In non-relativistic QM when one transforms to a different frame the wavefunction is also transformed:(adsbygoogle = window.adsbygoogle || []).push({});

(1) [tex] \psi ' = \psi e^{-\frac{imvx}{\hbar}-\frac{imv^{2}t}{2\hbar}} [/tex]

This looks a hell of a lot like:

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

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

The latter itself looks similar to:

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

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.

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# Another Schroedinger equation invariance question!

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