Uncertainty in some observable A

[CAF123]

Why do we define the uncertainty in some physical quantity $A$ as: $$\delta A = < \sqrt{<A^2> - <A>^2} > .$$ I know that it can be derived by computing the variance of $A$, but what is the physical meaning?
Thanks!

 

[cattlecattle]

Why do we define the uncertainty in some physical quantity $A$ as: $$\delta A = < \sqrt{<A^2> - <A>^2} > .$$ I know that it can be derived by computing the variance of $A$, but what is the physical meaning?
Thanks!

It has the same statistical meaning as usual variance. The deviation of any particular measurement from the expected value is A-<A>, the average value of this quantity is by definition zero. so in order to define variance, we square it (so it's no longer zero), and take the average. (followed by a square root to make dimensions right)

 

[jtbell]

The deviation of any particular measurement from the expected value is A-<A>, the average value of this quantity is by definition zero. so in order to define variance, we square it (so it's no longer zero), and take the average. (followed by a square root to make dimensions right)

Mathematically:

$$\Delta A = \sqrt{\langle (A - \langle A \rangle)^2 \rangle}$$

which can be shown to reduce to

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

 

[CAF123]

The deviation of any particular measurement from the expected value is A-<A>, the average value of this quantity is by definition zero.

Why is this zero? A is the measurement obtained from some experiment and <A> is the number obtained by making measurements on many different particles in the same quantum state (I,e described by the same wave function). Did I understand the meaning of these terms correctly?

so in order to define variance, we square it (so it's no longer zero), and take the average. (followed by a square root to make dimensions right)

Why does squaring zero not equal zero?

@jtbell - I proved the statement you gave.

Thanks!

 

[jtbell]

Why is this zero?

Half the measurements have a positive deviation (above the expectation value) and half have a negative deviation (below the expectation value). Their average is zero. More precisely, the average approaches zero as the number of measurements becomes larger and larger.

<A> is the number obtained by making measurements on many different particles in the same quantum state (I,e described by the same wave function).

I would say that $\langle A \rangle$ is the predicted or expected average of many measurements (hence the term "expectation value"), calculated from the wave function according to a certain recipe. The observed or experimental average is never exactly the expectation value, but it comes closer and closer to the expectation value as the number of measurements increases.

 

[Sonderval]

@CAF
I think you understood well and just misread: It is not
A-<A> that is zero, but its average.
Thus (A−⟨A⟩)² is not zero and its average isn't either.

 

[CAF123]

The observed or experimental average is never exactly the expectation value.

The above is true unless the wave function can be written as a product of two functions, one in $x$ and one in $t$, yes? In which case, the energy takes a definite value (the wavefunction models an energy eigenstate). I believe in general in this case, for a stationary state, I can write that $ΔA = 0$ and so every measurement of $A$ over lots of systems will give the same value, that value being the expected value.

 

[jtbell]

for a stationary state, I can write that $ΔA = 0$

That is true only if A = E (energy). If ΔE = 0, it's possible for Δx or Δp or ΔL or whatever to be nonzero.

 

[Sonderval]

@CAF
In general, if your system is in an eigenstate of the operator A, the measured value of A will always be the eigenvalue. There is no general conclusion to be drawn on any other operator B.