Uncertainty relation for non-simultaneous observation

Question

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 1

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.

Answer 2

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:

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

Answer 3

The answer is that time doesn't make a difference to the degree of uncertainty.

The point about Heisenberg's uncertainty principle is that when you make a measurement associated with one operator, A say, the wave function of the system becomes one of the eigenfunctions of A. If you then make a measurement corresponding to another operator, B say, that doesn't commute with A, the wave function of the system has to change to become an eigenstate of B, where the 'choice' of final eigenstate is uncertain. The probability of the wave function becoming a particular eigenstate of B depends upon the overlap between that eigenstate and the eigenstate of A that the system had previously.

In the previous paragraph, where I said 'If you then make a measurement' I did not impose any constraints on the time lag- you could wait as long as you liked. That is because the overlap between an eigenfunction of B and an eigenfunction of A is not time-dependent. Or, to put it another way, the expansion coefficients that allow you to express an eigenfunction of A in terms of some basis set are not time dependent.