0
$\begingroup$

I'm studying for an exam in quantum mechanics and tried to calculate the ground state and the first two excited states of two identical bosons (spin 0) in an infinite one dimensional potential well.

As I've understood, the state for two identical bosons should be $\frac{1}{\sqrt{2}}(\psi_1(x_1)\psi_2(x_2)+\psi_1(x_2)\psi_2(x_1))$

However, in the solutions manual the correct answer is $\psi_1(x_1)\psi_1(x_2)\\$ for the ground state $\frac{1}{\sqrt{2}}(\psi_1(x_1)\psi_2(x_2)+\psi_1(x_2)\psi_2(x_1))\\$ for the first excited state and $\psi_2(x_1)\psi_2(x_2)$ for the second excited state

How can the ground state and the second excited state be represented by the simple product and not on the same form as in the first excited state? They are still identical, right? Or does this have to do with symmetry? In that case, how should I think to get it right?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
0
$\begingroup$

The simple product $\psi_1(x_1)\psi_1(x_2)$ is already symmetric under $x_1 \leftrightarrow x_2$. So it is not necessary to symmetrize the wave function by adding in the term you get when you switch $x_1$ and $x_2$.

Improve this answer
$\endgroup$
4
  • $\begingroup$ Yes, I can see that. But why are we just having $\psi_1$s in the simple product and not both $\psi_1$ and $\psi_2$ as in the first excited state? $\endgroup$
    – Djamillah
    Commented Oct 22, 2014 at 20:17
  • $\begingroup$ because $\psi_1$ is the ground state, and the lowest energy two-particle state is where both particles are in their ground state. $\endgroup$
    – Brian Moths
    Commented Oct 22, 2014 at 20:28
  • $\begingroup$ Ok, so the ground state is defined as both particles in the ground state and the first state as one particle in the ground state and one in the first state? Then the third state should be one particle in the first state and one in the second state? $\endgroup$
    – Djamillah
    Commented Oct 22, 2014 at 21:25
  • $\begingroup$ You should make that into a new question. $\endgroup$
    – Brian Moths
    Commented Oct 22, 2014 at 21:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.