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Heisenberg's uncertainty relation in the Robertson-Schroedinger formulation is written as,

$$\sigma_A^2 \sigma_B^2 \geq \left|\frac{1}{2} \langle\{\hat A, \hat B\}\rangle -\langle \hat A\rangle\langle \hat B\rangle\right|^2+\left|\frac{1}{2}\langle[\hat A,\hat B]\rangle\right|^2 $$ where $\sigma_A^2 = \langle\psi|(\hat A-\langle \hat A \rangle)^2 |\psi\rangle$ and $\sigma_B^2 = \langle\psi|(\hat B-\langle \hat B \rangle)^2 |\psi\rangle$ calculated in the same state $\psi$ for both observables $\hat A$ and $\hat B$.

Now my question is what happens to the other side of the inequality if we calculate one variance for state $\psi(t)$ and then let the state evolve to $\psi(t+\delta t)$ and now calculate the other variance in the product. In other words, what is the QM lower limit of this product: $$ \langle {\psi(t)|(\hat A -\langle \hat A\rangle)^2|\psi(t)\rangle} ~\langle {\psi(t+\delta t)|(\hat B -\langle \hat B\rangle)^2|\psi(t+\delta t)\rangle} $$ for arbitrary $\delta t$ and $\psi(t)$ is evolving according to the time-dependent Schroedinger equation $$\hat H \psi(t)=i \hbar\frac{\partial \psi(t)}{\partial t}~?$$

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    $\begingroup$ I am surprised that this is construed as a "homework" or "off topic" problem. I have never seen asked before. Is the answer that obvious to you @Danu, Prahar, Jim, JamalS, Brandon Enright? The "RHS" must somehow depend on $\hat H$ I just don't know how, and down-voting does not solve the problem. The answer offered below is obviously and admittedly wrong. $\endgroup$ – hyportnex Nov 12 '14 at 0:15
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    $\begingroup$ Since $|\psi(t)\rangle=\exp\left(it\hat{H}/\hbar\right)|\psi(0)\rangle$, you probably could write $|\psi(t+\delta t)\rangle$ using this time-evolution operator. This probably would require you to know $[\hat{H},\,\hat{B}]$. $\endgroup$ – Kyle Kanos Nov 18 '14 at 3:37
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    $\begingroup$ In short, this can be zero. In the Heisenberg picture, it is possible for $B(0)$ to be canonically conjugate to $A(0)$, and then 'rotate' into $B(\Delta t)=A(0)$. Examples are easy to find with $x$ and $p$ in a harmonic oscillator, or $\sigma_x$ and $\sigma_y$ for a two-level system where $H\propto \sigma_z$. $\endgroup$ – Emilio Pisanty Nov 18 '14 at 12:20
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    $\begingroup$ Physical note: There are no "non-simultaneous observations" here. Either you measure a system in which case the quantum state will be altered and after some time you'll NOT have $|\psi(t+\delta t)\rangle$, or it won't, in which case you don't have an observation. Also note that the Robertson-Schrödinger uncertainty relation is not about simultaneous measurement, it is about variances of observables after preparation. $\endgroup$ – Martin Sep 24 '15 at 16:54
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    $\begingroup$ Technically, no, it is not an observarble, because of the state-dependent part $\langle A \rangle$, and the notation $\langle A \rangle$ might be ambiguous.in this case. $\endgroup$ – pppqqq Dec 5 '16 at 20:33
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Try to use the Heisenberg picture. It will become more evident what you are actually computing, namely the uncertainty relation between two obsevables $A(t_1)$ and $B(t_2)$. You can take as an exercise the harmonic oscillator and compute the uncertainty relation between $x(0)$ and $x(t)$, it will be nonzero for $t$ different from the period of the oscillator.

hint:

$$ x(t)=e^{iHt}x(0)e^{-iHt} $$

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