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**Prove the equation**

[itex]A\left|\psi\right\rangle = \left\langle A\right\rangle\left|\psi\right\rangle + \Delta A\left|\psi\bot\right\rangle[/itex]

where [itex] A[/itex] is a Hermitian operator and [itex] \left\langle\psi |\psi\bot\right\rangle = 0[/itex]

[itex]\left\langle A\right\rangle[/itex] = The expectation value of A.

[itex]\Delta A[/itex] = The standard deviation of A.

[itex]A\left|\psi\right\rangle = \left\langle A\right\rangle\left|\psi\right\rangle + \Delta A\left|\psi\bot\right\rangle[/itex]

where [itex] A[/itex] is a Hermitian operator and [itex] \left\langle\psi |\psi\bot\right\rangle = 0[/itex]

[itex]\left\langle A\right\rangle[/itex] = The expectation value of A.

[itex]\Delta A[/itex] = The standard deviation of A.

My attempt :

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

[itex]A\left|\phi_{n}\right\rangle = \lambda_{n}\left|\phi_{n}\right\rangle[/itex] , [itex]\left|\psi\right\rangle = \sum a_{n}\left|\phi_{n}\right\rangle[/itex]

so that [itex]A\left|\psi\right\rangle = \sum a_{n}\lambda_{n}\left|\phi_{n}\right\rangle[/itex]

and [itex]\left\langle A\right\rangle = \sum |a_{n}|^{2}\lambda_{n} [/itex]

[itex]\Delta A = \sqrt{\left\langle A^{2}\right\rangle - \left\langle A\right\rangle ^{2}}[/itex]

and I wrote [itex]\left|\psi\bot\right\rangle[/itex] as [itex]\left|\psi\bot\right\rangle = \sum b_{n}\left|\phi_{n}\right\rangle[/itex]

[itex]\sum a^{*}_{n}b_{n} = 0[/itex]

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

any ideas?

thank you! :)