How to calculate the quantum expectation of frequency of a particle?

Question

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?

EDIT 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.

Answer 1

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

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

References:

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

Answer 2

the expectation for time frequency is obtained replacing A with the Hamiltonian operator, for space frequency is obtained replacing A with the momentum operator

EDIT from your comments i think i have a better idea of the source of your confusion

first, frequency is not an operator; it is a parameter (a c-number); for instance, plane waves $ | \textbf{u}_k \rangle = e^{-i k x } $ have a single parameter $k$ called spatial-frequency (the x is not really a parameter of the state $ | \textbf{u}_k \rangle$ is just an artifact that appears only when you write the coefficients of this state in the basis of eigenstates of the position operator )

De Broglie had the insight of associating classical particle quantities like energy and momentum to temporal and spatial frequencies of plane waves. This means that $ | \textbf{u}_k \rangle$ represents a physical state of momentum $p = \hbar k$

if you consider the algebraic entity $ | \textbf{u}_k \rangle \langle \textbf{u}_k | $ and you accept the bra-ket convention, if you multiply this operator with an eigenfunction $ | \textbf{u}_j \rangle$ you will basically have zero if $k \neq j$ or $ | \textbf{u}_j \rangle$ if $k=j$

then, a correct way to write the momentum operator (which is the same for any operator, the only thing that will change is the form of the basis function) is like:

$$\textbf{P} = \sum_{k} { \hbar k | \textbf{u}_k \rangle \langle \textbf{u}_k |}$$

ANY state vector can be written in the following form

$$ | \Psi \rangle = \sum_{j} { \psi_{j} | \textbf{u}_k \rangle }$$

I leave you as an exercise to write now the expression $\langle \Psi | \textbf{P} | \Psi \rangle$ by replacing each term with the above sums

hint 1: what you want is to obtain an expression that looks like an average after all! but you need to identify the weight in the average

hint 2: the fact that $\textbf{P}$ is hermitian is equivalent to saying that 1) the scalar products $\langle \textbf{u}_j | \textbf{u}_k \rangle$ will be always zero if $j \neq k$ and 2) that the eigenvalues $\hbar k$ will always be real

Answer 3

Formally, $H$ is an observable since it is a Hermition operator. But physically, I don't quite see that we ever really measure energy directly, what we tend to measure is the momentum of a photon. But the same applies to frequency: we don't measure the frequency of a photon directly, we measure its momentum, deduce its energy, and apply Planck's law to get the frequency. Classically, we measure the frequency of a wave by trying to time many different peaks of the wave. But according to QM, this cannot be done since after trying to find even one peak, one has disturbed the photon and so cannot proceed to find the next peak (it no longer exists).

The Hamiltonian really is not an operator on the same footing as all the other observables.

This can be seen another way: if it were, then $t$ would be its canonical conjugate, but we already have had it explained that $t$ is a parameter, not an observable. This has often been seen as a defect in the relativistic invariance of QM, but I don't think it has ever succcessfully been removed (without moving to Quantum Field Theory instead).

So I would say that there isn't really any frequency operator in the sense of a quantum mechanical observable since $H$ plays such a special role apart from all the other observables.

Answer 4

Frequency and energy are synonyms in quantum mechanics, so the expected value of the frequency is equal to the expected value of the energy. This is true for angular frequency in radian units, in units where $\hbar=1$. The Hamiltonian operator is the frequency operator, and states of definite H are definite frequency, with frequency given by the DeBroglie relation.

There is no ambiguity or difficulty related to the formalism of nonrelativistic quantum mechanics distinguishing time from space.