Hermitian Operators

  • Thread starter sunsun
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  • #1
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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
 

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  • #2
dextercioby
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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.
 
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  • #3
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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 ) ^*[/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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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
dextercioby
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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?
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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