1. Dec 10, 2007

### naqo

Hi there, i have been studying a bit about QM, but ther's one fundamental question
about the wavefunction i can't understand: why is the wavef. defined complex? I mean,
couldn't one work from the begining with a real wave?

Thanks

2. Dec 10, 2007

### Ben Niehoff

One explanation is to consider a time-invariant potential, such as the infinite square well. Because the potential does not change over time, we know that the probability of finding an electron of a single energy eigenstate must also be constant in time. Therefore, we know that

$$\frac{\partial}{\partial t}|\psi|^2 = 0$$

Expanding in terms of the real and imaginary parts,

$$\frac{\partial}{\partial t}\left(\Re(\psi)^2 + \Im(\psi)^2\right) = 0$$

$$2\, \Re(\psi) \frac{\partial}{\partial t} \Re(\psi) + 2\, \Im(\psi) \frac{\partial}{\partial t} \Im(\psi) = 0$$

Now, if $\Im(\psi) = 0$, then this directly implies that

$$\frac{\partial}{\partial t} \Re(\psi) = 0$$

which would mean that $\psi$ is not changing at all! This is a bit problematic, because then $\psi$ is not a wave; it's just the square root of a probability distribution. It would have no frequency, because it would just be a fixed function in space.

The theory, if based on real waves only, would lose its ability to talk about the time dependence of energy eigenstates. Which means if a particle in an energy eigenstate ever had to interact with anything, the theory would be unable to model it. This would be a bit of a problem. You will run into a similar problem in situations with other conserved quantities; i.e., any situation where the probability distribution is unchanging in some space, such as momentum, time, etc.

If we allow the wavefunction to be complex, then it can have a constant squared magnitude, but still oscillate like a wave (by traversing a circle in the complex plane). This way, there is time-dependence even in time-invariant states, and we can model what happens when those states interact with some outside influence.

3. Dec 10, 2007

### akhmeteli

As I mentioned in some of my earlier posts, quantum mechanics does not necessarily needs complex wavefunctions (real wavefunctions may be enough). This is not just my personal opinion. Shroedinger demonstrated this for a Klein-Gordon field interacting with electromagnetic field. The reasons offered in the preceding post do not work there as the probability density does not equal \psi^2 in that case

4. Dec 11, 2007

### Gerenuk

Basically quantum mechanics works if you use complex wavefunctions and that's all we want.
One reason is that there are phenomena where for example three objects exist at some point individually with equal probability (amplitude), but if you combine them at this point (adding) they vanish (interfere destructively). That only works with complex numbers (exp(2pi*i*n/3) in this case).
Formally one could of course consider the real and imaginary part of the wavefunction. Then you have only real functions, but now two of them.

5. Dec 11, 2007

### akhmeteli

Again, this reasoning is based on the assumption that the probability density is \psi^2. This is not so, for example, for a Klein-Gordon field interacting with electromagnetic field (and this theory includes a lot of what quantum electrodynamics includes, the exceptions being such important phenomena as spin). In this case just one real field (not two) is enough to describe the charged Klein-Gordon field. This is achieved by choosing a gauge, as Schroedinger showed (there is a reference in one of my earlier posts).

6. Dec 11, 2007

### Gerenuk

Where can I read up about the Klein-Gordon field that has only one real values function? I've only heard about the 4 dimensional Dirac stuff so far.
Actually I didn't assume that the probability density is $$\psi\psi^*$$. I only assumed that it is zero for $$\psi=0$$ and it is non-zero for $$\psi\neq 0$$

7. Dec 11, 2007

### akhmeteli

E. Shroedinger, Nature, V. 169, P.538 (1952). There is also a discussion in my paper http://arxiv.org/abs/quant-ph/0509044
Actually, the assumption that the probability density is non-zero for \psi\neq 0 does not hold for the model I mentioned (you may check it in textbooks, in Shroedinger's short article, or look at expression (9) for the current in my paper; I am afraid I am too lazy to rewrite it here:-) )

8. Dec 11, 2007

### Peter Morgan

Hi Akmeteli,
Mathematically, you can certainly use a real Hilbert space for generating probabilities, but when you want to represent position and momentum as noncommuting observables (which if you're doing QM, you do; if you're doing not-QM, you might not), it's convenient to introduce complex numbers. The elementary paper on this that I like best is Leon Cohen, Foundations of Physics, Volume 18, page 983, 1988, but I'm afraid it's behind a paywall. I believe myself that the use of complex structure in quantum mechanics is not anything to worry about much, partly because you can always write a dimension n complex Hilbert space as a dimension 2n real Hilbert space. What you lose by going real, however, is the fundamental theorem of algebra (for which see wikipedia, say), which mathematically is a huge convenience.

9. Dec 11, 2007

### mendocino

I think it related to the phase of wavef.
You can't use one real number to represent wave

10. Dec 11, 2007

### akhmeteli

Sorry, I'm afraid I did not quite get it. Are you implying that one complex number can be used to represent a wave?
If, however, you meant "You can't use one real function to represent wave", I tend to disagree, as it seems you can use a sine or a cosine function to represent a wave.

11. Dec 11, 2007

### akhmeteli

I have no problems with complex numbers in quantum mechanics. However, the original poster asked:"why is the wavef. defined complex? I mean,
couldn't one work from the begining with a real wave?". I was just trying to answer that question, and my answer was "it might be possible". I would like to emphasize once again, I am not talking about using two real functions instead of one complex one. I might agree that "it is convenient to introduce complex numbers". However, whether something is more convenient or not, depends on what you want, on your agenda. The reason I am interested in formulating quantum mechanics in terms of real functions is such reformulation might be critical for interpretation of quantum mechanics. For example, it can lead to natural exclusion of the matter (the wavefunction). If you are interested in the Bohmian interpretation, the electromagnetic potential can play the role of the quantum potential after such reformulation.

12. Dec 11, 2007

### Xeinstein

You can use a sine or a cosine function to represent a wave with 0 or 90 degree phase, But you need both sine "and" cosine function to represent a wave with arbitrary phase

Last edited: Dec 11, 2007
13. Dec 11, 2007

### akhmeteli

I had in mind something like \sin(k x+\alpha) or \cos(k x+\alpha), where \alpha is an arbitrary phase.

14. Dec 12, 2007

### Xeinstein

Basically, what you did is just shift the origin,
But if you have many wave functions with different phases,
what can you do about them?
Are you gonna shift origins for all of them?
Then you end up with No common origin of them
I think what we want is just one common origin for all of them

The sine and cosine functions are just the basis vectors/function in Hilbert space
Then any function is just linear combination of the basis function

Last edited: Dec 12, 2007
15. Dec 12, 2007

### akhmeteli

I am afraid you've lost me. Of course, you may use the sine and cosine functions as a basis. Two things elude me, though. How your remarks relate to the statement that I disagreed ("You can't use one real function to represent wave") and how the situation with real numbers is significantly different from the situation you have when using complex numbers. Maybe, one more thing. What do you call "common origin" and why do we need it?

16. Dec 12, 2007

### Xeinstein

I think I should call it offset, so you add offset to sine or cosine function
My point is this, we need both sine "and" cosine as basis functions in Fourier transform/analysis, You can't do Fourier transform with just sine "or" cosine only for arbitrary functions

Last edited: Dec 12, 2007
17. Dec 12, 2007

### akhmeteli

I agree. Furthermore, I appreciate that the exponential function is more convenient than sines and cosines. Nevertheless, I believe that a real function, say, a sine, can represent a wave, contrary to what mendocino stated.

18. Dec 12, 2007

### reilly

There's no mystery about the fact that sin(kx) and cos(kx) are solutions to any linear wave equation -- k = wave number. But, the algebra and calculus of sin and cosine are somewhat cumbersome, not so for the algebra of complex exponentials. Complex variables bring a particularly convenient and powerful approach to dealing with wave functions, and many other topics as well.

So why complex wave functions? It's strictly pragmatism. We use complex variables, by active choice, because they work, and work well -- complex variables are used in virtually all branches of applied math, strictly for the sake of convenience. Complex variables provide some very powerful tools -- contour integration, for example -- for certain classes of mathematical problems, particularly those related to linear second order differential equations. Special functions and complex variables go hand-in-glove

Complex variables? No big deal, and with absolutely no conceptual bearing on physics.
Regards,
Reilly Atkinson

19. Dec 13, 2007

### Staff: Mentor

Can you replace the time-dependent Schrödinger Equation with a differential equation that does not have an $i$ in it, has a real (not complex) wave function as its solution, and describes all the phenomena that the SE and its solutions can?

20. Dec 13, 2007

### akhmeteli

Actually, yes.

Let me explain it (though you may have an impression that I am not explaining, but "caveating" it).

First, as I said, this is not my achievement and claim, but Shroedinger's (I gave a reference in this thread; sorry for "oe" instead of "o umlaut" in his name).

Second, saying "yes", I assume that the Klein-Gordon equation describes all the phenomena that the SE and its solutions can (actually, KGE is better than SE in the relativistic limit).

Third, the possible potentials are limited to such electromagnetic 4-potentials A^\mu that their source (the right-hand-side of the Maxwell equations) is the standard Klein-Gordon current (so we do not ignore the electromagnetic field generated by the Klein-Gordon particle, but why should we?) plus arbitrary conserved external currents (strictly speaking, Shroedinger considered a self-consistent theory, where the Klein-Gordon field interacts with electromagnetic field, but the generalization through addition of external conserved currents seems obvious). That does not look like a strong limitation.

Actually, the main idea is exremely straightforward (Shroedinger wrote: "That the wave function ... can be made real by a change of gauge is but a truism, though it contradicts the widespread belief about 'charged' fields requiring complex representation." And he did not have in mind a replacement of a complex function by two real ones!). Indeed, if you have a solution (\psi, A^\mu) of the Klein-Gordon-Maxwell equations (Klein-Gordon field interacting with electromagnetic field), then a gauge transform would give you an equivalent solution (\phi, B^\mu), where \phi is real, \phi^2=(\psi*)\psi, and the electromagnetic field is the same for A^\mu and B^/mu (Careful in thread https://www.physicsforums.com/showthread.php?p=830490#post830490 draw my attention to the fact that B^mu can have singularities, but, on the other hand, even the Coulomb potential is singular, let alone quantum field theory).