In quantum mechanics, operators can only be observables if the eigenfunctions they operate on have real eigenvalues. If they are complex, I am told that, surely, some observable of a physical system cannot be an imaginary number. This makes some innate sense to me, but I feel I am forced to acknowledge the "real-ness" of imaginary numbers. For instance, an eigenstate to the infinite square well cannot be described without complex time. But why are we forced to recognize the significance of complex numbers here and refuse it for something like an eigenvalue? When I'm doing math, real and complex numbers are just elements of a field, yet when we're dealing with real life observables, reals are then given some new preference. Why is this? It seems like we're using complex numbers when it's convenient, and then saying that they don't exist beyond mathematical trickery.

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    $\begingroup$ The day you invent a ruler that can measure $1+i\sqrt2$ meters let me know. $\endgroup$ Oct 23, 2018 at 22:29
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    $\begingroup$ But why is $1$ meters more real than $1 + i \sqrt{2}$ meters? If I want to invent a ruler than can measure $1 + i \sqrt{2}$ meters, I'll just make a tick mark somewhere along the ruler and write $1 + i \sqrt{2}$. We've decided how long $1$ meter is and $2$ meters and so on, but why is this any more real than $1 + i \sqrt{2}$ other than by our own convention? How is it innately more real? $\endgroup$
    – sangstar
    Oct 23, 2018 at 22:33
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    $\begingroup$ Any ruler induces a total ordering, but $\mathbb C$ does not admit such an ordering, so no such ruler may exist. So no: cannot invent a ruler that measures $1+i\sqrt2$ meters. $\endgroup$ Oct 23, 2018 at 22:36
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    $\begingroup$ You can use an imaginary number to represent a point in 2D, with magnitude and phase for example. These quantities can be added, subtracted etc. In QM the electron is constrained by an EM field which gives it wave properties ( like a guitar string) and we can only observe averages of behaviour (like energy level transitions) so the complex part helps with magnitude of these properties. $\endgroup$ Oct 23, 2018 at 22:52
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    $\begingroup$ More on complex numbers and physics: physics.stackexchange.com/q/11396/2451 , physics.stackexchange.com/q/76595/2451 , physics.stackexchange.com/q/82613/2451 $\endgroup$
    – Qmechanic
    Oct 24, 2018 at 8:54

5 Answers 5


First, note that physical measurements are fundamentally comparisons between some property of a system (i.e. the observable) and a set of reference quantities. Consider a physicist armed with a meter stick, a pendulum, and a mass balance - these constitute her references. Experimental questions might include

  • How many meter sticks away from me is that tree?
  • How many pendulum swings will it take for this rock to hit the ground if I drop it from the roof of my house?
  • How many cups of water (obviously held in a container) do I have to add to one arm of my balance to balance out this rock, which is sitting on the other arm?

From these questions, it should be clear that at the very least, we need to use a field such as $\mathbb{Q}$, as we need some notion of fractional parts. It's less clear, however, that we need more than this. After all, no direct physical measurement will ever yield an irrational number, because all measurements have finite resolution.

There are a number of reasons why we might come to regard $\mathbb{Q}$ as insufficient, but the one that immediately springs to mind for me is that $\mathbb{Q}$ is not complete. Imagine that our physicist constructs two identical cubic containers, each with side length $L\in\mathbb{Q}$. She fills both completely with water, and now sets out to construct a cubic container which would be just large enough to hold all of the water from the first two cubes.

If we restrict ourselves to observables in $\mathbb{Q}$, then this is not possible even in principle - such a cube would necessarily have side length $L\sqrt[3]{2}\notin\mathbb{Q}$. To avoid philosophical issues like this, as well as more technical mathematical details related to convergence, we should at the very least fill in the "holes" in $\mathbb{Q}$.

There is another requirement we have implicitly placed on our observables. If the idea of a measurement as a comparison to a reference is to hold water, then given any distances (or times, or masses, or whatever) $A,B,$ and $C$, I should be able to say the following:

  1. Either $A\leq B$ or $B\leq A$
  2. If $A\leq B$ and $B\leq A$ then $A=B$
  3. If $A\leq B$ and $B\leq C$, then $A\leq C$

This is the definition of a total order. The absence of such an ordering would mean that comparisons between observable quantities would not be universally well-defined, so that observables take their values in a totally ordered set is a natural requirement to impose.

$\mathbb{C}$ is an example of a field which cannot be equipped with a total order which respects its field structure. If lengths are allowed to be complex, then comparing the lengths of two objects (or their positions relative to some fixed point) become either ill-defined and/or inconsistent (e.g. if we add the same number to two equal lengths, the results may no longer be equal). This rules out $\mathbb{C}$ as a viable candidate for observables.

From the above arguments, we can conclude that if our observables take their values in some set $S$, then that set should be: (1) a field, (2) have no "holes" (i.e. be complete), and (3) equipped with a total order which is compatible with the field structure. If we take the philosophical considerations expressed above as axiomatic requirements for observables, then there is no freedom left for our choice - there is precisely one mathematical structure (up to isomorphism) which satisfies those requirements, and that is $\mathbb{R}$.

  • $\begingroup$ Ordering is not in general a requirement of measurement. Example: measure a two-state system such as electron spin. The only requirement is the ability to distinguish the two states. $\endgroup$
    – isometry
    Oct 26, 2018 at 21:42
  • $\begingroup$ @BruceGreetham There are no measurement devices which directly measure the state of a quantum mechanical system - they measure properties of a state (in this case, angular momentum or energy), which are ordered. I specifically did not bring up quantum mechanics because observables taking values in $\mathbb{R}$ is specifically baked in to the formalism, so it can be a bit circular. $\endgroup$
    – J. Murray
    Oct 26, 2018 at 22:00
  • $\begingroup$ OK, if you insist, then I change my example to any two-state classical measurement which has no particular ordering (black/white, male/female, in/out, above/below ... ) $\endgroup$
    – isometry
    Oct 26, 2018 at 22:24
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    $\begingroup$ @BruceGreetham I am talking about the definition of an observable, not a descriptor. They are not the same thing. In/out and above/below are descriptions which come down to measurement of relative position. Black/white and male/female are much too complicated to be reduced to the level of fundamental, direct physical observables like position, momentum, and energy. $\endgroup$
    – J. Murray
    Oct 26, 2018 at 22:42

I have a very different take on this than the other commenters: the reason why we take real numbers is because it is convenient, without being limiting.

On the other hand, expectations etc. are computed by taking weighted averages with real coefficients, so it should at least be a vector space over $\mathbb R$ if you don't want to change the interpretation of the wave function. This already shows that you won't gain anything by allowing more general values, you could always just replace that by a tuple of real numbers.

One could try to argue that physical quantities necessarily are real numbers, but that is only because we describe them to be. What we really work with is possible outcomes of measurement devices, like the mark saying $1 + i\sqrt2$ on the ruler in your comment. It doesn't matter if we assign to the spin states of an electron in the $z$-direction the values $\pm1$, $\{0,1\}$, $\pm\frac12$ or $\{\pi,i\}$. Neither does it matter if we express position in the plane by a coordinate pair in $\mathbb R^2$, a coordinate in $\mathbb C$, a pair of polar coordinates, etc. It is in interpreting these values that things can change, e.g. expectations, dynamics, but its validity is not affected.

An apparent issue with using complex numbers could be that the associated operator is not hermitian, but that doesn't make any real difference (if you want you could develop the same formalism using normal matrices, which have all properties you want that Hermitian matrices have, except it allows complex eigenvalues, but why would you), it just complicates things, and you could do the same with a pair of Hermitian matrices.

Summarizing, we don't want to mess with the interpretation of the wave function, so it must be possible to take positive real convex combinations of eigenvalues, so they must live in a real vector space. If you choose a real basis for such a vector space, every coordinate is an observable, so restricting to real numbers covers the general case.


After so many answers, perhaps nothing more remains to be said... However some points could be deepened a little.

E.g., J. Murray says

There are a number of reasons why we might come to regard $\Bbb Q$ as insufficient, but the one that immediately springs to mind for me is that $\Bbb Q$ is not complete.

I agree, and can give a more fundamental example than two cubes into one. I'm thinking of the law of free fall: $$s = {\textstyle{1 \over 2}}\,g\,t^2.$$ It is natural, maybe necessary, to require that for all positive values of $s$ a $t$ exists, and this cannot be accomodated within $\Bbb Q$.

And there is more, even if we remain in elementary mechanics. The requirement I issued for free fall should be extended to any law of motion $x=f(t)$, in this form (continuity):

If a body is found in $x=x_1$ at time $t=t_1$, and in $x=x_2$ at time $t=t_2$, for all $x_3\in[x_1,x_2]$ there exists $t_3$ such that $x_3=f(t_3)$.

This holds if $f:\Bbb R\to\Bbb R$, but not if $f:\Bbb Q\to\Bbb Q$.

On the other hand, it we think of experiments, $\Bbb Q$ is much more than required:

After all, no direct physical measurement will ever yield an irrational number, because all measurements have finite resolution.

To me, this shows that a complete field is a purely theoretical requirement. It is not experimental physics which needs so rich a structure. It is our theoretical model.

When it comes to complex numbers, my ideas are less clear. It is obvious that a dominant sector of modern physics has been built on $\Bbb C$ and that it's necessary. Think e.g. of polarization states of photons: it is a 2D space, but on $\Bbb C$. Otherwise we cannot explain how circular and elliptical polarizations exist, and can be obtained as superpositions of linear ones.

It is not true that after all $\Bbb C$ is isomorphic to ${\Bbb R}^2$. This holds for the vector space structure, but not for the field structure. Moreover, $\Bbb C$ owns an anti-automorphism (conjugation) which in QM acts as time reversal.

There is the argument of total ordering. I understand it, but am not entirely convinced. I would not exclude that a less stringent (topological?) requirement may suffice to establish a link between theory and measurements.


As an experimental physicist: the only possible measurements are into a one to one correspondence with the field of real numbers.

Start by counting the planets around the sun. The measurement is described by one real number, because it is a function of one parameter answering the question (how many).

Measure the distance from Boston to New York. It will be in a one to one correspondence with the real numbers, again answering a single question,( how long).

Then we come to the three variables (x,y,z) in space : they can always be reduced to one real number by taking the length of the vector. It is in special relativity that the necessity of differentiating between two types of variables, both mapping to the real numbers but independent as parameters, becomes important and the introduction of a complex number algebra simplifies matters. One could stick to four vectors and ignore the need for a square root of (-1) to end up with a real measurable number, but the algebra would really get impossible.

In physics the mathematical models used to predict measurable real numbers end up with a lot of square roots of (-1), and the algebra of complex number is simplifying and useful. This is particularly true for wave equations.

reals are then given some new preference. Why is this? It seems like we're using complex numbers when it's convenient, and then saying that they don't exist beyond mathematical trickery.

It is not mathematical trickery, it is mathematical simplification clarifying functional dependences which would be complicated just sticking with real number models. Quoting from an answer here

While we'll probably keep doing quantum mechanics in terms of complex realizations, one should keep in mind that the theory can be made manifestly real. This fact shouldn't really surprise anyone who has taken the bird's eye view instead of just looking throught the blinders of specific formalisms.


The field of complex numbersis isomorphic to the 2D vector plane, not to the reals. Thus you can use a complex ruler to measure distances and directions in 2D space. The information of that ruler will contain both, lenght and direction. The ruler cannot be a line though, it has to be 2D. Or you can have multiple linear rulers with different complex labels, and use each ruler only for a specific angle. For instance, the ruler used to measure at a 45 deg angle will be labeled $1+1i$ when the distance from the origin is $\sqrt 2$, etc.


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