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A system is entangled if and only if the Schmidt rank is greater than 1. Why does this $$\left[\frac{1}{\sqrt{2}}\left(\left|0\right\rangle+\left|1\right\rangle\right)\right]\otimes\left[\frac{1}{\sqrt{2}}\left(\left|0\right\rangle+i\,\left|1\right\rangle\right)\right] = \frac{\left|00\right\rangle+i\,\left|01\right\rangle+\left|10\right\rangle+i\,\left|11\right\rangle}{2}$$ have a Schmidt rank equal to one? or equivalently why is this not entangled? ($\otimes$ denotes the tensor product)

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I'm not sure why the down-vote was made... this seems like a pretty fundamental and important question, the only issue was the typesetting wasn't too impressive at first, but this was cleaned up. This isn't a "research-level" blog so I would appreciate some justification for the down-vote. – Squirtle Jun 5 '13 at 23:00
I can only guess why you were downvoted, but perhaps it's because your title is suggestive of a much more basic question. I've updated it to match the question more closely. Alternatively, perhaps it's because it's not clear what you're really confused about. Why can't you just calculate the Schmidt rank of the state, using the definition available in many resources, and find that it is 1? If you elaborate on what exactly you had trouble with, it would make a much better question. The advice in our homework policy may be helpful. – David Z Jun 5 '13 at 23:18
This is NOT homework.... I'm not taking any class... I don't think I've ever heard of being taught in the summer anyway. I'm just self-teaching and found computing it difficult. I tried to follow the approach used on wikipedia... not sure I understand it: – Squirtle Jun 5 '13 at 23:21
Did I ever say it is homework? No. I only said that the advice in our homework policy may be useful, and that's because it happens to be good advice for formulating all sorts of questions. (By the way, "snapping" at people who are trying to help you isn't giving us much incentive to keep doing so.) – David Z Jun 5 '13 at 23:26
No problem :-) By the way, we don't actually care whether a question is a homework question or not. We do have a homework tag, but what it means is a little different from "this question is a homework question" (and thus it gets applied, correctly, to some questions which are not homework questions). Just so you know. – David Z Jun 6 '13 at 0:27

Notice that the state you have written down is in the form $|u\rangle \otimes |v\rangle$. It is, by definition, an unentangled state because an state is said to be unentangled if and only if it can be written as a tensor product of states.

Now, the question becomes why its schmidt rank is equal to $1$. Well, consider the following orthonormal bases for the Hilbert space spanned by $\{|0\rangle, |1\rangle\}$ $$ B_1 = \{|u\rangle, |u'\}, \qquad B_2 = \{|v\rangle, |v'\rangle\} $$ where $$ |u\rangle = \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle), \qquad |u'\rangle = \frac{1}{\sqrt{2}}(|0\rangle - |1\rangle) $$ and $$ |v\rangle = \frac{1}{\sqrt{2}}(|0\rangle + i|1\rangle), \qquad |v'\rangle = \frac{1}{\sqrt{2}}(|0\rangle - i|1\rangle) $$ The Schmidt decomposition of $|u\rangle\otimes |v\rangle$ in these bases is $$ |u\rangle\otimes |v\rangle = 1\cdot|u\rangle\otimes |v\rangle + 0\cdot|u'\rangle\otimes |v\rangle + 0\cdot|u\rangle\otimes |v'\rangle+ 0\cdot|u'\rangle\otimes |v'\rangle $$ Notice that there is only one Schmidt (non-negative) coefficient in this decomposition, so the Schmidt rank of the state is $1$. Actually, the following fact is generally true

Fact. A state is unentangled if and only if its Schmidt rank is $1$.

So once you know that the Schmidt rank is 1, then you know that the state can be written as a tensor product of two states, and vice versa.

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@dustanalysis : With a state like $\psi = a|00> + b |01> + c |10> + d |11> $, the state is a pure state if $(ad - bc) = 0$ If $(ad - bc) \neq 0$, $\psi$ is an entangled state. – Trimok Jun 6 '13 at 11:12
@Trimok You probably mean the state is a separable (rather than pure) state if and only if $ad - bc = 0$. As it's written down, it's always pure... – SMeznaric Jun 6 '13 at 23:30
@SMeznaric : Aaaah!, sorry, you are perfectly right. One must read separable, and not pure !!!! – Trimok Jun 7 '13 at 8:23

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