In Quantum Mechanics one often by postulate has that the system is described by a Hilbert space $\mathscr{H}$ and observables are described by hermitian operators $A\in \mathfrak{A}(\mathscr{H})$ on the corresponding algebra. The possible values to be measured are the ones in the spectrum $\sigma(A)$ and we work with generalized eigenstates, so that for each $a\in \sigma(A)$ we have $|a\rangle$ the state with definite value for $A$ equal to $a$, which we assume normalized conveniently.

We can thus define projectors

$$\pi_a =|a\rangle\langle a|,$$

and we have:

$$\int_{\sigma(A)}\pi_a da=\int_{\sigma(A)}|a\rangle\langle a|da=\mathbf{1}$$

and they are also orthogonal, in the sense that $$\pi_a\pi_b=\delta_{ab}|a\rangle \langle b|.$$

Now, in the paper A Mini-Introduction To Information Theory by Witten, he turns this around. Working in the discrete and finite case for simplicity, he actually says that if we pick any collection $\pi_a$ of hermitian operators satisfying:

$$\sum_{a=1}^k \pi_a=\mathbf{1},\quad \pi_a^2=\pi_a,\quad \pi_a \pi_{a'}=0, \ a\neq a'$$

they determine a projective measurement.

Now, these are just abstract mathematical operators. How do they correspond to physical measurement?

I mean, take position. We know how to measure position. One puts detectors around, and when a particle is detected somewhere position has been measured. The same goes for momentum, energy, and so all.

In these cases, we have a physical measurement in mind, we create one abstract operator out of it. And when we say "after a measurement in position yielding $x$ the state is $|x\rangle$" we know what we are saying physically.

This is the opposite. We pick abstract operators first and want them to correspond to measurement.

So how to physically understand such a measurement, when it seems like a purely mathematicaly abstract concept? How does one actually "measure" what corresponds to $\pi_a$?

  • $\begingroup$ The direction from a given experiment to the related operator is far from trivial. But here, if you ask for the other way around, I think we are lost. Of course, on a formal level, one can consider all self-adjoint operators as describing "some" ideal measurement, but how to actually build it may be impossible to find out. $\endgroup$
    – Luke
    Jul 11 '18 at 9:43

A Hermitian operator $\pi$ satisfying $\pi^2 = 1$ is still an orthogonal projector, so this is not too exotic. The reason why Witten writes it this way is that it allows, for example, to measure:

Is my particle inside the interval $[0,1]$ or outside of it?

In that case $k=2$, $\pi_1 = \int_0^1 |x \rangle \langle x |\, \mathrm dx$ and $\pi_2 = 1 - \pi_1$.

Now, you were asking "given some projectors satisfying the conditions, what observable are we actually measuring"? Well, you have to find a Hermitian operator so that sums of its eigenspaces correspond to the projectors. That still leaves open the question what physical quantity actually corresponds to the Hermitian operator, but that question is also open in the postulates you are familiar with ;)

Finally note that "generalized measurement" usually means something more general than what you quoted. We want to describe also weak measurements and more. For that, there is the framework of generalized measurements where the $\pi$ are not required to be projectors (neither hermitianity nor idempotence is required), only that $\sum \pi^\ast \pi = 1$.


Note that if the projection operators $\pi_{\alpha}$ in question satisfy the postulates above that constitute the definition of an orthogonal projective measurement according to Witten, it is easy to prove that the operators $\pi^{'}_{\alpha}=U\pi_{\alpha}U^{\dagger}$ also constitute a projective measurement. Thus those properties alone do not define a particular set of observables on which we perform measurements, but rather the entirety of operators that perform a projection on a particular orthogonal basis (operators of the form $| \alpha\rangle\langle\alpha|$ with the states $|\alpha \rangle$ belonging to an orthonormal basis).

Witten references in his paper the E type of generalized measurement with the property $\sum_{s}E^{\dagger}_s E_{s}=1$ and now this equation is satisfied by any operators of the form $\pi_{\alpha\beta}=|\alpha\rangle\langle\beta|$, where now the states $|\alpha \rangle$ have to be part of a complete basis but no longer orthonormal or even orthogonal.

What would conclude this proof nicely, is a proof that these are the only types of operators that satisfy the above relations, but I couldn't think of a quick way to prove it.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.