The angular momentum operator is l r x p ir x 1 lz i l2...

The angular momentum operator is L = r x p = -iħr x ∇. 1. Lz = -iħ(∂/∂φ), L^2 = -ħ^2(sinθ)(∂/∂θ)(sinθ)(∂/∂θ) + (-ħ^2)(1/sinθ)(∂^2/∂φ^2). 2. If the eigenfunction of the hydrogen atom is Ψn,l,m, then: a. LzΨn,l,m = mħΨn,l,m b. L^2Ψn,l,m = l(l+1)ħ^2Ψn,l,m c. The results in a and b will not change if the Coulomb potential is replaced by another potential V(r).

Transcript

00:01 Okay, so we're going to define the angular momentum like this and we want to start with the canonical commutation relations.

00:12 The one between r and p, the components of the position and momentum vectors look like this.

00:19 The two i left off is that the components of r commute with each other and the components of p commute with each other.

00:30 Write those in a bit.

00:31 So we want to find out how lz commutes with x, y, and z.

00:42 So with x, the only factor in there that it doesn't commute with is the px.

00:51 So it commutes with x, excuse me, the py.

00:54 It commutes with py, it commutes with x and y, it doesn't commute with px.

01:05 So we get this and so we get that.

01:11 Lx with lz with x is ih bar y.

01:16 Lz with y similarly commutes with everything there except for py.

01:23 So there it is.

01:29 And then with z, z commutes with all of that.

01:36 So it's zero.

01:40 And then with px, lx with px, the only term that matters is the x term.

01:51 It's xpx times py.

01:54 That's going to give us ih bar py.

02:00 Lz with py comes out to be minus ih bar px.

02:19 And because actually we know that either x, y, z or px, py, pz form vectors that we expect these commutation relations.

02:35 These are the definition of vector operators is that they have these commutation relations with ls.

02:43 Okay, so that's the last one...

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