# A few operator questions

1. Mar 27, 2005

### 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.

Last edited by a moderator: Mar 27, 2005
2. Mar 28, 2005

### 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.

3. Mar 28, 2005

### HallsofIvy

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

4. Mar 28, 2005

### 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.

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

Last edited: Mar 28, 2005
5. Mar 28, 2005

### dextercioby

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

Daniel.

6. Mar 28, 2005

### 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.

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

Last edited: Mar 28, 2005
7. Mar 28, 2005

### 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.