- #1
JD_PM
- 1,125
- 156
- 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!
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!
Last edited: