This question is going to look a lot like a duplicate, but I've read dozens of related posts and they don't touch the subject. Here we go.

Why are observables represented by hermitian operators?

  1. Because then we'll measure real stuff, since the eigenvalues are real;

  2. Because hermitian operators (actually normal operators) have a complete set of eigenvectors, and we'll be sure the accessible states of our experiment will be modelled consistently;

Those are the two main reasons for the Physicists of the early XX century to have chosen hermitian operators as a mathematical model of what we can observe: mathematics proves they provide both the needed properties.

My question is of philosophical inclination: which one of the above requirements is more fundamental? I'm inclined to think the most important property is to be able to represent the system by the right model, which is II). Real eigenvalues come as a bonus. I've read that having real eigenvalues is more important, but I don't see why they couldn't be complex... Maybe Nature is tricking us some way and we only measure their real part. The problem for me would be describing the system with a model that doesn't mirror our accessible states, and is therefore flawed.

Maybe both are fundamental on the same level. Maybe the question is stupid. What do you think?


2 Answers 2


Hermitian operators (or more correctly in the infinite dimensional case, self-adjoint operators) are used not because measurements must use real numbers, but rather because we almost always decide to use real numbers.

As the OP mentions at one point, you might choose to use complex numbers to label a two-dimensional screen, and in that case you'll be able to use a so-called normal operator to represent the 2-dimensional observable. (Contrary to what Dirac thought, nothing whatsoever goes wrong here.)

It should not be too hard then to accept my next claim: You can use whatever measurement scale you want to measure a quantum observable! You can label pointer positions with items of fruit if you want to, and you can still build a perfectly legitimate observable.

There is no question that the reals and complexes have enormous advantages over other more arbitrary measurement scales (due to their rich internal structure which we capitalize on in the functional analysis), but the idea that real numbers are somehow endowed with a prestigious metaphysical status is baloney.

How to define an observable with any measurement scale you want to

Step 1. Set up a bunch of particle detectors

Step 2. Attach a label to each detector

The set of labels we'll denote by $\Omega$. Examples include: $$\Omega = \{0,1\},\mathbb{R},\mathbb{C},\{\heartsuit,\clubsuit,\diamondsuit,\spadesuit\}$$

Step 3. Write out the list of all possible events

By event, I mean a subset $\Delta$ of $\Omega$ that represents a possible question like "Did a detector in $\Delta$ fire?". We'll label the event structure $\Sigma$, e.g. $$\Sigma = \{\emptyset,\heartsuit,\clubsuit,\diamondsuit,\spadesuit,\heartsuit\clubsuit,\heartsuit\diamondsuit, \cdots,\heartsuit\clubsuit\diamondsuit\spadesuit\}$$

Step 4. Associate each event in $\Sigma$ with a projection operator

This is the hard bit, and there's no recipe for it. But you have to make sure that the family of projectors forms a Boolean algebra that perfectly mirrors the natural algebra of $\Sigma$.

We'll call the association $\sigma$, so that $\sigma:\Sigma\to\mathscr{P}(\mathscr{H})$.

And that's basically it! The object $\sigma$ (technically, a Projection Valued Measure on $\langle\Omega,\Sigma\rangle$) is a quantum observable. It contains all the probabilistic information you need to calculate the probability measure on your chosen measurement scale for any quantum state.

For example, suppose the state of the system is $\rho$ and you want the probability that a detector $\Delta\in\Sigma$ fires. The desired probability is just $p=tr[\rho\sigma(\Delta)]$.

What the heck does this have to do with Self-Adjoint operators?

Are you ready for the climax? Here it is...

IF you choose to use the measurement scale $\langle\mathbb{R},\mathscr{B}(\mathbb{R})\rangle$, THEN you will be able to build a self-adjoint operator which is precisely equivalent (in terms of the information it stores) to the PVM you constructed.

IF you choose a detection screen calibrated by $\langle\mathbb{C},\mathscr{B}(\mathbb{C})\rangle$, THEN repeat the above sentence replacing 'self-adjoint' with 'normal'.

(Neither of the above statements is obvious, by the way. They are famous results in Functional Analysis known as the Spectral Theorems.)

IF you choose to be a Fancy Nancy and use $\{\heartsuit,\clubsuit,\diamondsuit,\spadesuit\}$ for a measurement scale (with its power set for the event structure) THEN the fruits of your labors are more modest. In particular, you still get the answers to any questions you care to ask, but you don't get any neat operator to give you computational shortcuts. Instead you will forever be doing calculations like $p=tr[\rho\sigma(\heartsuit\clubsuit)]$.

I haven't even touched eigenvectors yet, but suffice it to say that they also do not have a fundamental status in the theory.

There is no doubt that we can learn something by reading the works of the great masters, but taking that work as the state of play can send you back a century. We've learned a lot since Einstein and Dirac.

  • 2
    $\begingroup$ Yes, I always though that statement of Dirac's to be uncharacteristically limp! I always wondered what he thought was wrong with pairs observables with common eigenvectors / invariant spaces such that one would give the real, the other imaginary part. I'm going to have a deck of cards always ready now in the lab whenever I do measurements now! Great answer. $\endgroup$ Commented Jun 27, 2016 at 0:29
  • 2
    $\begingroup$ @Physics-Footnotes Very nice answer! Can you provide some book or some other reference for this? Reading different places clears any leftover confusion. Thanks! $\endgroup$
    – user29978
    Commented Apr 3, 2020 at 10:57
  • $\begingroup$ I know this answer is old but does anyone know how $\rho$ looks in terms of these abstract labels like {♡,♣,♢,♠}? I have a hard time imaging how one would calculate $tr[\rho\sigma(\Delta)]$ without writing $\rho$ as a matrix. $\endgroup$ Commented Aug 8, 2022 at 12:26
  • $\begingroup$ @AccidentalTaylorExpansion What do you mean? Yes, $\rho$ is an operator, just like $\sigma(\ldots)$ is; the symbols are labels, and $\sigma(\ldots)$ is a projection operator. What the answer, neatly and correctly, points out is that you don't have to label these by real numbers, as "usually" done in QM. $\endgroup$ Commented Dec 19, 2023 at 13:04
  • $\begingroup$ I'd still be interested in a reference that explains this in slightly more detail. $\endgroup$
    – user56834
    Commented May 7 at 13:53

When in doubt go back to the masters. From Dirac's Principles of QM

When we make an Observation we measure some dynamical variable. It is obvious physically that the result of such a measurement must always be a real number, so we should expect that any dynamical. variable that we can measure must be a real dynamical variable. One might think one could measure a complex dynamical variable by measuring separately its real and pure imaginary Parts. But this would involve two measurements or two observations, which would be all right in classical mechanics, but would not do in quantum mechanics, where two observations in general interfere with one another-it is not in general permissible to consider that two observations can be made exactly simultaneously, and if they are made in quick succession the first will usually disturb the state of the System and introduce an indeterminacy that will affect the second. We therefore have to restrict the dynamical variables that we can measure to be real..

According to Dirac property I in your question is more fundamental. Since Hermitian matrices are normal property II is satisfied. He also explains why a real dynamical variable should be a Hermitian matrix by starting from simple assumptions in his book. The book is quite old and is available for free.

  • $\begingroup$ He's talking about measuring and therefore is already implying we have a model to apply measurement to. If we have a model then eigenvectors are complete. To assume mathematical exactness is already assume that the operators are normal. It still looks to me as II is more fundamental. $\endgroup$ Commented Jun 24, 2016 at 5:17
  • $\begingroup$ This is not a convincing argument to me. Maybe the imaginary part collapses also and is just never observed...something like an imaginary dimension that is inaccessible for measurement. Frequently we take the real part of systems and discard the imaginary; I see nothing here any different than wave function collapse hand-waiving. $\endgroup$
    – anon01
    Commented Jun 24, 2016 at 5:49
  • $\begingroup$ @ ConfusinglyCuriousTheThird I am not sure if I understand your comment correctly but if you can never observe the imaginary part and it has no influence on the real world, then I think Ockham's razor comes into the game. Why should we model this non influential variable at all? $\endgroup$
    – Noldig
    Commented Jun 24, 2016 at 10:03
  • $\begingroup$ @Noldig This has nothing to do with Ockham's Razor. There could plausibly be theories that predict real variables with unobserved imaginary artifacts. It is often the case that we use complex exponentials for phasors or vector fields for which we only care ab lout the evolution of the real part, and its not clear to me from the passage above why this couldn't be the case for observables in QM $\endgroup$
    – anon01
    Commented Jun 24, 2016 at 16:10

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