2
$\begingroup$

I was looking at the quantum mechanics book by Bransden and Joachain, specifically at the section about Galilean transformations, and I was trying to find out what they did here for the potential $V(\mathbf{r}, t) $.

What the books says, is the following:

Let frame $S'$ move with a constant velocity with respect to frame $S$, then $\mathbf{r}'=\mathbf{r}-\mathbf{v}t $.
The Schroedinger equation for a single particle of mass $m$ moving in a potential $V(\mathbf{r}, t) $ is
$$i\hbar \frac{\partial}{\partial t}\Psi(\mathbf{r},t) = \left[-\frac{\hbar}{2m} \nabla^2_\mathbf{r} + V(\mathbf{r},t)\right] \Psi(\mathbf{r},t)$$ Rewriting this in terms of $\mathbf{r}'$, we use
$\Psi(\mathbf{r},t) = \Psi(\mathbf{r}'+\mathbf{v}t,t)$
$\frac{\partial}{\partial t}\Psi(\mathbf{r},t)=\frac{\partial}{\partial t}\Psi(\mathbf{r}'+\mathbf{v}t,t) - \mathbf{v}\cdot\nabla_{\mathbf{r}'}\Psi(\mathbf{r}'+\mathbf{v}t,t)$
$\nabla_{\mathbf{r}}\Psi(\mathbf{r},t)=\nabla_{\mathbf{r}'}\Psi(\mathbf{r}'+\mathbf{v}t,t)$
to get
$$i\hbar \frac{\partial}{\partial t}\Psi(\mathbf{r}'+\mathbf{v}t,t) - i\hbar \mathbf{v}\cdot\nabla_{\mathbf{r}'}\Psi(\mathbf{r}'+\mathbf{v}t,t) = \left(-\frac{\hbar}{2m} \nabla^2_{\mathbf{r}'} + V(\mathbf{r}'+\mathbf{v}t,t)\right) \Psi(\mathbf{r}'+\mathbf{v}t,t)$$ This is not the form of a Schroedinger equation, so we use the unitary transformation
$$\Psi(\mathbf{r}'+\mathbf{v}t,t) = \exp\left[ i(m\mathbf{v}\cdot\mathbf{r}' + mv^2t/2)/\hbar\right]\Psi'(\mathbf{r}',t)$$ It is found that
$$i\hbar \frac{\partial}{\partial t}\Psi'(\mathbf{r}',t) = \left[-\frac{\hbar}{2m} \nabla^2_{\mathbf{r}'} + V'(\mathbf{r}',t)\right] \Psi'(\mathbf{r}',t)$$ where the potential in the $S'$ frame, $V'(\mathbf{r}',t)$, is defined as
$V'(\mathbf{r}' , t) = V(\mathbf{r}-\mathbf{v} t, t) $.

I left out some text to make it more compact. Here is a link to the section on Google books (might not work for some areas, including US).

Can anyone explain me how they arrived at $V'(\mathbf{r}' , t) = V(\mathbf{r}-\mathbf{v} t, t) $ ? It seems to me this should be $V'(\mathbf{r}' , t) = V(\mathbf{r}, t) $ since $V(\mathbf{r} , t) = V'(\mathbf{r}-\mathbf{v} t, t) = V'(\mathbf{r}',t)$.

Little side question: what is the physical reason $V$ has to vanish for Galilean covariance?

$\endgroup$
  • $\begingroup$ The link above doesn't send me to the section in Google Books; the resulting page says that there is "no preview available". This may be an issue of differing distribution rights here (USA) vs. where you are (Belgium?) You might get more people who can help you if you reproduce the pertinent points of the derivation in your question. $\endgroup$ – Michael Seifert Jun 4 '15 at 12:52
  • $\begingroup$ Alright, thanks for telling me. I'll see what I can do. $\endgroup$ – FatherNucleus Jun 4 '15 at 12:56
  • $\begingroup$ I added the derivation from the book. $\endgroup$ – FatherNucleus Jun 4 '15 at 13:51
  • $\begingroup$ How does one pronounce "Joachain"? $\endgroup$ – Harry Wilson Jun 4 '15 at 14:07
  • $\begingroup$ Probably in a French way $\endgroup$ – FatherNucleus Jun 4 '15 at 14:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.