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I am reading the following paper on mixed three-qubit states. It states that any three–qubit pure state can be written as (equation 1 of the paper)

$|\psi_{GHZ}\rangle=\lambda_0 |000\rangle+\lambda_1 e^{i\theta} |100\rangle+\lambda_2 |101\rangle+\lambda_3 |110\rangle+\lambda_4 |111\rangle$

and W class states have the form (equation 2 of the paper)

$|\psi_{W}\rangle=\lambda_0 |000\rangle+\lambda_1 |100\rangle+\lambda_2 |101\rangle+\lambda_3 |110\rangle$

However, later they use the following form for W states (page 2 of the paper)

$|\psi_{W}\rangle=\kappa_0 |000\rangle+\kappa_1\left(|100\rangle+ |010\rangle+ |001\rangle\right)$

I did not understand how equation 1 is obtained; also which one amongst equation 2 and the last equation is the "correct" representation of W-class states (or whether these two forms are equivalent to each other). Any help would be greatly appreciated.

glS
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Anindita Sarkar
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1 Answers1

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The GHZ form of the state follows from a bit of a manipulation after define what you mean by a GHZ state, remembering that all these parametrisations are "up to local unitaries". For me, the GHZ class of states is one that takes the form $$ \alpha|u_1\rangle|u_2\rangle|u_3\rangle+\beta|v_1\rangle|v_2\rangle|v_3\rangle, $$ so the first challenge is to see how we can reduce a state of this form, under the action of local unitaries, to the claimed standard form. Pick a convention for the global phase such that the $|000\rangle$ term is real. Start by looking at the two basis states $|111\rangle$ and $|011\rangle$. There must exist a rotation on the first qubit that can completely get rid of the $|011\rangle$ term, leaving the $|111\rangle$ term real. Now, we can apply the same trick using a single qubit rotation on the second qubit to cancel the $|010\rangle$ term out of the pair $|010\rangle$ and $\|000\rangle$. Finally, use the same trick to cancel the $|001\rangle$ term out of $|001\rangle$ and $|000\rangle$. You'll find the only basis states we have left are those from your statement of $|\psi_\text{GHZ}\rangle$, and the only one we've not had the freedom to make real has been the $|100\rangle$ term.

As for the correct representation of the W class, it's the first one. If you read the bit on page 2 more carefully, you'll see that because they're trying to maximise a specific overlap, they know that the state they need to achieve the maximum has certain symmetries that they've built into the description. It's perhaps a bit misleading to give it the same notation as the general version, but what they say seems clear enough.

DaftWullie
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