[Physics] How to calculate the quantum expectation of frequency of a particle

frequencyhamiltonianoperatorsquantum mechanics

I know how to calculate the expectation of $\langle \Psi|A|\Psi \rangle$
where the operator $A$ is the eigenfunction of energy, momentum or position, but I'm not sure how to perform this for a pure frequency.

In other words, what is the expectation of frequency?

Indeed, is there an expectation, and is it solved using a Fourier transform from position space into frequency space?

In response to the posts by lurscher and David Zaslavsky below, I think both may be right. Frequency f is a parameter and can be considered an operator.

a) For the plane wave there exists temporal and spatial frequencies which act as parameters. $|\textbf{u}_k\rangle = e^{-i k x }$

b) there is also an operator derived from the Hamiltonian ($E=hf$) where $$\hat{H} =- \frac{\hbar }{i}\frac{\partial }{\partial t}$$
Rearranging and inverting $i$
$$\hat{f} = \frac{H}{\hbar } = i \frac{\partial }{\partial t}$$

Intuitively this appears like an operator, it has the right units and it transforms into the frequency domain, however, it still requires that $\hat{f}$ be shown to be Hermitian and is an operator that is its own adjoint or satisfies a Poisson algebra. Also intuitively, we can observe a frequency, by taking a measurement.

Best Answer

I'm getting the impression that a good part of this question (and perhaps also this physics.SE question?) arises from a wrong presumption that time and position should be on equal footing in quantum mechanics. They are not. The position $\hat{\bf r}$ is an operator, while time $t$ is a parameter. (Notation: In the following boldface denotes a vector quantity, and a hat denotes an operator quantity.) In fact, Pauli's Theorem shows, under mild assumptions, that time cannot be an operator in quantum mechanics, cf. Ref. (1). See also this and this physics.SE questions.

In the position space Schroedinger representation of the Schroedinger picture, we have a wave function $\Psi({\bf r},t)=\left<{\bf r}|\Psi(t)\right>$, the position operator $\hat{\bf r}$ becomes multiplication with ${\bf r}$, and the momentum operator $\hat{\bf p}= \frac{\hbar}{i}{\bf\nabla}$ is differentiation wrt. position. Both $\hat{\bf r}$ and $\hat{\bf p}$ are Hermitian operators. The Hamiltonian $\hat{H}=H(\hat{\bf r},\hat{\bf p})$ is composed out of the operators $\hat{\bf r}$ and $\hat{\bf p}$ in such a way that it is Hermitian.

It is wrong to claim that the Hamiltonian $\hat{H}$ is $i\hbar \partial_{t}$. (If it was, for starters, the Schroedinger equation $i\hbar \partial_{t}\Psi({\bf r},t)=\hat{ H}\Psi({\bf r},t)$ would become an empty statement without any content.) This should be compared with the fact that (in the position space Schroedinger representation) the momentum operator $\hat{\bf p}= \frac{\hbar}{i}{\bf\nabla}$ is a differentiation wrt. position. In particular, time and position differentiations are, in this sense, not on equal footing.

Finally, the frequency operator may be defined as $\frac{1}{h}\hat{H}=\frac{1}{h}{H}(\hat{\bf r},\hat{\bf p})$, as David Zaslavsky observes in a comment to lurscher's answer. The corresponding expectation value is

$$\left<\Psi(t) \left|\frac{1}{h}\hat{H} \right| \Psi(t)\right> ~=~ \int\! d^{3}r \ \Psi^{*}({\bf r},t)\ \frac{1}{h}\hat{H}\Psi({\bf r},t). $$


(1) Wolfgang Pauli 1933, in Handbuch der Physik (Encyclopedia of Physics) (ed. S. Fluegge), vol. 5, pp.1-168, Springer Verlag, 1958.