# Hermitian Operators

1. Mar 14, 2007

### 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?

2. Mar 14, 2007

### dextercioby

It means to be included in its adjoint. By definition

$$A\subseteq A^{\dagger} \ \mbox{means that A is hermitian/symmetric} [/itex] As for the second part, i'm sure the question is ill posed, as there's no mentioning of domains for the operators. you can simplify it by assuming the involved operators are bounded, hence defined on all the Hilbert space. Last edited: Mar 14, 2007 3. Mar 14, 2007 ### StatMechGuy Typically in a quantum mechanics course, you can assume a basis for your operators that they will span. To prove that the above operators are Hermitian, you'd want to look at how the matrix elements transform: [tex]\langle n | \left ( \mathcal{O} | m \rangle \right ) = \left ( \langle n | \mathcal{O}^\dagger \right ) | m \rangle = \langle m | \left ( \mathcal{O}| n \rangle \right ) ^*$$
by definition. But if $$\mathcal{O}^\dagger = \mathcal{O}$$, what does that mean about the matrix elements?

4. Mar 14, 2007

### sunsun

Im sorry, I dont 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?

5. Mar 14, 2007

### dextercioby

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 $O\subseteq O^{\dagger\dagger}$