# What is the purpose of the imaginary portion of the wave function?

I recently watched this video. I'm trying to learn about the origin of the wave function and therefore understand its use in the Schrödinger Equation.

However at the end of the video I understood up to:

$$ψ = \cos(kx - ωt)$$

This would be the real part of the wave function wholly defined as:

$$\cos(αx) + i\sin(αx)$$

Or:

$$ψ = e^{iαx}$$

What I'm having trouble understanding is the imaginary part of the wave function. It was never explained why we needed the imaginary sine function added to the real part of the wave, or what $e$ is.

• In this particular case of a plane wave the magnitude of the wave function has to be the same everywhere. A real wave function, i.e. just the cosine part, doesn't do that. – CuriousOne Jun 2 '15 at 7:02
• @CuriousOne: I'm not sure that's true. A real cosine wave is a perfectly acceptable solution to the Schrodinger equation. It's just not the most general solution. – John Rennie Jun 2 '15 at 7:13
• @JohnRennie: That's why I prefaced the sentence with "in this particular case". It is likely that the introduction the OP saw was talking about the plane wave solution (as it probably should), in which case one would ask for a constant magnitude, I believe. One can, of course, make purely real wave functions, but then the magnitude wouldn't be constant, right? – CuriousOne Jun 2 '15 at 7:16
• @Serg: The real problem for the beginner is probably the question why the Schroedinger equation has to have an imaginary constant in it? Why can't we make a real equation? Maybe one should explore that and write it as two real components? Either I already forgot about it or it doesn't really tell us much... I have to think about that for a while or wait for someone who can wave a good looking hand. As for a real solution... I think it can exist, but it would have a complex energy, i.e. it's a decaying solution, which we don't look at until later. Even then it's a special case, though... – CuriousOne Jun 2 '15 at 7:21
• As a general hint: you can't start physics with quantum mechanics and expect to understand what is really going on. Quantum mechanics is deeply rooted in classical mechanics and the professional method of teaching it starts with a full course of classical mechanics. Before you have a solid understanding of the Lagrange and Hamiltonian forms of mechanics it's nearly impossible to understand why/how physicists came up with it. Everything we are doing here is handwaving. This is not how we teach it to students. – CuriousOne Jun 2 '15 at 7:38

## 1 Answer

It is misleading to consider the real and imaginary parts of the wave function separately. The wave function is a function of spacetime that returns a complex number. We interpret this as meaning that the wavefunction requires two components to describe it. You can think of this as an amplitude and a phase. However the split between the real and imaginary parts is arbitrary and can be changed by a coordinate transformation, so there is nothing special about about the real part or the imaginary part.

The wavefunction is not an observable, so the fact it is a complex quantity does not matter. Anything we can observe is given by acting on the wavefunction with a Hermitian operator, and these are guaranteed to return a real result.

• In the case of a photon, does it make sense to apply to the sin and the cos the two components electric field and magnetic field? – HolgerFiedler Jun 2 '15 at 7:23
• @HolgerFiedler: The Schroedinger equation is not relativistic, you don't get photons out of it this easily. – CuriousOne Jun 2 '15 at 7:26