- #1
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)?