# Does this wave function violate any rules of Quantum mechanics?

I am studying bosons and fermions in quantum mechanics. We learned that for a two-particle system with identical bosons, the wave function can have the form (assume symmetric spin) $$\frac{1}{\sqrt2} (\psi_{1} (x_{a}) \psi_{2} (x_{b}) + \psi_{1} (x_{b}) \psi_{2} (x_{a}))$$

where $$\psi_{1}, \psi_{2}$$ denote two single-particle states and $$x_{a}, x_{b}$$ denote the two bosons. Will the following also be a possible wave function (assume symmetric spin)? $$\frac{1}{\sqrt2} (\psi_{1} (x_{a}) \psi_{1} (x_{b}) + \psi_{2} (x_{a}) \psi_{2} (x_{b}))$$ This wave function is still symmetric under the exchange of the two bosons. However, each individual product wave function implies that the two bosons are at the same energy state.
Does this second wave function violate any rule, or can it describe the wave function of the system of two bosons?

• you have to edit the second line of formulas , there is a psi as a variable at the last and an extra parenthesis Jan 23 '21 at 5:07

• Going off the comment by @ZeroTheHero, can this wave function still describe the system of two bosons even though it's not a solution to the time-independent SE? I was thinking in analogous to a single particle wave function for the infinite square well, say $\frac{1}{\sqrt{3}}(|E_{1}\rangle+|E_{2}\rangle+|E_{3}\rangle)$, which describes a particle with equal probability of being in any of the three states. Jan 23 '21 at 5:23