I often read in books that an observable is represented by an Hermitean operator. But it is deceiving as operator isn't the observable.

As far as I've read the observable is denoted like $\langle \psi|\hat{x}|\psi\rangle$ which is equivalent of $\langle \psi |\hat{x} \psi\rangle$ or $\langle \hat{x}^\dagger\psi| \psi \rangle$. So I would say that an observable is represented by an inner product of a (1st) wavefunction with (2nd) an operator acted on a wavefunction. (If I look the second equation).

Is this even correct? What do you think about this?


2 Answers 2


Functionally speaking, an observable is a physical quantity with this property: you can design a physics experiment that measures the value of that quantity possessed by a particle. So, for example, you can measure the speed of an electron by checking the radius of motion as it goes through a magnetic field, you can measure a particle's spin with the Stern-Gerlach experiment, and so on. And, as it turns out, any quantity for which you can design these kinds of experiments corresponds to a Hermitian operator.

The inner product isn't the observable itself; rather, it's the expectation value of that observable. If I repeat the experiment I did many times over, the average value of the measurement will be $\langle\psi|\hat{x}|\psi\rangle$. This is why we say that, mathematically, the operator itself is the observable: it's the mathematical object that we use to make predictions regarding the actual measured numbers we get from experiments.

  • $\begingroup$ "The operator is the observable" might be better phrased by saying the observable is the physical quantity and the operator is a mathematical quantity. The operator represents the observable. $\endgroup$ Commented Nov 29, 2022 at 14:50

The above answer is correct but let me provide another one from the point of view of path integral and correlators. Besides canonical formulations of quantum mechanics, there is also the path integral formulation, where the central object is the partition "function" $Z$ given by $$ Z=\int e^{i\frac{S}{\hbar}} $$ where $S$ is the action of the model and $\hbar$ the Planck-constant. One can assign boundary conditions, which in canonical quantization will correspond to $\langle\psi|$ and $|\psi\rangle$, or generalization of that, such as traces. One can also insert functionals in the integrand, corresponding in canonical quantization to insertions of observables such as $\langle\psi|\mathcal{O}|\psi\rangle$. From the path integral perspective, the physical observables can be thought of as linearized deformations of the action, as in $$ e^{i\frac{S}{\hbar}+i\mathcal{O}}-e^{i\frac{S}{\hbar}}=ie^{i\frac{S}{\hbar}}\mathcal{O}+\dots $$ One important formula in the path integral approach to quantum mechanics is the Dyson-Schwinger equation which stems from $$ 0=\int \delta \left(e^{i\frac{S}{\hbar}}i\mathcal{O}\right) = \int e^{i\frac{S}{\hbar}} \left(-\frac{1}{\hbar}\delta S \mathcal{O} + i\delta\mathcal{O} \right) \implies i\hbar\delta\mathcal{O}\sim \delta S \mathcal{O} $$ where $\delta$ is a variation of the dynamical variables (the ones being integrated by the path integral). This means in particular that one is free to impose the equations of motion far from insertions, and so $\mathcal{O}$ is defined up to equations of motion. One can interpret this as follows: the observables of the theory correspond to linearized deformations of the action, up to field redefinitions. The fact that they are required to be hermitian means that they correspond to unitary linearized deformations of the action. Measurements of the expectation value of this observable are linear responses to this deformation, like in usual statistical mechanics.

  • $\begingroup$ Silly question: shouldn't your equation $\int e^{iS/\hbar}(-\hbar^{-1}\,\delta S + i\,\delta\mathcal{O})$ imply $i\hbar\,\delta\mathcal{O}\sim\delta S$? Not $\sim\delta S\,\mathcal{O}$? Or am I making a mistake somewhere (which may very likely be the case!)? $\endgroup$ Commented Nov 21, 2022 at 2:08
  • 1
    $\begingroup$ It is a typo, I corrected now $\endgroup$
    – Nogueira
    Commented Nov 29, 2022 at 14:29

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