130_notes.dvi

Recall the standard method of finding eigenvectors and eigenvalues:

Aψ=αψ

(A−α)ψ= 0

For spin^12 system we have, in matrix notation,

(

a 1 a 2

a 3 a 4

)(

χ 1

χ 2

)

=α

(

1 0

0 1

)(

χ 1

χ 2

)

⇒

(

a 1 −α a 2

a 3 a 4 −α

)(

χ 1

χ 2

)

= 0

For a matrix times a nonzero vector to give zero, the determinant of the matrix must be zero. This
gives the “characteristic equation” which for spin^12 systems will be a quadratic equation in the
eigenvalueα: ∣
∣
∣
∣

a 1 −α a 2

a 3 a 4 −α

∣

∣

∣

∣= (a^1 −α)(a^4 −α)−a^2 a^3 = 0

α^2 −(a 1 +a 4 )α+ (a 1 a 4 −a 2 a 3 ) = 0

whose solution is

α±=

(a 1 +a 4 )

4

±

√

(a 1 +a 4 )

2

2

−(a 1 a 4 −a 2 a 3 )

To find the eigenvectors, we simply replace (one at a time) each of the eigenvalues above into the
equation (
a 1 −α a 2
a 3 a 4 −α

)(

χ 1

χ 2

)

= 0

and solve forχ 1 andχ 2.

Now specifically, for the operatorA=Sx= ̄h 2

(

0 1

1 0

)

, the eigenvalue equation (Sx−α)χ= 0

becomes, in matrix notation,

̄h

2

(

0 1

1 0

)(

χ 1

χ 2

)

−

(

α 0

0 α

)(

χ 1

χ 2

)

= 0

⇒

(

−α ̄h/ 2

̄h/ 2 −α

)(

χ 1

χ 2

)

= 0

The characteristic equation isdet|Sx−α|= 0, or

α^2 −

̄h^2

4

= 0 ⇒ α=±

̄h

2

These are the two eigenvalues (we knew this, of course). Now, substituting α+ back into the
eigenvalue equation, we obtain
(
−α+ ̄h/ 2
̄h/ 2 −α+

)(

χ 1

χ 2

)

=

(

− ̄h/2 ̄h/ 2

̄h/ 2 − ̄h/ 2

)(

χ 1

χ 2

)

=

̄h

2

(

−1 1

1 − 1

)(

χ 1

χ 2

)

= 0