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To test the usual Robertson-Schrodinger uncertainty bound, we follow the methods delineated in the answer here by Timaeus. This method talks about non-interfering measurements. For two incompatible (or compatible) observables, A and B, we do not measure both on the same system.

However, if it turns out that for some experiments, it is necessary to make two subsequent measurements of two non-commuting operators, A and B, on the same system, then the Robertson-Schrodinger uncertainty relation cannot be used anymore. For those kinds of experiments, does there exist a separate, rigorous uncertainty relation like the Robertson-Schrodinger uncertainty relation?

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  • $\begingroup$ Do the observables have continuous spectra, like the traditional position or momentum observables? Then $\sigma_A$ is still nonzero even after $A$ is measured, because it can't be measured with infinite resolution, so the RS uncertainty relation can still be used to get $\sigma_B$. (Use the Heisenberg picture so that all unitary time-evolution is already built into the operators $A$ and $B$.) Or do the observables have discrete spectra? In that case you're right: the RS uncertainty relation is useless because $\sigma_A=0$ after a perfect measurement of $A$, so it says nothing about $\sigma_B$. $\endgroup$ Mar 28, 2021 at 15:47
  • $\begingroup$ @ChiralAnomaly Why can the RS uncertainty relation still be used in the case when $\sigma_A$ is non-zero? The derivation relies on the fact that the expectation values of the operators that go into $\sigma_{A,B}$ are evaluated over the same state. If one first measures $A$ and then $B$ then the expectation values that go into $\sigma_B$ will be evaluated over a different state than the state over which the expectation values that go into $\sigma_A$ are evaluated, right? $\endgroup$
    – user87745
    Mar 28, 2021 at 15:52
  • $\begingroup$ I am imagining an ensemble, in which all states are prepared in state $\psi$. On each copy, I first measure A at time t. At the next instant t+dt, suppose I measure B, on all the copies (which had already collapsed into some eigenstate of A, with known probabilities, due to the first measurement). I this sort of experiment will involve some other uncertainty principle. $\endgroup$ Mar 28, 2021 at 15:56
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    $\begingroup$ Does the "error-disturbance (uncertainty) relations" in Martin's answer have anything to do with my question? physics.stackexchange.com/questions/169730/… Also section 6.1 in the article plato.stanford.edu/entries/qt-uncertainty $\endgroup$ Mar 28, 2021 at 16:01
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    $\begingroup$ There appears to be a nice paper by Distler and Paban on this topic: Uncertainties in Successive Measurements. I am not sure how my answer relates to their results, but the paper looks interesting. I will try to study the paper in more detail when I get the time and append my answer accordingly. $\endgroup$
    – user87745
    Mar 29, 2021 at 15:51

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I don't have a prior familiarity with this problem, but I found it quite interesting so I tried to scrabble around a bit and it looks like one can get a marginally non-ugly relation for the uncertainties of the kind that OP is interested in.


Let's consider the density matrix $\tilde{\rho}$ of the system after the measurement of the first operator $A$. It is easy to see that

\begin{align} \tilde\rho&=\sum_a \mathbb{P}_a\rho\mathbb{P}_a \end{align}

where $\rho=\vert\psi\rangle\langle\psi\vert$ is the initial (pure state) density matrix of the system, and $\mathbb{P}_a$ are the projection operators corresponding to the eigensubspaces of the operator $A$.

The measurement of the operator $A$, performed over $\rho$, has the well-known uncertainty given by

\begin{align} \sigma_A^2 =\mathrm{Tr}\big[\rho\big(A-\langle A\rangle\big)^2\big] \end{align}

For a more detailed calculation leading to the above expression, starting from the familiar state-vector formulation, see this answer of mine which deals with the usual RS uncertainty principle in the density matrix formulation. In order to keep the following discussion self-contained, I will freely copy relevant parts of the linked answer.

Similarly, the measurement of the operator $B$, performed over $\tilde{\rho}$ would carry an uncertainty given by

\begin{align} \sigma_B^2 &=\mathrm{Tr}\big[\tilde\rho\big(B-\langle B\rangle\big)^2\big]\\ \end{align}

Notice that we are adopting a compact notation for the expectation value which can lead to confusion if one loses sight of the fact that $\langle B\rangle = \mathrm{Tr}(\tilde{\rho}B)$ whreas $\langle A\rangle = \mathrm{Tr}(\rho A)$.

Thus, we can write \begin{align} \sigma_A^2\sigma_B^2&=\mathrm{Tr}\big[\rho\big(A-\langle A\rangle\big)^2\big]\mathrm{Tr}\big[\tilde\rho\big(B-\langle B\rangle\big)^2\big]\\ \mathrm{(linearity\ of\ trace)}&=\sum_a\mathrm{Tr}\big[\rho\big(A-\langle A\rangle\big)^2\big]\mathrm{Tr}\big[\mathbb{P}_a\rho\mathbb{P}_a\big(B-\langle B\rangle\big)^2\big]\\ \mathrm{(cyclic\ property\ of\ trace)}&=\sum_a\mathrm{Tr}\big[\rho\big(A-\langle A\rangle\big)^2\big]\mathrm{Tr}\big[\rho\mathbb{P}_a\big(B-\langle B\rangle\big)^2\mathbb{P}_a\big]\\ \mathrm{(idempotent\ nature\ of\ }\mathbb{P}_a\mathrm{)}&=\sum_a\mathrm{Tr}\big[\rho\big(A-\langle A\rangle\big)^2\big]\mathrm{Tr}\big[\rho\mathbb{P}^2_a\big(B-\langle B\rangle\big)^2\mathbb{P}^2_a\big]\\ \end{align} Now, the Cauchy-Schwarz inequalities hold true for all inner products, and it can be shown that for Hermitian operators $X$ and $Y$, it takes the following form

\begin{align} \big|\text{Tr}\big(SX Y\big)\big|^2 \leq \text{Tr}\big(SX^2\big)\text{Tr}\big(SY^2\big) \end{align} where $S$ is a semi-positive definite matrix. Since $\rho$ is a semi-positive definite matrix and $A-\langle A\rangle$, and $\mathbb{P}_a(B-\langle B\rangle)\mathbb{P}_a$ are Hermitian operators, an application of this inequality yields \begin{align} \mathrm{Tr}\big[\rho\big(A-\langle A\rangle\big)^2\big]\mathrm{Tr}\big[\rho\mathbb{P}^2_a\big(B-\langle B\rangle\big)^2\mathbb{P}^2_a\big] &\geq \Big\vert\ \mathrm{Tr}\big[\rho\big(A-\langle A\rangle\big)\mathbb{P}_a\big(B-\langle B\rangle\big)\mathbb{P}_a\big]\ \Big\vert^2\\ \implies \sum_a\mathrm{Tr}\big[\rho\big(A-\langle A\rangle\big)^2\big]\mathrm{Tr}\big[\rho\mathbb{P}^2_a\big(B-\langle B\rangle\big)^2\mathbb{P}^2_a\big] &\geq \sum_a\Big\vert\ \mathrm{Tr}\big[\rho\big(A-\langle A\rangle\big)\mathbb{P}_a\big(B-\langle B\rangle\big)\mathbb{P}_a\big]\ \Big\vert^2 \\ \implies \sigma_A^2\sigma_B^2&\geq \sum_a\Big\vert\ \mathrm{Tr}\big[\rho\big(A-\langle A\rangle\big)\mathbb{P}_a\big(B-\langle B\rangle\big)\mathbb{P}_a\big]\ \Big\vert^2\\ \end{align} We take a pause from our mainline calculation and manipulate the terms appearing on the RHS in the expression above. Using the benefit of the hindsight, we write \begin{align} \zeta_a&\equiv\mathrm{Tr}\big[\rho\big(A-\langle A\rangle\big)\mathbb{P}_a\big(B-\langle B\rangle\big)\mathbb{P}_a\big] \\ &= \mathrm{Tr}\big[\rho\big(A\mathbb{P}_aB\mathbb{P}_a-A\mathbb{P}_a\langle B \rangle \mathbb{P}_a - \langle A\rangle\mathbb{P}_aB\mathbb{P}_a+\langle A\rangle \mathbb{P}_a\langle B\rangle\mathbb{P}_a\big)\big]\\ &=\mathrm{Tr}\big[\mathbb{P}_a\rho\mathbb{P}_a\big(AB-\langle A\rangle B- \langle B\rangle A + \langle A\rangle \langle B\rangle \big)\big]\\ \implies \zeta_a^* &= \mathrm{Tr}\big[\mathbb{P}_a\rho\mathbb{P}_a\big(BA-\langle A\rangle B - \langle B\rangle A + \langle A\rangle \langle B\rangle \big)\big]\\ \implies \mathrm{Im}(\zeta_a) &= \frac{1}{2i}\mathrm{Tr}\big(\mathbb{P}_a\rho\mathbb{P}_a[A,B]\big) \end{align} where we used the linearity of the trace, the cyclic property of the trace, and ${\mathrm{Tr}(A)^*}=\mathrm{Tr}(A^\dagger)$ along with the Hermiticity of the operators and the fact that $\mathbb{P}_a$ commutes with $A$ whenever necessary.

We now invoke the argument that modulus square of a complex number should be bigger than or equal to the square of its imaginary part. We can thus write \begin{align} \sigma_A^2\sigma_B^2&\geq \sum_a \Big\vert\ \mathrm{Im}(\zeta_a)\ \Big\vert^2 \\ \implies \sigma_A^2\sigma_B^2&\geq\frac{1}{4}\sum_a \Big\vert\ \mathrm{Tr}\big(\mathbb{P}_a\rho\mathbb{P}_a[A,B]\big)\ \Big\vert^2 \end{align} Or, in other words, $$\boxed{\sigma_A\sigma_B\geq\frac{1}{2}\sqrt{\sum_a \Big\vert\ \mathrm{Tr}\big(\mathbb{P}_a\rho\mathbb{P}_a[A,B]\big)\ \Big\vert^2}}$$ Notice that the traditional RS uncertainty relation in the form of density matrix reads \begin{align} \sigma_A\sigma_B&\geq\frac{1}{2} \Big\vert\ \mathrm{Tr}\big(\rho[A,B]\big)\ \Big\vert \end{align}

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  • $\begingroup$ The upshot of your answer is that subsequent measurements on the same system lead to a separate uncertainty inequality than the RS uncertainty inequality. Am I right? $\endgroup$ Mar 30, 2021 at 5:30

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