0
$\begingroup$

Suppose we have two entangled particles, A and B. If we measure particle A and then subsequently entangle particle B with a third particle, C, would the state of particle A influence the combined states of particles B and C? Specifically, would there be any correlation between the combined spins of B and C and the measured spin of A? Have there been any experiments conducted to explore or validate this scenario?

$\endgroup$
3
  • $\begingroup$ So you consider a tripartite system and ask for correlations/entanglement between the partition $A$ and ($B$ and $C$)? Your wording is a bit confusing here for me: "and then subsequently"?! $\endgroup$ Sep 20, 2023 at 10:01
  • $\begingroup$ Tobias Fünke, thank you for the answer. I 'll try to reformulate: Imagine a scenario where we start with two entangled particles, A and B. After taking a measurement of particle A, we then proceed to entangle particle B with a third particle, C. My primary question is: Would the measurement taken from particle A have any bearing on the combined states or correlations between particles B and C, particularly in terms of their combined spins? Are you aware of any experiments that have tested this particular setup? $\endgroup$
    – aestlist
    Sep 20, 2023 at 10:05
  • 1
    $\begingroup$ As you describe it: I'm not aware of any technique by which entangled B can become entangled with C. The exception is to perform a Bell State Measurement on A (which requires a fourth photon D, initially entangled with C). That of course is a normal entanglement swap. In such case, there is no direct connection between A and the combined spins of B & C. However, the Bell State Measurement outcome (of which A is a part) does indicate whether the B/C spins are correlated or anti-correlated. $\endgroup$
    – DrChinese
    Sep 20, 2023 at 14:48

1 Answer 1

2
$\begingroup$

There could be correlations between A and C (this is the realm of entanglement swapping) but there will be no way for A to affect the internal dynamics of the correlations between B and C. The term "subsequently" has no bearing here.

In math, assuming pure states, the most general way we can write the initial state is by using the Schmidt decomposition for A and B and considering C to be separable to begin: $$|\Psi\rangle=\left(\sum_i \psi_i |a_i\rangle\otimes |b_i\rangle\right) \otimes |\phi\rangle.$$ A measurement of A without communicating the result to B leaves the system in the mixed state $$\rho=\sum_i |\psi_i|^2 |b_i\rangle\langle b_i|\otimes |\phi\rangle\langle \phi|.$$ This this does not depend on A's measurement result, nothing you can do locally will change the correlations between B and C in a way that depends on A. You can always apply $U\rho U^\dagger$ to entangle the state between B and C, but this is independent from A.

Now if A performs a measurement and communicates the result "$i$" then B and C remain in the pure state $$|\psi^{(i)}\rangle=|b_i\rangle \otimes |\phi\rangle.$$ The structure of the correlations between B and C are the same regardless of the outcome "$i$" but the actual state depends on $i$; this is no different from any idea of entanglement without C being present. One can entangle the state using $U|\psi^{(i)}\rangle$, the result of which is no different from if one started with $|\psi^{(i)}\rangle$ and never knew that A existed. So yes, the measurement result of A affects the joint state of B and C because the measurement result of A affects the state of B - there is no other effect present that adds mysterious correlations between B and C. As mentioned at the outset, correlations between A and C are normally investigated, not induced correlations between B and C.

All of these results are the same if $U$ is performed before or after A's measurement.

$\endgroup$
7
  • $\begingroup$ "There could be correlations between A and C..." This is not possible if A is initially entangled with B, and later B and C become entangled (presumably via entanglement swapping). I'm sure you are aware of Monogamy of Entanglement. It states, in effect, that if A and B are maximally entangled, then there is no C which can be simultaneously entangled with A or B. If B becomes maximally entangled with C, then A cannot be entangled at all with B or C. I agree with everything you say about the order of measurements (and your comment about the word "subsequently"). $\endgroup$
    – DrChinese
    Sep 20, 2023 at 15:02
  • $\begingroup$ @DrChinese you are neglecting the annoying trivial case: when the unitary $U$ simply swaps B and C. Then it is C that becomes entangled with A while B becomes separable from AC. The swap unitary would yield $U|\Psi\rangle=\sum_i \psi_i |a_i\rangle\otimes|\phi\rangle\otimes|b_i\rangle$ $\endgroup$ Sep 20, 2023 at 16:55
  • $\begingroup$ @DrChinese the point of monogomy of entanglement, as you point out, is that A cannot become entangled with C except at the expense of losing entanglement with B. I am sure we agree overall $\endgroup$ Sep 20, 2023 at 16:56
  • $\begingroup$ I guess I have never seen the unitary case U mentioned (swapping B & C without a D). Can you point me to a paper where this operation is performed? Thanks. $\endgroup$
    – DrChinese
    Sep 20, 2023 at 23:10
  • $\begingroup$ @DrChinese the swap operation? For two qubits it's easy, e.g. see quantumcomputing.stackexchange.com/q/15129/15820 with a circuit diagram that can be run on any two-qubit device. Even a controlled version of this operation has been done, for the purpose of making the Fredkin gate and doing the swap test. Higher order versions that cycle through more than two qubits abound as well, e.g. arxiv.org/abs/2109.10006, arxiv.org/abs/2206.15405, arxiv.org/abs/2302.00705 $\endgroup$ Sep 21, 2023 at 2:42

Not the answer you're looking for? Browse other questions tagged or ask your own question.