Hermiticity of permutation operator

In summary, the conversation discusses the properties of a permutation operator on a two-variable complex function in a Hilbert space. The operator is verified to be linear and its square is shown to be the identity operator. The eigenvalues of the operator are found to be ±1, which implies that the operator is also Hermitian. The conversation then discusses the derivation of the eigenfunctions, with different approaches suggested.
  • #1
issacnewton
1,016
35
Hi

here's a problem I am having.

Consider the hilbert space of two-variable complex functions [tex]\psi (x,y)[/tex].

A permutation operator is defined by its action on [tex]\psi (x,y)[/tex] as follows.

[tex]\hat{\pi} \psi (x,y) = \psi (y,x) [/tex]

a) Verify that operator is linear and hermitian.

b) Show that

[tex]\hat{\pi}^2 = \hat{I}[/tex]

find the eigenvalues and show that the eigenfunctions of [tex]\hat{\pi}[/tex] are given by

[tex]\psi_{+} (x,y)= \frac{1}{2}\left[ \psi (x,y) +\psi (y,x) \right] [/tex]

and

[tex]\psi_{-} (x,y)= \frac{1}{2}\left[ \psi (x,y) -\psi (y,x) \right] [/tex]

I could show that the operator is linear and also that its square is unity operator I . I did
find out the eigenvalues too. I am having trouble showing that its hermitian and the
part b about its eigenfunctions.

Any help ?
 
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  • #2
After you already show that [itex]\hat{\pi}[/itex] is linear. Take a look at [itex]\hat{\pi}^2[/itex]. What is it? From this what can you infer about the possible eigenvalues for [itex]\hat{pi}[/itex]? What properties for the eigenvalues for an operator implies Hermiticity?
 
  • #3
Hi , eigenvalues are [tex]\pm 1[/tex] they are real. I know that hermitian operators have real eigenvalues but does it mean that if operator has real eigenvalues then its a hermitian
operator ( this is converse statement , so not necessarily true) ?
 
  • #4
IssacNewton said:
Hi , eigenvalues are [tex]\pm 1[/tex] they are real. I know that hermitian operators have real eigenvalues but does it mean that if operator has real eigenvalues then its a hermitian
operator ( this is converse statement , so not necessarily true) ?

If all the eigenvalues of an operators are real.

More explicitly, you are trying to show that:
[tex]\int_{-\infty}^{\infty} \psi(r)^{*} \phi(-r) dr = \int_{-\infty}^{\infty} \psi(-r)^{*} \phi(r) dr[/tex]
which is easy by simple change of variable.
 
Last edited:
  • #5
So since all eigenvalues of [itex]\hat{\pi}[/itex] are real , its hermitian. fine that's solved.
what about the second part about the eigenfunctions ? can you help ?
 
  • #6
Just apply the operator to the functions and show that you get an eigenvalue times the function.
 
  • #7
vela, yes that's one approach. but how do I derive these eigenfunctions in the first place ?
 
  • #8
IssacNewton said:
vela, yes that's one approach. but how do I derive these eigenfunctions in the first place ?

There are really no general approach for arbitrary operator. A good guess is a start.
 
  • #9
I suppose you could make a hand-waving argument that the Hilbert space can be written as a direct sum of subspaces, where each subspace is the span of Φ(x,y) and Φ(y,x) for a particular Φ(x,y). Then find the matrix representing [itex]\hat{\pi}[/itex] in one of these subspaces and diagonalize it. Blah, blah, blah...

It's more useful, though, to recognize this pattern of combining elements to achieve a certain symmetry.
 
  • #10
thanks fellas. makes some sense now...
 

Related to Hermiticity of permutation operator

1. What does it mean for a permutation operator to be Hermitian?

Hermiticity of a permutation operator means that it is equal to its own conjugate transpose. This means that the operator is its own inverse, and it maps a state to itself.

2. Why is the Hermiticity of a permutation operator important?

The Hermiticity of a permutation operator is important because it guarantees that the operator is unitary, meaning it preserves the norm of a vector. This is essential in quantum mechanics, where operators represent physical observables.

3. How is the Hermiticity of a permutation operator related to symmetry?

A permutation operator is Hermitian if and only if the underlying permutation has a certain symmetry. This symmetry is known as a transposition, where two elements are swapped. Therefore, Hermiticity of a permutation operator is closely related to the concept of symmetry in mathematics.

4. Can a permutation operator be Hermitian for any permutation?

No, not all permutations can have a Hermitian permutation operator. A permutation must have a certain symmetry, as mentioned before, in order for its operator to be Hermitian. For example, the identity permutation is always Hermitian since it does not change the state of a vector.

5. How is the Hermiticity of a permutation operator tested in practice?

In practice, the Hermiticity of a permutation operator can be tested by calculating its conjugate transpose and comparing it to the original operator. If the two are equal, then the operator is Hermitian. Additionally, the eigenvalues of a Hermitian operator are always real, so this can also be used to test for Hermiticity.

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