Quick Matrix Element Question using Hermitian Operator

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SUMMARY

The discussion centers on the manipulation of matrix elements involving Hermitian operators A and B in quantum mechanics. The key conclusion is that the expression <1|AB|2> can be rewritten as <2|BA|1> only under the condition that A and B are compatible observables and |1> and |2> are basis vectors of those observables. Additionally, the complex conjugate relationship is crucial, as <1|AB|2> = <2|BA|1>^*. The user is also exploring the relationship between momentum and position operators in the context of a harmonic oscillator.

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  • Understanding of Hermitian operators in quantum mechanics
  • Familiarity with matrix elements and their properties
  • Knowledge of eigenstates and observables
  • Basic concepts of harmonic oscillators and Hamiltonians
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  • Learn about the commutation relations for position and momentum operators
  • Explore the implications of compatible observables in quantum mechanics
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Quantum mechanics students, physicists working with quantum systems, and researchers studying operator theory and harmonic oscillators will benefit from this discussion.

starryskiesx
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Hi there,

This should be very simple...

If I have a state <1|AB|2> where A and B are Hermitian operators, can I rewrite this as <2|BA|1> ?

That would be, taking the complex conjugate of the matrix element and saying that A*=A and B*=B.

Thank you!
 
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Well, you need to remember to include the complex conjugation:

##\langle 1 | A B | 2 \rangle^* = \langle 2 | B^\dagger A^\dagger | 1 \rangle = \langle 2 | B A | 1 \rangle##

so ##\langle 1 | A B | 2 \rangle = \langle 2 | B A | 1 \rangle^*##
 
The_Duck said:
Well, you need to remember to include the complex conjugation:

##\langle 1 | A B | 2 \rangle^* = \langle 2 | B^\dagger A^\dagger | 1 \rangle = \langle 2 | B A | 1 \rangle##

so ##\langle 1 | A B | 2 \rangle = \langle 2 | B A | 1 \rangle^*##

Great, thank you for the help
 
not necessarily, you need to also consider that this is only true if A and B are compatible observables, and it is only a matrix element if |1> and |2> are basis vectors of those same compatible observables.
 
raymo39 said:
not necessarily, you need to also consider that this is only true if A and B are compatible observables, and it is only a matrix element if |1> and |2> are basis vectors of those same compatible observables.

So it appears I have a bigger problem!

|1> and |2> are eigenstates of the hamiltonian, my system is that of a harmonic oscillator. What I'm trying to prove is that <1|P|2> = -imw<1|X|2>, starting with the matrix element [P,H] where H is the hamiltonian. I thought I could do this just by swapping eigenstates so now I'm more stuck :)

So far I've tried two methods, one involving writing H = T + V for the harmonic oscillator and finding the commutation relation and the other working with H|1> = E1|1> etc.

Any suggestions?
 
raymo39 said:
not necessarily, you need to also consider that this is only true if A and B are compatible observables, and it is only a matrix element if |1> and |2> are basis vectors of those same compatible observables.
What The Duck did holds in general, for arbitrary state vectors |1> and |2>.
 
Fredrik said:
What The Duck did holds in general, for arbitrary state vectors |1> and |2>.

I agree, my post was in reply to the original poster. Guess i was late to the party
 

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