Two particle system

Thread Starter


Joined Oct 10, 2008
1. The problem statement, all variables and given/known data

Assume that we have a system where the two lowest one-particle states are \(\psi _{1} (r)\) with eigen energy E1 and \(\psi _{2} (r)\) with eigen energy E2. What is the lowest eigen energy E and the wave function \(\psi (r_{1},r_{2})\) for a two-particle system if

a) they are bosons
b) they are fermions

2. The attempt at a solution

a) The symmetric (boson) wave function becomes

\( \psi (r_{1},r_{2}) = \frac{1}{\sqrt{2}}[\psi _{1} (r1)\psi _{1} (r2) + \psi _{1} (r2)\psi _{1} (r1)] = \frac{2}{\sqrt{2}}[\psi _{1} (r1)\psi _{1} (r2)]\)

The energy is of course 2E1 because bosons can be in the same quantum state.

According to my book, the answer is \( \psi (r_{1},r_{2}) =\psi _{1} (r1)\psi _{1} (r2)\) only, so before proceeding with b) I want to know what I have done wrong.


Joined Jul 3, 2008
Think about the meaning of the wave function as a probablity amplitude. The square is a probablity distribution which must me normalized to one once integrated over space.


Joined Jul 3, 2008
So my answers are not really wrong?
I guess it depends on your point of view. I think the usual approach is to normalize the wave function so that, when you square it ,you have a true probability distribution that integrates to one over all space. I didn't work out the problem, but their answer has a properly normalized wave function that integrates to one. You can assume that the individual wave functions are properly normalized, and you can use that fact to normalize your answer. Then you should match.


Joined Nov 24, 2008
I think you may be assuming the states
are indentical, which would not be the general problem. If we assume that E1 < E2 then the lowest energy for a system with two bosons would be E1 + E1 which would imply each particle would be in state
. Work it out from here and see what you get.