Units of Wave function (Schrodinger equation)

[Avatrin]

Hi

I thought I knew the answer to this question until I encountered the following question:

What is the unit of R(r)?

We are of course talking about the radial part of the solution to Schrodinger's equation in spherical coordinates (i.e. [itex]\psi(r,\theta,\phi) = R(r)\Theta(\theta)\Phi(\phi)[/itex]).

I tried to look at the integral needed for normalization and just got more confused:

[itex]\int R(r)r^2 dr \int\Theta(\theta)sin\theta d\theta \int \Phi(\phi) d\phi = 1[/itex]

How should I approach problems like this?

 

[strangerep]

What is the unit of R(r)?

We are of course talking about the radial part of the solution to Schrodinger's equation in spherical coordinates (i.e. [itex]\psi(r,\theta,\phi) = R(r)\Theta(\theta)\Phi(\phi)[/itex]).

I tried to look at the integral needed for normalization and just got more confused:

[itex]\int R(r)r^2 dr \int\Theta(\theta)sin\theta d\theta \int \Phi(\phi) d\phi = 1[/itex]

How should I approach problems like this?

Well, "1" is dimensionless. Angles are dimensionless. "r" has dimensions of length (which I'll write as 'L').

Your normalization integral looks wrong. Maybe you meant (in cartesian coordinates) something like:
$$ \int \psi^*(x) \psi(x) d^3 x ~=~ 1 ~~~?$$(You only seem to have ##\psi## in your integral.)

##x## has dimensions L, so ##d^3 x## has dimensions ##L^3##. The whole integral must be dimensionless to match the right hand side, hence ##\psi(x)## must have dimensions ##L^{-3/2}##. Usually, the angular factors in the wavefunction would be dimensionless, so that should be enough to tell you the dimensions of the radial part -- if you write the spherical polar version of the integral correctly...

 

[Avatrin]

Yes, my integral is indeed wrong...

[itex] \int |R(r)|^2 r^2 dr \int |\Theta(\theta)|^2 sin\theta d\theta \int |\Phi (\phi)|^2 d\phi = 1[/itex]

However, the angles being dimensionless, while it makes sense; How, when the potential is not a function of only the radius, do you indicate that the probability density varies with the angle? This is the part that was confusing me the most. Of course, if we try to use some angular unit like the unit of distance, we encounter the problem [itex] \psi(r,\theta,\phi) = \psi(r,\theta + 2\pi,\phi + 2\pi)[/itex]

 

[vanhees71]

It's good practice to make the wave functions in a separation ansatz of their natural dimension as a probability-density amplitude, depending on the nature of the argument. In your case of separation in spherical coordinates thus, I'd choose to normalize them separately to unity, because then by integrating over one or two of the three coordinates you get directly probability densities for the other variables, i.e., the dimensions should be
$$[R]=\text{length}^{-3/2}, \quad [\Theta]=[\Phi]=1.$$

 

[strangerep]

How, when the potential is not a function of only the radius, do you indicate that the probability density varies with the angle? This is the part that was confusing me the most.

I don't understand where the confusion lies. The wave function ##\psi## involves the factors ##\Theta(\theta)\Phi(\phi)##, hence clearly depends on angle in general.