How to Calculate Uncertainty of an Operator with Known State

In summary, there are two ways to calculate the uncertainty of an operator: (1) using the formula \Delta\Omega2=<\Psi|(\Omega - <\Omega>)2|\Psi>, where the middle term is defined as the operator minus the expectation value multiplied by the identity operator, and (2) by summing the products of the probabilities of all the states with the states' deviation from the expected value squared. The second method can be seen as breaking down the state into a weighted average of eigenstates, applying the operator to each eigenstate separately, and then summing them all up according to their respective weights. Both methods are equivalent and do not require knowing all the probabilities of the states.
  • #1
Vaal
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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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  • #2
[tex]\Delta[/tex][tex]\Omega[/tex]2=<[tex]\Psi[/tex]|([tex]\Omega[/tex] - <[tex]\Omega[/tex]>)2|[tex]\Psi[/tex]> (hope that is legible)

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

I know I can also find [tex]\Delta[/tex][tex]\Omega[/tex]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 without having to know all the probabilities. Thanks.
 
  • #3
Vaal said:
[tex]\Delta[/tex][tex]\Omega[/tex]2=<[tex]\Psi[/tex]|([tex]\Omega[/tex] - <[tex]\Omega[/tex]>)2|[tex]\Psi[/tex]> (hope that is legible)

but I'm confused as to how the middle, ([tex]\Omega[/tex] -<[tex]\Omega[/tex]>) 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.

Vaal said:
I know I can also find [tex]\Delta[/tex][tex]\Omega[/tex]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 without having to know all the probabilities. Thanks.

I think that's exactly what the above is doing. Whenever you have [tex]\langle\Psi|\Omega|\Psi\rangle[/tex], 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 [tex]\Psi[/tex] 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.
 
  • #4
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.
 

1. What is "Uncertainty of an Operator"?

"Uncertainty of an Operator" refers to the concept in quantum mechanics of how accurately we can know the position and momentum of a particle at the same time.

2. How is "Uncertainty of an Operator" calculated?

"Uncertainty of an Operator" is calculated using the Heisenberg Uncertainty Principle, which states that the product of the uncertainty in position and momentum must always be greater than or equal to a specific value known as Planck's constant.

3. What does a high uncertainty of an operator value indicate?

A high uncertainty of an operator value indicates that there is a large range of possible values for the position and momentum of a particle, meaning that our knowledge of its exact location and movement is limited.

4. How does "Uncertainty of an Operator" relate to wave-particle duality?

The concept of "Uncertainty of an Operator" is closely related to the idea of wave-particle duality in quantum mechanics. This is because the uncertainty principle suggests that a particle cannot have a definite position and momentum at the same time, much like how a wave cannot have a definite wavelength and frequency at the same time.

5. Can the uncertainty of an operator be reduced?

The uncertainty of an operator cannot be reduced beyond a certain point due to the nature of quantum mechanics. However, by making measurements and observations, we can gain a better understanding of the particle's position and momentum, thus reducing the uncertainty to some extent.

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