Phy5645/Particle in Uniform Magnetic Field: Difference between revisions

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<math>
<math>
\begin{align}
\begin{align}
\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 [{\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 [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 ) \\
&=\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}
&=-\frac{eB}{2c}(-i\hbar)+\frac{eB}{2c}(i\hbar)=i\hbar \frac{eB}{c}
\end{align}
\end{align}
</math>
</math>


'''(b) The Hamiltonian for the system is
'''(b)''' The Hamiltonian for the system is


<math>
<math>
\begin{align}
\begin{align}
H&=\frac{1}{2m}\left (\mathbf{P}-\frac{e}{c}\mathbf{A}\right )^2 \\
\hat{H}&=\frac{1}{2m}\left (\hat{\mathbf{p}}-\frac{e}{c}\mathbf{A}\right )^2 \\
&=\frac{\Pi_{x}^{2}}{2m}+\frac{\Pi_{y}^{2}}{2m}+\frac{p_{z}^{2}}{2m}.
&=\frac{\hat{\Pi}_{x}^{2}}{2m}+\frac{\hat{\Pi}_{y}^{2}}{2m}+\frac{\hat{p}_{z}^{2}}{2m}.
\end{align}
\end{align}
</math>
</math>


If we label the first two terms as <math>\text{H}_{1}=\frac{\Pi_{x}^{2}}{2m}+\frac{\Pi_{y}^{2}}{2m}</math>, and the last one as <math>H_{2}=\frac{p_{z}^{2}}{2m}</math>, then we may write the Hamiltonian as <math>H=H_{1}+H_{2}\!</math>.
If we label the first two terms as <math>\hat{H}_{1}=\frac{\hat{\Pi}_{x}^{2}}{2m}+\frac{\hat{\Pi}_{y}^{2}}{2m}</math>, and the last one as <math>\hat{H}_{2}=\frac{\hat{p}_{z}^{2}}{2m}</math>, then we may write the Hamiltonian as <math>\hat{H}=\hat{H}_{1}+\hat{H}_{2}.</math> Using the identity,


<math>H_{1}=\frac{\Pi_{x}^{2}}{2m}+\frac{\Pi_{y}^{2}}{2m}</math>
<math>\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 ],</math>


<math>=\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}</math>
we may rewrite <math>\hat{H}_1</math> as


<math>=\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}</math>
<math>\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}.</math>


<math>=\frac{\Pi _{y}^{2}}{2m}+\frac{1}{2}m\left ({\frac{eB}{cm}} \right )^{2}\left ({\frac{c\Pi _{x}}{eB}} \right )^{2}</math>
If we now define the operators,


Then the Hamiltonian will look like <math>\text{H}_{1}=\frac{\Pi _{y}^{2}}{2m}+\frac{1}{2}m \tilde{w^{2}} \tilde{x^{2}}</math>  where <math> \tilde{w}= \left ({\frac{eB}{cm}} \right )</math>  and <math>\tilde{x}= \left ({\frac{c\Pi _{x}}{eB}} \right )</math>.
<math>\hat{a}=\sqrt{\frac{c}{2\hbar eB}}\left (\hat{\Pi}_x+i\hat{\Pi}_y\right )</math>


As we know, <math> \text{H}\Psi =E\Psi \!</math>
and


<math> \text{H=}\hbar \left ({\frac{eB}{cm}} \right )(n+\frac{1}{2}) + \frac{P_{z}^{2}}{2m} </math>
<math>\hat{a}^\dagger=\sqrt{\frac{c}{2\hbar eB}}\left (\hat{\Pi}_x-i\hat{\Pi}_y\right ),</math>


<math>\text{H=}\hbar \left ({\frac{eB}{cm}} \right )(n+\frac{1}{2})+\frac{\hbar ^{2}k^{2}}{2m}</math>
this becomes


So now we can write that;
<math>\hat{H}_1=\hbar\omega\left (\hat{a}^\dagger\hat{a}+\tfrac{1}{2}\right ),</math>


<math>\text{E}_{k,n}=\hbar\frac{eB}{cm}(n+\frac{1}{2})+\frac{\hbar ^{2}k^{2}}{2m}</math>
where <math>\omega=\frac{eB}{mc}.</math> This is just the Hamiltonian for a [[Harmonic Oscillator Spectrum and Eigenstates|harmonic oscillator]].  The contribution to the energy from this term is therefore


Back to [[Charged Particles in an Electromagnetic Field]].
<math>E_1=\left (n+\tfrac{1}{2}\right )\hbar\omega.</math>
 
The remaining part of the Hamiltonian, <math>\hat{H}_2,</math> is just that of a free particle in one dimension, and thus its contribution to the energy is just <math>E_2=\frac{\hbar^2k_z^2}{2m}.</math>  The total energy is then just
 
<math>E=\left (n+\tfrac{1}{2}\right )\hbar\omega+\frac{\hbar^{2}k_z^{2}}{2m},</math>
 
which is just the result that we obtained working in the Landau gauge, as expected.
 
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.

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