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 ${\hat {H}}={\hat {H}}_{1}+{\hat {H}}_{2}.$

${\hat {H}}_{1}={\frac {{\hat {\Pi }}_{x}^{2}}{2m}}+{\frac {{\hat {\Pi }}_{y}^{2}}{2m}}$

$={\frac {1}{2m}}\left({\frac {c\Pi _{x}}{eB}}\right)^{2}\left({\frac {e^{2}B^{2}}{c^{2}}}\right)+{\frac {\Pi _{y}^{2}}{2m}}$

$={\frac {\Pi _{y}^{2}}{2m}}+{\frac {1}{2m}}\left({\frac {m^{2}}{m^{2}}}\right)\left({\frac {e^{2}B^{2}}{c^{2}}}\right)\left({\frac {c\Pi _{x}}{eB}}\right)^{2}$

$={\frac {\Pi _{y}^{2}}{2m}}+{\frac {1}{2}}m\left({\frac {eB}{cm}}\right)^{2}\left({\frac {c\Pi _{x}}{eB}}\right)^{2}$

Then the Hamiltonian will look like ${\text{H}}_{1}={\frac {\Pi _{y}^{2}}{2m}}+{\frac {1}{2}}m{\tilde {w^{2}}}{\tilde {x^{2}}}$ where ${\tilde {w}}=\left({\frac {eB}{cm}}\right)$ and ${\tilde {x}}=\left({\frac {c\Pi _{x}}{eB}}\right)$.

As we know, ${\text{H}}\Psi =E\Psi \!$

${\text{H=}}\hbar \left({\frac {eB}{cm}}\right)(n+{\frac {1}{2}})+{\frac {P_{z}^{2}}{2m}}$

${\text{H=}}\hbar \left({\frac {eB}{cm}}\right)(n+{\frac {1}{2}})+{\frac {\hbar ^{2}k^{2}}{2m}}$

So now we can write that;

${\text{E}}_{k,n}=\hbar {\frac {eB}{cm}}(n+{\frac {1}{2}})+{\frac {\hbar ^{2}k^{2}}{2m}}$

Back to Charged Particles in an Electromagnetic Field.