46
$\begingroup$

Why can't we measure imaginary numbers? I mean, we can take the projection of a complex wave to be the "viewable" part, so why are imaginary numbers given this immeasurable descriptor? Namely with operators in quantum mechanics, why must measurable quantities be Hermitian, and consequently real?

$\endgroup$
4
  • 4
    $\begingroup$ This question (v2) seems to be about quantum observables & complex numbers. More on wave functions & complex numbers in this and this Phys.SE posts. More on physics & complex numbers in this Phys.SE post. $\endgroup$
    – Qmechanic
    Commented Oct 29, 2013 at 10:21
  • 1
    $\begingroup$ I remember Aharonov mentioning with weak measurements you could measure complex/imaginary values. He was somehow alluding to pertubation theory but I stayed ignorant. $\endgroup$
    – lalala
    Commented Jun 18, 2019 at 20:35
  • 2
    $\begingroup$ Who says we can't? You can measure AC electricity phase, for example. $\endgroup$ Commented Mar 6, 2020 at 14:20
  • $\begingroup$ I am not sure I understand what you are asking. Every quantum measurement is an energy, momentum, angular momentum and charge transfer between two systems. At most you can ask how von Neumann's Hilbert space description of quantum mechanics represents the ensemble averages/distributions of these measurements. $\endgroup$ Commented Oct 19, 2022 at 10:13

7 Answers 7

59
$\begingroup$

I) Well, one can identify a complex-valued observable with a normal operator

$$\tag{1} A^{\dagger}A~=~AA^{\dagger}.$$

A version$^1$ of the spectral theorem states that an operator $A$ is orthonormally diagonalizable iff $A$ is a normal operator.

Thus, normal operators are the only kind of operators that we can consistently extract measurements [i.e. eigenstates and (possibly complex) eigenvalues] from.

II) But notice that a normal operator

$$\tag{2} A~=~B+iC$$

can uniquely$^2$ be written as a sum of two commuting self-adjoint operators

$$\tag{3} B^{\dagger}~=~B, \qquad C^{\dagger}~=~C, \qquad [B,C]~=~0. $$

($B$ and $C$ are the operator analogue of decomposing a complex number $z=x+iy\in\mathbb{C}$ in real and imaginary part $x,y\in\mathbb{R}$.) Conversely, two commuting self-adjoint operators, $B$ and $C$, can be packed into a normal operator $(2)$. We stress that the commutativity of $B$ and $C$ precisely encodes the normality condition $(1)$.

Since the self-adjoint operators $B$ and $C$ commute, they can be orthonormally diagonalized simultaneously, i.e. the corresponding pair $(B,C)$ of real-valued observables may be measured simultaneously. This fact is consistent with the Heisenberg uncertainty principle applied to the operators $B$ and $C$.

We conclude that a normal operator does not lead to anything fundamentally new which couldn't have been covered by a commuting pair of standard real-valued observables, i.e. self-adjoint operators. For this reason, the possibility of using normal operators as complex observables is rarely mentioned when discussing the postulates of quantum mechanics.

For more on real-valued observables, see, e.g. this Phys.SE post and links therein.

--

$^1$ We will ignore subtleties with unbounded operators, domains, selfadjoint extensions, etc., in this answer.

$^2$ The unique formulas are $B=\dfrac{A+A^{\dagger}}{2}$ and $C=\dfrac{A-A^{\dagger}}{2i}$.

$\endgroup$
20
$\begingroup$

As a mathematical structure, the field of complex numbers does not admit an order relation which is an extension of the order we have in $\mathbb{R}$.

This means that there is absolutely no way of saying if $5+3i$ is bigger or smaller than $5+6i$ for example. We just know it is not equal and we have to stop here.

Therefore it is physically really hard (actually impossible) to compare "observables" having as eigenvalues complex numbers.

We could not tell anymore which particle has a bigger mass, a smaller energy and so on.

I believe that taking the real field as the primary field in which the measure results take values is just a matter of convenience. You could try to create a sort of quantum mechanics with complex eigenvalues, but then you could not fit experiments anymore and you model turns extremely less predictive.

Anyways, I have read in The Road to Reality by Penrose that some physicists considered as numerical fields somethinig like the cyclic $\mathbb{Z}_p$ with $p$ prime and extremely big. As it is not clear if this can lead to new physics, we just stick with $\mathbb{R}$.

That's it, as far as I understand the problem.

$\endgroup$
14
  • 1
    $\begingroup$ I don't understand your point, sorry. Could you make it clearer? I am saying that in the end, to fit experimental data, you will need "numbers" that can be compared to each other. From an experimental point of view, we are pretty sure that a $Z_0$ flying at $0.5 c$ has a far bigger energy than an electron flying at the same speed. Whatever the model I assume, the final theoretical prediction has to be comparable i.e has to take values on a field of numbers which has an order relation. However, sadly, all field extensions of $\mathbb{R}$ (the complex, quaternions, octonions) do not have it. $\endgroup$ Commented Oct 28, 2013 at 23:46
  • 6
    $\begingroup$ Well, the point is as follows: Say, you have a field $\vec E = \{0,1,2\}$. Foundamentally it is a vector. I guess, there is no problem in comparing it to some other field, say $\vec E'=\{2,1,0\}$. So, on foundamental level we are sometimes dealing with quantities, which are not ordered, hence not comparable in your sense. Why then do we need to be always able to compare observed, measured quantities to each other? What is different between observed and foundamental quantities, that one need to be ordered, while others don't? $\endgroup$ Commented Oct 28, 2013 at 23:52
  • 3
    $\begingroup$ This is all true. But then I read it as: sometimes we need to compare quantities, hence some measurements should be numbers. Then, why can't some other measurements be, say, vectors? I guess then your thesis statement is "Among all posible measurements we need to have those which produce reals, in order to be able to compare them with each other". $\endgroup$ Commented Oct 29, 2013 at 1:36
  • 1
    $\begingroup$ Oh wow only now I fully understand what you say! Yes, exactly. In principles the measure results could be reals and "other things". But we must have at least the reals, and we shall never compare the "other things". Now I see. Thanks for sharing your opinions! $\endgroup$ Commented Oct 29, 2013 at 1:39
  • 1
    $\begingroup$ A small remark: I think you are confusing finite fields $\mathbb F_p = \mathbb Z/p\mathbb Z$, which are cyclic as groups, but don't have a norm compatible with the algebraic structure, and the p-adic fields $\mathbb Q_p$ (and their ring of integers $\mathbb Z_p$) which are harder to define, but which are complete normed fields in which calculus can be done. $\endgroup$
    – doetoe
    Commented Oct 29, 2013 at 2:07
13
$\begingroup$

Imaginary numbers can be represented as pairs of real numbers. You can also make a device, which mixes the measurement outcomes of two reals on hardware level to produce complex "amplitude" and "phase" as outcomes, which you further might call as measuring a complex number.

More generally, any measurement is eventually reading off the values on the indicators of your instruments. Those are numbers, hence reals. However, they can also be sets (arrays) of reals, as is the case for cameras, for example. So, perhaps, most general statement would be that one can measure quantities, which are expressible as a set of real numbers.

$\endgroup$
3
$\begingroup$

Since the complex numbers are quite abstract term, and do not have physical representation, it is still possible to see them, although only imaginary part or only real part of some measurement is not going to give full information. Only full complex number represents a full information.

Now, I said it is possible to really measure imaginary and real values. While it is not quantum mechanics, QA modulation is a good example how you can really measure imaginary and real parts of a signal.

$\endgroup$
2
$\begingroup$

At least complex energy and imaginary time are used in quantum mechanics. Complex energy to describe non-stationary processes. Imaginary time is used in Landau Lifshits’s book, Volume 3, Problem 3 to paragraph 77. At the same time, such words are used: “The imaginary value of a moment in time $\tau_0$ corresponds to the classical impracticability of the process” $$W=\exp\left[-2\mathrm{Im}\left(\int_{\tau}^{\tau_0}\frac{4F^2}{\Omega^2}\sin^2\Omega u\mathrm{d}u+\tau_0\right)\right]$$ But to use the complex eigenvalues of quantum mechanics operators, it is necessary to use non-self-adjoint operators, and then the eigenvalues can turn out to be complex. In complex space, the energy and momentum operators are general operators, non-self-adjoint. $$\hat H=\sum_{k=1}^3 -\frac{\hbar^2}{2m}\frac{\partial^2 }{\partial z_k^2},z_k=\mathrm{Re}\,z_k+i \mathrm{Im}\,z_k$$ $$\hat p_r=-i\hbar(\frac{\partial }{\partial r}+\frac{1}{r})$$ The eigenfunction of the radial part of the momentum operator in three-dimensional space is equal to $\psi=exp(ip_r r/\hbar)/r$. The eigenvalues of these operators can be complex. When writing a solution for the wave function in the complex plane there is a problem. When using real space, there is a non-damping solution only with real coordinates. Similarly, in a complex plane with a complex eigenvalue, a non-damping solution exists at a certain phase of the complex coordinate. We need to think about the physical meaning of the complex solution. In hydrodynamics, the physical meaning of the imaginary part is the standard deviation. In quantum mechanics, apparently also. You need to measure the constant term described by the real part and the variable one, the fading away term described by the imaginary part. Complex energy and momentum describes the localization in time and in space, respectively, energy and momentum. The imaginary part of the complex energy value is determined from the lifetime of the system. The imaginary part of the impulse is determined from the known complex value of the energy and equations $E^2=p^2c^2+m^2c^4$.

$\endgroup$
1
$\begingroup$

Well "imaginary numbers" are just that; imaginary. We made them up to allow for the square root of a negative number. Likewise, we made up complex numbers.

The problem is NOT with the imaginary, or complex number. The difficulty is with the interpretation that we CHOOSE to use for an imaginary or complex number.

For example in AC electrical circuit problems, we consider multiplying by -1 to be the equivalent of a 180 degree rotation of a vector that represents say a Voltage. It is then just a small jump to suggest multiplying by the square root of -1 (i or j) to be equivalent to rotating the vector through half of 180, or 90 degrees, since multiplying twice by i (or j) gives us the same rotation of twice 90 degrees or 180 degrees.

So the mathematics, is nothing more than a convention for describing a 90 degree vector rotation, or a combination of two vectors at right angles. It's no more mysterious than the Germans using a capital letter for ALL nouns. It's just the way they do it.

So the REALITY of imaginary, or complex numbers, is nothing more than our definition of how WE HUMANS interpret them or their use. There is NO universal mysterious reason at all.

$\endgroup$
4
  • 7
    $\begingroup$ Yes, but, we "made up" the negative numbers to allow an answer to, e.g., "what is the number that, when 3 is added equals 2" and we "made up" the rational numbers to allow an answer to , e.g., "what is the number that, when multiplied by 2 equals 1". By that reckoning, these numbers are "imaginary" too. $\endgroup$ Commented Oct 29, 2013 at 2:12
  • $\begingroup$ """...Yes, but, we "made up" the negative numbers...""" So if I understand you, after we made up negative numbers (we did), you now wish to rename them "imaginary numbers" and also rename rational numbers too ?? Given that we made them all up (everything in mathematics), why would you now want to rename everything, to "imaginary", simply because we made them up; what's the point ?? $\endgroup$
    – user26165
    Commented Oct 29, 2013 at 5:28
  • 6
    $\begingroup$ The point is that complex numbers are made up in exactly the same way as all other types of numbers: by defining the mathematical rules that govern them. Your first sentence suggests that "imaginery" in "imaginery numbers" should be interpreted in the usual English sense of the word. $\endgroup$
    – MSalters
    Commented Oct 29, 2013 at 14:09
  • $\begingroup$ Well it is very common to use words in the English language, to have the meaning that is conventionally assigned to them, by usage and dictionary definitions. Now why would I use the word "imaginary" to have some different meaning; what would be the purpose ? So you too evidently, would prefer to call imaginary numbers, by some other name. Did you catch this in my answer: """...So the REALITY of imaginary, or complex numbers, is nothing more than our definition of how WE HUMANS interpret them or their use...."" ? You should answer OP's question. $\endgroup$
    – user26165
    Commented Oct 30, 2013 at 5:53
0
$\begingroup$

Actually you can measure imaginary numbers by measuring separately the real and the imaginary part. However this is only possible in classical mechanics. In quantum mechanics measuring the two parts simultaneously is not possible, because the first measurement would necessarily change the outcome of the second measurement, as Dirac explains nicely in his book:

"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 alright 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 indeterminancy that will affect the second." (P.A.M Dirac, The principles of quantum mechanics, §10, p.35)

$\endgroup$
6
  • $\begingroup$ Dear asmaier, it is usually frown upon to directly copy-paste identical answers. (The problem is if everybody start to copy-paste identical answers en mass.) In general in such situations, please consider one of the following options: (i) Delete three of your answers. (ii) Flag for duplicate posts and delete three of your answers. (iii) If you think the four posts are not duplicates, then personalize each answer to address the four different specific questions. $\endgroup$
    – Qmechanic
    Commented Nov 2, 2013 at 23:14
  • $\begingroup$ Dear Qmechanic, isn't is also frown upon to copy-paste identical comments? ;-) However I admit, that my answers were too similar. So I tried to follow your suggestion (iii) and personalized my answers to address the specific question in a better way. However I still believe the quote from Dirac is very relevant and important, so I will refer to it in every answer. $\endgroup$
    – asmaier
    Commented Nov 3, 2013 at 14:49
  • $\begingroup$ Dear asmaier, Dirac's quote fits perfectly in this answer, because OP's post is here about quantum observables & complex numbers. Dirac's quote is not about wave functions & complex numbers, nor more generally about physics & complex numbers, so please remove the quote from the other three answers. The four answers are still too identical. $\endgroup$
    – Qmechanic
    Commented Nov 3, 2013 at 15:37
  • $\begingroup$ Dear Qmechanic, I believe otherwise and I'm willing to take the risk and let the crowd decide if my answers are relevant or not for the questions. I don't care about the reputation points. If I get any, I promise I will donate them to Dirac. $\endgroup$
    – asmaier
    Commented Nov 3, 2013 at 15:52
  • $\begingroup$ Well, if it were just about your posts, I would look the other way. The problem is that it is not just about your posts. It's about principles. The problem is if everybody start to quadruple their answers. Note that the 4 posts are already linked, so a serious student of the subject would not miss you answer even if it is only posted one time. $\endgroup$
    – Qmechanic
    Commented Nov 3, 2013 at 18:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.