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, 2014 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, 2014 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 $H\propto \sigma_z$. $\endgroup$ Nov 18, 2014 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, 2015 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, 2016 at 20:33

3 Answers 3


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.

  • $\begingroup$ Thank you, I have been waiting for this answer but it also brings up another question: let us assume an ensemble of identically prepared systems, we measure one parameter, say, position of one system and sometime later we measure the conjugate variable, say, momentum of another system in the ensemble. Would this potentially lower "lower bound" than that of the HUP help us refine further what a trajectory might mean in QM? $\endgroup$
    – hyportnex
    Nov 7, 2019 at 21:05
  • $\begingroup$ I cannot understand how such sort of coupled and retarded measures could help refine a trajectory. $\endgroup$ Nov 7, 2019 at 21:27

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.


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


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.

  • $\begingroup$ Let me rephrase: Imagine you have two identical systems prepared in the same quantum mechanical state characterized by the wave function $\psi$. Now you measure $\hat A$ for one system at time $t$ while measure the other for $\hat B$ at time $t+\delta t$. What is the lower limit of the product of the two variances as function of $\delta t$? We know the answer for $\delta t =0$. $\endgroup$
    – hyportnex
    Oct 4, 2019 at 12:24
  • $\begingroup$ Without doing the maths, I would say that it would still make no difference. The uncertainty relation is time independent for a reason. None of the expectation values involved vary by time, and none of the expansion coefficients vary by time. The functions are normalised in a time independent way. In short, time plays no role in the calculation. The two states you mention are indistinguishable, from a measurement perspective, regardless of the time at which measurements are taken. $\endgroup$ Oct 4, 2019 at 13:34
  • $\begingroup$ you should read Prof Moretti's answer above $\endgroup$
    – hyportnex
    Nov 7, 2019 at 20:59

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