- #1
maverick280857
- 1,774
- 5
Hi,
I'm teaching myself quantum mechanics (so this isn't homework). I came across the following question:
\Phi(p) = A\Theta\left[\frac{\hbar}{d}-|p-p_{0}|\right]
I have to find the constant of normalization, \psi(x), and the coordinate space wave function \psi(x,t) in the limit \frac{h/d}{p_{0}} << 1.
I started by finding A:
\int_{-\infty}^{\infty}\frac{|\Phi(p)|^2}{2\pi\hbar}dp = 1
This gives A = \sqrt{\pi d}.
Now,
\psi(x) = \frac{1}{2\pi\hbar}\int_{-\infty}^{\infty}dp'\Phi(p')e^{-ip'x/\hbar}
which gives
\psi(x) = \sqrt{\frac{d}{\pi}}e^{ip_{0}x/\hbar}\frac{\sin(x/d)}{x}
(There may be an algebraic error here..)
My problem is: how do I find \psi(x,t)? I am not sure how to proceed here.
I'm teaching myself quantum mechanics (so this isn't homework). I came across the following question:
\Phi(p) = A\Theta\left[\frac{\hbar}{d}-|p-p_{0}|\right]
I have to find the constant of normalization, \psi(x), and the coordinate space wave function \psi(x,t) in the limit \frac{h/d}{p_{0}} << 1.
I started by finding A:
\int_{-\infty}^{\infty}\frac{|\Phi(p)|^2}{2\pi\hbar}dp = 1
This gives A = \sqrt{\pi d}.
Now,
\psi(x) = \frac{1}{2\pi\hbar}\int_{-\infty}^{\infty}dp'\Phi(p')e^{-ip'x/\hbar}
which gives
\psi(x) = \sqrt{\frac{d}{\pi}}e^{ip_{0}x/\hbar}\frac{\sin(x/d)}{x}
(There may be an algebraic error here..)
My problem is: how do I find \psi(x,t)? I am not sure how to proceed here.
