Question is pretty simple, here is the first part.

Consider a spin-1/2 particle. I have two polariser kind of device which detects their spin along the axis of polariser. (I am using polariser in a general sense. You can assume two Stern Gerlach Apparatus which somehow let's you measure the spin state successively) Now, if we put our two polarisers one after other, and would like to study what would it give after these two operations. This successively measuring or operation is denoted by a multiplication of operators in linear algebra. Like if we measure A first then B, total operation on your state is BA. Good.

Now the second part.

All physical operations in Quantum Mechanics are Hermitian Operators as given in the postulates. The converse is also true? All Hermitian Operators can be associated to a physical operation but that physical operation may not be very insightful.

Therefore, these A and B, which are polarising actions (maybe pauli spin X and Y matrices) are physically realisable and are Hermitian indeed.

But when we apply them successively, that is BA, is not a Hermitian matrix and hence not physically realisable. But in the first paragraph, we discussed the exact situation where we applied BA.

So, these questions come out from this

  1. What would we see after applying X and Y on a particle in an experiment? We have read series of Stern Gerlach operations many times in books, so it should be possible, right?

  2. Why do we impose hermiticity on observables still, when there are many cases of non-conservative systems having Non-Hermitian operators. Because BA is realisable and also non-commuting (this is an imposition from my side, otherwise the question wouldn't have existed).


2 Answers 2


This successively measuring or operation is denoted by a multiplication of operators in linear algebra. Like if we measure A first then B, total operation on your state is BA. Good.

No. You're confusing measurement - which is a projection - with the action of an operator. To be explicit: measuring the spin along $+\hat z$ (for instance) is represented by \begin{align} \hat \Pi_{z,+}=\vert +;z\rangle\langle +;z\vert \mapsto \left(\begin{array}{cc} 1&0\\ 0&0\end{array}\right)\, ,\tag{1} \end{align} whereas measuring the spin along $+\hat x$ is represented by the matrix \begin{align} \hat\Pi_x= \frac{1}{2}\left(\vert +;z\rangle- \vert -;z\rangle\right)\left(\langle +;z\vert- \langle -;z\vert\right) \mapsto \frac{1}{2} \left(\begin{array}{cc} 1&-1\\ -1&1 \end{array}\right)\, .\tag{2} \end{align} (Perfect) measurements (the simplest ones anyways) are irreversible and indeed the matrix realizations in (1) and (2) do not have an inverse, and have eigenvalues $1$ or $0$: the eigenvalue $1$ is for positive outcome (i.e. the state is indeed along $+\hat z$ or $+\hat x$, depending on the case), while the eigenvalue $0$ corresponds to a negative outcome.

These eigenvalues are certainly different from those of the hermitian operators $\sigma_z$ or $\sigma_x$, which are $\pm \hbar/2$ in both cases.

Measuring $\hat A$ then $\hat B$ in sequence is NOT the same as measuring $\hat A\hat B$, as you can verify by applying $\hat \Pi_x\hat \Pi_z$ to any state. Indeed, as you point out, $\hat A\hat B$ is in general not hermitian so it would not always be possible to come up with a method to measure $\hat A\hat B$ as a product.

  • $\begingroup$ Just wanted to add from the discussion I had with my professor. The confusion stems from the commutation relation, which is often depicted in the picture I had in my mind. That commuting observables don't change the eigenstates. This is correct in their context but the measuring and preparing of a state called quantum tomography contain, projecting on an observable's eigenvectors, as you mentioned. And consecutive measurement is formalised using projection on density operator we would get after each measurement. Very different from measuring something called AB, [A,B] which maybe not realised. $\endgroup$
    – Prabhat
    Jun 23, 2022 at 8:52

In this specific case BA = XY ~ Z, the Pauli Z matrix. There is a global phase $i$ on this term, but the product is still Hermitian.

Yes, measuring X and Y with the Stern-Gerlach (SG) experiment is possible - It is not different from the regular SG experiment where you first measure Z and then X. Non-eigenstates are measured in the exact same way - we expect the same result.

In general, non-hermitian operators can have complex expectation values. We insist all operators must be Hermitian because we never measure complex numbers.

There are examples of non-hermitian operators in physics, but these still necessarily have real eigenvalues.

Non-conserving operators (non-unitary operators e.g. depolarizing noise) are generated by non-hermitian operators. The operations just appear non-unitary because we do not have enough information about the whole system (the thermal bath in the case of depolarizing noise).


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