Phy5645/Particle in Uniform Magnetic Field

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An electron moves in magnetic field which is in the z direction, ${\overrightarrow {B}}=B{\hat {z}}$, and the Landau gauge ${\overrightarrow {A}}=({\frac {-By}{2}},{\frac {Bx}{2}},0)$

• Evaluate $\left[{\Pi _{x},\Pi _{y}}\right]$
• Using the hamiltonian and commutation relation obtained in a), obtain the energy eigenvalues.

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• According to the Ladau gauge, ${\text{A}}_{x}={\frac {-By}{2}}{\text{ A}}_{y}={\frac {Bx}{2}}{\text{ A}}_{z}=0$

$\left[{\Pi _{x},\Pi _{y}}\right]=\left[{\left({P_{x}-{\frac {e}{c}}A_{x}}\right),\left({P_{y}-{\frac {e}{c}}A_{y}}\right)}\right]$

$=\left[{\left({P_{x}+{\frac {eBy}{2c}}}\right),\left({P_{y}-{\frac {eBx}{2c}}}\right)}\right]$

$=\left\lbrace {\left[{P_{x},P_{y}}\right]-\left[{P_{x},{\frac {eBx}{2c}}}\right]+\left[{{\frac {eBy}{2c}},P_{y}}\right]-\left[{{\frac {eBy}{2c}},{\frac {eBx}{2c}}}\right]}\right\rbrace$

${\text{= -}}{\frac {eB}{2c}}(-i\hbar )+{\frac {eB}{2c}}(i\hbar )$ ${\text{=}}i\hbar {\frac {eB}{c}}$

• The hamiltonian fot the system is;

$H={\frac {({\overrightarrow {P}}-{\frac {e{\overrightarrow {A}}}{c}})^{2}}{2m}}$

$={\frac {\left({P_{x}-{\frac {e}{c}}A_{x}}\right)^{2}}{2m}}+{\frac {\left({P_{y}-{\frac {e}{c}}A_{y}}\right)^{2}}{2m}}+{\frac {\left({P_{z}-{\frac {e}{c}}A_{z}}\right)^{2}}{2m}}$

${\text{=}}{\frac {\Pi _{x}^{2}}{2m}}+{\frac {\Pi _{y}^{2}}{2m}}+{\frac {P_{z}^{2}}{2m}}$

If we define first two terms as ${\text{H}}_{1}={\frac {\Pi _{x}^{2}}{2m}}+{\frac {\Pi _{y}^{2}}{2m}}$, and the last one as ${\text{H}}_{2}={\frac {P_{z}^{2}}{2m}}$, The hamiltonian will be ${\text{H=H}}_{1}+H_{2}$.

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

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

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

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

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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