Harmonic Oscillator in an Electric Field: Difference between revisions

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The Hamiltonian of the system is:
The Hamiltonian of the system is


<math>H=\frac{P^2}{2m}+\frac{1}{2}m\omega ^2r^2-eE_{0}x</math>
<math>H=\frac{p^2}{2m}+\tfrac{1}{2}m\omega^2r^2-eE_{0}x.</math>


we seprate the Hamiltonian (<math>H=H_{x}+H_{y}+H_{z} f</math>) where
We may seprate the Hamiltonian into three terms, <math>H=H_{x}+H_{y}+H_{z},\!</math> where
   
   
<math>H_{x}=\frac{p_{x}^{2}}{2m}+\frac{1}{2}m\omega ^2x^2-eE_{0}x</math>
<math>H_{x}=\frac{p_{x}^{2}}{2m}+\tfrac{1}{2}m\omega^2x^2-eE_{0}x,</math>


<math>H_{y}=\frac{p_{y}^{2}}{2m}+\frac{1}{2}m\omega ^2y^2 </math>
<math>H_{y}=\frac{p_{y}^{2}}{2m}+\tfrac{1}{2}m\omega^2y^2,</math>


<math>H_{z}=\frac{p_{z}^{2}}{2m}+\frac{1}{2}m\omega ^2z^2</math>
and


Notice that <math>H_{x} ,H_{z}</math>are identical to the Hamiltonian of the one dimensional harmonic oscillator, so we can write the wave function
<math>H_{z}=\frac{p_{z}^{2}}{2m}+\tfrac{1}{2}m\omega^2z^2.</math>  


<math>\psi (x,y,z)=\psi _{1}(x)\psi _{2}(y)\psi _{3}(z)</math>, where
Note that each of these terms depends on only one coordinate, and that, in fact, <math>H_y\!</math> and <math>H_z\!</math> are each the Hamiltonian of a one-dimensional harmonic oscillator.  In fact, if we "complete the square" in <math>H_x,\!</math> we will find that it is also a one-dimensional harmonic oscillator, but with a shifted center.  Let us, in fact, do this:


<math>\psi _{2}(y)</math>, and
<math>H_x=\frac{p_{x}^{2}}{2m}+\tfrac{1}{2}m\omega^2\left (x^2-\frac{2eE_{0}}{m\omega^2}x\right )=\frac{p_{x}^{2}}{2m}+\tfrac{1}{2}m\omega^2\left (x-\frac{eE_{0}}{m\omega^2}\right )^2-\frac{e^2E_0^2}{2m\omega^2}</math>
<math>\psi _{3}(z)</math> are the wave functions of the one dimensional harmonic oscillator:
<math>\psi (x,y,z)=\psi _{1}(x)\psi _{2}(y)\psi _{3}(z)</math>


<math>\psi _{2}(y)=\frac{1}{\sqrt{\pi \lambda 2 ^{n_{2}}n_{2}!}}H_{n_{2}}e^{\frac{-y^{2}}{2\lambda ^{2}}}</math>
We may now easily write down the solution.  If we take <math>\psi(x,y,z)=X(x)Y(y)Z(z),\!</math> then


<math>\psi _{3}(z)=\frac{1}{\sqrt{\pi \lambda 2 ^{n_{3}}n_{3}!}}H_{n_{3}}e^{\frac{-z^{2}}{2\lambda ^{2}}}</math>
<math>X(x)=\frac{1}{2^{n_1}n_1!}\left (\frac{m\omega}{\pi\hbar}\right )^{1/4}\exp\left [-\frac{m\omega}{2\hbar}\left (x-\frac{eE_0}{m\omega^2}\right )^2\right ]H_{n_1}\left [\sqrt{\frac{m\omega}{\hbar}}\left (x-\frac{eE_0}{m\omega^2}\right )\right ],</math>
<math>\lambda =\sqrt{\frac{\hbar}{m\omega }},</math> The equation of the<math>\psi _{1}(x)</math> is


<math>-\frac{\hbar^{2}}{2m}\frac{\partial^2\psi _{1}(x)}{\partial x^2}+\frac{m\omega ^{2}}{2}x^{2}\psi _{1}(x)-eE_{0}(x)\psi _{1}(x)=E_{1}\psi _{1}(x)</math>
<math>Y(y)=\frac{1}{2^{n_2}n_2!}\left (\frac{m\omega}{\pi\hbar}\right )^{1/4}e^{-m\omega y^2/2\hbar}H_{n_2}\left (\sqrt{\frac{m\omega}{\hbar}}y\right ),</math>


changing variables to <math>\xi =\frac{x}{\lambda }-\frac{eE_{0}}{\sqrt{\hbar m \omega }}</math>
and


<math>\frac{\partial^2 \psi _{1}}{\partial \xi ^2}+(\frac{2E_{1}}{\hbar \omega })\frac{(eE_{0})^{2})}{\sqrt{\hbar m\omega ^{3}}})\psi _{1}-\xi ^{2}\psi _{1}=0</math>
<math>Z(z)=\frac{1}{2^{n_3}n_3!}\left (\frac{m\omega}{\pi\hbar}\right )^{1/4}e^{-m\omega z^2/2\hbar}H_{n_3}\left (\sqrt{\frac{m\omega}{\hbar}}z\right ).</math>


we obtain the diffrential equation for a one dimensional harmonic oscillator with the solution
The energy may simply be written as <math>E=E_x+E_y+E_z,\!</math> where <math>E_x,\!</math> <math>E_y,\!</math> and <math>E_z\!</math> are the contributions to the energy from each of the harmonic oscillators.  These are


<math>\psi _{1}(\xi )=\frac{1}{\sqrt{\pi \lambda 2 ^{n_{1}}n_{1}!}}H_{n_{1}}e^{\frac{-\xi ^{2}}{2\lambda ^{2}}}</math>
<math>E_x=\left (n_1+\tfrac{1}{2}\right )\hbar\omega-\frac{e^2E_0^2}{2m\omega^2},</math>


<math>\psi _{1}(\xi )=\frac{1}{\sqrt{\pi \lambda 2^{n_{1}}n_{1}!}}H_{n_{1}}(x) exp[-\frac{1}{2}(\frac{x}{\lambda }-\frac{eE_{0}}{\sqrt{\hbar m\omega ^{3}}})^{2}]
<math>E_y=\left (n_2+\tfrac{1}{2}\right )\hbar\omega,</math>
(E_{1})_{n_{1}}=(n_{1}+\frac{1}{2})\hbar \omega -\frac{(eE_{0})^{2}}{2m\omega ^{2}}</math>


The quantization condition in this case is
and
<math>\frac{(2E_{1})}{\hbar\omega }+\frac{(eE_{0})^2}{\hbar m\omega ^{3}}=2n_{1}+1</math>
so the energy eigenvalues are
<math>(E_{1})_{n_{1}}=(n_{1}+\frac{1}{2})\hbar \omega -\frac{(eE_{0})^{2}}{2m\omega ^{2}}</math>


In conclusion,the wave functions are <math>\psi (x,y,z)=\psi _{1}(x)\psi _{2}(y)\psi _{3}(z)</math>
<math>E_z=\left (n_3+\tfrac{1}{2}\right)\hbar\omega.</math>


<math>E_{n_{1},n_{2},n_{3}}=E_{n_{1}}+E_{n_{2}}+E_{n_{3}}=(n_{1}+n_{2}+n_{3}+\frac{3}{2})\hbar \omega -\frac{(eE_{0})^{2}}{2m\omega ^{2}}</math>
The total energy is thus
 
<math>E=\left (n_{1}+n_{2}+n_{3}+\tfrac{3}{2}\right )\hbar\omega-\frac{e^2E_{0}^{2}}{2m\omega^{2}}.</math>


Back to [[Analytical Method for Solving the Simple Harmonic Oscillator]]
Back to [[Analytical Method for Solving the Simple Harmonic Oscillator]]

Revision as of 16:18, 9 August 2013

The Hamiltonian of 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 H=\frac{p^2}{2m}+\tfrac{1}{2}m\omega^2r^2-eE_{0}x.}

We may seprate the Hamiltonian into three terms, 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 H=H_{x}+H_{y}+H_{z},\!} 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 H_{x}=\frac{p_{x}^{2}}{2m}+\tfrac{1}{2}m\omega^2x^2-eE_{0}x,}

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 H_{y}=\frac{p_{y}^{2}}{2m}+\tfrac{1}{2}m\omega^2y^2,}

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 H_{z}=\frac{p_{z}^{2}}{2m}+\tfrac{1}{2}m\omega^2z^2.}

Note that each of these terms depends on only one coordinate, and that, in fact, 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 H_y\!} 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 H_z\!} are each the Hamiltonian of a one-dimensional harmonic oscillator. In fact, if we "complete the square" in 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 H_x,\!} we will find that it is also a one-dimensional harmonic oscillator, but with a shifted center. Let us, in fact, do this:

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 H_x=\frac{p_{x}^{2}}{2m}+\tfrac{1}{2}m\omega^2\left (x^2-\frac{2eE_{0}}{m\omega^2}x\right )=\frac{p_{x}^{2}}{2m}+\tfrac{1}{2}m\omega^2\left (x-\frac{eE_{0}}{m\omega^2}\right )^2-\frac{e^2E_0^2}{2m\omega^2}}

We may now easily write down the solution. If we take 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 \psi(x,y,z)=X(x)Y(y)Z(z),\!} then

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 X(x)=\frac{1}{2^{n_1}n_1!}\left (\frac{m\omega}{\pi\hbar}\right )^{1/4}\exp\left [-\frac{m\omega}{2\hbar}\left (x-\frac{eE_0}{m\omega^2}\right )^2\right ]H_{n_1}\left [\sqrt{\frac{m\omega}{\hbar}}\left (x-\frac{eE_0}{m\omega^2}\right )\right ],}

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 Y(y)=\frac{1}{2^{n_2}n_2!}\left (\frac{m\omega}{\pi\hbar}\right )^{1/4}e^{-m\omega y^2/2\hbar}H_{n_2}\left (\sqrt{\frac{m\omega}{\hbar}}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 Z(z)=\frac{1}{2^{n_3}n_3!}\left (\frac{m\omega}{\pi\hbar}\right )^{1/4}e^{-m\omega z^2/2\hbar}H_{n_3}\left (\sqrt{\frac{m\omega}{\hbar}}z\right ).}

The energy may simply be written 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 E=E_x+E_y+E_z,\!} 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 E_x,\!} 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_y,\!} 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 E_z\!} are the contributions to the energy from each of the harmonic oscillators. These are

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_x=\left (n_1+\tfrac{1}{2}\right )\hbar\omega-\frac{e^2E_0^2}{2m\omega^2},}

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_y=\left (n_2+\tfrac{1}{2}\right )\hbar\omega,}

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 E_z=\left (n_3+\tfrac{1}{2}\right)\hbar\omega.}

The total energy is thus

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_{1}+n_{2}+n_{3}+\tfrac{3}{2}\right )\hbar\omega-\frac{e^2E_{0}^{2}}{2m\omega^{2}}.}

Back to Analytical Method for Solving the Simple Harmonic Oscillator

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