Is Q Self-adjoint Given Certain Conditions?

[noblegas]

Homework Statement

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

Homework Equations

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?

 

[noblegas]

is my question to esoteric

 

[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.

 

[noblegas]

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 z

(\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.

shouldn't I plug in \phi_2 as well ?

 

[kuruman]

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

 

[noblegas]

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

<br /> ((\varphi+\chi),Q(\varphi+\chi))=q(\varphi+\chi), (\varphi+\chi))=q*(\varphi+\chi), (\varphi+\chi)), q is some number.
So now what?

 

[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.

 

[noblegas]

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.

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
<br /> (\phi,Q\phi)=(Q\phi,\phi)<br />,then (\varphi,Q\varphi)+(\chi,Q(\chi)=Q(\varphi,\varphi)+Q(\chi,\chi)?

 

[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?

 

[noblegas]

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?

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)

 

[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.

 

[noblegas]

<br /> (\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)?<br />

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

 

[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.

 

[noblegas]

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.

<br /> (Q(\varphi+\chi),(\varphi+\chi))=((Q\varphi,(\varphi+\chi))+(Q\chi,(\varphi+\chi)=<br /> ((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) ?