Why is the square modulus of the wave function used in quantum mechanics?

[lalbatros]

The wave function is a probability amplitude.
So far so good.
This means that it doesn't give us the probability directly and it can be a complex number.
Taking the square modulus of it, gives rise to interference. That's very good.

But why the square, why not the fourth power or any even function, like cosh(|phi|)?

Simplicity is the winner, but why?

Michel

 

[Rach3]

I think it reduces to this: if [tex]\psi[/tex] is normalized, then so must be any [tex]\hat{U} \psi[/tex] for unitary U, and the only general way to have the unitarily-invariant statement [tex]A(\psi)=1[/tex] is something like [tex]A(\psi)\equiv \psi^{\dagger}\psi \rightarrow \psi^{\dagger} (U^{\dagger} U) \psi = \psi^{\dagger}\psi[/tex]. So the '2' comes from the number of times U enters in the unitarity definition.

 

[Mickey]

The squared modulus of [tex]\psi[/tex] is equal to [tex]\psi[/tex] times its complex conjugate.

[tex]|\psi|^{2} = \psi\bar{\psi}[/tex]

The complex conjugate allows us to make a function that can be expressed in terms of both the real and imaginary parts of [tex]\psi[/tex]. If instead we treated [tex]\psi[/tex] as a single complex parameter, then it would be one-dimensional. Complex conjugation, combined with other operations, doesn't require a mapping to be conformal either.

 

[loom91]

The vector formalism requires the probability density to be the inner product of the unit vector and its dual, which when translated to the position basis gives us phi times its conjugate, that is mod phi squared.

 

[Rach3]

The vector formalism requires the probability density to be the inner product of the unit vector and its dual, which when translated to the position basis gives us phi times its conjugate, that is mod phi squared.

So in other words, the question of why [tex]|\psi(x)|^2[/tex] reduces to that of why [tex]\psi^{\dagger}\psi[/tex] (or in Dirac notation, [tex]\langle \psi | \psi \rangle [/tex]). Which is still an interesting question (see my previous post). More generally yet, why do normed Hilbert spaces correctly represent the physics, as opposed to some other algebraic system? I'm not sure if there's a more fundamental principle involved here. Anyway the reality of it is very convenient.

 

[vanesch]

I think it reduces to this: if [tex]\psi[/tex] is normalized, then so must be any [tex]\hat{U} \psi[/tex] for unitary U, and the only general way to have the unitarily-invariant statement [tex]A(\psi)=1[/tex] is something like [tex]A(\psi)\equiv \psi^{\dagger}\psi \rightarrow \psi^{\dagger} (U^{\dagger} U) \psi = \psi^{\dagger}\psi[/tex]. So the '2' comes from the number of times U enters in the unitarity definition.

Well, I will make some publicity for my little paper where I show that you can use very well some other rules, and that |psi|^2 is a postulate which does not follow from the unitary alone:

http://www.arxiv.org/abs/quant-ph/0505059

(at least, under the condition that there are only a finite number of possible measurement outcomes, which is always the case in a real measurement).

 

[Rach3]

In case any lurkers are confused, the basic idea here is that the QM structure is in vector spaces and linear algebra, and wavefunctions [tex]\psi(x)[/tex] and [tex]\phi(p)[/tex] are only a particular representations of this; the rules of wavefunctions derive from the more fundamental rules of inner products of vectors.