How Does a Galilean Transformation Affect the Wave Function?

[doggydan42]

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.

 

[nrqed]

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.

Watch out, you are using the same symbols for different things, which leads to difficulties. Let's call $E_i,p_i$ the initial energy en momentum. When you do your boost, the $mv$ and $1/2 mv^2$ appearing in your $f(x,t)$ are not in general equal to the initial values so you must use different symbols to distinguish them. Also, why would you expect $E'=E$? The kinetic energy is different in different frames.

 

[doggydan42]

Watch out, you are using the same symbols for different things, which leads to difficulties. Let's call $E_i,p_i$ the initial energy en momentum. When you do your boost, the $mv$ and $1/2 mv^2$ appearing in your $f(x,t)$ are not in general equal to the initial values so you must use different symbols to distinguish them. Also, why would you expect $E'=E$? The kinetic energy is different in different frames.

So if I use separate p's, then in the formula:
$$-mvx + \frac{1}{2}mv^2t+px-\frac{p^2}{2m}t = (p-mv)x + (\frac{1}{2}mv^2-\frac{p^2}{2m})t$$
After applying the galilean tranformation and some simplification, I arrived at
$$(p-mv)x' + (pv-\frac{1}{2}mv^2-\frac{p^2}{2m}) = (p-mv)x' - \frac{1}{2m}(p-mv)^2t'$$
So that gives $p' = p-mv$ and $E' = \frac{1}{2m}(p-mv)^2$

 

[nrqed]

So if I use separate p's, then in the formula:
$$-mvx + \frac{1}{2}mv^2t+px-\frac{p^2}{2m}t = (p-mv)x + (\frac{1}{2}mv^2-\frac{p^2}{2m})t$$
After applying the galilean tranformation and some simplification, I arrived at
$$(p-mv)x' + (pv-\frac{1}{2}mv^2-\frac{p^2}{2m}) = (p-mv)x' - \frac{1}{2m}(p-mv)^2t'$$
So that gives $p' = p-mv$ and $E' = \frac{1}{2m}(p-mv)^2$

Perfect! Do you see that this is indeed what one would expect?

 

[doggydan42]

Perfect! Do you see that this is indeed what one would expect?

Yes, I noticed it is what you expect for the momentum and energy with a galilean tranformation. Thank you!