Entanglement never at first order

When two initially uncorrelated quantum systems interact through a weak coupling, no entanglement is generated at first order in the coupling constant. This is a useful and very easy to prove fact that I haven’t seen pointed out anywhere, although I assume someone has. I’d love a citation reference if you have one.

Suppose two systems \mathcal{A} and \mathcal{B} evolve under U = \exp(- i H t) where the Hamiltonian coupling them is of the form

(1)   \begin{align*} H=H_A + H_B + \epsilon H_I, \end{align*}

with H_A = H_A \otimes I_B and H_B = I_A \otimes H_B as usual. We’ll show that when the systems start out uncorrelated, \vert \psi^0 \rangle = \vert \psi_A^0 \rangle \otimes \vert \psi_B^0 \rangle, they remain unentangled (and therefore, since the global state is pure, uncorrelated) to first order in \epsilon. First, note that local unitaries cannot change the entanglement, so without loss of generality we can consider the modified unitary

(2)   \begin{align*} U' = e^{+i H_A t} e^{+i H_B t} e^{-i H t} \end{align*}

which peels off the unimportant local evolution of \mathcal{A} and \mathcal{B}. Then the Baker–Campbell–Hausdorff formula gives

(3)   \begin{align*} U' = e^{+i H_A t} e^{+i H_B t} e^{-i (H_A + H_B) t} e^{-i \epsilon H_I t}  e^{Z_2} e^{Z_3} \cdots \end{align*}

where the first few Z‘s are given by

(4)   \begin{align*} Z_2 &= \frac{(-i t)^2}{2} [H_A+H_B,\epsilon H_I] \\ Z_3 &= \frac{(-i t)^3}{12} \Big( [H_A+H_B,[H_A+H_B,\epsilon H_I]]-  [\epsilon H_I,[H_A+H_B,\epsilon H_I]] \Big) \\ Z_4 &= \cdots. \end{align*}

The key feature here is that every commutators in each of the Z‘s contains at least one copy of \epsilon H_I, i.e. all the Z‘s are at least first order in \epsilon. That allows us to write

(5)   \begin{align*} U' = e^{-i \epsilon H'_I t} \big(1 + O(\epsilon^2) \big) \end{align*}

for some new H'_I that is independent of \epsilon. Then we note just that a general Hamiltonian cannot produce entanglement to first order:

(6)   \begin{align*} \rho_A &= \mathrm{Tr}_B \left[ U' \vert \psi^0 \rangle \langle \psi^0 \vert {U'}^\dagger \right] \\ &=  \vert \psi'_A \rangle \langle \psi'_A \vert + O(\epsilon^2) \end{align*}

where

(7)   \begin{align*} \vert \psi'_A \rangle &= \left( I - i \epsilon t \langle \psi^0_B  \vert H_I' \vert  \psi^0_B \rangle \right) \vert \psi^0_A \rangle . \end{align*}

This is potentially a very important (negative) result when considering decoherence detection of very weakly coupled particles. If the coupling is so small that terms beyond first order are negligible (e.g. relic neutrinos), then there is no hope of being sensitive to any decoherence.

Of course, non-entangling (unitary) effect may be important. Another way to say this result is: Two weakly coupled systems act only unitarily on each other to first order in the coupling constant.… [continue reading]

Wavepacket spreading produces force sensitivity

I’m still trying to decide if I understand this correctly, but it looks like coherent wavepacket spreading is sufficient to produce states of a test-mass that are highly sensitive to weak forces. The Wigner function of a coherent wavepacket is sheared horizontally in phase space (see hand-drawn figure). A force that perturbs it slightly with a small momentum shift will still produce an orthogonal state of the test mass.


The Gaussian wavepacket of a test mass (left) will be sheared horizontally in phase space by the free-particle evolution governed by H=p^2/2m. A small vertical (i.e. momentum) shift by a weak force can then produce an orthogonal state of the test mass, while it would not for the unsheared state. However, discriminating between the shifted and unshifted wavepackets requires a momentum-like measurement; position measurements would not suffice.

Of course, we could simply start with a wavepacket with a very wide spatial width and narrow momentum width. Back when this was being discussed by Caves and others in the ’80s, they recognized that these states would have such sensitivity. However, they pointed out, this couldn’t really be exploited because of the difficulty in making true momentum measurements. Rather, we usually measure momentum indirectly by allowing the normal free-particle (H=p^2/2m) evolution carry the state to different points in space, and then measuring position. But this doesn’t work under the condition in which we’re interested: when the time between measurements is limited.The original motivation was for detecting gravitational waves, which transmit zero net momentum when averaged over the time interval on which the wave interacts with the test mass. The only way to notice the wave is to measure it in the act since the momentum transfer can be finite for intermediate times.[continue reading]

Decoherence Detection FAQ—Part 1: Dark matter

[Updated 2016-7-2]

I’ve submitted my papers (long and short arXiv versions) on detecting classically undetectable new particles through decoherence. The short version introduces the basic idea and states the main implications for dark matter and gravitons. The long version covers the dark matter case in depth. Abstract for the short version:

Detecting Classically Undetectable Particles through Quantum Decoherence

Some hypothetical particles are considered essentially undetectable because they are far too light and slow-moving to transfer appreciable energy or momentum to the normal matter that composes a detector. I propose instead directly detecting such feeble particles, like sub-MeV dark matter or even gravitons, through their uniquely distinguishable decoherent effects on quantum devices like matter interferometers. More generally, decoherence can reveal phenomena that have arbitrarily little classical influence on normal matter, giving new motivation for the pursuit of macroscopic superpositions.

This is figure 1:

MZ2_cropped
Decoherence detection with a Mach-Zehnder interferometer. System \mathcal{N} is placed in a coherent superposition of spatially displaced wavepackets \vert N_{L} \rangle and \vert N_{R} \rangle that each travel a separate path and then are recombined. In the absence of system \mathcal{E}, the interferometer is tuned so that \mathcal{N} will be detected at the bright port with near unit probability, and at the dim port with near vanishing probability. However, if system \mathcal{D} scatters off \mathcal{N}, these two paths can decohere and \mathcal{N} will be detected at the dim port 50% of the time.

Below are some FAQs I have received.

Won’t there always be momentum transfer in any nontrivial scattering?

For any nontrivial scattering of two particles, there must be some momentum transfer.  But the momentum transfer can be arbitrarily small by simply making the mass of the dark particle as tiny as desired (while keeping its velocity fixed).  … [continue reading]