Is <A> Always Zero for Anti-Hermitian Operators in Real Functions?

AI Thread Summary
The discussion revolves around proving that the expectation value <A> is zero for anti-Hermitian operators A when applied to any real function f. Participants clarify that if (Af)* = -Af, then <f|A|f> must equal zero, leading to the conclusion that the integral of fAf results in a purely imaginary value. The conversation highlights the importance of the completeness relation and the symmetric range of integration in this context. There is also mention of the potential lack of clarity in the textbook regarding the problem's requirements. Ultimately, the consensus is that the proof aligns with the properties of anti-Hermitian operators as outlined in the textbook.
mrquantum
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I'm stuck on a question in atkins molecular quantum mechanics 4e (self test 1.9).

If (Af)* = -Af, show that <A> = 0 for any real function f.

I think you are expected to use the completeness relation sum,s { |s><s| = 1.

I'm sure the answer is simple but I'm stumped.
 
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What are A and f?
 
I'm assuming A is any real or complex operator for which the relation holds and f is any real function.
 
And what is <f> ?

In QM <f> would be the expectation value of an operator f sandwitched between two states. But I guess you have something in mind like ∫ dx f(x)
 
Sorry, my mistake. Need to show that the expectation value of the operator A is zero. <f|A|f> = 0
 
Ok: If

(Af)^\ast = -Af

show that

\langle f|A|f\rangle = 0

for any real function f.
 
Yes.
 
Does that make sense?
 
I think something is missing.

What I can show immediately is

a = \int dx \, fAf = \int dx \, f (-Af)^\ast = -a^\ast

and therefore that a must be purely imaginary.

Note that I ommited the range of integration which must be symmetric [-L,L]
 
  • #10
http://www.kinetics.nsc.ru/chichinin/books/spectroscopy/Atkins05.pdf

That's a link to the textbook. Self-test 1.9. The question doesn't really specify anything else.
 
  • #11
As I said, something is missing. In the context of Atkins 1.9 only hermitean operators a considered.

1) so by the singe line above \text{Re}\,a = 0
2) by hermiticity \text{Im}\,a = 0

Therefore a=0
 
  • #12
Why don't you try rereading the problem yourself and then state the problem statement here again with all the relevant information? You might notice you've made mistakes or overlooked some important info.
 
  • #13
Thanks vela. This isn't actually a homework assignment, I'm just working my way through the textbook. I got the same result as Tom, (Af)* = (Af*)* = -Af = -Af* => a = -a* => a = 0 given hermiticity. This result is actually stated in the book in one of the worked-through examples. I was wondering if somebody knew how to solve the problem in the context of the textbook i.e. using matrix notation and the completeness relation. Do you think the textbook intends that the problem be solved that way? I know atkins mqm4 has got a reputation for not being overly clear.
 
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