Proving Properties of Hermitian and Anti-Hermitian Operators

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Homework Help Overview

The discussion revolves around proving properties of Hermitian and anti-Hermitian operators in the context of quantum mechanics and functional analysis. Participants are exploring definitions and relationships involving operators, particularly focusing on the hermitian conjugate and its implications for eigenvalues.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to prove that the hermitian conjugate of a hermitian conjugate returns the original operator. Questions arise regarding the conditions under which this holds true, particularly in relation to densely defined linear operators on separable Hilbert spaces. There is also discussion about the eigenvalues of anti-hermitian operators and their properties.

Discussion Status

The discussion is ongoing, with participants providing insights and references to literature. Some participants have raised concerns about the correctness of certain statements and definitions, suggesting a need for further clarification and exploration of the underlying assumptions.

Contextual Notes

There are mentions of specific definitions and conditions necessary for the properties being discussed, including the distinction between self-adjoint and hermitian operators. Participants note the importance of understanding the implications of these definitions in the context of quantum mechanics.

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 [tex](A_{op}^t)^t=A_{op}[/tex] Where [tex]A_{op}[/tex] 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, [tex]I_{op}^t=-I_{op}[/tex] 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 [tex]A_{op}[/tex], but I'm still having trouble with the second one.
 
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It's straightforward to see that for an eigenvector of [itex]\hat{A}[/itex] corresponding to a spectral value [itex]\lambda[/itex]

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

Therefore the spectral value [itex]\lambda[/itex] is either 0 or purely imaginary...

Daniel.
 
To show that the hermitian conjugate of the hermitian conjugate of A is A itself, use the definition of "hermitian conjugate"!
 
For this operatorial equality

[tex]\hat{A}=\left(\hat{A}^{\dagger}\right)^{\dagger}[/tex] (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

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

and

[tex]\hat{A}|\psi\rangle=\left(\hat{A}^{\dagger}\right)^{\dagger}|\psi\rangle,\forall |\psi\rangle \in \mathcal{D}_{\hat{A}}[/tex] (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 [itex]\hat{A}[/itex] this operatorial inclusion holds true
[tex]\hat{A}\subset\left(\hat{A}^{\dagger}\right)^{\dagger}[/tex] (4)

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

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

Daniel.
 
DISCLAIMER!

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

The correct relation is

[tex]\hat{A}\subseteq \left(\hat{A}^{\dagger}\right)^{\dagger}[/tex] (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

[tex]\hat{A}=\hat{A}^{\dagger}[/tex] (the operator is selfadjoint) (2)
[tex]\hat{A}^{\dagger}=\left(\hat{A}^{\dagger}\right)^{\dagger}[/tex] (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".
 
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I addition to post #6,i would recommend 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...:eek:

Daniel.
 

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