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Consider the quantum system $\mathcal{B}(\mathbb{C}^d\otimes\mathbb{C}^d)$ and $|\psi\rangle=\frac{1}{\sqrt{d}}\sum_{i=0}^{d-1}|i,i\rangle$ be the (standard) maximally entangled state. Consider the state

$\rho_\lambda=\lambda \frac{\mathbb{I}_{d^2}}{d^2}+(1-\lambda)|\psi\rangle\langle\psi|.$

Now for some values of $\lambda$ this state is entangled (example $\lambda=0$ it is $|\psi\rangle\langle\psi|$) and hence its entanglement can be detected by partial transpose operation.

Can $\rho(\lambda)$ be an entangled state which is positive under partial transpose (known in literature as PPT entangled state) for some values $\lambda$? My intuition tells me that this is the case. However, I was told (without reference) that this is not the case and for the values of $\lambda$ we get only separable states or not PPT entangled states. I could not find the corresponding paper. May be I am not giving the proper string for searching. Advanced thanks for any suggestion, reference or comment.

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The summation limits should be 0 to d-1 or 1 to d. –  Emilio Pisanty Mar 15 '13 at 16:31
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The correct search string is "isotropic state", see e.g. quantiki.org/wiki/Isotropic_state –  Norbert Schuch Mar 16 '13 at 22:41
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1 Answer

up vote 2 down vote accepted

To answer this question, we will first compute the values of $\lambda$ for which $\rho(\lambda)$ is PPT and separately compute the values for which it is entangled.

Let $T$ be the transpose map, such that the partial transpose map may be written as $(\mathbb{I}\otimes T)$, where $\mathbb{I}$ is the identity on $\mathbb{C}^d$. One can show that the partial transpose maps the standard maximally entangled state into the SWAP operator

$$(\mathbb{I}\otimes T)|\psi\rangle\langle\psi|=\frac{1}{d}W,$$

where $W=\sum_{i,j}|i\rangle\langle j|\otimes|j\rangle\langle i|$. For reference, you can take a look at John Watrous' excellent lecture notes. The SWAP operator has states with eigenvalue $-1$, let's call one of them $|w\rangle$. We then have

\begin{align} \langle w|(\mathbb{I}\otimes T)\rho(\lambda)|w\rangle&=\lambda\langle w|\frac{\mathbb{I}}{d^2}|w\rangle+(1-\lambda)\langle w|W|w\rangle\\ &=\frac{\lambda}{d^2}-\frac{(1-\lambda)}{d}. \end{align}

We want this expression to be positive, which gives us the condition $$\lambda\geq\frac{1}{1+d}.$$ On the other hand, we can calculate the maximum overlap $\langle\psi|\rho_s|\psi\rangle$ that a separable state $\rho_s$ can have with $|\psi\rangle$, such that if the overlap of $\rho(\lambda)$ is greater than this maximum, we know that $\rho(\lambda)$ is entangled. It can be shown (see for example this review) that in our case this maximum is precisely $\frac{1}{d}$. Therefore, $\rho(\lambda)$ is entangled whenever

\begin{align} \langle\psi|\rho(\lambda)|\psi\rangle&\geq\frac{1}{d}\\ \Rightarrow\frac{\lambda}{d^2}+(1-\lambda)&\geq\frac{1}{d}, \end{align}

which gives the condition

$$\lambda\geq\frac{d^2-d}{d^2-1}.$$

However, you can quickly check that both conditions cannot be met simultaneously, so there is no value of $\lambda$ for which $\rho(\lambda)$ is entangled and PPT.

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