Question

How do we know that the atomic orbital angular momentum magnitude is `sqrt(l(l+1))ℏ`, where `ℏ = h//2pi`?

This is asked for @Saran V.

Answer 1

What we will show is that `overbrace(hatL^2)^("operator")Y_(l)^(m_l)(theta,phi) = overbrace(l(l+1)ℏ^2)^"eigenvalue"Y_(l)^(m_l)(theta,phi)`.

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Consider a spherical harmonic wave function `Y_(l)^(m_l)(theta,phi)`, where `l` is the orbital angular momentum quantum number and `m_l` is the magnetic quantum number.

Its general formula for a hydrogen atom is very difficult to use to obtain eigenvalues, so we will take some example wave functions:

`Y_(0)^(0)(theta,phi) = Y(s) = 1/2sqrt(1/pi)`
`Y_(1)^(0)(theta,phi) = Y(p_z) = 1/2sqrt(3/pi)costheta`
`Y_(2)^(0)(theta,phi) = Y(d_(z^2)) = 1/2sqrt(5/pi)(3cos^2theta - 1)`

The orbital angular momentum is `L`, and the known eigenvalue relates to `hatL^2`, the squared orbital angular momentum operator.

That is, the value of `l` arises when we operate on `Y` using the `hatL^2` operator:

`hatL^2 = -ℏ^2[1/(sintheta)(del)/(del theta)(sintheta (del)/(del theta)) + 1/(sin^2theta)(del^2)/(delphi^2)]`

where `theta` is the angle from the `z` axis down to the `xy` plane, and `phi` is the azimuthal angle, counterclockwise from the `x` axis on the `xy` plane.

![https://upload.wikimedia.org/]()

What we expect is to get eigenvalues of:

`color(green)(hatL^2)Y(s) = overbrace(color(green)(0ℏ^2))^("eigenvalue")Y(s)`, since `0(0+1)ℏ^2 = 0`

`color(green)(hatL^2)Y(p_z) = overbrace(color(green)(2ℏ^2))^("eigenvalue")Y(p_z)`, since `1(1+1)ℏ^2 = 2ℏ^2`

`color(green)(hatL^2)Y(d_(z^2)) = overbrace(color(green)(6ℏ^2))^"eigenvalue"Y(d_(z^2))`, since `2(2+1)ℏ^2 = 6ℏ^2`

where `|L|` then corresponds to the square root of the eigenvalue of `hatL^2`.

And we would see that `s` orbitals (and hydrogen atom) have no orbital angular momentum (eigenvalue 0), and they are therefore spheres.

Let's take the most nontrivial example, as if it works out, there is no way it could be a coincidence. Plug it in first:

`color(blue)(hatL^2) Y(d_(z^2)) = -ℏ^2[1/(sintheta)(del)/(del theta)(sintheta (delY(d_(z^2)))/(del theta)) + 1/(sin^2theta)(del^2Y(d_(z^2)))/(delphi^2)]`

`= -ℏ^2[1/(sintheta)(del)/(del theta)(sintheta (del)/(del theta)(1/2sqrt(5/pi)(3cos^2theta - 1))) + cancel(1/(sin^2theta)(del^2)/(delphi^2)(1/2sqrt(5/pi)(3cos^2theta - 1)))^(0)]`

Now what we want to do is operate on the function, and get the SAME function back when we are all done.

The first derivative of `3cos^2theta - 1` with respect to `theta` is `-6sinthetacostheta`, and the second derivative with respect to `phi` is `0`:

`= -ℏ^2[1/(sintheta)(del)/(del theta)(1/2sqrt(5/pi)sintheta (-6sinthetacostheta))]`

Now we simplify this a bit:

`= 6ℏ^2cdot 1/2sqrt(5/pi)[1/(sintheta)(del)/(del theta) (sin^2thetacostheta)]`

The first derivative of `sin^2thetacostheta` with respect to `theta` is `-sin^3theta + 2sinthetacos^2theta`, giving:

`= 6ℏ^2cdot 1/2sqrt(5/pi)[1/cancel(sintheta)(-sin^(cancel(3)^(2))theta + 2cancel(sintheta)cos^2theta)]`

`= 6ℏ^2cdot 1/2sqrt(5/pi)[(-sin^2theta + 2cos^2theta)]`

`= 6ℏ^2cdot 1/2sqrt(5/pi)[(-sin^2theta + 2cos^2theta + cos^2theta - cos^2theta)]`

`= 6ℏ^2cdot 1/2sqrt(5/pi)[3cos^2theta - (sin^2theta + cos^2theta)]`

`= color(blue)(6ℏ^2) cdot overbrace(1/2sqrt(5/pi)[3cos^2theta - 1])^(Y(d_(z^2))`

And our eigenvalue is indeed `6ℏ^2`.

If you did this on other `Y_(l)^(m_l)(theta,phi)` functions, you would find the pattern `l(l+1) ℏ^2` works. Taking the square root gives the magnitude of the orbital angular momentum, `|L|`.

Therefore,

`color(blue)(|L| = sqrt(l(l+1)) ℏ)`

You should try this on `Y_(0)^(0)` and `Y_(1)^(1)` to get the idea.