Proving a squared Hermitian operator is Hermitian

6 replies · 3K views · last reply
kehler
Messages
103
Reaction score
0

Homework Statement


Typeset problem statement: particle with wave function Psi and Hermitian operator A-hat; part (a) asks to show <A^2> = integral from -infinity to infinity of (A-hat Psi)* (A-hat Psi) dx.



Homework Equations


Text defining a Hermitian operator: an operator Θ̂ is Hermitian if the integral of ψm*(Θ̂ψn) dx equals the integral of (Θ̂ψm)*ψn dx.



The Attempt at a Solution


I've gone round in circles doing this! I started of by writing it as an integral of (psi* x A_hat2 x psi) w.r.t dx, then using the equation above but I keep coming back at my original equation after flipping it around. Are there any tricks in particular that I should know when dealing with squares of operators? Or any assumptions that I need to make? Please help :(
 
It may help to insert a "resolution of the identity" (as Shankar calls it). The operators can be treated in terms of their "matrix elements" on the Hilbert space, which is the space of the functions, Ψ, on which the operators act.

Are you familiar with the Dirac (bra-ket) notation? I would start with Dirac notation, and then simply insert the appropriate identity resolutions.
 
What's a resolution of the identity? I don't have the Shankar text. We use Griffiths in class... No unfortunately I'm not too familiar with Dirac Notation :(. I know roughly what it is but I'm not confident in using it just yet. So far we've just been taught the integral form. I think we're just meant to manipulate the equation till we get to be the required function but I've tried that several times without getting anywhere
 
OK. The first thing to do is to express the expectation value explicitly. Then, split A2 into factors of A. Then, use your definition for Hermiticity in a clever way.
 
I've done the first two steps. It's the third that I'm having trouble with.
Handwritten derivation: <A^2> equals the integral from minus infinity to infinity of Psi star times A_hat times A_hat Psi dx, then equals the integral of (A_hat A_hat Psi) star times Psi dx.

What do I do from here? :s
 
You didn't really do the second step, then. Think of the action of each A independently. That is, think of the action of the first A on the wavefunction as producing a new wavefunction, and then the other A acts on this new wavefunction.
 
Ahh ok I see :). Thanks. Gosh it seems really easy now.