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

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Homework Help Overview

The discussion revolves around the properties of anti-Hermitian operators in quantum mechanics, specifically focusing on the expectation value of such operators when applied to real functions. The original poster is attempting to demonstrate that the expectation value equals zero for any real function f, as posed in a self-test question from a quantum mechanics textbook.

Discussion Character

  • Conceptual clarification, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore the definitions of the operator A and the function f, questioning the nature of the expectation value and its calculation. There is discussion about the implications of the anti-Hermitian property and the completeness relation in the context of the problem.

Discussion Status

Some participants have provided insights into the mathematical implications of the anti-Hermitian property, noting that the expectation value must be purely imaginary. Others express uncertainty about the completeness of the problem statement and suggest that additional context from the textbook may be necessary for a full understanding.

Contextual Notes

There is mention of the need for a symmetric range of integration and the potential oversight of relevant information in the problem statement. The original poster clarifies that this is not a formal homework assignment but rather a self-study effort, indicating a desire for deeper understanding rather than a straightforward solution.

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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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