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Tranformation of the wave function

  • #1
166
18

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.
 
Last edited:

Answers and Replies

  • #2
nrqed
Science Advisor
Homework Helper
Gold Member
3,605
205

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.
 
  • #3
166
18
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##
 
  • #4
nrqed
Science Advisor
Homework Helper
Gold Member
3,605
205
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?
 
  • #5
166
18
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!
 

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