Harmonic Oscillator in an Electric Field: Difference between revisions

From PhyWiki
Jump to navigation Jump to search
No edit summary
No edit summary
 
(2 intermediate revisions by the same user not shown)
Line 1: Line 1:
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


Back to [[Analytical Method for Solving the Simple Harmonic Oscillator]]
<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#Problem|Analytical Method for Solving the Simple Harmonic Oscillator]]

Latest revision as of 13:34, 18 January 2014

The Hamiltonian of the system is

We may seprate the Hamiltonian into three terms, where

and

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

We may now easily write down the solution. If we take then

and

The energy may simply be written as where and are the contributions to the energy from each of the harmonic oscillators. These are

and

The total energy is thus

Back to Analytical Method for Solving the Simple Harmonic Oscillator

Retrieved from "http://wiki.physics.fsu.edu/index.php?title=Harmonic_Oscillator_in_an_Electric_Field&oldid=15803"