- #1

- 2

- 0

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

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- #1

- 2

- 0

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

- #2

- 13,309

- 2,406

It means to be included in its adjoint. By definition

[tex] 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.

[tex] 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:

- #3

- 223

- 2

[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 ) ^*[/tex]

by definition. But if [tex]\mathcal{O}^\dagger = \mathcal{O}[/tex], what does that mean about the matrix elements?

- #4

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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

- 13,309

- 2,406

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

[tex] (O+O^{\dagger})^{\dagger}\supseteq O^{\dagger}+O^{\dagger\dagger} \supseteq O^{\dagger}+O [/tex] ,

since [itex] O\subseteq O^{\dagger\dagger} [/itex]

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