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

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This is asked for @Saran V.

This is asked for @Saran V.

##### 1 Answer

What we will show is that

Consider a **spherical harmonic wave function**

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 **operator**.

That is, the value of

#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.

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

#= -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

#= 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

If you did this on other

Therefore,

#color(blue)(|L| = sqrt(l(l+1)) ℏ)#

You should try this on