# Why Quantum Mechanics is Complex

1. Apr 24, 2015

### Zerkor

Why do Complex Numbers arise in Quantum Mechanics' computations? What kind of physical significance do they carry?

Someone told me to read this paper:
W E Baylis, J Hushilt, and Jiansu Wei, Why i?, American Journal of Physics 60 (1992), no. 9, 788–797.
But I found it difficult for me to understand completely. I need some simplified explanation.

2. Apr 24, 2015

### ddd123

3. Apr 24, 2015

### kith

4. Apr 24, 2015

### ChrisVer

Well, I am not sure how used you are in Classical Mechanics (waves) and/or Electromagnetism?

As a general setup, complex numbers never arise in describing physical observables not even in QM. In QM the observables are the eigenvalues of Hermitian operators , and Hermitian operators have real-valued eigenvalues.
In classical mechanics you also use complex numbers. For example when you describe the solution to the problem:
$\frac{d^2 x}{dt^2} + \gamma \frac{dx}{dt} + cx =0$
Harmonic oscillator with damping... in fact complex numbers can be used anytime you come across trigonometric functions thanks to the Euler's formula ($e^{\pm ia}= \cos (a) \pm i \sin (a)$). Except for this mechanical problem which complex numbers can help you solve, you can see them in electrodynamics, since the mechanical differential equations also apply in electromagnetism (in similar forms, maybe with different physical reasonings/explanations). In wave equations you use them to find easily the solutions which afterwards you take their real parts, and so on...

So nothing is special about complex numbers. They just make your life easier.
Now QM use a lot the wave mechanics, so it's not a big deal that you have complex numbers playing a role in your calculations. But anything that you can actually observe and use for testing the theory/experimenting, is absolutely real: It's the same reason you can never measure a vector in an experiment as a vector, but only its components that are "real", a complex number can be seen as a vector $\vec{z}= \begin{pmatrix} r \cos \theta \\ r \sin \theta \end{pmatrix}$ on the complex plane, since any complex number can be written as $z= r e^{i \theta}$. There is a small difference in inner products, but that's why instead of using the transpose you use the dagger (complex conjugate of the transpose).

Can a complex number carry physical significance? Yes, it depends on the context. For a fast classical example, if you take the refractive index of a wave to be complex $n= a + i b$, then your wave will not only receive refraction but the imaginary part will give attenuation. Both you can measure as $a$ and $b$ seperately, but you won't measure some $i$ in there...

5. Apr 24, 2015

### bhobba

6. Apr 29, 2015

### Zerkor

Thanks a lot for your help. I got it.

7. Apr 29, 2015

### Zerkor

Thanks a lot :)

8. May 21, 2015

### andresB

I would like to hear the opinion of informed people on this article that seems to defy the opinion that complex number are important in QM

http://arxiv.org/pdf/0912.3996.pdf

9. May 21, 2015

### ChrisVer

I'm sleepy right now...but can you expand in Fourier series an operator? because p is an operator and not a function.

10. May 21, 2015

### bhobba

I had a quick look. Don't know what the issue is. Its pretty conventional as far as I can see - he simply writes one equation in a non complex form - that's not an issue.

It talks about Schroedinger's acceptance of complex numbers in his famous equation. The following explains that Schroedinger made an error in deriving it, and complex numbers are essential in the correct derivation:
http://arxiv.org/pdf/1204.0653.pdf

As Feynman sorted out in his path integral formulation, unless you use complex numbers, nearby paths will bot cancel and you will not get the principle of least action.

Thanks
Bill

Last edited: May 21, 2015
11. May 24, 2015