What Is the Probability of Finding the Perturbed Oscillator in Its Ground State?

Demon117
Messages
162
Reaction score
1

Homework Statement


I showed earlier this semester that in the presence of a "constant force", F_{o}, i.e. V=-Fx, that the eigenvalues for the Harmonic oscillator are shifted by

\frac{F^{2}}{2m\omega^{2}}

from the "unperturbed" case. It was also discussed that x\rightarrow x-\frac{F}{m\omega^{2}}.

An oscillator is initially in its ground state (n=0). At t=0, a perturbation V is suddenly applied. What is the probability of finding the system in its (new) ground state for t>0, i.e. find |a_{o}|^{2}.


Homework Equations


For this |a_{n}|^{2}=|\int \Phi^{*}_{n}(x)\Psi_{o}(x)dx|^{2} over all space.

The Attempt at a Solution



For t>0, the state of the system is \Psi(x,t)=\sum a_{n}exp(-i(\frac{E_{n}}{\hbar})t)\Phi_{n}(x). Here \Phi_{n}(x) is an eigenvector of H. And the coefficients a_{n} are obtained by expanding \Psi_{o}(x), the ground state of H_{o}, in terms of \Phi_{n}(x).

I also know that the basis states \Phi_{n}(x) as well as \Psi_{o}(x) are Hermite polynomials.

With that in mind my assumption would simply be to integrate the following:

|a_{o}|^{2}=|\int \Phi_{o}(x) \Psi_{o}(x) dx|^{2} =|\int 1*1 dx|^{2}

If I integrate this over all space I end up with a probability that goes to infinity. . . .Maybe I am missing something as far as Hermite polynomials go. . . or maybe I have the wrong idea about this problem. Any suggestions would be helpful.
 
Last edited:
Physics news on Phys.org
Hi matumich, sorry I'm not sure if I can help but if you could so kindly explain to me where you got the relavant equation for |a_n|^2 and for psi(x,t) I would greatly appreciate it. Also, could you explain to me why the last line is equal to |integral 1*1 dx|^2?
 
Hi matumich, sorry I'm not sure if I can help but if you could so kindly explain to me where you got the relavant equation for |a_n|^2 and for psi(x,t)
.

Well I made several mistakes. The equation |a_{n}|^{2}=|∫Φ^{∗}_{n}(x)Ψ_{o}(x)dx|^{2} was given in class. The coefficients a_{n} are obtained by expanding Ψ_{o}(x), the ground state of H_{o}, in terms of Φ_{n}(x). From this it follows that the probability of finding the system in some state (t>0) is given by that integral.


I would greatly appreciate it. Also, could you explain to me why the last line is equal to |integral 1*1 dx|^2?

This was just purely as mistake and since the eigenfunctions Φ_{n}(x) correspond to the state once the perturbation is applied they can be expanded in terms of their basis elements, which in this case are the Hermite polynomials. Similarly, the unperturbed wave function Ψ_{o}(x) can be expressed in terms of the Hermite polynomials. Therefore we have:

Φ_{n}(x) = C exp(-\frac{1}{2}\alpha^{2}(x-\frac{F}{m \omega^{2})^{2})H_{n}(x)

&

Ψ_{o}(x) = Cexp(-\frac{1}{2}\alpha^{2} x^{2})H_{o}(x) = exp(-\frac{1}{2}\alpha^{2} x^{2})

The probability would be along the lines of

|a_{o}|^{2}=|∫C exp(-\frac{1}{2}\alpha^{2}(x-\frac{F}{m \omega^{2})^{2})exp(-\frac{1}{2}\alpha^{2} x^{2})dx|^{2}

After normalizing both wave functions you can integrate this and find the appropriate probability of finding the system in the ground state for t>0.
 
Hello everyone, I’m considering a point charge q that oscillates harmonically about the origin along the z-axis, e.g. $$z_{q}(t)= A\sin(wt)$$ In a strongly simplified / quasi-instantaneous approximation I ignore retardation and take the electric field at the position ##r=(x,y,z)## simply to be the “Coulomb field at the charge’s instantaneous position”: $$E(r,t)=\frac{q}{4\pi\varepsilon_{0}}\frac{r-r_{q}(t)}{||r-r_{q}(t)||^{3}}$$ with $$r_{q}(t)=(0,0,z_{q}(t))$$ (I’m aware this isn’t...
Hi, I had an exam and I completely messed up a problem. Especially one part which was necessary for the rest of the problem. Basically, I have a wormhole metric: $$(ds)^2 = -(dt)^2 + (dr)^2 + (r^2 + b^2)( (d\theta)^2 + sin^2 \theta (d\phi)^2 )$$ Where ##b=1## with an orbit only in the equatorial plane. We also know from the question that the orbit must satisfy this relationship: $$\varepsilon = \frac{1}{2} (\frac{dr}{d\tau})^2 + V_{eff}(r)$$ Ultimately, I was tasked to find the initial...
Back
Top