The wavefunction of a two-particle system (both Fermions and Bosons possible): $$ \psi_\pm(x_1,x_2) = \sqrt{\frac{1}{2}}[\psi_n(x_1)\psi_m(x_2) \mp \psi_m(x_1)\psi_n(x_2)] $$

And a simpler position operator in order to simplify the computation (harmonic oscillator):

$$ x = \sqrt{\frac{\hbar}{2m\omega}}\left(a_+ + a_- \right) $$

where $a_+$ and $a_-$ are the creation and respectively the annihilation operators.

We now know that the distance (squared) is given by:

I simplify $\psi_n$ with $n$ and $\psi_m$ with $m$:

$$\langle(x_1 - x_2)^2 \rangle_\pm = \langle x^2 \rangle_n + \langle x^2 \rangle_m -2\langle x \rangle_n\langle x \rangle_m \mp |\langle x \rangle_{nm}|^2$$

for the last part we have: \begin{align} \langle x \rangle_{nm} &= \sqrt{\frac{\hbar}{2m\omega}}( \sqrt{m+1}\langle n|m+1\rangle\ + \ \sqrt{m}\langle n|m-1\rangle)\\ & = \sqrt{\frac{\hbar}{2m\omega}}( \sqrt{m+1}\delta_{n,m+1} + \ \sqrt{m}\delta_{n,m-1} ) \tag{1}\\ & = \sqrt{\frac{\hbar}{2m\omega}}( \sqrt{n}\delta_{n,m+1} + \ \sqrt{m}\delta_{n+1,m} ) \tag{2}\end{align}

My question is quite a simple one: what happens between line $(1)$ and line $(2)$ ? What properties are used to carry out that computation step?

  • $\begingroup$ That's a property of the Kronecker Delta right? $\endgroup$ – Sounak Sinha Jun 16 at 23:06
  • $\begingroup$ @SounakSinha, really? just that? could you please link me to it or give me an example? $\endgroup$ – Mattia Jun 16 at 23:10
  • $\begingroup$ The Kronecker Delta will click only if m+1 and n are equal (I'm referring to the first expression within the brackets), so writing m+1 is equivalent to writing n in front of the Delta function. Same holds for the second part. $\endgroup$ – Sounak Sinha Jun 16 at 23:13
  • $\begingroup$ For example $a\delta_{a2}=2\delta_{a2}$, LHS=RHS for all integral values of $a$. $\endgroup$ – Sounak Sinha Jun 16 at 23:16
  • $\begingroup$ @SounakSinha, yes... that's kinda obvious right now, I'm sorry you to read though the premise, thanks a lot!!! $\endgroup$ – Mattia Jun 16 at 23:22

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