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Wavefunctions of fermions and bosons

  • #1
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Homework Statement


Consider two noninteracting particles p and q each with mass m in a cubical box od size a. Assume the energy of the particles is

[tex] E = \frac{3 \hbar^2 \pi^2}{2ma^2} + \frac{6\hbar^2 pi^2}{2ma^2} [/tex]

Using the eigenfunctions
[tex] \psi_{n_{x},n_{y},n_{z}} (x_{p},y_{p},z_{p}) [/tex]
and
[tex] \psi_{n_{x},n_{y},n_{z}} (x_{q},y_{q},z_{q}) [/tex]

write down the two particle wave functions which could describe the system when the particles are
a) distinguishable, spinless bosons
b) identical, spinless bosons
c) identical spin-half fermions in a symmetric spin state
d) identical spin half fermions in an antisymmetric spin state

Homework Equations


For a cube the wavefunction is given by

[tex] \psi_{n_{x},n_{y},n_{z}} = N \sin\left(\frac{n_{x}\pi x}{a}\right)\sin\left(\frac{n_{y}\pi y}{a}\right)\sin\left(\frac{n_{z}\pi z}{a}\right) [/tex]

[tex] E = \frac{\hbar^2 \pi^2}{2ma^2} (n_{x}^2 + n_{y}^2 +n_{z}^2) [/tex]

The Attempt at a Solution


for the fermions the wavefunction mus be antisymmetric under exhange
c) [tex] \Phi^{(A)}(p,q) = \psi^{(A)} (r_{p},r_{q}) \chi^{(S)}_{S,M_{s}}(p,q) [/tex]
where chi is the spin state

so since the energy is 3 E0 for the first particle there possible value nx,ny,nz are n=(1,1,1) and the second particle n'=(1,1,2).

we could select
[tex] \Psi^{(A)} (x_{p},x_{q},t) = \frac{1}{\sqrt{2}} (\psi_{n}(x_{p})\psi_{n'}(x_{q} - \psi_{n}(x_{q})\psi_{n'}(x_{p}) \exp[-\frac{i(E_{n} + E_{n'})t}{\hbar} [/tex]
A means it is unsymmetric
the spin state chi could be
[tex] \chi^{(S)}_{1,1}(p,q) = \chi_{+}(p) \chi_{-}(q) [/tex]
S means it is symmetric

For d it is similar but switched around

for a) and b) i have doubts though
For a) the bosons must be distinguishable so we could have WF like this

[tex] \Psi_{1} (r_{p},r_{q},t) = \psi_{n}(r_{p})\psi_{n'}(r_{q}) \exp[-\frac{i(E_{n} + E_{n'})t}{\hbar} [/tex]
for one of the particles. Under exchange this would be symmetric.

b) if the bosons are identical then we simply have to construct a wavefunction that is smmetric like we did in part c for the fermions.

Is this right? Please help!

thanks for any and all help!
 
Last edited:

Answers and Replies

  • #2
1,444
2
bump :biggrin:
 
  • #3
1,444
2
by the way [tex] \chi= \chi_{S,M_{S}} [/tex]

and
[tex] \chi_{+} [/tex] when Ms = +1/2
and
[tex] \chi_{-} [/tex] when Ms = -1/2
 
  • #4
cepheid
Staff Emeritus
Science Advisor
Gold Member
5,192
36
Hi stunner5000pt,

I'm just learning this stuff myself, but I'll try to help. Your answer to part c looks correct. The spatial part of the wf was chosen to be antisymmetric, since the spin state is given to be symmetric, so that the *overall* wf is antisymmetric.

for a) and b) i have doubts though
For a) the bosons must be distinguishable so we could have WF like this

[tex] \Psi_{1} (r_{p},r_{q},t) = \psi_{n}(r_{p})\psi_{n'}(r_{q}) \exp[-\frac{i(E_{n} + E_{n'})t}{\hbar} [/tex]
for one of the particles. Under exchange this would be symmetric.
No it wouldn't be! But that's okay! :smile: Because the particles are meant to be distinguishable. So I think your answer is correct...an acceptable wavefunction for a two-particle system is the product of the individual one-particle wavefunctions, *if* you know that one is in the state n, and
the other in the state n', because you are able to tell the difference between them.

b) if the bosons are identical then we simply have to construct a wavefunction that is smmetric like we did in part c for the fermions.
Yes, you do have to construct a wavefunction that is symmetric. But no, it's not like part c, because in part c, you constructed a wf that was antisymmetric. :tongue2: Not only that, but your bosons are spinless, so you'd only have a spatial part to your wavefunction, and it would have to be symmetric on its own.

Hope this helps
 

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