State of qubits: $\frac{1}{2} (| 111 \rangle + | 010 \rangle + | 101 \rangle + | 000 \rangle)$
Are the first and second qubits of this register entangled with each another?
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1$\begingroup$ What are your ideas? $\endgroup$– RammusNov 15, 2021 at 9:48
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$\begingroup$ I found info about 2 qubits here quantumcomputing.stackexchange.com/questions/2263/… but I don't understand it quite well $\endgroup$– Lily SandersNov 15, 2021 at 9:54
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$\begingroup$ My hint would be to first try to reduce the problem to a two-qubit problem. And then you can try to check whether the two-qubit system is entangled, for instance you could use the PPT condition to check entanglement. $\endgroup$– RammusNov 15, 2021 at 9:58
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2 Answers
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Perhaps some of your confusion is coming from the phrasing of the question. Without more detail, I'm not sure I even necessarily know what it means to say two qubits within a three qubit state are entangled (should you be tracing out the third qubit and looking at the entanglement of the mixed state??).
This suggests that, to be answerable, this question should actually be a special case, perhaps of the form where two of the qubits are entangled and one of them is separable. If you can identify which qubit is separable, you'll be able to apply your bipartite entanglement knowledge to answering the question.
So, what I suggest is to take each qubit in turn, and try factoring out that specific qubit. For instance, if I take the first qubit, $$ |0\rangle(|00\rangle+|10\rangle)+|1\rangle(|11\rangle+|01\rangle). $$ Since the two terms in brackets are different, the first qubit is not separable from the other two. You could continue to do the same thing for each of the other two qubits. Or, perhaps you'll notice something about the above equation if you try simplifying it a little bit, which might push you in the right direction (I'm quite deliberately not giving you the answer).
answered Nov 15, 2021 at 10:45
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If you consider only the first and the second qubits then they are not entangled in the sense that knowing the state of one qubit is not sufficient to make any prediction on the state of the other qubit, because there is equal probability of either possibility. The same can be said about the pair of the second qubit and the third qubit. But the first qubit and the third qubit are entangled, which means if you know the state of one of them then the state of the other qubit would be determined.
Here is a Qiskit circuit for the same:
You can see that by measuring the first qubit, the state of second qubit has equal probability of being either $|0\rangle$ or $|1\rangle$.
