What is the relation between maximally entangled and maximally mixed?

Question

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.

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.

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

1.

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 $$\text{Tr}_B(\rho) = \rho_A \stackrel{?}{=} \frac{1}{2}I\,,\tag{4}$$ $$\text{Tr}_A(\rho) = \rho_B \stackrel{?}{=} \frac{1}{2}I\,.\tag{5}$$

2.

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?

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

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^{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 $B$.}