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The Bell state is $$|\beta(a,b)\rangle = \frac{1}{\sqrt{2}}\sum_{k=0}^{1}(-1)^{ka}|k,k \oplus b\rangle\,.$$

How do I get to a general Bell state given an initial state $|ab\rangle$ here $a,b = 0, 1$?

A beam-splitter gate $B=\frac{1}{\sqrt{2}}[I + iX]$ and a $CNOT$ gate is necessary, but I might require another one more which I can't seem to figure out.

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2 Answers 2

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Starting from any of the four basis states, Hadamard + controlled-not will produce each of the four Bell states. So, expressed in terms of the gates you need, you just need to add $X$ to either qubit in order to be able to prepare each of those basis states.

Alternatively, once you've produced one Bell state, you can convert it into any of the other three by choosing to apply one of $X$, $Y$ or $Z$ to a single qubit (this is essentially the content of superdense coding).

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  • $\begingroup$ Thank you for your reply. Just to be sure I understand you correctly, you are saying that B gate and CNOT does nothing to put me closer to the state I am wanting to construct? I'm asking because it contradicts a lecture questions that the class was asked to think about and that both B and CNOT gates are necessary. Note: It's a beam splitter B gate not a H gate. $\endgroup$
    – Physkid
    Nov 16, 2023 at 8:43
  • $\begingroup$ No, I'm not saying that at all. But your question is perhaps not sufficiently clear about your constraints. Are you saying that you have been told you have to include the $B$ gate in your construction? $\endgroup$
    – DaftWullie
    Nov 16, 2023 at 9:36
  • $\begingroup$ Have you considered an $S$ gate? $\endgroup$
    – DaftWullie
    Nov 16, 2023 at 9:37
  • $\begingroup$ I managed to solve it with an $S$ and Z gate just before your reply. Timely! $\endgroup$
    – Physkid
    Nov 16, 2023 at 9:40
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I think I nailed it.

The 4 Bell states are

$$\frac{1}{\sqrt{2}}\big [|00\rangle + |11\rangle\big],\quad \frac{1}{\sqrt{2}}\big[|01\rangle + |10\rangle\big],\quad \frac{1}{\sqrt{2}}\big[|00\rangle - |11\rangle\big],\quad \frac{1}{\sqrt{2}}\big[|01\rangle - |10\rangle\big]$$

The solution: $$\bigg(CNOT \cdot (Z \otimes I) \cdot (S \otimes I) \cdot (B \otimes I)\bigg)|ab\rangle$$

Take initial state $|11\rangle$ as an example. This gives $$\frac{1}{\sqrt{2}}[i|01\rangle -i|10\rangle]$$

Then, observe:

$$\frac{1}{\sqrt{2}}[i|01\rangle -i|10\rangle] = e^{i\phi}\bigg[\frac{1}{\sqrt{2}}(|01\rangle - |10\rangle)\bigg]$$ for $\phi = \frac{\pi}{2}$ so that $e^{i\phi}\cos(\phi) + i\sin(\phi) = i$.

The global phase may be omitted as it is physically insignificant.

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    $\begingroup$ The better way of expressing your final line is that you can remove an irrelevant global phase of $i$. (Just looking at the mod-square of probabilities doesn't tell you enough.) $\endgroup$
    – DaftWullie
    Nov 16, 2023 at 9:46

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