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One-dimensional HO in squeeze state

  1. Dec 2, 2014 #1
    1. The problem statement, all variables and given/known data
    A paticle in one-dimensional harmonic potential $$H=\frac{p^2}{2m}+\frac 1 2 kx^2$$ is at tme ##t=0## in squeeze state for which we know $$(z\frac{x}{x_0}+i\frac{p}{p_0})|\psi >=0$$ where ##z\in \mathbb{C}## so that ##Re(z)>0##, ##x_0=\sqrt{\frac{\hbar }{m\omega }}##, ##x_0p_0=\hbar ## and ##\omega ^2=\frac k m##.

    a)What is expected value of particle position and it's uncertainty? Hint: Write the equation above in coordinate presentation.
    b) A particle in squeeze state at ##t=0## stays in that state even for ##t>0##. So $$(z(t)\frac{x}{x_0}+i\frac{p}{p_0})|\psi >=0$$ Prove that and find ##z(t)## (which is complex).
    HINT:
    Show ##(z\frac{x(t)}{x_0}+i\frac{p(t)}{p_0})|\psi >=0## where ##x(t)=e^{-i\frac H \hbar t}xe^{i\frac H \hbar t}## and ##p(t)=e^{-i\frac H \hbar t}pe^{i\frac H \hbar t}## and solve differential equations for ##\dot x (t)## and ##\dot p (t)##.
    c)At time ##t=0## the particle is in ground state of hamiltonian ##H##. At ##t=0## we change the potential so that Hamiltonian for ##t>0## is $$\tilde{H}=\frac{p^2}{2m}+\frac 1 2 k\tilde x^2$$ How do expected value of particle's position and uncertainty change as a function of time?

    2. Relevant equations


    3. The attempt at a solution
    a)$$(z\frac{x}{x_0}+i\frac{p}{p_0})|\psi >=(z\frac{x}{x_0}+i\frac{-i\hbar \frac{\partial }{\partial x}}{p_0})|\psi >=0$$ $$z\frac{x}{x_0}\psi (x)+\frac{\hbar}{p_0} {\psi}'(x)=0$$ And after solving this differential equation and normalization of ##\psi (x)## we should get something like $$\psi (x)=(\frac{zm\omega}{\pi \hbar})^{1/4}e^{-\frac{zm\omega}{2\pi \hbar}x^2}$$ Than $$<x>=(\frac{zm\omega}{\pi \hbar})^{1/2}\int _{-\infty}^{\infty}xe^{-\frac{zm\omega}{\pi \hbar}x^2}dx=(\frac{\pi \hbar}{zm\omega})^{3/2}$$ and $$<x^2>=(\frac{zm\omega}{\pi \hbar})^{1/2}\int _{-\infty}^{\infty}x^2e^{-\frac{zm\omega}{\pi \hbar}x^2}dx=(\frac{\pi \hbar}{zm\omega})^{2}\frac{\sqrt \pi}{2}$$ Finally $$\delta _x^2=<x^2>-<x>^2$$
    b) $$\dot x (t)=\frac i \hbar [H,x]=\frac i \hbar (\frac{1}{2m}[p^2,x^]+\frac 1 2 k[x^2,x])=\frac{p(t)}{m}$$ and very similar for $$\dot p(t)=-kx(t)$$ If derive the second equation again, I can combine the both $$\ddot p(t)+\frac k m p(t)=0$$ which brings me to $$p(t)=Ae^{i\omega_0 t}+Be^{-i\omega_0 t}$$ Using the first equation also $$x(t)=\frac 1 m (\frac{1}{i\omega}Ae^{i\omega t}-\frac{1}{i\omega}Be^{-i\omega t})+C$$ Now I am not really sure how to find those constants ##A##, ##B## and ##C##. ??

    c) No idea how to start. :/
     
  2. jcsd
  3. Dec 7, 2014 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
     
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