quantum mechanics - Uncertainty relation for non-simultaneous observation

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 $$\hat H \psi(t)=i \hbar\frac{\partial \psi(t)}{\partial t}~?$$

Answer

The requested lower limit is zero already for $X$ and $P$ as I am going to prove.

Let us consider the Fourier-Plancherel transform $F: L^2(\mathbb{R},dx)\to L^2(\mathbb{R},dx)$, formally for integrable functions (otherwise a further extension is necessary)

$$(F\psi)(x) = \frac{1}{(2\pi)^{1/2}} \int_{\mathbb R} e^{ixy} \psi(y) dx$$ It is clear that if $\psi$ is very concentrated around the value $p_0/\hbar$, then $F\psi$ tends to approach $$const.\frac{e^{ip_0x/\hbar}}{(2\pi)^{1/2}}$$ In other words:

$F$ transforms approximated eigenvectors of the position operator $X$ to approximated eigenvectors of the momentum operator $P$.

from now on I set $\hbar=1$ for the sake of semplicity.

It is known that the spectrum of the unitary operator $F$ is made of four elements $\pm 1, \pm i$. The eigenvectors are nothing but the Hermite functions, but the details are not relevant here. From the spectral theorem of unitary operators we can therefore write down

$$F = 1 P_1 -1 P_{-1} + i P_{i} -i P_{-i}$$ where $P_\lambda$ is the orthogonal projector onto the eigenspace of $F$ with eigenvalue $\lambda \in \{\pm 1, \pm i\}$. We can re-write the previous spectral decomposition as $$F = e^{i0} Q_0 +e^{i\pi/2} Q_{\pi/2} + e^{i\pi} Q_{\pi} + e^{i3\pi/2} Q_{3/2} = e^{iH}$$ where we have introduced the selfadjoint operator $$H = 0 Q_0 +(\pi/2) Q_{\pi/2} + \pi Q_{\pi} + (3\pi/2) Q_{3/2}$$ with obviously $$Q_0 := P_1\:,\quad Q_{\pi/2}:= P_{i}\:, \quad Q_{\pi}:= P_{-1}\:, \quad Q_{3\pi/2}:= P_{-i}\:.$$ $H$ has pure point spectrum made of the four eigenvalues $0, \pi/2, \pi, 3\pi/2$.

Let us finally consider the time evolutor $U_t = e^{-itH}$. According to the definitions above, it reads

$$U_t = e^{-i0t} Q_0 +e^{-it\pi/2} Q_{\pi/2} + e^{-it\pi} Q_{\pi} + e^{-i3t\pi/2} Q_{3/2}\:.$$ As a consequence: $$U_0 =I \quad \mbox{and}\quad U_{-1} = F\:.$$

This discussion permits to prove that the requested lower limit for

$$\sigma^{(t)2}_X \sigma^{(t+\delta t)2}_P$$ is zero.

It is sufficient to set $t=0$ and $\delta t =-1$ and referring to a state $\psi$ at $t=0$ which is sufficiently concentrated around $p_0/\hbar$, so that $\sigma^{(0)2}_X$ can be made as small as wanted. With that choice $\sigma^{(-1)2}_P$ is the standard deviation of $F\psi$ which is arbitrarily close to an eigenvector of $P$ so that, in turn, also $\sigma^{(-1)2}_P$ tends to vanish. The product

$$\sigma^{(0)2}_X \sigma^{(-1)2}_P$$ can be made as small as wanted choosing $\psi$ arbitrarily concentrated around $p_0/\hbar$.

ADDENDUM. I was a bit sloppy on this point, but the fact that $F\psi$ approaches a normalized eigevector of the momentum as $\psi$ approaches a a normalized eigenvector of $X$ easily follows form the spectral theorem using the fact that the spectral measure of $P$ and that of $X$ are bijectively related through the Fourier transform.