Physics GRE Discussion Forums

Hey guys,
So I got a bunch of flash cards from "Case Western Reserve" University, courtesy of a recommendation from someone on this forum ages ago, and one of the flash cards is asking to write down the wave functions for a free particle with periodic boundary conditions. I've never heard of periodic boundary conditions. Is this something I should be aware of? I looked in griffiths index and online, didn't really find alot of info...
Any help would be appreciated.

Ari

That's solid state physics. Could show up, but not likely to be vital. It's not really QM, at least in the sense of QM tested on the PGRE. You might find something in the E&M book...

for reference, the wave function (1D) in a period potential is given by $$\Psi(x) = e^{ikx}U(x)$$, where U(x) is a function with the same period as the potential, and $$k = (2\pi/L)*n$$, which is a result of the boundary condition.

You might recognize bits of this from your studies of diffraction...

bfollinprm wrote:for reference, the wave function (1D) in a period potential is given by $$\Psi(x) = e^{ikx}U(x)$$, where U(x) is a function with the same period as the potential, and $$k = (2\pi/L)*n$$, which is a result of the boundary condition.

This is Bloch's theorem, not boundary conditions

AriAstronomer wrote:I've never heard of periodic boundary conditions

It's OK. They will not appear on the PGRE for at least 10-15 years. But you can read about them in Chapter 8, Ashcroft and Mermin "Solid state physics", if you want.

Sounds like a particle on a ring [periodic boundary conditions, no potential]:
http://physchem.ox.ac.uk/~hill/tutorial ... index.html

Bloch's theorem does still apply, but with U(x)=1.

kangen558 wrote:Bloch's theorem does still apply, but with U(x)=1.

no. Boundary conditions are far more strong statements, than Bloch's theorem with $$U(x) \equiv 1$$.

physicsworks wrote:

kangen558 wrote:Bloch's theorem does still apply, but with U(x)=1.

no. Boundary conditions are far more strong statements, than Bloch's theorem with $$U(x) \equiv 1$$.

Perhaps I've misunderstood. I was only commenting that the eigenstates for the particle on a ring satisfy Bloch's Theorem with U(x)=constant. The periodic BCs will quantize the momentum. Am I missing something?

for periodic boundary conditions:
For example if your region is bet. x=0 to L, for any point after L, lets say for L+n,
f(L+n)=f(n)
for bloch theorem :
the function needs to be multiplied with exp(ikL) after moving L(assuming L is the period of the potential)
f(L+n)=f(n)exp(ikL)
so they are not the same. make sure that you have a periodic potential not a periodic boundary before using bloch theorem.

lol. I think all this confusion is a pretty good indicator of how important this topic is for the PGRE (not very).

Haha perfect. That was the answer I wanted to hear.