# Uncertainty in some observable A

1. Dec 14, 2012

### 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!

2. Dec 14, 2012

### cattlecattle

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)

3. Dec 14, 2012

### Staff: Mentor

Mathematically:

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

which can be shown to reduce to

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

4. Dec 15, 2012

### CAF123

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?

Why does squaring zero not equal zero?

@jtbell - I proved the statement you gave.

Thanks!

5. Dec 15, 2012

### Staff: Mentor

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.

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.

6. Dec 15, 2012

### 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.

7. Dec 15, 2012

### CAF123

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.

8. Dec 15, 2012

### Staff: Mentor

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.

9. Dec 15, 2012

### 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.