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## Homework Statement

$$\Psi = Ae^{\frac{i}{\hbar}(px-\frac{p^2}{2m}t)}$$

where ##p = \hbar k## and ##E = \hbar \omega = \frac{p^2}{2m}## for a nonrelativistic particle.

Find ##\Psi'(x',t')##, E' and p', under a galilean tranformation.

## Homework Equations

$$\Psi'(x',t') = f(x,t)\Psi(x,t)$$

where,

$$f(x,t) = e^{\frac{i}{\hbar}(-mvx + \frac{1}{2}mv^2t)}$$

For the galilean tranformation, ##x' = x - vt## and ##t' = t##

## The Attempt at a Solution

When multiplying f and ##\Psi##, the terms in the exponents will add. I am looking for the terms in the exponent that are multiplied by ##\frac{i}{\hbar}##. Since both f and ##\Psi## have ##\frac{i}{\hbar}## in the exponent, this can be factored out. So I only need to add the terms in the exponents ignoring the ##\frac{i}{\hbar}##. This gives

$$-mvx + \frac{1}{2}mv^2t+px-\frac{p^2}{2m}t$$

Since ##E= \frac{p^2}{2m} = \frac{1}{2}mv^2##, the expression becomes

$$-mvx +px$$

From the galilean tranformations, ##p=-mv##, so the expression becomes

$$2px = 2p(x'+vt') = 2px' + 2pvt'$$

I would then expect ##p' = 2p## and ##E' = 2pv##. I initially expected E' to be E and p' to be ##-p##. I know my calculated result is wrong, but I cannot figure out where I made a mistake using the tranformation.

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