Phy5645/Particle in Uniform Magnetic Field

(a) In the symmetric gauge, $A_{x}=-{\tfrac {1}{2}}By,$ $A_{y}={\tfrac {1}{2}}Bx,$ and $A_{z}=0.\!$

${\begin{aligned}\left[{{\hat {\Pi }}_{x},{\hat {\Pi }}_{y}}\right]&=\left[{\hat {p}}_{x}-{\frac {e}{c}}A_{x},{\hat {p}}_{y}-{\frac {e}{c}}A_{y}\right]=\left[{\hat {p}}_{x}+{\frac {eB}{2c}}{\hat {y}},{\hat {p}}_{y}-{\frac {eB}{2c}}{\hat {x}}\right]\\&=\left(\left[{\hat {p}}_{x},{\hat {p}}_{y}\right]-\left[{\hat {p}}_{x},{\frac {eB}{2c}}{\hat {x}}\right]+\left[{\frac {eB}{2c}}{\hat {y}},{\hat {p}}_{y}\right]-\left[{\frac {eB}{2c}}{\hat {y}},{\frac {eB}{2c}}{\hat {x}}\right]\right)\\&=-{\frac {eB}{2c}}(-i\hbar )+{\frac {eB}{2c}}(i\hbar )=i\hbar {\frac {eB}{c}}\end{aligned}}$

(b) The Hamiltonian for the system is

${\begin{aligned}{\hat {H}}&={\frac {1}{2m}}\left({\hat {\mathbf {p} }}-{\frac {e}{c}}\mathbf {A} \right)^{2}\\&={\frac {{\hat {\Pi }}_{x}^{2}}{2m}}+{\frac {{\hat {\Pi }}_{y}^{2}}{2m}}+{\frac {{\hat {p}}_{z}^{2}}{2m}}.\end{aligned}}$

If we label the first two terms as ${\hat {H}}_{1}={\frac {{\hat {\Pi }}_{x}^{2}}{2m}}+{\frac {{\hat {\Pi }}_{y}^{2}}{2m}}$, and the last one as ${\hat {H}}_{2}={\frac {{\hat {p}}_{z}^{2}}{2m}}$, then we may write the Hamiltonian as ${\hat {H}}={\hat {H}}_{1}+{\hat {H}}_{2}.$ Using the identity,

${\hat {A}}^{2}+{\hat {B}}^{2}=\left({\hat {A}}-i{\hat {B}}\right)\left({\hat {A}}+i{\hat {B}}\right)-i\left[{\hat {A}},{\hat {B}}\right],$

we may rewrite ${\hat {H}}_{1}$ as

${\hat {H}}_{1}={\frac {1}{2m}}\left({\hat {\Pi }}_{x}-i{\hat {\Pi }}_{y}\right)\left({\hat {\Pi }}_{x}+i{\hat {\Pi }}_{y}\right)+{\frac {\hbar eB}{2mc}}.$

If we now define the operators,

${\hat {a}}={\sqrt {\frac {c}{2\hbar eB}}}\left({\hat {\Pi }}_{x}+i{\hat {\Pi }}_{y}\right)$

and

${\hat {a}}^{\dagger }={\sqrt {\frac {c}{2\hbar eB}}}\left({\hat {\Pi }}_{x}-i{\hat {\Pi }}_{y}\right),$

this becomes

${\hat {H}}_{1}=\hbar \omega \left({\hat {a}}^{\dagger }{\hat {a}}+{\tfrac {1}{2}}\right),$

where $\omega ={\frac {eB}{mc}}.$ This is just the Hamiltonian for a harmonic oscillator. The contribution to the energy from this term is therefore

$E_{1}=\left(n+{\tfrac {1}{2}}\right)\hbar \omega .$

The remaining part of the Hamiltonian, ${\hat {H}}_{2},$ is just that of a free particle in one dimension, and thus its contribution to the energy is just $E_{2}={\frac {\hbar ^{2}k_{z}^{2}}{2m}}.$ The total energy is then just

$E=\left(n+{\tfrac {1}{2}}\right)\hbar \omega +{\frac {\hbar ^{2}k_{z}^{2}}{2m}},$

which is just the result that we obtained working in the Landau gauge, as expected.

Back to Charged Particles in an Electromagnetic Field.