eigenstates ofLzas a basis for our states and operators. Ignoring the (fixed) radial part of the
wavefunction, our state vectors forℓ= 1 must be a linear combination of theY 1 m
ψ=ψ+Y 11 +ψ 0 Y 10 +ψ−Y 11whereψ+, for example, is just the numerical coefficient of the eigenstate.
We will write our3 component vectorslike
ψ=
ψ+
ψ 0
ψ−
.
The angular momentum operators are therefore 3X3 matrices. Wecan easilyderive(see section
18.11.1)the matrices representing the angular momentum operators forℓ= 1.
Lx=̄h
√
2
0 1 0
1 0 1
0 1 0
Ly=√ ̄h
2 i
0 1 0
−1 0 1
0 −1 0
Lz= ̄h
1 0 0
0 0 0
0 0 − 1
(1)
The matrices must satisfy the samecommutation relationsas the differential operators.
[Lx,Ly] =i ̄hLzWe verify this with anexplicit computation of the commutator.(see section 18.11.2)
Since these matrices represent physical variables, we expect them to beHermitian.That is, they
are equal to their conjugate transpose. Note that they are alsotraceless.
As an example of the use of these matrices, let’s compute anexpectation valueofLxin the matrix
representation for the general stateψ.
〈ψ|Lx|ψ〉 = (ψ∗ 1 ψ 2 ∗ ψ∗ 3 )
̄h
√
2
0 1 0
1 0 1
0 1 0
ψ 1
ψ 2
ψ 3
=
̄h
√
2(ψ∗ 1 ψ 2 ∗ ψ∗ 3 )
ψ 2
ψ 1 +ψ 3
ψ 2
=
̄h
√
2(ψ∗ 1 ψ 2 +ψ 2 ∗(ψ 1 +ψ 3 ) +ψ∗ 3 ψ 2 )18.3 Eigenvalue Problems with Matrices
It is often convenient to solveeigenvalue problemslikeAψ=aψusing matrices. Many problems
in Quantum Mechanics are solved by limiting the calculation to a finite, manageable, number of
states, then finding the linear combinations which are the energy eigenstates. The calculation is
simple in principle but large dimension matrices are difficult to work with byhand. Standard