130_notes.dvi

(Frankie) #1

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
Free download pdf