# Complex or real wave function?

Discussion in 'Physics & Math' started by computAI, Nov 12, 2022.

1. ### computAIRegistered Member

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23
It is not clear what complex wave function means in Schrödinger, Pauli, Dirac equations. Is it always two-component (complex), or can it be real, or are both variants possible in different situations?

For example, how to understand:

-i · h/(2·π) · ∂ψ/∂t = h^2/(8·π^2·m) · div grad ψ

(for simplicity in absence of potential multiplied by function).

The imaginary unit “i” simply shows that quantum operator is used instead of classical derivative, or function must be divided into two components: ψ = ψ1 + i · ψ2

and then in reality there are two equations

(~ symbol means is proportional with a constant multiplier).

In this case, the question arises how this relates to de Broglie equation, because it turns out to be

or ∂^2ψ/∂t^2~rot rot ψ for different kinds of waves.

Or is function real (should be, or can be)?

If Maxwell's equation is written as one formula, there are two components, electric field and magnetic, but instead of squared nabla single nabla (curl) is used, and this is consistent as de Broglie wave.

Do Pauli and Dirac equations follow the same principle as Schrödinger equation with respect to the complexity of function, or there are differences?

3. ### exchemistValued Senior Member

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12,071
I don't understand your difficulty. Any wavelike equation is complex. That arises from the fact that it models a system with periodic character in which phase is important. You get just the same thing in electrical AC theory.

The important thing is that when one applies any of the operators corresponding to observable properties, the outcome is real.

5. ### computAIRegistered Member

Messages:
23
Yes, sum of the squares of wavefunction two components is real and represents probability to encounter an electron (charge and mass density also).
But anyway, I am confused that Maxwell equation is compatible with de Broglie, and of Schrödinger is not. Double time derivative corresponds to "nabla in the fourth power". Have it to be so?

7. ### exchemistValued Senior Member

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12,071
You can't use Schrödinger's equation for a massless entity such as a photon, because mass appears in it. It is simply not appropriate for a photon. Maxwell's equations, however, don't involve mass.

Why are you considering a double time derivative? The time derivative in Schrödinger's equation is a single derivative.

8. ### computAIRegistered Member

Messages:
23
Thank you for your answer. I thought if function is called "wave", it has to be consistent in some way with de Broglie formula: second time derivative is proportional to squared nabla (div grad or curl curl, depending on is the wave longitudinal or transverse). But now I see "wave" does mean only continuous distribution of something in the space and is not really a wave process in this case.

9. ### exchemistValued Senior Member

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12,071
Yup, see also my response on the other forum where you have posed this. I understand you a bit better now. In its more general time-dependent form, it's actually a diffusion equation. However, for stable bound states, e.g. an electron in an atomic orbital, which is how it was originally used, the time independent version applies. This is equivalent to a standing wave. So in such cases the distinction between a wave and diffusion vanishes! So I think that's why it's called a wave equation, for historical reasons.

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