Harmonic Oscillator in an Electric Field

Source: "Theory and problems of quantum mechanics", Schaum, chapter 5

consider a particle with charge e moving under three dimensional isotropic harmonic potential l

$V(r)={\frac {1}{2}}m{\omega }^{2}{r}^{2}$in an electric field $E=E_{0}(x)$ Find the eigen states and eigen values of the patricle

the Hamiltonian of the system is:

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

we seprate the Hamiltonian ($H=H_{x}+H_{y}+H_{z}f$) where

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

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

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

Notice that $H_{x},H_{z}$are identical to the Hamiltonian of the one dimensional harmonic oscillator, so we can write the wave function

$\psi (x,y,z)=\psi _{1}(x)\psi _{2}(y)\psi _{3}(z)$, where

$\psi _{2}(y)$, and $\psi _{3}(z)$ are the wave functions of the one dimensional harmonic oscillator: $\psi (x,y,z)=\psi _{1}(x)\psi _{2}(y)\psi _{3}(z)$

$\psi _{2}(y)={\frac {1}{\sqrt {\pi \lambda 2^{n_{2}}n_{2}!}}}H_{n_{2}}e^{\frac {-y^{2}}{2\lambda ^{2}}}$

$\psi _{3}(z)={\frac {1}{\sqrt {\pi \lambda 2^{n_{3}}n_{3}!}}}H_{n_{3}}e^{\frac {-z^{2}}{2\lambda ^{2}}}$ $\lambda ={\sqrt {\frac {\hbar }{m\omega }}},$ The equation of the$\psi _{1}(x)$ is

$-{\frac {\hbar ^{2}}{2m}}{\frac {\partial ^{2}\psi _{1}(x)}{\partial x^{2}}}+{\frac {m\omega ^{2}}{2}}x^{2}\psi _{1}(x)-eE_{0}(x)\psi _{1}(x)=E_{1}\psi _{1}(x)$

changing variables to $\xi ={\frac {x}{\lambda }}-{\frac {eE_{0}}{\sqrt {\hbar m\omega }}}$

${\frac {\partial ^{2}\psi _{1}}{\partial \xi ^{2}}}+({\frac {2E_{1}}{\hbar \omega }}){\frac {(eE_{0})^{2})}{\sqrt {\hbar m\omega ^{3}}}})\psi _{1}-\xi ^{2}\psi _{1}=0$

we obtain the diffrential equation for a one dimensional harmonic oscillator with the solution

$\psi _{1}(\xi )={\frac {1}{\sqrt {\pi \lambda 2^{n_{1}}n_{1}!}}}H_{n_{1}}e^{\frac {-\xi ^{2}}{2\lambda ^{2}}}$

$\psi _{1}(\xi )={\frac {1}{\sqrt {\pi \lambda 2^{n_{1}}n_{1}!}}}H_{n_{1}}(x)exp[-{\frac {1}{2}}({\frac {x}{\lambda }}-{\frac {eE_{0}}{\sqrt {\hbar m\omega ^{3}}}})^{2}](E_{1})_{n_{1}}=(n_{1}+{\frac {1}{2}})\hbar \omega -{\frac {(eE_{0})^{2}}{2m\omega ^{2}}}$

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

In conclusion,the wave functions are $\psi (x,y,z)=\psi _{1}(x)\psi _{2}(y)\psi _{3}(z)$

$E_{n_{1},n_{2},n_{3}}=E_{n_{1}}+E_{n_{2}}+E_{n_{3}}=(n_{1}+n_{2}+n_{3}+{\frac {3}{2}})\hbar \omega -{\frac {(eE_{0})^{2}}{2m\omega ^{2}}}$