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I would like to create a quantum system with the gates for a W state where each qubit is at a different location. Entanglement distribution has been proven in several research articles. I'm new to this space and interested:

  • if three qubits W-state can be implemented where each qubit has a different location?
  • if in this case, measuring q1 will fix the measurement of q2 and q3 (if one is 1 the others are 0) and only one measurement of the distributed qubits will be possible?
  • if this would also be possible for n qubits W-state over n locations?
Mark Spinelli
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TimW
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    you can (and are encouraged to) edit the post to include further details clarifying the question. Also, please note that questions should be as laser-focused as possible and ask about an individual issue. You can open separate posts to ask separate questions – glS May 31 '21 at 13:58
  • Looking at your updated link, there are three 2-qubit gates used in the circuit. These gates require the qubits acting on them to be "local" to each other; otherwise no entanglement is possible. Note that as soon as the middle gate acts on q2, q2 does not need to stick around and wait for the other two gates to do their work, as the circuit is done with q2. – Mark Spinelli May 31 '21 at 14:27

2 Answers2

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Recall that the W state may be defined as:

$$\vert W\rangle=\frac{1}{\sqrt 3}(\vert 001\rangle+\vert 010\rangle+\vert 100\rangle).$$

Given three qubits, initially to prepare such a state local operations (wherein at least two of the three qubits are at the same location) will need to be performed. Depending on your background, see, for example, this question for some circuits that can prepare such states.

However, once prepared each of the three qubits may "go their own way", and measurement on any one of the three qubits will collapse the other two. For example measuring the rightmost qubit to be in the state $\vert 1\rangle$ will collapse the first two to be in the state $\vert 00\rangle$. All three qubits can be lightyears apart.

And certainly one can define the generalized $\vert W_n\rangle$ state on $n$ qubits, as a uniform superposition over the $n$ one-hot basis states. For example $\vert W_4\rangle$ would be defined as:

$$\vert W_4\rangle=\frac 1 2(\vert 0001\rangle+\vert 0010\rangle+\vert 0100\rangle+\vert 1000\rangle).$$

In this pairs of each of the four qubits will need to be local to each other in order to prepare the state, but then afterwards each of the four can be at a different location.

Take note that once one qubit is measured the other qubits are irrevocably collapsed. If you want to repeat the experiment you will need to bring the qubits back together again to re-prepare the state.

Mark Spinelli
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You may be interested in this paper Effcient quantum algorithms for GHZ and W states, and implementation on the IBM quantum computer. The paper provides general method how to prepare $n$-qubit W state.

Concerning your question on how place the state's qubits to different location, you can prepare W state on qubits from $q_{1}$ to $q_n$ and then use SWAP gates to exchange positions of some qubits or use CNOT gate with control on some qubit in original W state and target on some other qubit originally in state $|0\rangle$ (i.e. the CNOT works as fan-out).

Martin Vesely
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  • Thanks for the answer. Do you think such an implementation is already possible for a larger number of n (e.g. 30)? (see also my other question [link] (https://quantumcomputing.stackexchange.com/questions/17746/quantum-based-lottery-using-w-state-and-spatial-separation) – TimW Jun 01 '21 at 11:56
  • @TimW: Thank you for accepting my answer. I am afraid that in current NISQ the 30 qubits is too much and in the end you will get completely decoherent result. However, the algorithm is designed for any number of qubits, so you can try to run it on simulator (IBM Q simulator of general quantum processors offers 32 qubits). – Martin Vesely Jun 01 '21 at 12:38