Mixing
Maximally mixed states for more than 1 qubit
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
For 1 qubit, the maximally mixed state is I/2.
So, for two qubits, I assume the maximally mixed state is the maximally mixed state is I/4?
Which is:
$1/4 (|00rangle langle 00| + |01rangle langle 01| + |10rangle langle 10| + |11rangle langle 11|)$
Why is this state more mixed than the following, for instance?
$1/2 (|00rangle langle 00| + |11rangle langle 11|)$
Also, does this generalize to higher dimensions similarly?
quantum-entanglement density-matrix mixed-state
asked 6 hours ago
Tinkidinki
1725
add a comment |Â
up vote
1
down vote
favorite
For 1 qubit, the maximally mixed state is I/2.
So, for two qubits, I assume the maximally mixed state is the maximally mixed state is I/4?
Which is:
$1/4 (|00rangle langle 00| + |01rangle langle 01| + |10rangle langle 10| + |11rangle langle 11|)$
Why is this state more mixed than the following, for instance?
$1/2 (|00rangle langle 00| + |11rangle langle 11|)$
Also, does this generalize to higher dimensions similarly?
quantum-entanglement density-matrix mixed-state
asked 6 hours ago
Tinkidinki
1725
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
For 1 qubit, the maximally mixed state is I/2.
So, for two qubits, I assume the maximally mixed state is the maximally mixed state is I/4?
Which is:
$1/4 (|00rangle langle 00| + |01rangle langle 01| + |10rangle langle 10| + |11rangle langle 11|)$
Why is this state more mixed than the following, for instance?
$1/2 (|00rangle langle 00| + |11rangle langle 11|)$
Also, does this generalize to higher dimensions similarly?
quantum-entanglement density-matrix mixed-state
asked 6 hours ago
Tinkidinki
1725
For 1 qubit, the maximally mixed state is I/2.
So, for two qubits, I assume the maximally mixed state is the maximally mixed state is I/4?
Which is:
$1/4 (|00rangle langle 00| + |01rangle langle 01| + |10rangle langle 10| + |11rangle langle 11|)$
Why is this state more mixed than the following, for instance?
$1/2 (|00rangle langle 00| + |11rangle langle 11|)$
Also, does this generalize to higher dimensions similarly?
quantum-entanglement density-matrix mixed-state
quantum-entanglement density-matrix mixed-state
asked 6 hours ago
Tinkidinki
1725
asked 6 hours ago
Tinkidinki
1725
asked 6 hours ago
Tinkidinki
1725
asked 6 hours ago
Tinkidinki
1725
asked 6 hours ago
Tinkidinki
1725
1725
add a comment |Â
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
4
down vote
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.
answered 5 hours ago
Norbert Schuch
74019
add a comment |Â
up vote
1
down vote
The Von Neumann entropy of $1/2 (|00rangle langle 00| + |11rangle langle 11|)$ 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.
answered 5 hours ago
Craig Gidney
2,857119
add a comment |Â
up vote
1
down vote
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.
answered 5 hours ago
glS
3,261337
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
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.
answered 5 hours ago
Norbert Schuch
74019
add a comment |Â
up vote
4
down vote
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.
answered 5 hours ago
Norbert Schuch
74019
add a comment |Â
up vote
4
down vote
up vote
4
down vote
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.
answered 5 hours ago
Norbert Schuch
74019
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.
answered 5 hours ago
Norbert Schuch
74019
edited 1 hour ago
answered 5 hours ago
Norbert Schuch
74019
answered 5 hours ago
Norbert Schuch
74019
answered 5 hours ago
Norbert Schuch
74019
74019
add a comment |Â
add a comment |Â
up vote
1
down vote
The Von Neumann entropy of $1/2 (|00rangle langle 00| + |11rangle langle 11|)$ 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.
answered 5 hours ago
Craig Gidney
2,857119
add a comment |Â
up vote
1
down vote
The Von Neumann entropy of $1/2 (|00rangle langle 00| + |11rangle langle 11|)$ 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.
answered 5 hours ago
Craig Gidney
2,857119
add a comment |Â
up vote
1
down vote
up vote
1
down vote
The Von Neumann entropy of $1/2 (|00rangle langle 00| + |11rangle langle 11|)$ 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.
answered 5 hours ago
Craig Gidney
2,857119
The Von Neumann entropy of $1/2 (|00rangle langle 00| + |11rangle langle 11|)$ 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.
answered 5 hours ago
Craig Gidney
2,857119
answered 5 hours ago
Craig Gidney
2,857119
answered 5 hours ago
Craig Gidney
2,857119
answered 5 hours ago
Craig Gidney
2,857119
2,857119
add a comment |Â
add a comment |Â
up vote
1
down vote
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.
answered 5 hours ago
glS
3,261337
add a comment |Â
up vote
1
down vote
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.
answered 5 hours ago
glS
3,261337
add a comment |Â
up vote
1
down vote
up vote
1
down vote
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.
answered 5 hours ago
glS
3,261337
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.
answered 5 hours ago
glS
3,261337
answered 5 hours ago
glS
3,261337
answered 5 hours ago
glS
3,261337
answered 5 hours ago
glS
3,261337
3,261337
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fquantumcomputing.stackexchange.com%2fquestions%2f4594%2fmaximally-mixed-states-for-more-than-1-qubit%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
