Proving Properties of Hermitian and Anti-Hermitian Operators

[vsage]

I think these questions may be so simple that I don't know how to properly prove them, but I am really stumped on how to show especially the first one:

Show that (A_{op}^t)^t=A_{op} Where A_{op} is an operator and 't' is my way of saying hermitian conjugate

Secondly, I'm not too sure at all how to show this:

Show that the anti-hermitian operator, I_{op}^t=-I_{op} has at most one real eigenvalue

Any hints would be greatly appreciated.

Edit: oops! Hidden in the annals of the internet I found the appropriate definition for A_{op}, but I'm still having trouble with the second one.

 

[dextercioby]

It's straightforward to see that for an eigenvector of \hat{A} corresponding to a spectral value \lambda

\lambda=\langle \psi|\hat{A}|\psi\rangle =\left(\langle\psi|\hat{A}^{\dagger}|\psi\rangle\right)^{*}=(-\lambda)^{*}

Therefore the spectral value \lambda is either 0 or purely imaginary...

Daniel.

 

[HallsofIvy]

To show that the hermitian conjugate of the hermitian conjugate of A is A itself, use the definition of "hermitian conjugate"!

 

[dextercioby]

For this operatorial equality

\hat{A}=\left(\hat{A}^{\dagger}\right)^{\dagger} (1)

to hold,for a densly defined linear operator A,then the 2 conditions from the definition of operatorial equality on a separable Hilbert space must be met

\mathcal{D}_{\hat{A}}=\mathcal{D}_{\left(\hat{A}^{\dagger}\right)^{\dagger}} (2)

and

\hat{A}|\psi\rangle=\left(\hat{A}^{\dagger}\right)^{\dagger}|\psi\rangle,\forall |\psi\rangle \in \mathcal{D}_{\hat{A}} (3)


Unfortunately,there's no densly defined linear operator on a separable Hilbert space for which (1) (and implicitely (2) & (3)) to hold.

It can be proved that for an arbitrary densly defined linear operator \hat{A} this operatorial inclusion holds true
\hat{A}\subset\left(\hat{A}^{\dagger}\right)^{\dagger} (4)

Since \hat{A}^{\dagger} is a closed operator,we use (4) to assert that \left(\hat{A}^{\dagger}\right)^{\dagger} is a closed extension of the operator \hat{A}.

Ergo,equality (1) doesn't hold true for any densly defined linear operator on a separable Hilbert space...

Daniel.

--------------------------------------------
NOTE:Red=Incorrect statements.Relation (4) doesn't hold for essentially self adjoint densly defined linear operators.See disclaimer in post #6.

 

[dextercioby]

What I'm saying is that u can't prove something which is incorrect...

Daniel.

 

[dextercioby]

DISCLAIMER!

My professor of QM was wrong (and it's my fault i didn't check other sources on functional analysis)...

The correct relation is

\hat{A}\subseteq \left(\hat{A}^{\dagger}\right)^{\dagger} (1)

Among the operators for which the equality holds,one finds operators which are called ESSENTIALLY SELFADJOINT densly defined linear operators.They form a subset of the set of selfadjoint (not hermitian/symmetric) linear operators acting on a separable Hilbert space for which their adjoint is selfadjoint.They satisfty the relations

\hat{A}=\hat{A}^{\dagger} (the operator is selfadjoint) (2)
\hat{A}^{\dagger}=\left(\hat{A}^{\dagger}\right)^{\dagger} (its adjoint is selfadjoint) (3)

The relation (1) coupled with the relations (2) & (3) prove that for most of the selfadjoint operators,their adjoint is not selfadjoint,but hermitean/symmetric.

See [1] for a proof that H-atom Hamiltonian (Dirac formulation-Schrödinger picture-coordinate representation) is essentially selfadjoint...

As a comment to post #5,u need to prove the relation (1) and show that the equality limit holds true,only under certain conditions (u need to find these conditions);for example,if the operator is essentially selfadjoint,then it verifies the equality limit.

Daniel.

--------------------------------------------------------------
[1]J.Prugoveçki:"Quantum Mechanics in Hilbert Space".

 

[dextercioby]

I addition to post #6,i would reccomend you to read Akhiezer & Glazman's "The Theory of Linear Operators in Hilbert Space",2nd.ed,Dover,1993.

In section #22,he proves that for a bounded linear ooperator defined on the whole Hilbert space the equality limit holds.On page #80 he mentions the strict inclusion,though he mentions that there are operators for which the equality relation holds...

Daniel.