# One-dimensional HO in squeeze state

1. Dec 2, 2014

### skrat

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. Dec 7, 2014