# Another Schroedinger equation invariance question!

1. Oct 10, 2004

### Kane O'Donnell

In non-relativistic QM when one transforms to a different frame the wavefunction is also transformed:

(1) $$\psi ' = \psi e^{-\frac{imvx}{\hbar}-\frac{imv^{2}t}{2\hbar}}$$

This looks a hell of a lot like:

(2) $$\psi ' = \psi e^{-i\frac{<p>x+<E>t}{\hbar}$$

where <p> and <E> are obviously for a free particle.

The latter itself looks similar to:

(3) $$\psi '(x) = \psi(x+<x>) e^{-\frac{i<p>x}{\hbar}}$$

where <x> and <p> are for the wavefunction psi. Indeed, for the wavefunction psi' the expectation values of position and momentum are zero (simple calculation) - this seems to suggest that (3) is also a transform from a stationary frame ('watching a particle go past') to the particle's frame. (3) has no time dependence explicitly but we can guess that it is the usual phase evolution factor with E = <E> for a free particle, making (2) and (3) very similar indeed.

I can't quantify the connection precisely. I guess <x> = -vt with v the relative frame velocity. What *is* the exponential term besides calling it a 'kinematic transformation'? Does it simply subtract energy and momentum from the wavefunction's 'internal information' (*very* imprecise wording, relax!)?

Cheers,

Kane O'Donnell

PS - (1) is derived when one requires that the Schroedinger equation be Galilean invariant, by assuming that psi has the form Kpsi' and solving for K. You have to assume K satisfies the TDSE for a free particle and that the potential in the TDSE for psi is a Galilean invariant.

2. Oct 11, 2004

### lalbatros

Hello

I don't see clearly what you are looking for, but I was surprised that it is not straightforward to come to equation (1). Could you explain a little bit more? What it means exactly and how you derived it.

The derivation of equation (1) should normally contain everything needed for an interpretation of this result. Could you post here?

If I have well understood, equation (1) for psi' should be the solution of the SE in the 'moving frame' as a function of the solution psi for the SE in the stationnary frame. I call the 'moving frame', the frame where the potential is time dependent: V(x+vt)). Is that the real meaning ?

I tried to derive equation (1) myself, but I could not do better than checking that it works for a noninterracting plane wave exp(i/hb(p.x-E.t)).

I hope I could go further later.

Michel

3. Oct 11, 2004

### Kane O'Donnell

Hi,

Thanks for your reply. What you do (I took my starting point from Jackson's Classical Electrodynamics 3rd Ed chapter 11) is suppose that the TDSE holds in a frame S' (the wavefunction $$\psi '$$ must then represent the whole SYSTEM in S'). The form is (in 1D obviously):

$$\frac{-\hbar^2}{2m}\frac{\partial^{2}\psi'}{\partial x'^{2}} + V(x')\psi' = i\hbar\frac{\partial\psi'}{\partial t'}$$​

Then you use $$x' = x-vt, t' = t$$ and the differential forms to transform the equation back into the S frame, and you get:

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

Notice that this is *almost* the Schroedinger equation - we *almost* have Galilean invariance. The extra term on the right is very suggestive of an extra momentum. I try to think about it like this - the wavefunction contains all the information we can know for the system. When we shift frames so that we are watching the 'system' go past us, we should measure extra momentum. This suggests we need to make a kinematic transform to the wavefunction.

We try: $$\psi' = K\psi$$ (not the same K as I last posted! ) and sub that in to the first Schroedinger equation. After much algebra (you have to expand everything using the product rule and then transform frames again) you get:

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

To get to the latter you also have to use the fact that $$\psi'$$ obeys the TDSE and that V is Lorentz invariant itself. I am unsure what the consequence of this last assumption has for the generality the TDSE Galilean invariance.

Anyway, the left hand factor of the left hand side is obviously the TDSE for a plane wave. K being a plane wave of sorts would be very good, because then the *probability distribution* of $$\psi$$ and $$\psi'$$ would be unchanged by the frame transform. Indeed, if we assume K is a plane wave, then we are done, because the left hand side goes to zero and since the derivative of psi' isn't zero we can set the left hand factor of the right hand side to zero, solve for K, sub into the TDSE for a plane wave, get the time dependence and we have the form of K.

Substituting -v for v gives the inverse version of K which appears in equation (1) in my original post.

So, easy! My question is just whether the interpretation of 'factoring in a plane wave adds momentum that we'd need to see in the moving frame' is a VALID physical interpretation. One can *always* multiple a wavefunction by a phase factor without affecting the probability density. I have derived a particular phase factor here so that different observers will measure the same probability density but different momenta. Is this valid? Are different observers supposed to see the same probability density? Or is the idea that the Schroedinger equation is Galilean invariant not true? There is very little discussion of the point.

(There has been one paper in AJP, but I haven't gotten around to reading it yet)

Is that more clear?

Regards,

Kane

Last edited: Oct 11, 2004
4. Oct 14, 2004

### lalbatros

Hello

Here are the connexions I can imagine to explain the form of the Galilean Invariance for the Schrodinger equation.

The gauge transformation reads as follows, assuming x'=x-Vt and t'=t :

$$\Psi'(x',t') \;\;= \;\;\Psi(x'+Vt',t')\;\;\;e^\frac{mVx - \frac{mV^2}{2}t}{i\hbar} \;\;= \;\;\Psi(x'+Vt',t')\;\;\;e^\frac{mVx' + \frac{mV^2}{2}t'}{i\hbar}$$

This form of the gauge factor is connected to basic needs.
Indeed, it ensures that:

• the probability density is invariant with respect to the Galilean transformation

• the average momentum is transformed according to the usual classical rule, indeed:

$$P'\Psi'(x',t') = -i\hbar \frac{\partial}{\partial x'}\;\;\Psi'(x',t')$$

yields by substitution and quantum average:

$$p' = p - mV$$

• the average energy is tranformed according to the usual classical rules, indeed:

$$H'\Psi'(x',x') = i\hbar \frac{\partial}{\partial t'}\;\;\Psi'(x',t')$$

yields by substitution and quantum average:

$$E' = E + \frac{mV^2}{2} - pV$$

• the phase of the gauge factor transforms exactly as the action of classical mechanics, indeed the action

$$S' = \int \left(\frac{mv'^2}{2} - U(x'+Vt)\right) dt = \int \left(\frac{mv^2}{2} - U(x) + \frac{mV^2}{2} - mvV\right) dt$$

or

$$S' = S + \frac{mV^2}{2}t - mVx$$

this tranformation matches the gauge factor since:

$$\psi = e^{-\frac{S}{i\hbar}}$$

Going through all this was interresting for me because it really goes back to the fundamentals of the Schrodinger equation and how Schrodinger presented it in his 1927 paper. In this paper, the links to classical mechanics are completely detailled. I will read it sometimes hopefully.

5. Oct 14, 2004

### Kane O'Donnell

That's much of what I suspected, so it's just a 'compare to the classical case and if it works that's good enough' scenario.

Kane

6. Oct 15, 2004

### lalbatros

Indeed, there were no reason that the transformation of momentum and energy would differ from the classical limit.

However, there is a llink between the SE and the principle of least action. This link was embedded by Schrodinger in the SE (see 1927 paper). It is also the basis of the Bohm interpretation of QM.

It is then no surprise that the jauge factor is most directly linked to the way the action is modified under a Galilean transformation.

This reflects the fact that QM extend CM.