# Uncertainty of an Operator

Given the state and an operator I know the uncertainty of this operator can be calculated via

(see next post latex is being weird, sorry)

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$$\Delta$$$$\Omega$$2=<$$\Psi$$|($$\Omega$$ - <$$\Omega$$>)2|$$\Psi$$> (hope that is legible)

but I'm confused as to how the middle, ($$\Omega$$ -<$$\Omega$$>) is defined. Isn't this an operator minus a scalar?

I know I can also find $$\Delta$$$$\Omega$$2 by summing the the products of the probabilities of all the states with the states deviation from the expected value squared, but I thought there was a way to do this with out having to know all the probabilities. Thanks.

$$\Delta$$$$\Omega$$2=<$$\Psi$$|($$\Omega$$ - <$$\Omega$$>)2|$$\Psi$$> (hope that is legible)

but I'm confused as to how the middle, ($$\Omega$$ -<$$\Omega$$>) is defined. Isn't this an operator minus a scalar?
The scalar is multiplied by the identity operator. Then you can subtract them, and things work out like you'd expect.

I know I can also find $$\Delta$$$$\Omega$$2 by summing the the products of the probabilities of all the states with the states deviation from the expected value squared, but I thought there was a way to do this with out having to know all the probabilities. Thanks.

I think that's exactly what the above is doing. Whenever you have $$\langle\Psi|\Omega|\Psi\rangle$$, you can envision breaking down the state into a weighted average of eigenstates of the operator. Then you know that the operator's effect on each eigenstate will just be multiplying it by the eigenvalue, so that allows you to turn the calculation into a weighted average of eigenvalues. For the above expression, I believe you can do the same thing: break down $$\Psi$$ into weighted eigenstates, then apply the operator to each eigenstate separately to get the eigenvalue, subtract the expectation value from it, square it, and then sum them all up according to the original weights on the states.

Yeah, I thought that might be that case but I wasn't sure. Thanks.

The two definitely are pretty much equivalent, I just wasn't quite seeing how so, thanks again.