We use the raising operator (See section 10) equation for an energy eigenstate.
A†un=√
n+ 1un+1Now simply compute the matrix element.
A†ij=〈i|A†|j〉=√
j+ 1δi(j+1)Now this Kronecker delta puts us one off the diagonal. As we have it set up, i gives the row and j
gives the column. Remember that in the Harmonic Oscillator we start counting at 0. For i=0, there
is no allowed value of j so the first row is all 0. For i=1, j=0, so we have an entry forA† 10 in the
second row and first column. All he entries will be on a diagonal from that one.
A†=
√0 0 0 0 ...
1 0 0 0 ...
0
√
2 0 0 ...
0 0
√
3 0 ...
0 0 0
√
4 ...
18.10.3Harmonic Oscillator Lowering Operator
We wish to find the matrix representing the 1D harmonic oscillator lowering operator. This is similar
to the last section.
The lowering operator (See section 10) equation is.
Aun=√
nun− 1Now we compute the matrix element from the definition.
Aij=〈i|A|j〉=√
jδi(j−1)A=
0
√
1 0 0 0 ...
0 0
√
2 0 0 ...
0 0 0
√
3 0 ...
0 0 0 0
√
4 ...
This should be the Hermitian conjugate ofA†.
18.10.4Eigenvectors ofLx
We will do it as if we don’t already know that the eigenvalues arem ̄h.
Lxψ=aψ
0 1 0
1 0 1
0 1 0
ψ 1
ψ 2
ψ 3
=
√ 2 a
̄h
ψ 1
ψ 2
ψ 3
≡b
ψ 1
ψ 2
ψ 3