Hermitian Operators: Meaning & Showing Properties

[sunsun]

1.What does it mean for an operator to be hermitian?

Note: the dagger is represented by a '
2. How do I show that for any operator ie/ O' that O + O' , i(O-O') and OO' are hermitian?

Thanks in advanced

 

[dextercioby]

It means to be included in its adjoint. By definition

A\subseteq A^{\dagger} \ \mbox{means that A is hermitian/symmetric} [/itex]<br /> <br /> As for the second part, I&#039;m sure the question is ill posed, as there&#039;s no mentioning of domains for the operators. you can simplify it by assuming the involved operators are bounded, hence defined on all the Hilbert space.

 

[StatMechGuy]

Typically in a quantum mechanics course, you can assume a basis for your operators that they will span. To prove that the above operators are Hermitian, you'd want to look at how the matrix elements transform:

\langle n | \left ( \mathcal{O} | m \rangle \right ) = \left ( \langle n | \mathcal{O}^\dagger \right ) | m \rangle = \langle m | \left ( \mathcal{O}| n \rangle \right ) ^*
by definition. But if \mathcal{O}^\dagger = \mathcal{O}, what does that mean about the matrix elements?

 

[sunsun]

Im sorry, I don't really get that.

How would I go around starting to answer the Q2? I know that O' = O
But how would I show that O+O' is hermitian?

 

[dextercioby]

Im sorry, I don't really get that.

How would I go around starting to answer the Q2? I know that O' = O
But how would I show that O+O' is hermitian?

I'll let you figure out the domain issues, but

(O+O^{\dagger})^{\dagger}\supseteq O^{\dagger}+O^{\dagger\dagger} \supseteq O^{\dagger}+O ,

since O\subseteq O^{\dagger\dagger}