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operator $\hat A$ is a mathematical rule that when applied to a ket $\hat A|\phi\rangle$ transforms it into another ket $\hat A|\phi '\rangle $ and too for bra.

$\langle \phi| \hat A|\phi\rangle$ for short $\langle\hat A\rangle$

  1. what is the square of item $\langle\hat A\rangle^2$ ?
  2. what is the expectation value item $\langle\hat A^2\rangle$?
  3. what is difference between them$\Delta A=\sqrt{\langle\hat A^2\rangle-\langle\hat A\rangle^2}$?
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    cross-post: http://physics.stackexchange.com/q/24207/24512012-04-22

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Uncertainty Relation Between Two Operators

An interesting application of the commutator algebra is to derive a general relation giving the uncertainties product of two operator $\hat A$ and $\hat B$

if $\langle\hat A\rangle$ and $\langle\hat B\rangle$ be the expectation valus of to hermitian operators $\hat A$ and $\hat B$ with respect to a normalized state vector $|\psi\rangle$. that is, $\langle \psi| \hat A|\psi\rangle=\langle\hat A\rangle$ and $\langle \psi| \hat B|\psi\rangle=\langle\hat B\rangle$

The uncertainties $\delta A$ and $\delta B$ are defined by:

$\delta A=\sqrt {\langle\hat A^2\rangle - \langle\hat A\rangle^2}$, $\delta B=\sqrt {\langle\hat B^2\rangle - \langle\hat B\rangle^2}$, $\delta A \delta B\ge \frac {1}{2}|\langle [\hat A, \hat B] \rangle|$

leads to Heisenberg Uncertainty Relations

$\Delta x \Delta p_x \ge \frac {1}{2}|\langle [\hat X, \hat P_x] \rangle|=\frac {1}{2} |\langle i\hbar \hat I \rangle|=\frac {1}{2} \hbar$