Homework Help: 1D Harmonic Oscillator in a Constant Electric Field

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1. Apr 4, 2015

Entanglement717

1. The problem statement, all variables and given/known data
Hello,
I'm just curious as to whether I'm going about solving the following problem correctly...
Problem Statement:
A particle mass m and charge q is in the ground state of a one -dimensional harmonic oscillator, the oscillator frequency is ω_o.
An electric field ε_o is suddenly turned on. Show that immediately afterwards the probability the particle is in the new ground state is given by exp(-q^2ε_o^2/2mω_o^3)

3. The attempt at a solution

I know I have to take the inner product of the old wavefunction with the new wavefunction, and I know that the new potential of the harmonic oscillator is going to be given by
H = p^2/(2m) + (1/2)mw^2x^2 - ekx
where V = -integral(E) = -kx

So my question is how does the wavefunction change and why?
and also my teacher hinted that we would have to complete the square and I don't see any equations that require completing the square in my attempt at a solution so I'm know something is horribly wrong...
Any help would be appreciated :)
thanks!

2. Apr 4, 2015

TSny

The wavefunction of the particle immediately after the field is turned on will be the same as before the field is turned on.

Follow the hint and complete the square for the expression (1/2)mw2x2 - ekx. Then step back and compare the Hamiltonian with the field to the Hamiltonian without the field.

3. Apr 4, 2015

Entanglement717

Hey TSny thanks!
Ok after completing the square the Hamiltonian become H = p2/2m + 1/2mω2(x - ek/mω2)2
and the wavefunction has to be different then the usual harmonic oscillator since x is replaced by (x - ek/mω2)2 in the groundstate wavefunction

So the probability that the particle will be found in the ground state is the square of the magnitude of the wavefunction perviously found but that leads to an ugly integral that isn't giving me the solution of exp(-q^2ε_o^2/2mω_o^3)

the integral I'm getting is (mω/πħ)1/2∫e(-mω/ħ)(x - (ek/mω2)2 dx

is this correct?

4. Apr 4, 2015

Entanglement717

oh and I guess I replaced qξo with eK in my notation

5. Apr 4, 2015

TSny

When completing the square there will be an additional constant term. But this just shifts the energy by a constant amount, so I think it's ok not to worry about it.

Before the field is switched on, the particle is just sitting in the ground state of the original Hamiltonian. Just after the field is switched on, the wavefunction of the particle is still the same. However, this wavefunction is no longer the ground state of the new Hamiltonian. You need to consider the "overlap" of the wavefunction of the particle with the wavefunction corresponding to the ground state of the new Hamiltonian.

6. Apr 4, 2015

Entanglement717

Cool got it! ^-^
Thank you! :)