# Homework Help: Self-adjoint operator problem

1. Sep 21, 2009

### noblegas

1. The problem statement, all variables and given/known data

Show that if the operator Q satifies

$$(\phi,Q\phi)=(Q\phi,\phi)$$ for all $$\phi$$, then Q is self-adjoint , that is

$$(\varphi,QX)=(Q\varphi,X).$$ Consider the functions

$$\phi_1=\varphi+X$$,$$\phi_2=\phi+i*X$$

Note: X is NOT a matrix. Could not find the latex code for the curvy X so i just typed X

2. Relevant equations

3. The attempt at a solution

$$(\phi,Q\phi)=(Q\phi,\phi)$$ =$$(\phi_1,Q\phi_1)=(Q\phi_1,\phi_1)=(\varphi+X),Q(\varphi+X),$$ (Q\phi,\phi)=$$(\phi_2,Q\phi_2)=(Q\phi_1,\phi)=(\varphi+X)i,Q(\varphi+X)i$$ ? Am I off in the right direction?

2. Sep 22, 2009

### noblegas

is my question to esoteric

3. Sep 22, 2009

### kuruman

It is hard to sort out what you are doing because your Latex code is mangled up. I would start with what is known, namely

$$((\varphi+\chi),Q(\varphi+\chi))=(Q(\varphi+\chi),(\varphi+\chi))$$

and develop each side separately, i.e.

Left side =
Right side =

Set the two sides equal and use that

$$(\chi,Q\chi)=(Q\chi,\chi);\ (\varphi,Q\varphi)=(Q\varphi,\varphi)$$

to get some terms to cancel. After that you should be able to see how to finish. BTW "curvy X" is \chi.

4. Sep 22, 2009

### noblegas

shouldn't I plug in $$\phi_2$$ as well ?

5. Sep 22, 2009

### kuruman

Yes, that is part 2. Do this one first to see how it works, then do the other one following the same procedure.

6. Sep 22, 2009

### noblegas

$$((\varphi+\chi),Q(\varphi+\chi))=q(\varphi+\chi), (\varphi+\chi))=q*(\varphi+\chi), (\varphi+\chi))$$, q is some number.
So now what?

7. Sep 23, 2009

### kuruman

Are you given that Qφ = qφ where q is some number? It does not appear so. You need to distribute Q and get four terms on each side.

8. Sep 23, 2009

### noblegas

no, it is given in a section of my textbook but not in the problem.$$(\varphi+\chi),Q(\varphi+\chi)=(\varphi+\chi),Q(\varphi)+Q(\chi)=(\varphi,Q\varphi)+(\chi,Q(\chi)$$ given that
$$(\phi,Q\phi)=(Q\phi,\phi)$$,then $$(\varphi,Q\varphi)+(\chi,Q(\chi)=Q(\varphi,\varphi)+Q(\chi,\chi)?$$

9. Sep 23, 2009

### kuruman

What does your textbook say the definition of the notation

$$(\varphi,Q\varphi)$$

is? Is there any reason to believe that

$$(\varphi,Q\varphi)=Q(\varphi,\varphi) ?$$

The parentheses with the comma in between them mean something. What is it?

10. Sep 23, 2009

### noblegas

it is the notation for the inner product. I think that I can only move Q out side when its a constant, but Q is a matrix. my book says: $$(\varphi,Q\varphi)=(Q\varphi,\varphi)$$

11. Sep 23, 2009

### kuruman

Right. So you cannot move Q outside the parentheses. It is a matrix and it needs a vector to operate on. This means that Q needs to be inside the parentheses either on the left or on the right side of the comma. What you wrote in posting #8 is incorrect. Try again.

12. Sep 23, 2009

### noblegas

$$(\varphi+\chi),Q(\varphi+\chi)=(\varphi+\chi),Q(\varphi)+Q(\chi)=(\varphi,Q\varphi)+(\chi,Q(\chi)= (\varphi,Q\varphi)+(\chi,Q(\chi)=(Q\varphi,\varphi )+(Q\chi,\chi)=(Q\varphi+Q\chi,\varphi+\chi)=(Q(\varphi+\chi),(\varphi+\chi)?$$

therefore$$(\varphi+\chi),Q(\varphi+\chi))=(Q(\varphi+\chi),(\varphi+\chi))$$ and is self-adjointed?

Last edited: Sep 23, 2009
13. Sep 23, 2009

### kuruman

One more time. Start with the left side and split it into two terms

$$((\varphi+\chi),Q(\varphi+\chi))=((\varphi+\chi),Q\varphi)+((\varphi+\chi),Q\chi)$$

the last result can be split into four terms

$$=(\varphi,Q\varphi)+(\chi,Q\varphi)+(\varphi,Q\chi)+(\chi,Q\chi)$$

Now do the same with the right side which is

$$(Q(\varphi+\chi),(\varphi+\chi))$$

then set the two sides equal.

14. Sep 23, 2009

### noblegas

$$(Q(\varphi+\chi),(\varphi+\chi))=((Q\varphi,(\varphi+\chi))+(Q\chi,(\varphi+\chi)= ((Q\varphi,(\varphi+\chi))+(Q\chi,(\varphi+\chi)=(Q\varphi,\varphi)+((Q\varphi,\chi)+(Q\chi, \varphi)+(Q\chi,\chi)=(\varphi,Q\varphi)+((\varphi,Q\chi)+(\chi, Q\varphi)+(\chi,Q\chi) ?$$