• Support PF! Buy your school textbooks, materials and every day products Here!

Operators and expectation values.

  • Thread starter mrquantum
  • Start date
  • #1
14
0
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.
 
Last edited:

Answers and Replies

  • #2
tom.stoer
Science Advisor
5,766
161
What are A and f?
 
  • #3
14
0
I'm assuming A is any real or complex operator for which the relation holds and f is any real function.
 
  • #4
tom.stoer
Science Advisor
5,766
161
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)
 
  • #5
14
0
Sorry, my mistake. Need to show that the expectation value of the operator A is zero. <f|A|f> = 0
 
  • #6
tom.stoer
Science Advisor
5,766
161
Ok: If

[tex](Af)^\ast = -Af[/tex]

show that

[tex]\langle f|A|f\rangle = 0 [/tex]

for any real function f.
 
  • #8
14
0
Does that make sense?
 
  • #9
tom.stoer
Science Advisor
5,766
161
I think something is missing.

What I can show immediately is

[tex] a = \int dx \, fAf = \int dx \, f (-Af)^\ast = -a^\ast[/tex]

and therefore that a must be purely imaginary.

Note that I ommited the range of integration which must be symmetric [-L,L]
 
  • #11
tom.stoer
Science Advisor
5,766
161
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 [itex]\text{Re}\,a = 0[/itex]
2) by hermiticity [itex]\text{Im}\,a = 0[/itex]

Therefore [itex]a=0[/itex]
 
  • #12
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,631
1,267
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
14
0
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.
 
Last edited:

Related Threads on Operators and expectation values.

  • Last Post
Replies
1
Views
4K
  • Last Post
Replies
2
Views
2K
Replies
6
Views
1K
  • Last Post
Replies
12
Views
14K
  • Last Post
Replies
1
Views
3K
  • Last Post
Replies
3
Views
3K
  • Last Post
Replies
4
Views
10K
Replies
1
Views
5K
Replies
2
Views
459
Replies
1
Views
644
Top