computer utilities are readily available to help solve this problem.
A 11 A 12 A 13 ...
A 21 A 22 A 23 ...
A 31 A 32 A 33 ...
... ... ... ...
ψ 1
ψ 2
ψ 3
...
=a
ψ 1
ψ 2
ψ 3
...
Subtracting the right hand side of the equation, we have
A 11 −a A 12 A 13 ...
A 21 A 22 −a A 23 ...
A 31 A 32 A 33 −a ...
... ... ... ...
ψ 1
ψ 2
ψ 3
...
= 0.
For the product to be zero, thedeterminantof the matrix must be zero. We solve this equation
to get the eigenvalues.
∣ ∣ ∣ ∣ ∣ ∣ ∣
A 11 −a A 12 A 13 ...
A 21 A 22 −a A 23 ...
A 31 A 32 A 33 −a ...
... ... ... ...
∣ ∣ ∣ ∣ ∣ ∣ ∣
= 0
- See Example 18.10.4:Eigenvectors ofLx.*
The eigenvectors computed in the above example show that the x axis is not really any different than
the z axis. Theeigenvaluesare + ̄h, 0, and− ̄h, the same as for z. The normalizedeigenvectors
ofLxare
ψ+ ̄(xh)=
1(^21)
√ 2
1
2
ψ(0 ̄xh)=
√^1
2
0
−√^12
ψ−(xh ̄)=
− 1(^21)
√ 2
− 1
2
.
These vectors, and anyℓ= 1 vectors, can be written in terms of the eigenvectors ofSz.
We can check whether theeigenvectors are orthogonal,as they must be.
〈ψ0 ̄h|ψ+ ̄h〉=( 1
√
2∗
0 −√^12∗)
1(^21)
√
2
1
2
= 0
The others will also prove orthogonal.
Shouldψ(+ ̄xh)andψ(−z ̄h)be orthogonal?
NO. They are eigenvectors ofdifferenthermitian operators.
The eigenvectors may be used to compute the probability or amplitude of a particular measurement.
For example, if a particle is in a angular momentum stateχand the angular momentum in the x
direction is measured, the probability to measure + ̄his
P+ ̄h=∣
∣
∣〈ψ
(x)
+ ̄h|χ〉∣
∣
∣
2