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I am reading Introduction to Quantum Computing by Kaye, Laflamme, and Mosca. Here is a question I am struggling with:

Exercise 3.5.6: Prove that the transpose map, which maps $\rho \mapsto \rho^{T}$ is positive1, but not completely positive2.

Now, positivity is defined in terms of the inner product, i.e. $\rho$ is positive iff $\forall v\in \mathcal{H}$, $\left<v, \rho v\right> \ge 0$, but "transpose" is defined in terms of an operation on matrices. So I was able to get this under the assumption that $\rho^{T} = \rho^{*}$, but not under the weaker assumption the $\rho$ is simply positive. Is this true even if I don't assume $\rho^{T} = \rho^{*}$?

As for showing that the transpose map is not completely positive, I frankly do not know what I'm doing and am asking for any and all help you can give me. My attempt is given below, though it is not worth reading:

Let $\rho\otimes \gamma$ be a positive map, so that $\forall u\otimes v \in \mathcal{H}_A \otimes \mathcal{H}_{B}$ $$ \left<u\otimes v, \rho u \otimes \gamma v\right> := \left<u, \rho u\right>\left< v, \gamma v \right> \ge 0. $$ Then the transpose map tensored with the identity takes $\rho \otimes \gamma$ to $\rho^{T}\otimes \gamma$, and we have (again assuming $\rho^{T}=\rho^{*}$) $$ \left<u\otimes v, \rho u \otimes \gamma v\right> := \left<u, \rho^{T} u\right>\left< v, \gamma v \right> = \overline{\left< u, \rho u\right>}\left<v, \gamma v \right> =\frac{\overline{\left< u, \rho u\right>}}{\left< u, \rho u\right>}\left<u, \rho u\right>\left< v, \gamma v \right>. $$ So $\frac{\overline{\left< u, \rho u\right>}}{\left< u, \rho u\right>}$ must be $\ge 0$. . .(at this point I am stumped.)

1 "Positive" in this context means "maps positive operators to positive operators.

2 They define completely positive as follows: A map is completely positive iff it is positive, and in addition, when tensored with the identity operation, they still map positive operators to positive operators.

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The transposing map for $2\times 2$ matrices is discussed on this Wikipedia page‌​. –  Qmechanic Apr 4 '12 at 16:29

1 Answer 1

up vote 2 down vote accepted

Since it's a homework question, I won't give you the full answers, but hints towards their resolution.


You don't have to assume hermiticity of $\rho$. To show it, you just need to look at $\left<\psi\middle|\rho\middle|\psi\right>^T$ and notice that the set of all $|\psi^*\rangle$ is the same as the set of all $|\psi\rangle$.

2.Non complete-positivity

By restricting yourself to vector of the form $u\otimes v$, you essentially restrict yourself to separable states. This being a book about quantum information, looking at entangled states can help.

The Hilbert state of interest is $\mathcal H_A\oplus\mathcal H_B$ (with a 'oplus' in the middle not 'otimes' ) and contains vectors of the form $\alpha u\otimes v + \beta u'\otimes v'$ which cannot be decomposed under product form.

A way to prove non-complete-positivity, is to exhibit a counter example. Take for $\rho$ the density matrix of any of the Bell states, and apply transpose on one particle and the identity on the other, and check whetther the resulting operator is still positive.

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Dr. Grosshans, I thank you for your time. –  Nick Thompson Mar 12 '12 at 20:09

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