# States with minimum energy for electrons in mag field with nonzero Lz

1. Sep 8, 2014

### vdweller

1. The problem statement, all variables and given/known data

For an electron in a uniform magnetic field, say $B\hat{z}$ with no angular momentum, the Hamiltonian can be expressed as $\hat{H}=\frac{1}{2m}\Big(\hat{p}_x^2+\frac{mω^2}{2}\hat{x}^2\Big)+\frac{1}{2m}\Big(\hat{p}_y^2+\frac{mω^2}{2}\hat{y}^2\Big)$

Which is equivalent to two separate harmonic oscillators.

Now the ground state is $Ψ=Ne^{\frac{mω}{\hbar}r^2}$

where $r=\sqrt{x^2+y^2}$ and $ω=ω_B/2=eB/2mc$

which yields energy equal to $\hbarω$

Now there are other states with nonzero angular momentum $L_z$ which yield the same energy. Those states are $Ψ_n=Nr^ne^{inφ}e^{\frac{mω}{\hbar}r^2}$

(n is not the quantum number)

The question is to prove that $\hat{H}Ψ_n=\frac{\hbarω_B}{2}Ψ_n$

since $Ψ_n$ is also an eigenstate of the Hamiltonian.

My question is, how do we prove that? I tried using the Hamiltonian with polar coordinates but I can't seem to get to the result.

Is there any other way (using the creation/annihilation operators)?

2. Sep 15, 2014

### Greg Bernhardt

I'm sorry you are not finding help at the moment. Is there any additional information you can share with us?

3. Sep 16, 2014

### nrqed

Did you try simply staying in cartesian coordinates and applying the hamiltonian?

4. Sep 16, 2014

### vdweller

Yes, it's still a mess.

By the way, the correct ground state is $Ψ=Ne^{-\frac{mω}{ℏ}r^2}$ (forgot a minus). I can't see the edit button any more...

5. Sep 16, 2014

### nrqed

Sorry, I did not notice your $\phi$ dependence on the one you must check.
Then you must use polar coordinates (you are working in two dimensions, right?).
You need $\nabla^2$ in polar coordinates. Once you reexpress the hamltonian in polar coordinates, it should be easy to check.