# Schroedinger Equation - Galilean Invariance

1. Sep 21, 2004

### Kane O'Donnell

Hi All,

I'm new to this forum. I'm a third-year undergrad Physics Major in Australia, about to go on to Honours, very exciting project in Helium atom detection.

To the point. My 3rd year Special Rel project is an investigation of the development of relativistic QM (RQM). I have to prove all the key components of the development, for example that the TDSE is Gal but not Lorentz invariant, that the lack of invariance arises from the unequal treatment of the time/momentum operators, etc.

Anyway, what I want to ask is - everytime I go to prove Galilean invariance for the 1D TDSE with an arbitrary potential, I get an extra term appearing in the transformed equation of the form vp^, where p^ is the momentum operator and v is the relative velocity of the frame. How can this be considered "of the same form" as the original if that term is in there?

I'm not sure if I'm going about it the right way - the method I have used is to start in the S' frame and transform backwards to the S frame, ie Psi(x', t') goes to Psi(x-vt, t), etc, and using the chain rule for the partial derivatives.

I'd like to be as rigorous as possible, but I can't seem to find references where the proof is actually done and not left as a student exercise!

Thanks,

Kane

2. Sep 21, 2004

### Tom Mattson

Staff Emeritus
Hi, and welcome to PF!

I've only got a second right now, but I can give you a reference: See Jackson's Classical Electrodynamics, 2ed, Chapter 11. There is a short explanation of the Galilean invariance of the Schrodinger equation there. It turns out that you also have to do a transformation on the wavefunction to recover the form of the SE under Galilean transformations. I'll post mathematical details later today.

3. Sep 21, 2004

### Kane O'Donnell

Classical Electrodynamics

Thanks, our library has that reference, I'll check it out.

Kane

4. Sep 27, 2004

### Tom Mattson

Staff Emeritus
Sorry I didn't get back to this. How did you make out? Did you find the reference to be understandable?

5. Sep 27, 2004

### Kane O'Donnell

Yes, thanks, I found the reference. It was fairly brief (just a footnote) but I did want to be able to prove this thing myself.

Kane

6. Sep 29, 2004

### Kane O'Donnell

Damnit. I'm still a bit stuck with deriving the form of the wavefunction transformation. That is, I've assumed that:

$$\psi = K\psi^{\prime}$$

Then it can be shown that one of the governing equations for K (in 1D) is:

$$(\frac{-\hbar^{2}}{2m}\frac{\partial^{2}K}{\partial x^{2}}-i\hbar \frac{\partial K}{\partial t})\psi^{\prime}=(\frac{\hbar^{2}}{m}\frac{\partial K}{\partial x} -i\hbar vK)\frac{\partial\psi^{\prime}}{\partial x}$$

It can be checked that this is correct by using the given version of K from Jackson and substituting in above.

The Schroedinger equation transformed from the S' frame back to the S frame is:

$$\frac{-\hbar^{2}}{2m}\frac{\partial^{2}\psi^{\prime}}{\partial x^{2}} +V\psi^{\prime} = i\hbar\frac{\partial \psi^{\prime}}{\partial t} +i\hbar v\frac{\partial \psi^{\prime}}{\partial x}$$

where the extra term on the right is the momentum term I was babbling on about previously. (v is the relative velocity of the frames)

However, I'm not quite sure where to proceed. I notice that that first factor in the left hand side of the governing equation is just the Schroedinger expression in K, but I can't justify saying that K must be a solution to the Schroedinger equation by itself. If that were the case, then we could set the left hand side in that equation to zero and then the right hand side yields a differential equation in K that can be solved quite easily to give part of the answer, which can then be substituted into the Schroedinger equation for K to give a second differential equation for K to get the other part.

The only problem is I can't justify the assumption that K is a solution to Schroedinger's equation!

Any ideas?

Kane