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Harmonic oscillator in electric field

  1. Jan 15, 2014 #1
    1. The problem statement, all variables and given/known data
    Potential energy of electron in harmonic potential can be described as ##V(x)=\frac{m\omega _0^2x^2}{2}-eEx##, where E is electric field that has no gradient.
    What are the energies of eigenstates of an electron in potential ##V(x)##? Also calculate ##<ex>##.

    HINT: Use ##(x-a)^2=x^2-2ax+a^2##


    2. Relevant equations



    3. The attempt at a solution

    I am sorry, to say, but I have no idea how start here.

    I know that if there were no electric field, the energies would be ##E_n=\hbar \omega (n+1/2)##. Since there is Electric field, I assume I have to solve ##\hat{H}\psi =E\psi ## for ##\hat{H}=\hat{T}+\hat{V}##... but, how on earth can i do that?
     
  2. jcsd
  3. Jan 15, 2014 #2

    TSny

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    The hint is to "complete the square" in the expression ##V(x)=\frac{m\omega _0^2x^2}{2}-eEx##.
    That is, can you write it as ##V(x) = b(x-a)^2 - c## for certain constants ##a##, ##b## and ##c##?
     
  4. Jan 16, 2014 #3
    Ok, I got that but I can't see how this makes my life any easier..

    ##V(x)=\frac{m\omega _0^2x^2}{2}-eEx=(\sqrt{\frac{m}{2}}\omega _0x-\frac{eE}{\sqrt{2m}\omega _0})^2-\frac{e^2E^2}{2m\omega _0^2}##
     
  5. Jan 16, 2014 #4

    TSny

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    Try to factor out something inside the parentheses so that the x has a coefficient of 1 inside the parentheses. Then define a new variable in terms of x and solve the problem in the new variable.
     
  6. Jan 16, 2014 #5
    Hmm,

    ##V(x)=(\sqrt{\frac{m}{2}}\omega _0x-\frac{eE}{\sqrt{2m}\omega _0})^2-\frac{e^2E^2}{2m\omega _0^2}##

    ##V(x)=(\sqrt{\frac{m}{2}}\omega _0x-\frac{eE\sqrt{m}\omega_0}{\sqrt{2}m\omega _0^2})^2-\frac{e^2E^2}{2m\omega _0^2}## and finally

    ##V(x)=(\frac{m}{2})^{1/4}\omega_0^{1/2}(x-\frac{eE}{m\omega _0^2})^2-\frac{e^2E^2}{2m\omega _0^2}##

    now let's say ##u=x-\frac{eE}{m\omega _0^2}## than ##V(u)=(\frac{m}{2}\omega_0^2)^{1/4}u^2-\frac{e^2E^2}{2m\omega _0^2}##

    Now ##\hat{H}\psi = E_n \psi ##

    ##\hat{V}\psi=((\frac{m}{2}\omega_0^2)^{1/4}\hat{u}^2-\frac{e^2E^2}{2m\omega _0^2})\psi =E_n \psi##

    Are the energies of eigenstates than ##W_n=(\hbar \omega(n+1/2)+\frac{e^2E^2}{2m\omega _0^2})(\frac{m}{2}\omega_0^2)^{-1/4}##?
     
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