Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
The summation limits should be 0 to d-1 or 1 to d. – Emilio Pisanty Mar 15 '13 at 16:31
The correct search string is "isotropic state", see e.g. – Norbert Schuch Mar 16 '13 at 22:41
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


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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.