Expectation value of a product of operators

  • Thread starter ryanwilk
  • Start date
  • #1
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Just to check something:

If A and B are operators and B|a> = 0, does this imply that <a|AB|a> = 0 ?

Or can you not split up the operators like <a|A (B|a>) ?

Thanks.
 

Answers and Replies

  • #2
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Yes. The meaning of AB|a> is to first operate with B, then with A - but the first operation produces a zero.
 
  • #3
Matterwave
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Yes, you are correct to assume <a|AB|a>=0; however, be sure to remember that, it's not necessarily true that <a|BA|a>=0 (if A and B don't commute, and if B is not hermitian, then this may not equal 0)
 
  • #4
dextercioby
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The Dirac notation makes sense, iff AB seen as an operator from D(B) to Ran(A) is self-adjoint. That's why you're permitted to convert the normal scalar product into the <|AB|> notation.
 

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