0
$\begingroup$

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?

$\endgroup$
1
  • 1
    $\begingroup$ you have to edit the second line of formulas , there is a psi as a variable at the last and an extra parenthesis $\endgroup$
    – anna v
    Jan 23 '21 at 5:07
1
$\begingroup$

You're correct that the second wavefunction is symmetric under particle exchange. But the wavefunctions won't solve the two-particle Schrodinger equation with the correct eigenvalues.

$\endgroup$
3
  • $\begingroup$ Could the wave function be thought of as the superposition of two two-particle states, each with equal probability? Or can it just not describe a system of two identical bosons at all? $\endgroup$
    – ferris
    Jan 23 '21 at 3:50
  • $\begingroup$ It is indeed a superposition of two product states. What should be intuitive, however, is that whatever the composite state is, it should have energy eigenvalues as a sum of those of the individual eigenstates. For example, for an exciton, the wavefunction is written just as a product of individual electron and hole(both fermions) states. See for example pdfs.semanticscholar.org/529d/… $\endgroup$ Jan 23 '21 at 4:58
  • $\begingroup$ 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. $\endgroup$
    – ferris
    Jan 23 '21 at 5:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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