Question

Evaluate the commutator [lx,ly,ly]

Answer 1

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Answer:

Angular momentum Algebra

We have

[Lx, Ly] = ihL¯ z ,

[Ly, Lz] = ihL¯ x , and

[Lz, Lx] = ihL¯ y .

These angular momentum commutation relations and their consequences are extraor-

dinarily important in understanding three-dimensional quantum mechanics in general and

atoms and molecules in particular.

Eigenvalue spectrum

The algebra of commutation relations can be used to obtain the eigenvalue spectrum.

First note that since Lx and Ly do not commute they cannot be simultaneously diagonal-

ized and do not share an orthonormal basis set of simultaneous eigenfunctions.

We choose to use as a basis the eigenkets of Lz. We note that the magnitude of the

angular momentum is related to the squared angular momentum, i.e., L~ ·L~ and the latter

turns out to be a useful operator. It is easy to see that [L

2

, L~ ] = 0. We verify one of the

three equations:

[L

2

, Lx] = [L

2

x + L

2

y + L

2

z

, Lx] = [L

2

y

, Lx] + [L

2

z

, Lx]

= Ly [Ly, Lx] + [Ly, Lx]Ly + Lz [Lz, Lx] + [Lz, Lx]Lz

= −ihL¯ yLz − ihL¯ zLy + ihL¯ zLy + ihL¯ yLz = 0 . (1)

Please verify [L

2

, Lz] = 0 to get some practice with such manipulations. Therefore, since

L

2 ≡ L

2

x + L

2

y + L

2

z

commutes with Lz they can be diagonalized simultaneously and we

can determine common eigenfunctions. This we do below.

Let |λ, µi be a normalized eigenfunction of L

2 with eigenvalue λ h¯

2

and of Lz with

eigenvalue µh¯, i.e.,

L

2

|λ, µi = ¯h

2

λ |λ, µi (2)

Lz |λ, µi = µh¯ |λ, µi (3)

We wish to find λ and µ.

The trick is to define raising and lowering operators (as in the solution to the harmonic

oscillator problem) by

L± ≡ Lx ± iLy . (4)

As is usual when we encounter new operators we determine their commutation relations

with known operators! We note that

[ Lz L± ] = ± hL¯ ± (5)

1

Answer 2

Answer:

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Explanation:

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