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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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    Is this a question about mathematics, or about understanding quantum mechanics? If mathematics, then what kind of answer to "what is ..." do you expect? The expressions are what they are -- which other description than that do you imagine getting?2012-04-22
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    @Henning Makholm what ever you like explain it.2012-04-22
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    The point is I don't think there _is_ any explanation that doesn't just repeat the expression you want explained. $\langle \hat A\rangle^2$ is the number $\langle \hat A\rangle$ multiplied by itself. That seems to be all there is to say about it, absent any knowledge of $\hat A$ or $\phi$.2012-04-22
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    @Henning Makholm i need mathematical proof For Example $|z|+|z'|\ge |z+z'|$ mathematical proof is: $(xy'-x'y)^2\ge 0$ some thing like this.2012-04-22
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    You need mathematical proof of _what_? "Multiplied by itself" is what "$^2$" _means_. That's not a matter of proof, it is just how the notation works.2012-04-22
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    @Henning Makholm proof that there is difference between then them2012-04-22
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    Well, if we take $\hat A$ to be the identity operator on $\mathbb R^1$ and $\phi$ to be $2$, then $\langle \phi|\hat A|\phi\rangle^2=16$ but $\langle \phi|\hat A^2|\phi\rangle=4$. That looks pretty different.2012-04-22
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    @Henning Makholm assume $\hat A=complex\,number$2012-04-22
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    $1$ is a perfectly cromulent complex number!2012-04-22
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    @Henning Makholm and $\Delta A$ will be complex too. isn't that ture!?2012-04-22
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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

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