4...$$

(a) what is the lowest possible value for the energy E in the cubical box? show that there is only one quantum state corresponding to this energy (that is, only one choice of the set $n_x,n_y,n_z$).

(b) What is the second-lowest energy E? Show that this value of energy is shared by three distinct quantum states. Given the probability function $|\psi|^2$ for each of these states, could they be distinguuished from one another?

(c) let $n^2=n_x^2+n_y^2+n_z^2$ be an integer proportional to a given permitted energy. List the degeneracies for energies corresponding to $n^2=3, 6, 9, 11, 12$. Can you find a value of $n^2$ for which the energy level is sixfold degenerate?
(a) the lowest possible value for $E$ is
$$ E=\frac{h^2}{8ma^2}(1^2+1^2+1^2)=\frac{3h^2}{8ma^2}$$
where the quantum state corresponding to this energy is $(n_x,n_y,n_z)=(1,1,1)$.

(b)the second lowest energy $E$ is
$$ E=\frac{h^2}{8ma^2}(2^2+1^2+1^2)=\frac{3h^2}{4ma^2}$$
shared by three distinct quantum states $(n_x,n_y,n_z)=(2,1,1)$ or $(1,2,1)$ or $(1,1,2)$.

They cannot be distinguished from one another because $$|\psi|^2=sin^2(\frac{n_x\pi x}{a})sin^2(\frac{n_y\pi x}{b})sin^2(\frac{n_z\pi x}{c})$$ where $a=b=c$, and the three quantum states have identical probabilities.

(c)
$n^2=3\Rightarrow (1,1,1)$

$n^2=6\Rightarrow (2,1,1), (1,2,1), (1,1,2)$

$n^2=9\Rightarrow (2,2,1), (2,1,2), (1,2,2)$

$n^2=11\Rightarrow (3,1,1), (1,3,1), (1,1,3)$

$n^2=12\Rightarrow (2,2,2)$

When $n_x\neq n_y\neq n_z$, the number of degeneracies are given by the number of permutations of $n_x, n_y, n_z$.
$$3!=3\times2\times1=6$$

 

[mfb]

They cannot be distinguished from one another because $$|\psi|^2=sin^2(\frac{n_x\pi x}{a})sin^2(\frac{n_y\pi x}{b})sin^2(\frac{n_z\pi x}{c})$$ where $a=b=c$, and the three quantum states have identical probabilities.

Is that really true everywhere in the box?

For the six-fold degeneracy the problem asks for an explicit n² as answer.

 

[PeroK]

They cannot be distinguished from one another because $$|\psi|^2=sin^2(\frac{n_x\pi x}{a})sin^2(\frac{n_y\pi x}{b})sin^2(\frac{n_z\pi x}{c})$$ where $a=b=c$, and the three quantum states have identical probabilities.

This should be $|\psi(x, y, z)|^2$.

 

[mfb]

Ah good catch, didn't notice that the typo made it harder to see the difference.

 

[docnet]

Is that really true everywhere in the box?

For the six-fold degeneracy the problem asks for an explicit n² as answer.

No! I believe I made a mistake. $\psi$ is not the same everywhere in the box, and can be distinguished from one another as I work out below.##

This should be $|\psi(x, y, z)|^2$.

I realize that now. My work was careless.

Let me try this again, this time aware of my error.

(b)the second lowest energy $E$ is
$$ E=\frac{h^2}{8ma^2}(2^2+1^2+1^2)=\frac{3h^2}{4ma^2}$$
shared by three distinct quantum states $(n_x,n_y,n_z)=(2,1,1)$, $(1,2,1)$ and $(1,1,2)$.

The distinct quantum states can be distinguished from one another because the probability densities for the three states are given by
$$|\psi_1(x,y,z)|^2=sin^2(\frac{2\pi x}{a})sin^2(\frac{\pi y}{a})sin^2(\frac{2\pi z}{a})$$
$$|\psi_2(x,y,z)|^2=sin^2(\frac{\pi x}{a})sin^2(\frac{2\pi y}{a})sin^2(\frac{\pi z}{a})$$
$$|\psi_3(x,y,z)|^2=sin^2(\frac{\pi x}{a})sin^2(\frac{\pi y}{a})sin^2(\frac{2\pi z}{a})$$
The three product terms in $\psi_i(x,y,z)$ are independent in x,y, and z. So, quantum operators that contain partial derivatives with respect to x, y, and z can give different values on $\psi_i(x,y,z)$. For example, we can identify $\psi_1(x, y, z)=sin(\frac{2\pi x}{a})sin(\frac{\pi y}{a})sin(\frac{\pi z}{a})$ using the three momentum operators on $\psi_1(x, y, z)$
$$\hat{P_x}\psi(x,y,z)=-i\hbar\Big[\frac{\partial}{\partial x}sin(\frac{2\pi x}{a})sin(\frac{\pi y}{a})sin(\frac{\pi z}{a})\Big]=-i\hbar\frac{2\pi}{a}\Big[cos(\frac{2\pi x}{a})sin(\frac{\pi y}{a})sin(\frac{\pi z}{a})\Big]$$
$$\hat{P_y}\psi(x,y,z)=-i\hbar\Big[sin(\frac{2\pi x}{a})\frac{\partial}{\partial y}sin(\frac{\pi y}{a})sin(\frac{\pi z}{a})\Big]=-i\hbar\frac{\pi}{a}\Big[sin(\frac{2\pi x}{a})cos(\frac{\pi y}{a})sin(\frac{\pi z}{a})\Big]$$
$$\hat{P_z}\psi(x,y,z)=-i\hbar\Big[sin(\frac{2\pi x}{a})sin(\frac{\pi y}{a}) \frac{\partial}{\partial z}sin(\frac{\pi z}{a})\Big]=-i\hbar\frac{\pi}{a}\Big[sin(\frac{2\pi x}{a})sin(\frac{\pi y}{a}) cos(\frac{\pi z}{a})\Big]$$
We can work out the expectation values for momentum, and we will find that $\psi_1(x,y,z)$ has a two-fold larger momentum in the x direction than in y or z directions.

Likewise, the kinetic energy operators $\hat{T_j}$ operating on $\psi_i(x,y,z)=sin(\frac{2\pi x}{a})sin(\frac{\pi y}{a})sin(\frac{\pi z}{a})$ have different eigenvalues. It is worth mentioning the three quantum states correspond to a single expectation value for position because the density functions $\rho=|\psi_i(x,y,z)|^2$ are symmetrically distributed about the center of the cube. In similar terms, we find there is a single expectaion value for energy because the Hamiltonian contains partial derivatives with respect to all three components x, y, and z in a symmetric manner, and because we defined the three wavefunctions $\psi_i(x,y,z)$ to be degenerate quantum states.

edited for wording and explanations

 

[mfb]

You are far overthinking this. Consider what happens at e.g. x=a/2 in the first case and how that differs from the second and third case.

 

[docnet]

You are far overthinking this. Consider what happens at e.g. x=a/2 in the first case and how that differs from the second and third case.

I see what you mean. For $x=\frac{a}{2}$ the probability density vanishes to $0$ for all $y$ and $z$ for $\psi_1(x,y,z)$, and that is not the case for $\psi_2(x,y,z)$ or $\psi_3(x,y,z)$. However, I thought we were only allowed to use operators to test our hypothesis.

 

[PeroK]

However, I thought we were only allowed to use operators to test our hypothesis.

I don't know what that means. The wave functions are different, with different probability densities for finding the particle, so the states are distinguishable.

 

[docnet]

I don't know what that means. The wave functions are different, with different probability densities for finding the particle, so the states are distinguishable.

oh no... I spent too long on this problem.