Proving Hermiticity for the Product of Two Hermitian Operators

  • Thread starter Trance
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Hello everybody, long time reader, first time poster.
I've searched the forums extensively (and what seems like 60% of the entire internet) for anything relevant and haven't found anything, please point me in the right direction if you've seen this before!



Homework Statement


Show that even though Q and R are Hermitian operators, QR is only Hermitian if [Q,R]=0



Homework Equations


[Q,R] = QR - RQ
[Q,R]* = R*Q*-Q*R*


The Attempt at a Solution


I've managed to prove, using the definition of Hermiticity that:

∫f(i(QR-RQ))g dτ = ∫g(i(QR-RQ))*f dτ

For two random functions, f and g, and i= Sqrt(-1)

But I'm not sure how relevant this is... I don't have a strong enough intuition to see whether or not this immediately proves anything. Any help would be greatly appreciated!
 

Answers and Replies

  • #2
dextercioby
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For bounded linear operators acting on a separable Hilbert space, you have that, if [itex] Q=Q^{\dagger} ,\, R=R^{\dagger} [/itex], then [itex] (QR)^{\dagger} = R^{\dagger}Q^{\dagger} = RQ [/itex] and it is equal to QR, if Q and R commute. For unbounded operators, the proof is much more complicated, since the concept of commutativity is harder to define.
 
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