For 1 qubit, the maximally mixed state is $\frac{\mathrm{I}}{2}$.

So, for two qubits, I assume the maximally mixed state is the maximally mixed state is $\frac{\mathrm{I}}{4}$? Which is:

$\frac{1}{4} (|00\rangle \langle 00| + |01\rangle \langle 01| + |10\rangle \langle 10| + |11\rangle \langle 11|)$

Why is this state more mixed than the following, for instance?

$\frac{1}{2} (|00\rangle \langle 00| + |11\rangle \langle 11|)$

Also, does this generalize to higher dimensions similarly?


4 Answers 4


The Von Neumann entropy of $1/2 (|00\rangle \langle 00| + |11\rangle \langle 11|)$ is one bit. For $I/4$ it's two bits of entropy instead.

The entropy of states that only have entries on the diagonal of a density matrix is very easy to compute in general, because you just treat the entries as probabilities and compute the Shannon entropy. The Shannon entropy is maximized when all the probabilities are equal, which is why 0.25 four times beats 0.5 twice with 0 twice.


For two probability distributions, there is a clear notion how to say which one is more mixed: $\vec p$ is more mixed than $\vec q$ if it can be obtained from $\vec p$ by a mixing process, this is, a stochastic process described by a doubly stochastic matrix (i.e. one which preserved the flat distribution).

Birkhoff's theorem relates this to a concept called majorization, which introduces a partial order on the space of probability distributions.

The same concept generalized to mixed states, allowing us to say which mixed state is more mixed -- for instance, one can establish an order by using the majorization condition on the eigenvalues, and then use a Birkhoff's theorem to prove that one can be converted into the other by a quantum "mixing map" (a unital channel).

This is explained in detail e.g. in http://michaelnielsen.org/papers/majorization_review.pdf, or also in the book of Nielsen and Chuang.

Specifically, this yields that the state with all eigenvalues equal (or equivalently the flat probability distribution) is most mixed.

To relate this to the quantification of mixedness through entropy mentioned in the other answers, the connection comes from the fact that if a state $\rho$ is more random than another state $\sigma$ in the above sense -- i.e., if $\sigma$ can be transformed into $\rho$ by mixing, or equivalently the eigenvalues of $\sigma$ majorize those of $\rho$ -- then the entropy of $\rho$ is larger than the one of $\sigma$. This property (monotonicity under majorization) is known as Schur-concavity, a property shared e.g. by all Renyi-entropies.


For a general $d$-dimensional system, the maximally mixed state is the one described by the normalised identity matrix: $$\rho=I/d.$$ In the specific case of a two-qubit system, this reduces to the first state you write.

Why is this state more mixed than the following

This is a bit hard to answer, as it depends on what your current understanding/intuition of "more mixed" is. One possible answer is that it is "more mixed" because it represents a state associated with a higher uncertainty, as quantifiable for example via the von Neumann entropy. You can also have a look at the answers to a similar question on physics.SE for more details.


In the sense of the entropy, the maximum entropy from a "$d$" dimensional density matrix is just $\log_2 d$ if you measure wrt two degrees of freedom. This is more than $\log_22=1 \ \text{bit}$, and hence more information. This is because we always compare the Quantum Information with the Classical Information as over a "bit". However, if we measure the information of a $d$ dimensional system over an "information basis" of "$d$" itself we will see that the maximum is $\log_d d$ which will be equivalent to $1$bit again in the sense of comparison of maximally mixed states and maximum information.

The intuitive answer as to why the higher dimensional system has more entropy as a bound is simply because it has more parameters to be identified and measured to describe it completely. This means, conversely; for two systems $M$ and $N$ with Hilbert Space dimesnions $m$ and $n$ ($m\geq n$), $M$ will have more information if both are initially completely mixed. Because more mixedness$\implies$ more uncertainty before measurement$\implies$ more surprise after a measurement. This is the intuition behind mixedness of a state.


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