# Hermitian operator- Problem

Prove the equation

$A\left|\psi\right\rangle = \left\langle A\right\rangle\left|\psi\right\rangle + \Delta A\left|\psi\bot\right\rangle$

where $A$ is a Hermitian operator and $\left\langle\psi |\psi\bot\right\rangle = 0$

$\left\langle A\right\rangle$ = The expectation value of A.
$\Delta A$ = The standard deviation of A.

My attempt :

I tried to write $\left|\psi\right\rangle$ as a superposition of the eigenfuncion of the operator $A$ and used the fact that it's a Hermitian operator

$A\left|\phi_{n}\right\rangle = \lambda_{n}\left|\phi_{n}\right\rangle$ , $\left|\psi\right\rangle = \sum a_{n}\left|\phi_{n}\right\rangle$

so that $A\left|\psi\right\rangle = \sum a_{n}\lambda_{n}\left|\phi_{n}\right\rangle$

and $\left\langle A\right\rangle = \sum |a_{n}|^{2}\lambda_{n}$

$\Delta A = \sqrt{\left\langle A^{2}\right\rangle - \left\langle A\right\rangle ^{2}}$

and I wrote $\left|\psi\bot\right\rangle$ as $\left|\psi\bot\right\rangle = \sum b_{n}\left|\phi_{n}\right\rangle$

$\sum a^{*}_{n}b_{n} = 0$

I don't know how to go on from here...
any ideas?

thank you! :)

dextercioby
Homework Helper
Hmm, using that the s-adj A has a purely discrete spectrum is not ok, it's not assumed in the problem. And I think you left out a square. I get

$$A\psi = \langle A \rangle \psi + (\Delta A)^2 \psi_{\perp}$$

under the simplifying assumptions $\psi_{\perp} \in D(A)$ and $\mbox{Ran}(A)\subset D(A)$.

dextercioby
Homework Helper
The first step I found to resolution is to calculate:

$$\langle \psi,(A-\langle A\rangle ) \psi\rangle$$

What do you get and what do you conclude ?

I didn't miss the squar.. this is the question.
you get wrong units if you put the squar there ...

The first step I found to resolution is to calculate:

$$\langle \psi,(A-\langle A\rangle ) \psi\rangle$$

What do you get and what do you conclude ?

this yields zero .. i didn't understand how it can help me

thank you

dextercioby
Homework Helper
If it yields zero, then the vector in the rhs of the scalar product is perpendicular on psi, so it's psi perp times a non-zero constant.

I don't understand something -
there are lots of functions that can be perpendicular to psi

e.g (euclidean space)- the vector z (cartesian coordinates) , x is prep to z and y is prep to z but also the superposition (x+y) is prep to z.

so by writing psi prep do u mean to the superposition of all the functions that prep to psi or it's enough to choose only one?

dextercioby