Quantum physics - Symmetrizer operator

[vabite]

Hi everyone.

I am studying 'identical particles' in quantum mechanics, and I have a problem with the properties of the Symmetrizer (S) and Antisymmetrizer (A) operators.

S and A are hermitian operators. Therefore, for what I know, their set of eigenkets must constitute a basis of the space ket.
However, the set of eigenkets of S (A) contains only symmetric (antisymmetric) kets. Given an arbitrary ket (not necessarily symmetrical or antisymmetrical), I don't think it is always possible to write it as linear combination of only symmetrical (antisymmetrical) kets. Therefore, I can't understand how the eigenkets of S (A) can constitute a basis of the space ket.

 

[George Jones]

Try a 4-dimensional example: Consider the state space V spanned by {|1>|1>, |1>|2>, |2>|1>, |2>|2> }. What are the eigenvalues of S? What are the corresponding eigenspaces?

 

[vabite]

I think in this case the eingevalue is 1 (4 times degenerate), and the eigenkets
|1>|1>
(1/√2) (|1>|2>+|2>|1>)
(1/√2) (|2>|1>+|1>|2>)
|2>|2>
So.. 4 eigenkets for a 4D space. Does this mean that every ket can be obtained as a linear combination of symmetric (antisymmetric) kets?

 

[George Jones]

I think in this case the eingevalue is 1 (4 times degenerate), and the eigenkets
|1>|1>
(1/√2) (|1>|2>+|2>|1>)
(1/√2) (|2>|1>+|1>|2>)
|2>|2>

Addition is commutative, so your middle two kets are the same.

 

[vabite]

Err... right. I did not solve the eigenvalue equation, but I can't see a 4th eigenket of S...

I guess the missing one to form a basis of the ket space is:
(1/√2) (|1>|2>-|2>|1>)
that is an eigenket of A (and thus not of S).

 

[George Jones]

Err... right. I did not solve the eigenvalue equation, but I can't see a 4th eigenket of S...

I guess the missing one to form a basis of the ket space is:
(1/√2) (|1>|2>-|2>|1>)
that is an eigenket of A (and thus not of S).

Are you sure this isn't an eigenket of S?

 

[vabite]

S [(1/√2) (|1>|2>-|2>|1>)]=
(1/2)(P12+P21) [(1/√2) (|1>|2>-|2>|1>)] =
(1/2) {[(1/√2) (|1>|2>-|2>|1>)] + [(1/√2) (|2>|1>-|1>|2>)]} =
0
...

At this point, the only thing I can imagine is that
(1/√2) (|1>|2>-|2>|1>)
is an eigenket of S with eigenvalue 0.

 

[George Jones]

S [(1/√2) (|1>|2>-|2>|1>)]=
(1/2)(P12+P21) [(1/√2) (|1>|2>-|2>|1>)] =
(1/2) {[(1/√2) (|1>|2>-|2>|1>)] + [(1/√2) (|2>|1>-|1>|2>)]} =
0
...

At this point, the only thing I can imagine is that
(1/√2) (|1>|2>-|2>|1>)
is an eigenket of S with eigenvalue 0.

Yup.

S [(1/√2) (|1>|2>-|2>|1>)] = 0 [(1/√2) (|1>|2>-|2>|1>)].

Things are sort of reversed for A.

 

[vabite]

Great: thanks!