Proof of Hermitian Operator: f,g,h Functions of x

[KeyToMyFire]

Homework Statement

If <h|Qh> = <Qh|h> for all functions h, show that <f|Qg> = <Qf|g> for all f and g.

f,g, and h are functions of x
Q is a hermitian operator

Hints: First let h=f+g, then let h=f+ig

Homework Equations

<Q>=<Q>*

Q(f+g)= Qf+Qg

The Attempt at a Solution

(All integral are with respect to x, and go from -∞ to +∞)

<(f+g)|Q(f+g)> = <Q(f+g)|(f+g)>

∫(f*+g*)(Qf+Qg) = ∫(Qf*+Qg*)(f+g)

∫f*Qf + ∫g*Qg + ∫f*Qg + ∫g*Qf = ∫(Qf*)f + ∫(Qg*)g + ∫(Qf*)g + ∫(Qg*)f

<f|Qf> + <g|Qg> + <f|Qg> + <g|Qf> = <Qf|f> + <Qg|g> + <Qf|g> + <Qg|f>

<f|Qg> + <g|Qf> = <Qf|g> + <Qg|f>

And then I'm stuck.

 

[dextercioby]

Again I don't see the relevance of this problem with <Advanced Physics>.
The problem is mathematically ill-posed. A correct formulation would be

"Let A be a densly defined linear operator acting on a complex separable Hilbert space. Show that if

$$\langle h,Ah\rangle = \langle Ah,h\rangle, ~ ~\forall h\in D(A)$$

then A is symmetric. "

The solution to the correctly formulated problem is trivial, once you use the definition of a symmetric operator.

 

[KeyToMyFire]

Again I don't see the relevance of this problem with <Advanced Physics>.
The problem is mathematically ill-posed. A correct formulation would be

"Let A be a densly defined linear operator acting on a complex separable Hilbert space. Show that if

$$\langle h,Ah\rangle = \langle Ah,h\rangle, ~ ~\forall h\in D(A)$$

then A is symmetric. "

The solution to the correctly formulated problem is trivial, once you use the definition of a symmetric operator.

____"Again"? I've never posted a question on this site before, let alone an irrelevant one. If other people do so I don't see why that's held against me. I felt this problem was relevant to <Advanced Physics> because it comes straight from quantum mechanics homework.
____As to the mathematics being "ill posed" or "incorrectly formulated", it was copied directly from the textbook. With the admitted exclusion that "h is in hilbert space", though seeing as it was in parathesis I felt it was implied.
____And I'm glad you find the solution trivial, but seeing as it is the question I asked, I clearly do not. So thank you failing to address my question, other than to critisize that way it was asked, and being entirely helpful.

 

[dextercioby]

Letting aside the mathematical rigor, "If <h|Qh> = <Qh|h> for all functions h, show that <f|Qg> = <Qf|g> for all f and g", is this the whole text of your problem ? Are there any other assumptions to be made ?

 

[KeyToMyFire]

Word for word (with "hats" on the Q's)

"Problem 3.3 Show that if <h|Qh> = <Qh|h> for all functions h (in Hilber space), then <f|Qg> = <Qf|g> for all f and g (i.e., the two definitions of "hermitian" -- Equations 3.16 and 3.17 -- are equivalent). Hint: First let h=f+g, and then let h=f+ig."



Then on the previous page:

"Thus operators representing observables have the very special property that:

<f|Qf>= <Qf|f> for all f(x) [3.16]

We call such operators hermitian.

Actually most books require an ostenibly stronger condition:

<f|Qg> = <Qf|g> for and f(x) and all g(x) [3.17]

But as it turns out, in spite of appearances, that this is perfectly equivalent to my definition (Equation 3.16), as you will prove in Problem 3.3. So use which ever you like. The essential point is that the hermitian can be applied to the first member of an inner product or to the second . . . "

 

[Oxvillian]

You're halfway there - you've done the h=f+g part. Now you just have to do the h=f+ig part and subtract the two answers

Note that it was never necessary to write out the integrals because you were always going to come back to the abstract notation anyway.

I don't see the relevance of this problem with <Advanced Physics>

I think it's appropriate here because the mathematics in question is traditionally taught in introductory quantum mechanics courses, albeit without the rigor you seem to enjoy :yuck: