In Wikipedia I read the following:

"A multipartite state ${\displaystyle |\psi \rangle }$ of a system is called absolutely maximally entangled if for any bipartition ${\displaystyle A|B} $ of ${\displaystyle S}$, the reduced density operator is maximally mixed ${\displaystyle \rho _{A}=\rho _{B}=I/d}$."

Unfortunately there was no reference for that. For more studying I need some references about the above content.


1 Answer 1


What is a maximally mixed state?

Intuitively, think of a maximally mixed state as a quantum state where all possible measurement outcomes are equally likely to occur at random upon measurement. If your state is not 'maximally' mixed, then you will not get all possible outcomes with equal probability. There will be some bias.

Example for single qubit states:

$$\text{Maximally mixed state} = \begin{bmatrix} 0.5 & 0\\ 0 & 0.5 \end{bmatrix}\,.\tag{1}$$

Here, the probability of getting $|0\rangle$ is the same as the probability of getting $|1\rangle$, which is 0.5 each.

$$\text{Mixed, but not 'maximally' mixed state} = \begin{bmatrix} 0.3 & 0\\ 0 & 0.7 \end{bmatrix}\,.\tag{2}$$

Here, the probability of getting $|0\rangle$ is $0.3$, and the probability of getting $|1\rangle$ is $0.7$. There is a bias.

What is an entangled state?

Given a system with two parts, let's say subsystem $A$ and subsystem $B$, where $A$ and $B$ are entangled, then if you only measure $A$, then even without measuring $B$, you can know with 100% certainty what is state the of system $B$. Similarly, by only measuring $B$, you can conclude what state your system $A$ would be in, without measuring $A$ explicitly.

What is a 'maximally' entangled state?

Given a system with two parts, let's say subsystem $A$ and subsystem $B$, if you ignore the system $B$ and only measure the system $A$, your perception of the $A$ upon many measurements would be that $A$ is maximally mixed state. Similarly, if you ignore system $A$ and only measure $B$, your perception of $B$ will be that it is a maximally mixed state, i.e. you get all possible outcomes with equal probability. Along with this, things written above for entangled states also remain true.

However, if your states are not 'maximally' entangled, then ignoring the system $B$ and only measuring system $A$ will not give you all possible measurement outcomes with equal probability. There will be some bias.

It would be beneficial to cement your understanding if you try some exercises yourself.


Let $$|\psi\rangle = \frac{1}{\sqrt{2}}\big(|0_A0_B\rangle + |1_A1_B \rangle \big)\,.\tag{3}$$

This is a maximally entangled state. To check this, take the partial trace$^1$ of $\rho = |\psi \rangle \langle \psi |$ and see if you get a density matrix which has an equal probability distribution, i.e., verify that $$\text{Tr}_B(\rho) = \rho_A \stackrel{?}{=} \frac{1}{2}I\,,\tag{4}$$ $$\text{Tr}_A(\rho) = \rho_B \stackrel{?}{=} \frac{1}{2}I\,.\tag{5}$$


Do a simillar analysis for $$|\psi\rangle = \sqrt{0.3}|0_A0_B\rangle + \sqrt{0.7}|1_A1_B \rangle \,\tag{6}.$$

What's your conclusion?

As for the resources, Mark Wilde's Quantum Information Theory textbook, Sections 3.5, 3.6 and 3.7, should be helpful regarding this.

1: Taking the partial trace over $B$ is an operation, which means that you ignore system $B$ and only see what would be the state of the system $A$, i.e., see what is the probability distribution for measurement on $A$, ignoring $B$.

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    $\begingroup$ +1. I just added equation numbers because even if you're not referring to them in your answer, other people would rather be able to say "Eq. 5 in this answer by FDGod" rather than "the second question between 'verify that' and 'Do a similar analysis', in this answer by FDGod". $\endgroup$ Nov 7, 2023 at 19:25
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    $\begingroup$ Thank you, and I 100% agree, that writing equation numbers is useful in general, always. We should start implementing this throughout the site from now, as much as possible. $\endgroup$
    – FDGod
    Nov 7, 2023 at 22:07
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    $\begingroup$ To implement it throughout the site automatically, all we need is to add the lines from here to the MathJax script. However, that proposal was not greatly supported, so it would probably be supported even less over here. $\endgroup$ Nov 7, 2023 at 23:51
  • $\begingroup$ That's sad. I hope it could get implemented, along with support for \ket{}, \bra{} and \braket{} commands. $\endgroup$
    – FDGod
    Nov 8, 2023 at 5:09
  • $\begingroup$ If it's sad and you hope it can get implemented, the first step would be to support the Meta proposal to which I provided the link, and then this one too. For the latter one, it's a bit funny that my answer got a negative comment from someone who 3 years later changed his mind and said he now supports it. I think if it doesn't get approved there, where the request was made more than 10 years ago, on the network's second largest site, then proposing it here on QCSE would very unlikely be fruitful. $\endgroup$ Nov 8, 2023 at 21:34

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