# Eigenstates of the momentum operator

1. Dec 27, 2012

### aaaa202

For the free particle the solution to the SE are eigenstates of the momentum.
You get something like:
ψ = Aexp(ik(x-vt)) + Bexp(-ik(x+vt)) , where k is a constant
And my book then says that first term represents a wave travelling to the right and the second a wave travelling to the left. But I have a problem with the second statement, because doesn't the minus sign on the second term, that is the minus sign in front of (x+vt), change anything. Certainly the waves are not identical in shap because of that minus sign?

2. Dec 27, 2012

### Staff: Mentor

Eigenstates of the momentum are solutions, but not the only solutions - every superposition of them is a solution, too.

Your equation can be written as ψ = Aexp(ikx-ikvt) + Bexp(ik(-x)-ikvt)), which is just a substitution x -> -x, so the wave is mirrored.

3. Dec 27, 2012

### aaaa202

but I dont understand then. Isn't x+vt the mirror of x-vt, not as you say -x+vt mirror of x-vt.

4. Dec 27, 2012

### Staff: Mentor

Just put them into the SE to see which signs you need :).

5. Dec 27, 2012

### aaaa202

what? I already put them in the SE and the general solution is:

ψ(x,t) = Aexp(ik(x-vt)) + Bexp(-ik(x+vt))

What I dont understand is just that the form a wave travelling to the left is for me exp(ik(x+vt)) not exp(-ik(-x+vt)) - what is the essential difference between these two waveforms?

6. Dec 27, 2012

### Staff: Mentor

They have a different relation between phase and x-coordinate, while their time-evolution of the phase is the same (just determined by k and v).
All this is multiplied with i and in an exponential - it is just a phase.

7. Dec 27, 2012

### aaaa202

so you say its just a phasefactor? That would make sense I guess but if you multiply
exp(-ik(x+vt)) by a phasefactor exp(i) you get exp(k(x+vt)) - clearly that is not what you wanted is it?

8. Dec 27, 2012

### Staff: Mentor

If you multiply exp(-ik(x+vt)) by exp(i) the result is exp(-ik(x+vt) + i) = exp(-ik(x+vt-1/k))
Which is equivalent to a shift by 1/k in x-direction or 1/(kv) in time.

The whole exp()-part is just a phase.

9. Dec 27, 2012

### aaaa202

man I still dont get it. My multiplication of the exponential was wrong but I still dont see the answer to why:

exp(-ik(x+vt)) is the same as exp(ik(x-vt)) except for a phase.

10. Dec 27, 2012

### jmcelve

Well, you agree that, in general, complex exponentials with the same coefficients differ only by a phase factor, yes? If so, then surely you can see how this is simply a particular case of that general rule.

11. Dec 27, 2012

### aaaa202

but the the function in the exponential is not the same! x-vt≠x+vt

12. Dec 27, 2012

### jmcelve

Sure, and that's the whole point. There is a *phase* by which they differ, though the *amplitudes* of these particular plane waves are equal. In particular, their phases differ such that they propagate in opposite directions (+x and -x).

13. Dec 27, 2012

### aaaa202

ugh I see but normally you have exp(ik(x-vt)) and exp(ik(x+vt)) where you in this case instead of the last term have exp(ik(-x+vt)) - my question is really just what difference this -x instead of x does.