- #1
Henk
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I'm having some difficulties with the following problem:
Consider two (spinless)free bosons in a box of volume V with periodic boundary conditions. Let the momenta of the bosons be p and q.
a) Write down the normalized wavefunction for p is not equal to q and p = q.
\Psi_{pq}(r1,r2)
I thought since they are bosons Y has to be symmetric thus:
\Psi_{pq}(r1,r2) = \frac{1}{\sqrt{2}}(\varphi_{p}(r1)\varphi_{q}(r2)+ \varphi_{p}(r2)\varphi_{q}(r1))
Where
\varphi_{p}(r1)\varphi_{q}(r2) = \frac{1}{(2\pi)^3}(e^(i(p \cdot r1))(e^(i(q \cdot r2))
and
\varphi_{p}(r2)\varphi_{q}(r1) = \frac{1}{(2\pi)^3}(e^(i(p \cdot r2))(e^(i(q \cdot r1))
For p=q this means:
\Psi_{pq}(r1,r2) = \frac{1}{\sqrt{2}} \frac{1}{(2\pi)^3}(2e^(i(k \cdot (r1+r2)))
b) Show that for p is not equal to q:
\Psi_{pq}(r,r)|^2 > |\Psi_{pp}(r,r)|^2
But I don't know how to do this.
Consider two (spinless)free bosons in a box of volume V with periodic boundary conditions. Let the momenta of the bosons be p and q.
a) Write down the normalized wavefunction for p is not equal to q and p = q.
\Psi_{pq}(r1,r2)
I thought since they are bosons Y has to be symmetric thus:
\Psi_{pq}(r1,r2) = \frac{1}{\sqrt{2}}(\varphi_{p}(r1)\varphi_{q}(r2)+ \varphi_{p}(r2)\varphi_{q}(r1))
Where
\varphi_{p}(r1)\varphi_{q}(r2) = \frac{1}{(2\pi)^3}(e^(i(p \cdot r1))(e^(i(q \cdot r2))
and
\varphi_{p}(r2)\varphi_{q}(r1) = \frac{1}{(2\pi)^3}(e^(i(p \cdot r2))(e^(i(q \cdot r1))
For p=q this means:
\Psi_{pq}(r1,r2) = \frac{1}{\sqrt{2}} \frac{1}{(2\pi)^3}(2e^(i(k \cdot (r1+r2)))
b) Show that for p is not equal to q:
\Psi_{pq}(r,r)|^2 > |\Psi_{pp}(r,r)|^2
But I don't know how to do this.
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