Phy5645/Particle in Uniform Magnetic Field: Difference between revisions

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


Back to [[Charged Particles in an Electromagnetic Field]].
Back to [[Charged Particles in an Electromagnetic Field#Problem|Charged Particles in an Electromagnetic Field]].

Latest revision as of 13:36, 18 January 2014

(a) In the symmetric gauge, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A_{x}=-\tfrac{1}{2}By,} Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A_{y}=\tfrac{1}{2}Bx,} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A_{z}=0.\!}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} \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{align} }

(b) The Hamiltonian for the system is

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} \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{align} }

If we label the first two terms as Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{H}_{1}=\frac{\hat{\Pi}_{x}^{2}}{2m}+\frac{\hat{\Pi}_{y}^{2}}{2m}} , and the last one as Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{H}_{2}=\frac{\hat{p}_{z}^{2}}{2m}} , then we may write the Hamiltonian as Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{H}=\hat{H}_{1}+\hat{H}_{2}.} Using the identity,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{H}_1} as

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{a}=\sqrt{\frac{c}{2\hbar eB}}\left (\hat{\Pi}_x+i\hat{\Pi}_y\right )}

and

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{a}^\dagger=\sqrt{\frac{c}{2\hbar eB}}\left (\hat{\Pi}_x-i\hat{\Pi}_y\right ),}

this becomes

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{H}_1=\hbar\omega\left (\hat{a}^\dagger\hat{a}+\tfrac{1}{2}\right ),}

where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \omega=\frac{eB}{mc}.} This is just the Hamiltonian for a harmonic oscillator. The contribution to the energy from this term is therefore

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E_1=\left (n+\tfrac{1}{2}\right )\hbar\omega.}

The remaining part of the Hamiltonian, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{H}_2,} is just that of a free particle in one dimension, and thus its contribution to the energy is just Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E_2=\frac{\hbar^2k_z^2}{2m}.} The total energy is then just

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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.