So I had this Problem when I had to learn about classical electromagnetism: Why is it, that we use complex numbers when calculating stuff, but in the end only the real part is important (for example for the electric field and hence for the poynting vector)?

On the other hand in quantum mechanics the imaginary part has a role to play, as the expected value depends on it.

In Griffith's Introduction To Quantum Mechanics I read:

"Incidentally, in electrodynamics we would write the azimuthal function in terms of sines und cosines, instead of [complex] exponentials, because electric potentials must be real. In quantum mechanics there is no such constraint..."

So potentials in quantum mechanics don't have to be real. But why not? And electric potentials have to be real. Why is that? And if the electric potentials have to be real, why then work with complex numbers and in the end just forget about the imaginary part? Or on a different level: What is the role imaginary numbers play in physics? I never understood the complex numbers. After some time I could calculate with them, but I never really got what they MEAN. Can somebody help?

  • $\begingroup$ What does Griffiths use a complex quantum mechanical potential for in a book about electrodynamics (I never liked the book, so I don't have a copy)? Can you give a longer citation? That one wouldn't use complex numbers in electrodynamics is complete nonsense, for sure, even for the electromagnetic potential. I bet he uses them just a couple chapters later himself. $\endgroup$
    – CuriousOne
    Commented Jun 16, 2016 at 8:26
  • 3
    $\begingroup$ Griffiths is referring to the wavefunction not to any potential. $\endgroup$ Commented Jun 16, 2016 at 9:28
  • $\begingroup$ Related: physics.stackexchange.com/q/8062/2451 , physics.stackexchange.com/q/17168/2451 and links therein. $\endgroup$
    – Qmechanic
    Commented Jun 16, 2016 at 11:39

2 Answers 2


Short (but cryptic) answer: complex numbers arise in quantum mechanics because we would like find solutions to the differential equation $$\frac{\partial}{\partial x}f(x) = cf(x)$$ which don't blow up as $x\to \pm\infty$.

Long answer:

Fundamentally, the shift from classical mechanics to quantum mechanics is replacing functions (observables) and numbers (states) by matrices (quantum observables) and vectors (quantum states). The first examples of this (say, in Griffiths) is where the vectors are complex valued functions and the "matrices" are differential operators which act on the space of such functions. As in classical mechanics, in order to make sense of measured quantities, we would like the results of measurements to be real, which is analogous to requiring that the eigenvalues of the matrices which correspond to our quantum observables be real (in other words, the matrices must be Hermitian). As long as the final, measured answer is real there is no reason to exclude the use of complex numbers.

However, having no reason to exclude something is very different from having a reason for it! So why do the complex numbers appear in quantum mechanics? This can probably be tied, at least in part, to the relationship between the position and momentum operators. The position operator is not hard to explain. If we take the wavefunction $\psi$ (really, the square of its norm, $|\psi|^2$) as a probability distribution telling us where a particle is likely to be, then the integral $$\langle x \rangle = \int x\,|\psi(x)|^2dx$$ measures the expected value of the position of the particle. However, the momentum operator $\hat{p} = i\frac{\partial}{\partial x}$ is more difficult to interpret. One possibility is that you know that the momentum of the particle (wave) should be related to the derivative of the wavefunction. But then you would check that, due to the minus sign acquired when doing integration by parts, the first derivative operator $\frac{\partial}{\partial x}$ is not Hermitian (this is the part where we secretly require that our wavefunctions not blow up as $x\to \pm\infty$). In fact, it is as far from having real eigenvalues as possible, all its eigenvalues are purely imaginary! Of course, we can fix this by multiplying by $i$, but we have had to introduce complex numbers into our theory in order to do so!

Furthermore, this definition of $\hat{p}$ seems almost inevitable. Once you know that the position operator and momentum operator must have commutator equal to a constant $$[\hat{x},\hat{p}] = c$$ you have pretty much forced $\hat{p}$ to be a first order differential operator. The Hermitian condition then forces us to introduce the $i$ for the reasons mentioned above. So we can trace the presence of complex numbers in our theory back to the requirement that $\hat{x}$ and $\hat{p}$ have a non-zero commutation relation. Why this is necessary also has an interesting answer, but I will end my ramblings here!

  • $\begingroup$ I always thought it just boiled down to an easy way of avoiding writing sin() and cos() terms all the time. That won't work here? $\endgroup$
    – uhoh
    Commented Jun 18, 2016 at 4:08
  • $\begingroup$ Maybe the difference lies in the fact that if you want to solve the equation $\frac{\partial^2}{\partial x^2} f(x) = -c^2 f(x)$ (and require that the solution doesn't blow up at infinity) then either plane waves $e^{\pm icx}$ or sines and cosines can be used as a basis of the space of solutions. However, if you want to solve the differential equation I mentioned in my answer, then only the plane waves work. Whereas the second order equation appears in some classical situations involving waves, I think the first order version is less common outside of quantum mechanics. $\endgroup$ Commented Jun 18, 2016 at 4:26
  • $\begingroup$ OK, can you help me out a bit - where is $f^\prime=cf$ used in QM exactly? It's been a while. $\endgroup$
    – uhoh
    Commented Jun 18, 2016 at 4:32
  • 1
    $\begingroup$ This is just the eigenvalue equation for the momentum operator. Plane waves (forgetting about square integrability conditions for now) are the momentum eigenstates. $\endgroup$ Commented Jun 18, 2016 at 4:46
  • $\begingroup$ OK ya that's right. I'll go back and read now. Thanks! $\endgroup$
    – uhoh
    Commented Jun 18, 2016 at 4:53

Imaginary numbers are very useful for calculations, but as their name suggests - they are imaginary, not real. So if you measure something in the real world, you can only expect to get real numbers. This is true for both electrodynamics and QM, expectation values of quantum mechanics observables (i.e. measurements) will always turn out real.

Imaginary numbers are useful in describing waves, and arise naturally in quantum mechanics.

  • 2
    $\begingroup$ Your point about expectation values is relevant of course, but just because imaginary numbers are used in calculations, does not make them imaginary. I mean, following your logic, do negative numbers exist? Of course they do not. You can't have a negative number of apples. Yet by introducing them, all of a sudden we can use them to solve problems we were never ever able to solve before, or we can solve them in a much simpler way. But are they "real"? Do they "exist"? No, they are just tools that help us solve real life problems. But would you say they are imaginary? $\endgroup$
    – user129412
    Commented Jun 16, 2016 at 8:24
  • $\begingroup$ Well, yeah, I might have exaggerated a bit, but you can have negative numbers as natural results of measurements. Not measurements of apples, of course, but anything with a direction works. $\endgroup$
    – milo
    Commented Jun 16, 2016 at 8:37
  • $\begingroup$ In a sense, but that only works with a point of reference. Is it still real if you need to specify with respect to what it is negative? $\endgroup$
    – user129412
    Commented Jun 16, 2016 at 8:43
  • $\begingroup$ And you count apples in reference to 0 apples. 3 apples are -2 if you reference to 5 apples. Are they still real? $\endgroup$
    – milo
    Commented Jun 16, 2016 at 8:47
  • 1
    $\begingroup$ In any case the point is that the question of what is real or what is not is perhaps not so set in stone, or even important to consider for that matter. Your argument is correct of course, that expectation values should correspond to what we can measure classically, which you could link to Ehrenfests theorem. And classically, we do not measure complex valued observables. $\endgroup$
    – user129412
    Commented Jun 16, 2016 at 9:10

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