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What is the Schmidt decomposition of the tripartite states$|GHZ\rangle=\frac{1}{\sqrt 2}[|000\rangle+|111\rangle]$ or $|W\rangle=\frac{1}{\sqrt 3}[|001\rangle+|010\rangle+|100\rangle]$? Are these same as that of their canonical forms?

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    $\begingroup$ Schmidt decomposition doesn't have a three particle equivalent. I mean you may not always write a 3 particle state $|\psi \rangle $ as $\sum{\alpha_i |\phi_i\rangle \otimes |\chi_i \rangle \otimes |\xi_i \rangle}$ $\endgroup$ – Ali Jul 18 '13 at 10:04
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    $\begingroup$ @Ali I'm not talking about any generalized tri-qubit state but only about GHZ and W state. $\endgroup$ – Prabir Das Jul 18 '13 at 12:46
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In this reference , it is stated that the sufficient as well as necessary condition for the existence of Schmidt decomposition for a tripartite system is that the partial inner product of a basis of any one of the subsystems (belonging to a Hilbert space of smaller dimension) with the state of the composite system gives a disentangled basis.

[EDIT] (Following Peter Shor commentary, the criteria was not correctly applied)

The GHZ state is already obviously Schmidt-decomposed. If we choose one of the basis $|0_1\rangle |1_1\rangle$, or $|0_2\rangle |1_2\rangle$, or $|0_3\rangle, |1_3\rangle$ and take the inner product, we find a separable state : For instance, the inner product with $|0_1\rangle$ gives $\sim$ $|0_20_3\rangle$ which is a separable basis).

The W state cannot be Schmidt-decomposed :

First, remember that a state $a|00\rangle + b|01\rangle + c|10\rangle + d|11\rangle$ is separable if and only if the discriminant $ \Delta = ad -bc=0$.

Now, we take a general basis for the first subsystem $\alpha |0_1\rangle + \beta |1_1\rangle$, $\beta |0_1\rangle - \alpha |1_1\rangle$ with $\alpha^2 + \beta^2 =1$, and take the inner product with the W-state. So, we get, for the inner products :

$\alpha (|0_21_3\rangle + |1_20_3\rangle) + \beta |0_20_3\rangle$ and $\beta (|0_21_3\rangle + |1_20_3\rangle) - \alpha |0_20_3\rangle$

The two discriminants are $ - \alpha^2$ and $ - \beta^2$.

So, the inner-product resulting states are separable only if $\alpha =0$ and $\beta =0$ , which is impossible.

Because of the symmetry of the W state, the demonstration is the same for a general basis of the $2nd$ and $3rd$ sub-system.

So, applying the criteria, the $W$ state is not Schmidt-decomposable

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    $\begingroup$ You are using the criterion incorrectly. You have to show that there is no basis which satisfies the criterion in the W system. For example, if we take the inner product of $\frac{1}{\sqrt{2}}\left(|0_1 \rangle + |1_1\rangle\right)$ with the GHZ state, you get $\frac{1}{2}\left(|0_2 0_3\rangle+ |1_2 1_3\rangle\right)$, which is entangled. So to show a state is not Schmidt-decomposable, it's not enough to just choose one state, take the inner product, and find it's entangled. This is what you've done in your answer. $\endgroup$ – Peter Shor Jul 19 '13 at 11:38
  • $\begingroup$ @PeterShor : Thanks. I have made an edit to the answer. Hope it is correct. $\endgroup$ – Trimok Jul 19 '13 at 18:54

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