Schroedinger Equation - Galilean Invariance

In summary: Well, from what you've said it sounds like you're trying to prove that the wavefunction in the S' frame is the same as the wavefunction in the S frame after applying a Galilean transformation. However, that term you're getting that appears in the transformed equation- the momentum term- doesn't seem to fit that description. If you're trying to prove that the wavefunction in the S' frame is the same as the wavefunction in the S frame, then you need to prove that the momentum term in the transformed equation is the same as the momentum term in the original equation. From what you've said, it seems like you're not able to do that.
  • #1
Kane O'Donnell
Science Advisor
124
0
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
 
Physics news on Phys.org
  • #2
Kane O'Donnell said:
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?

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
Classical Electrodynamics

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

Kane
 
  • #4
Sorry I didn't get back to this. How did you make out? Did you find the reference to be understandable?
 
  • #5
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
Damnit. I'm still a bit stuck with deriving the form of the wavefunction transformation. That is, I've assumed that:

[tex] \psi = K\psi^{\prime} [/tex]

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

[tex](\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} [/tex]

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:

[tex]\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}[/tex]

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
 

What is the Schrödinger equation?

The Schrödinger equation is a mathematical equation that describes how quantum particles behave over time. It is a fundamental equation in quantum mechanics and is used to predict the probability of finding a particle in a certain location at a certain time.

Why is the Schrödinger equation important?

The Schrödinger equation is important because it allows us to understand and predict the behavior of quantum particles, which are the building blocks of matter. It has been extensively tested and has been found to accurately describe the behavior of particles at the microscopic level.

What is Galilean invariance?

Galilean invariance is a principle in physics that states that the laws of motion should be the same for all observers in uniform motion. This means that the laws of physics should not change depending on the frame of reference used by an observer.

How is the Schrödinger equation related to Galilean invariance?

The Schrödinger equation is derived from the principles of quantum mechanics, which are based on the assumption of Galilean invariance. This means that the Schrödinger equation is valid in all inertial reference frames and is consistent with the principles of Galilean invariance.

Are there any limitations to the Schrödinger equation's Galilean invariance?

Yes, the Schrödinger equation does not hold true in situations where particles are moving at speeds close to the speed of light. In these cases, the principles of special relativity must be taken into account and the Schrödinger equation must be modified to be consistent with these principles.

Similar threads

  • Classical Physics
Replies
13
Views
3K
Replies
5
Views
1K
Replies
1
Views
843
  • High Energy, Nuclear, Particle Physics
2
Replies
38
Views
3K
  • Special and General Relativity
Replies
30
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
2K
  • Advanced Physics Homework Help
Replies
8
Views
2K
  • Classical Physics
Replies
17
Views
22K
  • Special and General Relativity
Replies
7
Views
1K
  • Electromagnetism
Replies
5
Views
4K
Back
Top