Expectation value of a product of operators

[ryanwilk]

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.

 

[EmpaDoc]

Yes. The meaning of AB|a> is to first operate with B, then with A - but the first operation produces a zero.

 

[Matterwave]

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)

 

[dextercioby]

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.