We would expect that the form of the TDSE is preserved under a galilean transformation. Let us see whether we can arrive at this result starting from the equation in one frame, and performing the transformations.
We'll work in 1D for simplicity. Let us take a frame of reference with co-ordinates and , and another frame with co-ordinates and such that is moving at a constant velocity relative to . The co-ordinates are related by a galilean boost like so:
The potential energy is given by in , and by in , which are related in the following way:
In , the TDSE has the form;
where, is the wavefunction in . On the other hand, in frame , the wavefunction may have a different form and if the TDSE is galilean covariant, we would expect it to satisfy the equation:
Since the jacobian of this transformation is 1, we know that the probability density at a specific point in space must be the same in and ;
We can now find the transformation of the derivatives under the this galilean boost:
We still do not know the form of f(x', t'), but if the TDSE is galilean covariant, the consistency of the equations in the two frames will uniquely determine a form of this function, proving that the equation is indeed covariant under this transformation.
We begin with the TDSE in the frame :
Substituting the relations between the new derivatives and wavefunctions:
We will now drop the labels (prime) for ease of writing, since all quantities are represented in terms of which can just as well be written as the variables .
To preserve the form of the TDSE in frame F, the last two terms must vanish for arbitrary .
The second equation gives us that . This sets . Substituting this in equation (1) gives us:
This gives us the expression for :
So we see that if in frame satisfies the equation:
Then in frame satisfies the equation:
where the transforms like so:
The existence of a solution for imposing this form invariance of the TDSE indicates the galilean covariance of this equation.