130_notes.dvi

The last equality is satisfied only ifχ 1 =χ 2 (just write out the two component equations to see
this). Hence the normalized eigenvector corresponding to the eigenvalueα= + ̄h/2 is

χ(+x)=

1

√

2

(

1

1

)

.

Similarly, we find for the eigenvalueα=− ̄h/2,

χ(−x)=

1

√

2

(

1

− 1

)

.

18.10.11 Eigenvectors ofSyfor Spin^12

To find the eigenvectors of the operatorSywe follow precisely the same procedure as we did forSx
(see previous example for details). The steps are:

1. Write the eigenvalue equation (Sy−α)χ= 0


2. Solve the characteristic equation for the eigenvaluesα±


3. Substitute the eigenvalues back into the original equation


4. Solve this equation for the eigenvectors

Here we go! The operatorSy= ̄h 2

(

0 −i

i 0

)

, so that, in matrix notation the eigenvalue equation

becomes (
−α −i ̄h/ 2
i ̄h/ 2 −α

)(

χ 1

χ 2

)

= 0

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

α^2 −

̄h^2

4

= 0 ⇒ α=±

̄h

2

These are the same eigenvalues we found forSx(no surprise!) Pluggingα+back into the equation,
we obtain (
−α+ −i ̄h/ 2
i ̄h/ 2 −α+

)(

χ 1

χ 2

)

=

̄h

2

(

− 1 −i

i − 1

)(

χ 1

χ 2

)

= 0

Writing this out in components gives the pair of equations

−χ 1 −iχ 2 = 0 and iχ 1 −χ 2 = 0

which are both equivalent toχ 2 =iχ 1. Repeating the process forα−, we find thatχ 2 =−iχ 1.
Hence the two eigenvalues and their corresponding normalized eigenvectors are

α+= + ̄h/ 2 χ

(y)

+ =

1

√

2

(

1

i

)

α−=− ̄h/ 2 χ(−y)=

1

√

2

(

1

−i

)