|Jan30-04, 03:05 PM||#1|
Symmetric, antisymmetric and parity
Problem 5.5 In David Griffiths “Introduction to Quantum Mechanics” says:
Imagine two non interacting particles, each of mass m, in the infinite square well. If one is in the state psin and the other in state psim orthogonal to psin, calculate < (x1 - x2) 2 >, assuming that (a) they are distinguishable particles, (b) they are identical bosons, (c) they are identical fermions.
(a) a2 [1/6 – 1/(2*pi2)(1/n2 + 1/m2)
(b) The answer to (a) - (128*a2*m2n2) / pi4(m2 - n2) 4
But this last term is present only when m,n have opposite parity.
(c) The answer to (a) plus the term added in (b) with the same stipulation as in (b)
What does this mean? It seams to be saying that all three particles would have the same separation unless their states have opposite parity. Is this correct? Bosons and Fermions would have the same separation unless their states have odd parities? I never heard of this before, how does this work?
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|Feb1-04, 12:31 AM||#2|
The way to solve the problem goes as follows:
if the particles are distinguishable, then the wave function is psi_n(x1) psi_m(x2). If they are fermions, then you have to use the antisymmetric form: 1/sqrt(2) (psi_n(x1) psi_m(x2) - psi_n(x2)psi_m(x1))
and if they are bosons, then you have to use the symmtric form:
1/sqrt(2) (psi_n(x1) psi_m(x2) + psi_n(x2)psi_m(x1))
This for the technical part. Now the interpretational part is more difficult, and the effects are called "exchange effects" But in general, fermions tend to "repulse" eachother and bosons tend to "attract" eachother. But not with a term in the hamiltonian, but purely through these exchange effects.
|Feb2-04, 07:07 PM||#3|
In case anyone's interested...........
I finally found the answer to this question in Quantum Physics of Atoms molecules etc. by Eisberg and Resnick on page 315.
I'd try to explain it but I don't think I can pull it off.
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