# Prove Eigenvalues of an (operator)^2 are real and positive

Q: Using Dirac notation, show that if A is an observable associated with the operator A then the eigenvalues of A^2 are real and positive.

Ans: I know how to prove hermitian operators eigenvalues are real:
A ket(n) = an ket(n)
bra(n) A ket(n) = an bra(n) ket(n) = an
[bra(n) A ket(n)]* = an* bra(n) ket(n) = an*
[bra(n) A(dager) ket(n)] =[bra(n) A ket(n)] = an*
therefore, an*=an

But, I don't know how to treat an operator^2 and start this question.

DrClaude
Mentor
Q: Using Dirac notation, show that if A is an observable associated with the operator A then the eigenvalues of A^2 are real and positive.

Ans: I know how to prove hermitian operators eigenvalues are real:
A ket(n) = an ket(n)
bra(n) A ket(n) = an bra(n) ket(n) = an
[bra(n) A ket(n)]* = an* bra(n) ket(n) = an*
[bra(n) A(dager) ket(n)] =[bra(n) A ket(n)] = an*
therefore, an*=an

But, I don't know how to treat an operator^2 and start this question.
What is the relation between an observable and Hermetian operators?

Can you think of a way to rewrite the operator in the expression ##\hat{A}^2 | n \rangle##?

Relationship: it's a postulate of QM that every dynamical observable is represented by a linear hermitian operator.

A*A ket(n) = ? but not sure what will be on the right hand side

DrClaude
Mentor
Relationship: it's a postulate of QM that every dynamical observable is represented by a linear hermitian operator.
Correct.

A*A ket(n) = ? but not sure what will be on the right hand side
Yes, which you can write as
$$\hat{A}^2 | n \rangle = \hat{A} \left( \hat{A} | n \rangle \right)$$
What can you say about the contents of the parenthesis?

it equals =an ket(n)
than A(an ket(n)) = an A ket(n) = an (an ket (n))= an^2 ket(n)

is that correct way to do it?

DrClaude
Mentor
it equals =an ket(n)
than A(an ket(n)) = an A ket(n) = an (an ket (n))= an^2 ket(n)

is that correct way to do it?
Yes. Now you can use the fact that A is Hermitian to complete the answer to the problem.

Thanks
So for hermitian A= A(dager), but would u write (A^2)* = A*A* = A*(an* ket(n)) ....

DrClaude
Mentor
Thanks
So for hermitian A= A(dager), but would u write (A^2)* = A*A* = A*(an* ket(n)) ....
No, that's not what I meant. You should use the proof you gave in the OP (I don't think you need the put the proof itself in the answer, but rather use what the proof says about Hermetian operators).

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