Hermitian operator- Problem

[Raz91]

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]

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]

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 ?

 

[Raz91]

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

 

[Raz91]

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]

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.

 

[Raz91]

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]

Yes, Psi_perp is an arbitrary vector perpendicular to the psi already chosen. You get from ⟨ψ,(A−⟨A⟩)ψ⟩ = 0 that C psi_perp = A psi - <A> psi. All you need to show is that C is related to the standard deviation in square.