Computing a wave function through a (non-relativistic) propagator

  • #1
JD_PM
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Homework Statement
Given the non-relativistic propagator of a one-dimensional free particle



\begin{equation*}
K(x,t;x',0)=\left( \frac{m}{2 \pi i \hbar t} \right)^{1/2} \exp\left( \frac{-m(x-x')^2}{2i \hbar t}\right)
\end{equation*}


And an initial wave function (which happens to be a gaussian)

\begin{equation*}
\psi(x,0) = \left( \frac{2b}{\pi}\right)^{1/4} \exp( -b x^2)
\end{equation*}

Compute the wave function ##\Psi(x,t)## of the system
Relevant Equations
N/A
We know that the non-relativistic propagator describes the probability for a particle to go from one spatial point at certain time to a different one at a later time.

I came across an expression (lecture notes) relating ##\Psi(x,t)##, an initial wave function and the propagator. Applying the given information we get

\begin{align*}
&\Psi(x,t) = \int_{-\infty}^{\infty} dx' \psi(x',0) K(x,t;x',0) \\
&= \left( \frac{2b}{\pi}\right)^{1/4}\left( \frac{m}{2\pi i\hbar t}\right)^{1/2} \int_{-\infty}^{\infty} dx' \exp(-bx'^2) \exp\left( \frac{-m(x-x')^2}{2i \hbar t}\right) \\
&= \left( \frac{2b}{\pi}\right)^{1/4}\left( \frac{m}{2\pi i\hbar t}\right)^{1/2} \int_{-\infty}^{\infty} dx' \exp\left( -bx'^{2} - \frac{mx'^{2}}{2i \hbar t} + \frac{mx'}{i \hbar t} - \frac{mx^2}{2i \hbar t} \right)
\end{align*}

Does this make sense so far?

If yes, let us deal with the computation of the integral. I was thinking of using the complete-the-square trick. Doing so, our integral becomes$$\int_{-\infty}^{\infty} dx' \exp\left( \left[x'\sqrt{\frac{-m}{2i \hbar t}-a} + \frac{m}{i \hbar t\sqrt{\frac{-m}{2i \hbar t}-a}}\right]^2 - \frac{mx^2}{2i \hbar t} -\frac{m}{(i \hbar t)^2 \left(\frac{-m}{2i \hbar t}-a\right)} \right)$$

Before continuing with such method: is there an easier way to solve it?

Any help is appreciated

Thank you! :biggrin:
 
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  • #2
I fear you just have to go on.

An alternative way is to work it out in the momentum representation first, but there you again encounter Gauß integrals of comparable complexity. It's not so difficult per se but a lot to write ;-).
 
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  • #3
Thank you @vanhees71

It is good to know that the approach is OK.
 

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