One-dimensional HO in squeeze state

[skrat]

Homework Statement

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?

Homework Equations

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. :/

 

[mark726]

a) Your approach looks correct, but I think you made a mistake in the calculation of <x^2>. It should be <x^2>=(pi*hbar/zmomega)^2/2 instead of (pi*hbar/zmomega)^2*sqrt(pi)/2. Also, your expression for delta_x^2 is incorrect. It should be delta_x^2=<x^2>-<x>^2=(pi*hbar/zmomega)^2/4 instead of (pi*hbar/zmomega)^2.

b) To find the constants A, B and C, you can use the initial conditions given in the problem (z>0, x0p0=hbar, etc.) and solve for them. Remember that A and B must be complex conjugates of each other.

c) To find the expected value of the particle's position and uncertainty as a function of time, you can use the time evolution of the wave function given by the Schrodinger equation. The initial state of the particle is the ground state of the original Hamiltonian, so you can use the time evolution operator to find the wave function at time t, and then use it to calculate the expected value and uncertainty.