Harmonic Oscillator in an Electric Field

The Hamiltonian of the system is

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

We may seprate the Hamiltonian into three terms, $H=H_{x}+H_{y}+H_{z},\!$ where

$H_{x}={\frac {p_{x}^{2}}{2m}}+{\tfrac {1}{2}}m\omega ^{2}x^{2}-eE_{0}x,$

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

and

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

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

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

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

$X(x)={\frac {1}{2^{n_{1}}n_{1}!}}\left({\frac {m\omega }{\pi \hbar }}\right)^{1/4}\exp \left[-{\frac {m\omega }{2\hbar }}\left(x-{\frac {eE_{0}}{m\omega ^{2}}}\right)^{2}\right]H_{n_{1}}\left[{\sqrt {\frac {m\omega }{\hbar }}}\left(x-{\frac {eE_{0}}{m\omega ^{2}}}\right)\right],$

$Y(y)={\frac {1}{2^{n_{2}}n_{2}!}}\left({\frac {m\omega }{\pi \hbar }}\right)^{1/4}e^{-m\omega y^{2}/2\hbar }H_{n_{2}}\left({\sqrt {\frac {m\omega }{\hbar }}}y\right),$

and

$Z(z)={\frac {1}{2^{n_{3}}n_{3}!}}\left({\frac {m\omega }{\pi \hbar }}\right)^{1/4}e^{-m\omega z^{2}/2\hbar }H_{n_{3}}\left({\sqrt {\frac {m\omega }{\hbar }}}z\right).$

The energy may simply be written as $E=E_{x}+E_{y}+E_{z},\!$ where $E_{x},\!$ $E_{y},\!$ and $E_{z}\!$ are the contributions to the energy from each of the harmonic oscillators. These are

$E_{x}=\left(n_{1}+{\tfrac {1}{2}}\right)\hbar \omega -{\frac {e^{2}E_{0}^{2}}{2m\omega ^{2}}},$

$E_{y}=\left(n_{2}+{\tfrac {1}{2}}\right)\hbar \omega ,$

and

$E_{z}=\left(n_{3}+{\tfrac {1}{2}}\right)\hbar \omega .$

The total energy is thus

$E=\left(n_{1}+n_{2}+n_{3}+{\tfrac {3}{2}}\right)\hbar \omega -{\frac {e^{2}E_{0}^{2}}{2m\omega ^{2}}}.$

Back to Analytical Method for Solving the Simple Harmonic Oscillator