How to Calculate Degenerate Energy Levels in an Infinite Square Well?

[Hertz]

Homework Statement

Given \Psi(x, y, z)=(2/L)^{3/2}sin(\frac{n_x\pi x}{L})sin(\frac{n_y\pi y}{L})sin(\frac{n_z\pi z}{L}), calculate the first few energy levels and tell which are degenerate.

The Attempt at a Solution

I don't have much of an attempt to be honest.. What I've done so far is find the Energy Operator \hat{E}=i\hbar \frac{\partial}{\partial t}, however, I quickly realized this wouldn't help because my wave function is not time dependent. Next, I realized that this particle has no potential energy (this is an infinite well problem), so E=\frac{p^2}{2m}, and p=\frac{\hbar}{\lambda}. However, I also don't know how to calculate lambda and I'm not even sure that I'm headed in the right direction >.<

I've seen other formulas for energy levels of a particle. Of course, my teacher went through this in class (unfortunately the class has trouble keeping up with him). The problem is, I don't know if those formulas apply to this specific wave function, and furthermore, I should be able to do this from scratch myself, so that's my aim. Any help is appreciated.

edit-
From a few things I posted above:
E=\frac{\hbar^2}{2m\lambda^2}

Since there are multiple wavelengths, (in each of the three degrees of freedom), can I say:
E=\sum{\frac{\hbar^2}{2m\lambda^{2}_{i}}}

 

[George Jones]

E=\frac{p^2}{2m}, and p=\frac{\hbar}{\lambda}.

Do you know of any other way that p is expressed in quantum theory?

 

[Hertz]

Do you know of any other way that p is expressed in quantum theory?

The only other ways that I know of (or that I can think of) to express p involve the velocity, and I wouldn't know how to calculate the "velocity" of the particle in this example :\

edit-
Oh and the momentum operator. Is this what you're referring to?
\hat{p}=-i\hbar \Delta^{2}

edit2-
I realize my symbol is upside down. Not worth the effort right now! Haha

 

[George Jones]

In your course, you haven't seen expressed as an operator?

 

[Hertz]

In your course, you haven't seen expressed as an operator?

Yes I do remember that. Interesting. How come the momentum operator applies here but the energy operator doesn't? (I will see if I can get the solution now)

Spoiler

edit-
Using the momentum operator, I found the total energy to be:
E=C(n_x^2+n_y^2+n_z^2)\Psi^2

But as you can see, this still depends on x, y and z because it still depends on Psi. In the other energy equations I've seen it's a total energy, not just the energy at a point.. Does this mean I need to integrate this energy over the volume of the box? Is there an easier way?

edit2-
Blah! I just realized momentum is not a Laplacian, it's a gradient >.<! I've got some erasing in my near future. The question still stands though.. Is there an easier way? The gradient squared will still return a function of x, y, and z..

edit3-
Disregard previous edits. I've been experimenting with operators a lot and I've realized that I don't understand them very well!

I'm looking at the derivation on wikipedia of the momentum operator in one dimension
HERE

My first question is:
They used the plane wave solution of the Shrodinger equation to derive it. So does it apply to any wave function or just the plane wave? (Plane wave is U = 0 right?)

Second question:
I'm looking at:
\frac{\partial \Psi}{\partial x} = i \frac{p}{\hbar} \Psi
Solving for p, I get p=\frac{-i\hbar\frac{\partial \Psi}{\partial x}}{\Psi}
So it's clear from this that the momentum is:
p=\frac{\hat{p}\Psi}{\Psi}
Is this true for energy as well? If I want to find energy do I apply the energy operator and then divide by the wavefunction? Is it like this for all operators? What's the general rule?

 

[vela]

Not all states will satisfy $\hat{p}\psi = p\psi$ where $p$ is some constant. Those that do are called momentum eigenstates. They have a definite momentum, namely $p$, so it makes sense to talk about THE momentum of such a state.

Say you have $\psi(x) = e^{ikx}$. If you apply $\hat{p}$ to it, you get
$$\hat{p}\psi(x) = -i\hbar \frac{\partial}{\partial x} e^{ikx} = (\hbar k) \psi(x).$$ This result tells you that $\psi(x)$ is a momentum eigenstate, and the constant $\hbar k$ multiplying $\psi(x)$ at the end is the momentum associated with that state. On the other hand, if you had the state $\phi(x) = \sin(kx)$, applying $\hat{p}$ yields
$$\hat{p}\phi(x) = i\hbar k \cos(kx).$$ Clearly, the righthand side is not a multiple of $\phi(x)$, so $\phi(x)$ isn't a momentum eigenstate, and the state doesn't have a definite momentum.

The time-independent Schrodinger equation
$$\hat{H}\psi(x) = E\psi(x)$$ is an eigenvalue equation with the energy E as the eigenvalue. You want to come up with the appropriate operator $\hat{H}$, which is called the Hamiltonian, and apply it to the given wave function to verify that it's indeed an energy eigenstate and to identify the eigenvalue E.

 

[vanhees71]

Although there is some danger in confusing you even more, I must stress that for the infinite-square-well problem there is no momentum operator defined. This is a pretty subtle point that is often overlooked, and I think it's even wrong in some textbooks.

The point is that an operator that is representing an observable has to be not only Hermitean but also self-adjoint, i.e., if there was a momentum operator, defined to generate spatial translations, in position representation it would have the form \hat{p}=-\mathrm{i}/\hbar \vec{\nabla}. The space of wave functions is L^2([0,L]^3) with the boundary conditions \psi(0)=\psi(L)=0.

Further, if \hat{p} was self-adjoint it should have a real spectrum, but the wave functions would be
u_{\vec{p}}(\vec{x})=N \exp(\mathrm{i} \vec{x} \cdot \vec{p}/\hbar),
but for no value of \vec{p} you can fulfill the boundary conditions, i.e., there are no (generalized) eigenvectors within the Hilbert space of your problem, and thus \hat{p} cannot be a self-adjoint operatator.
This argument does not hold for the Hamilton operator of this problem, which indeed is
\hat{H}=-\hbar^2 \frac{\Delta}{2m}.
This operator is indeed selfadjoint, and a complete set of eigenfunctions is given by the functions in the OP, which fullfills both
\hat{H} u_{n_x,n_y,n_z}(\vec{x})=E(n_x,n_y,n_z) u_{n_x,n_y,n_z}(\vec{x}), \quad E(n_x,n_y,n_z) =\frac{\hbar^2 \pi^2}{2m}(n_x^2+n_y^2+n_z^2), \quad (n_x,n_y,n_z) \in \mathbb{N}^3
and the boundary conditions of the infinite-square-well. This means that \hat{H} is a valid operator to represent an observable, and since it's also positive definite, it's a good candidate for a Hamiltonian.