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

[vdweller]

Homework Statement

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)?

 

[Greg Bernhardt]

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

 

[nrqed]

Homework Statement

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)?

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

 

[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...

 

[nrqed]

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...

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.