Consider a simple harmonic oscillator; the position operator is $\hat{x}=(a^\dagger+a)/\sqrt{2}$ and the momentum operator is $\hat{p}=-i(a-a^\dagger)/\sqrt{2}$.

One may verify that the eigenstates of $\hat{x}$ and $\hat{p}$ are $$\left|x\right>\propto e^{\sqrt{2}~xa^\dagger-(a^\dagger)^2/2}\left|0\right>$$ $$\left|p\right>\propto e^{ip\sqrt{2} ~a^\dagger+(a^\dagger)^2/2}\left|0\right>.$$ My question is: how do I verify that the position eigenstates and momentum eigenstates are orthogonal themselves, and that $$\left<x|p\right>\propto e^{ipx} ~~?$$ I'm not able to calculate this inner product using the commutator of $a$ and $a^\dagger$.

  • $\begingroup$ @knzhou what do you mean by that? Yes I want a direct approach. $\endgroup$
    – Nahc
    Commented Feb 14, 2017 at 4:13
  • 2
    $\begingroup$ is this correct? what is $z$ in your first exponential? Moreover, is $p$ in your second exponential an operator or a number? $\endgroup$ Commented Feb 14, 2017 at 7:08
  • $\begingroup$ BCH formula and friends are what you want to commute exponentials of operators. What's the conceptual question here? What have you tried? $\endgroup$
    – ACuriousMind
    Commented Feb 14, 2017 at 12:58
  • $\begingroup$ @ACuriousMind Hi, of course I know I can use BCH formula to do that calculation. The problem is that when you use BCH formula, you need to calculate infinite number of commutators, and because the $a^2$ and $(a^\dagger)^2$, they don't vanish after a few order. $\endgroup$
    – Nahc
    Commented Feb 14, 2017 at 15:33
  • 1
    $\begingroup$ Without an i your normalization of your p is toxically antihermitean, and prevents your from simple implementation of adjoint symmetries involved. Your combinatorics would be vastly expedited by utilizing the isomorphism $a^\dagger \to y$ and $a \to \partial_y$, after you normalize your a's properly, provided you understand translation operators. $\endgroup$ Commented Feb 14, 2017 at 16:17