I've been reading through a derivation of the wavefunctions and energy levels for the quantum harmonic oscillator. It defines

$$\hat R^\pm=\frac{1}{\sqrt{2}}[\hat p \pm \mathrm{i}\omega \hat q]$$ in mass-weighted coordinates $q=x\sqrt{\mu},$ such that $\hat p= -\mathrm{i}\hbar(\mathrm d/\mathrm dq)$ and $\hat q$ is the position operator.

It then asserts that that $\langle \psi|\hat R^+\hat R^-|\psi\rangle$ must be real and non-negative, since $\hat R^+$ and $\hat R^-$ are hermitian conjuagtes. I know that $(\hat A\hat B)^\dagger = \hat B^\dagger \hat A^\dagger$ and so $\hat R^+\hat R^-$ is a hermitian operator meaning that the bra-ket must be real, but I can't see why it should be non-negative. Could anyone shed some light on this?


1 Answer 1


Let $\hat{R}^-|\psi\rangle = |\psi'\rangle$ (a new ket; we don't care what it is). If you take the adjoint of this equation you get

$$\langle\psi'| = \langle\psi|(\hat{R}^-)^\dagger = \langle\psi|\hat{R}^+$$

and hence

$$\langle\psi|\hat{R}^+\hat{R}^-|\psi\rangle = \langle\psi'|\psi'\rangle$$

which by definition of the inner product must be real and non-negative.

  • $\begingroup$ Where you've taken the adjoint of the equation, is that a valid step? It looks like you're saying that the conjugate of psi dash is equal to the conjugate of psi multiplied by the R+ operator, but what is it operating on? $\endgroup$ Sep 18, 2017 at 13:52
  • 2
    $\begingroup$ Operators don't need to operate on anything. Your post has a couple examples of standalone operators. In this case you could think of it as operating on the bra to the left, though. If you think in terms of matrices, then bra times operator is no different from operator times ket. $\endgroup$ Sep 18, 2017 at 13:55

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