5
$\begingroup$

What is meant by maximally mixed states? Does this mean that there are partially mixed states? For example, consider $\rho_{GHZ} = \left| {GHZ} \right\rangle \left\langle {GHZ} \right|$ and $\rho_W = \left| {W} \right\rangle \left\langle {W} \right|$, where $$\left| {GHZ} \right\rangle = {1 \over {\sqrt 2 }}\left( {\left| {000} \right\rangle + \left| {111} \right\rangle } \right)$$ $$\left| W \right\rangle = {1 \over {\sqrt 3 }}\left( {\left| {100} \right\rangle + \left| {010} \right\rangle + \left| {001} \right\rangle } \right).$$

Are they maximally mixed?

$\endgroup$
4
  • $\begingroup$ perhaps you meant to ask about maximally entangled states? neither of those examples are maximally mixed (or mixed at all). $\endgroup$
    – forky40
    Feb 10 at 5:19
  • $\begingroup$ @forky40 The following sentence was mentioned in PHYSICAL REVIEW A 102, 052227 (2020): "...and the subsystem A is maximally mixed, such as the GHZ state." I don't know what is meant by maximally mixed? $\endgroup$
    – Bekaso
    Feb 10 at 5:25
  • $\begingroup$ I edited my question, please take a look. $\endgroup$
    – Bekaso
    Feb 10 at 5:59
  • 1
    $\begingroup$ take a look at these: physics.stackexchange.com/q/170318 , and quantumcomputing.stackexchange.com/q/4594/9858 $\endgroup$
    – KAJ226
    Feb 10 at 6:51

3 Answers 3

10
$\begingroup$

The maximally mixed state is a quantum state whose density matrix is proportional to the identity matrix. Physically, it may be interpreted as a uniform mixture of states in an orthonormal basis.

The density matrix of a maximally mixed state is full rank. On the other hand, the density matrix of a pure state, such as $|GHZ\rangle$ or $|W\rangle$, is rank one. Therefore, no pure state is maximally mixed. That said, tracing out two subsystems of $|GHZ\rangle$ does leave the other one in the maximally mixed state since

$$ \mathrm{tr}_{23}|GHZ\rangle\langle GHZ|=\frac{|0\rangle\langle 0|}2+\frac{|1\rangle\langle 1|}2 = \frac{I}{2}. $$

However, this is not the case for $|W\rangle$, because

$$ \mathrm{tr}_{23}|W\rangle\langle W|=\frac{2|0\rangle\langle 0|}3+\frac{|1\rangle\langle 1|}3 $$

is not a multiple of the identity matrix.

$\endgroup$
6
$\begingroup$

If you want to describe one part of a many-qubit state, such as a GHZ state, then usually, that one qubit cannot be said to be in a state $|\phi\rangle$. That only works if the overall state is separable. If the overall state is entangled, the single qubit is in a mixed state. Mathematically, this is described by a density matrix instead of a state vector.

Indeed, there is a great range of mixedness of mixed states, as can be measured by the purity. But maximally mixed states are those that have the maximum allowed mixedness, and can be written in the form $I/2$. One qubit from a GHZ state is maximally mixed. 1 qubit from a W-state is not maximally mixed.

$\endgroup$
4
$\begingroup$

Indeed, there are partially mixed states, although this terminology is seldom used. One can define measures of mixedness to make this quantitative. The best one is the von Neumann entropy, defined as $$ S(\rho) = -\operatorname{tr}(\rho \log\rho). $$ A maximally mixed state is then the ones that achieve the maximal von Neumann entropy $S(\rho) = \log(d)$, where $d$ is the dimension of the quantum state. As the other answers have pointed out, this is achieved only by the state $\rho = \frac1d I$. Pure states are then "minimally mixed", having von Neumann entropy 0, and any state with von Neumann entropy between 0 and $\log(d)$ will be a "partially mixed state".

One could also define "mixedness" as $$m(\rho) = 1-\operatorname{tr}(\rho^2),$$ with similar properties. It doesn't have much meaning, but sometimes this measured is used because it's easy to compute.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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