# Two-Particle System

I thought that the general composite wave function for Identical Bosons is: $$\label{} \psi_{+}(r_1,r_2)=A[\psi_{a}(r_1)\psi_b(r_2)+\psi_b(r_1)\psi_a(r_2)]$$

but I stumbled upon an example in Grifith's QM 2nd ed for the infinite square well where $\psi_{11}= \frac{2}{a}sin(\frac{\pi x_1}{a})sin(\frac{\pi x_2}{a})$

which, I think, is missing the second term of the first equation. Can someone explain where my misunderstanding is?

• Which example? Which page? Commented Dec 14, 2014 at 23:52
• Without knowing anything of the context, this is of that form, for $\psi_a = \psi_b = \sin\left({\pi\,\cdot\,\over a}\right)$ and $A = \frac1a$. Commented Dec 14, 2014 at 23:58
• This particular example is on page 205-206 Commented Dec 15, 2014 at 0:49

For identical Bosons, the rule is that the wavefunction must be symmetric under the exchange of the positions of the two particles, and that is all. The form you have given satisfies this property explicitly, exchanging $r_1$ and $r_2$ results in the same exact wavefunction; and it is independent of your choice of$\psi_a$ and $\psi_b$ functions.

But, that does not change the original rule: The test to apply is to exchange the variables of the two particles, and the wavefunction should stay exactly the same. If it does, we are just fine.

Your example of $\psi_{11}$ obeys this exactly, $\psi_{11}(x_1, x_2)$ is exactly the same as $\psi_{11}(x_2, x_1)$. So no problems there.

If it looks like it is missing the second term, we can write the whole thing as follows:

$\psi_{11}(x_1, x_2) = \frac{1}{a}(\sin(\frac{\pi x_1}{a}) \cdot \sin(\frac{\pi x_2}{a}) + \sin(\frac{\pi x_2}{a}) \cdot \sin(\frac{\pi x_1}{a}))$

Here, you can identify:

$A = \frac{1}{a}$,

$\psi_a(x) = \sin(\frac{\pi x}{a})$,

$\psi_b(x) = \sin(\frac{\pi x}{a})$,

Turns out, the whole apparent missing term problem occurs because in this case $\psi_a = \psi_b$, so the terms just add together.

Take the original expression you have with $\psi_+$ and try putting the ground state wavefunctions in for both particles. By doing so, and by recalling a trig identity, you should see that the two are equivalent.

• I'm don't see how a trig identity would make the two equivalent. It seems to be off by a a factor of 1/2 from my math. I think my misunderstanding is due to the normalization constant which I can't seem to get. Commented Dec 15, 2014 at 2:02

The normalization constant in this case -- where $\psi_a=\psi_b$ -- is indeed 1/2. You can show this by normalizing the first equation. If you then plug in the ground states for the infinite square well for both particle, it simplifies to the second equation!