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

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vdweller
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Homework Statement



For an electron in a uniform magnetic field, say [itex]B\hat{z}[/itex] with no angular momentum, the Hamiltonian can be expressed as [itex]\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)[/itex]

Which is equivalent to two separate harmonic oscillators.

Now the ground state is [itex]Ψ=Ne^{\frac{mω}{\hbar}r^2}[/itex]

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

which yields energy equal to [itex]\hbarω[/itex]

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

(n is not the quantum number)

The question is to prove that [itex]\hat{H}Ψ_n=\frac{\hbarω_B}{2}Ψ_n[/itex]

since [itex]Ψ_n[/itex] 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)?
 
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vdweller said:

Homework Statement



For an electron in a uniform magnetic field, say [itex]B\hat{z}[/itex] with no angular momentum, the Hamiltonian can be expressed as [itex]\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)[/itex]

Which is equivalent to two separate harmonic oscillators.

Now the ground state is [itex]Ψ=Ne^{\frac{mω}{\hbar}r^2}[/itex]

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

which yields energy equal to [itex]\hbarω[/itex]

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

(n is not the quantum number)

The question is to prove that [itex]\hat{H}Ψ_n=\frac{\hbarω_B}{2}Ψ_n[/itex]

since [itex]Ψ_n[/itex] 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?
 
Yes, it's still a mess.

By the way, the correct ground state is [itex]Ψ=Ne^{-\frac{mω}{ℏ}r^2}[/itex] (forgot a minus). I can't see the edit button any more...
 
vdweller said:
Yes, it's still a mess.

By the way, the correct ground state is [itex]Ψ=Ne^{-\frac{mω}{ℏ}r^2}[/itex] (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.