Phy5645/Particle in Uniform Magnetic Field

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

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

(b) The Hamiltonian for the system is

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

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

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

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

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

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

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;

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

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