Wave Function Units: |ψ(x,t)| - Dimensionless?

[tomlry]

|ψ(x,t)|² is the probability density.

Probabilities are dimensionless.

What would the SI units of |ψ(x,t)| be, then? Also dimensionless?

 

[jtbell]

In order to get a (dimensionless) probability, you have to integrate the probability density. What else do you have inside the integral besides the probability density, and what are its units?

 

[tomlry]

Oh, I see. |ψ|² is a function and you will only get a probability if this function is integrated.

I only have a wave function for a free particle

ψ(x,0)=Ae^{-(x-x₀)2/4a2}e^ilx,

where a and l are constants that I don't know the dimensions of.

 

[jtbell]

$$P(a < x < b) = \int^b_a {|\psi|^2 dx}$$

What are the units of dx, and what units does $|\psi|^2$ therefore have to have, in order that the result be dimensionless?

 

[tomlry]

The unit of dx is length, so that means |ψ|² has to be 1/length for the probability density to become dimensionless.

|ψ| must then be [length]^-1/2.

Thank you :)

 

[Jazzdude]

The unit of dx is length, so that means |ψ|² has to be 1/length for the probability density to become dimensionless.
|ψ| must then be [length]^-1/2.

This argument can be found regularly but it's not quite correct. The right answer is that the wave function is entirely unit agnostic and can very well be treated as unitless.

The reason for this is that a quantum state is really a ray in the hilbert space, or in other words an element of the C-projective space constructed from the hilbert space.

Rays are invariant under scalar multiplication and therefore also the choice of units. Probabilities in quantum theory are also really calculated using a quotient with the wavefunction in both numerator and denominator, as in

$$ p(X) = \frac{\int_X |\psi(x)|^2 dx}{\int_\mathbb{R} |\psi(x)|^2 dx} $$

and any choice of units for the integration variable or wavefunction cancels.

If you use the assumption of a normalized wavefunction you don't have a quotent but just the numerator. This does not change the argument however, because you just take the quotient out of the probability calculation and cancel the units during normalization.

So to sum up, it does not matter which unit you assign to the wave function as it will cancel once you calculate probabilities. So to make your life easier you can just omit any kind of unit.