# Operators Uncertainty

$\hat A$ is an operator. The uncertainty on $\hat{A}$, $\Delta A$ is defined by:

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

what is difference between $\langle\hat A^2\rangle$ and $\langle\hat A\rangle^2$ that leads to Uncertainty Relation between two Operators?

more details: $$\langle\hat A^2 \rangle=\langle\psi|\hat A^2|\psi \rangle$$ What is the name of difference between absolute value of these two complex conjugates

• The meaning and significance of the notation don't change for different letters used as the operator, so I have removed the redundant $\hat{B}$. Also note the layout differences implied by inline math (single $s) and block formatted math (double $s) and the need to uses braces {} to indicate the range of effect for operators like \sqrt. Apr 21 '12 at 19:16
• thank you but when "that leads to Uncertainty Relation between two Operators?" Oops! where is the second operator:D
– user8784
Apr 21 '12 at 19:31
• Well, you can put $\hat B$, back in but I think that you are mixing up two ideas here. Without writing a commutator between $\hat A$ and $\hat B$, you don't know anything about their relationship. Apr 21 '12 at 19:36
• "you don't know anything about their relationship" thank you! for your opinion.
– user8784
Apr 21 '12 at 19:41

1. $\langle\hat A\rangle$ is the expectation value of $\hat A$.

2. $\langle\hat A\rangle^2$ is the square of item 1.

3. $\langle\hat A^2\rangle$ is the expectation value of $\hat A^2=\hat A \hat A$.

Item 2 and 3 do not have to be equal.

If $\hat A$ is selfadjoint, then it is possible to show

1. that the expectation value $\langle\hat A\rangle~\in~\mathbb{R}$ is real, and

2. that $\langle\hat A^2\rangle ~\geq~ \langle\hat A\rangle^2$.

• thank you "2. $\langle\hat A^2\rangle ~\geq~ \langle\hat A\rangle^2$." so when $\langle\hat A^2\rangle \approx \langle\hat A\rangle^2$. $\Delta A$ should be equal zero
– user8784
Apr 21 '12 at 19:21
• could you please Explain with equation that what is difference between expectation value of: 1. $\langle\hat A\rangle$ 2. $\langle\hat A^2\rangle$ regards that $\langle\psi|\hat A|\psi \rangle=\langle\psi|\psi' \rangle$
– user8784
Apr 25 '12 at 20:16

Although Qmechanics's answer is formally complete and correct, there is a more intuitive formulation of this identity that makes it self evident. Consider the operator B which is A minus its expectation value in some state.

$$B = A - \langle A\rangle$$

Then the expectation value of B is zero in the same state (obviously--- it has been shifted to make it so). The expected value of $B^2$ can be nonzero--- it is a measure of the spread in B in state $\psi$. It is positive, as you can see by the definition of matrix multiplication (or by "inserting the identity in a basis")

$$\langle B^2 \rangle = \sum_i \langle |B|i\rangle\langle i|B\rangle$$

The last thing on the right is the sum of positive quatities of the form $c^*c$. If you now reexpress the expectation value of $B^2$ in terms of A,

$$\langle B^2 \rangle = \langle (A-\langle A\rangle)^2\rangle = \langle A^2\rangle - 2 \langle A\langle A\rangle \rangle + \langle A\rangle^2 = \langle A^2\rangle - \langle A\rangle^2$$

This manipulation justifies this thing.