How to show a density matrix is in a pure/mixed state? Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern) Announcing the arrival of Valued Associate #679: Cesar Manara Unicorn Meta Zoo #1: Why another podcast?Density matrices for pure states and mixed statesHow do density matrices act on $mathcalH_A$?How is measurement modelled when using the density operator?How do we derive the density operator of a subsystem?Density matrix after measurement on density matrixPartial Trace over a complicated looking stateNon-uniqueness of pure states ensemble decompositionWhy is a density operator defined the way it's defined?Computing von Neumann entropy of pure state in density matrixQuantum teleportation with “noisy” entangled state
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How to show a density matrix is in a pure/mixed state?
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)
Announcing the arrival of Valued Associate #679: Cesar Manara
Unicorn Meta Zoo #1: Why another podcast?Density matrices for pure states and mixed statesHow do density matrices act on $mathcalH_A$?How is measurement modelled when using the density operator?How do we derive the density operator of a subsystem?Density matrix after measurement on density matrixPartial Trace over a complicated looking stateNon-uniqueness of pure states ensemble decompositionWhy is a density operator defined the way it's defined?Computing von Neumann entropy of pure state in density matrixQuantum teleportation with “noisy” entangled state
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$begingroup$
Say we have a single qubit with some density matrix, for example lets say we have the density matrix $rho=beginpmatrix3/4&1/2\1/2&1/2endpmatrix$. I would like to know what is the procedure for checking whether this state is pure or mixed.
I know that a pure state is one which can be written $rho=|psiranglelanglepsi|$, and that a mixed state is one which can be written $rho=sum_k=1^Np_k|psi_kranglelanglepsi_k|$. But I have two problems. Firstly I'm not sure what the $p_k$ is, I know it's called the weight function but I don't understand it significance, how to find it, or its role mathematically. Secondly I don't see how it is possible to determine from these definition whether the sate I mentioned above is pure or mixed. How can we? Or perhaps there's some method that doesn't use the definitions directly?
Could anyone please clear up this problem for me ?
quantum-state quantum-information mathematics density-matrix matrix-representation
$endgroup$
add a comment |
$begingroup$
Say we have a single qubit with some density matrix, for example lets say we have the density matrix $rho=beginpmatrix3/4&1/2\1/2&1/2endpmatrix$. I would like to know what is the procedure for checking whether this state is pure or mixed.
I know that a pure state is one which can be written $rho=|psiranglelanglepsi|$, and that a mixed state is one which can be written $rho=sum_k=1^Np_k|psi_kranglelanglepsi_k|$. But I have two problems. Firstly I'm not sure what the $p_k$ is, I know it's called the weight function but I don't understand it significance, how to find it, or its role mathematically. Secondly I don't see how it is possible to determine from these definition whether the sate I mentioned above is pure or mixed. How can we? Or perhaps there's some method that doesn't use the definitions directly?
Could anyone please clear up this problem for me ?
quantum-state quantum-information mathematics density-matrix matrix-representation
$endgroup$
$begingroup$
Is it possible that you instead mean the density matrix $rho=beginpmatrix3/4&1/2\1/2&1/4endpmatrix$?
$endgroup$
– Mithrandir24601♦
29 mins ago
add a comment |
$begingroup$
Say we have a single qubit with some density matrix, for example lets say we have the density matrix $rho=beginpmatrix3/4&1/2\1/2&1/2endpmatrix$. I would like to know what is the procedure for checking whether this state is pure or mixed.
I know that a pure state is one which can be written $rho=|psiranglelanglepsi|$, and that a mixed state is one which can be written $rho=sum_k=1^Np_k|psi_kranglelanglepsi_k|$. But I have two problems. Firstly I'm not sure what the $p_k$ is, I know it's called the weight function but I don't understand it significance, how to find it, or its role mathematically. Secondly I don't see how it is possible to determine from these definition whether the sate I mentioned above is pure or mixed. How can we? Or perhaps there's some method that doesn't use the definitions directly?
Could anyone please clear up this problem for me ?
quantum-state quantum-information mathematics density-matrix matrix-representation
$endgroup$
Say we have a single qubit with some density matrix, for example lets say we have the density matrix $rho=beginpmatrix3/4&1/2\1/2&1/2endpmatrix$. I would like to know what is the procedure for checking whether this state is pure or mixed.
I know that a pure state is one which can be written $rho=|psiranglelanglepsi|$, and that a mixed state is one which can be written $rho=sum_k=1^Np_k|psi_kranglelanglepsi_k|$. But I have two problems. Firstly I'm not sure what the $p_k$ is, I know it's called the weight function but I don't understand it significance, how to find it, or its role mathematically. Secondly I don't see how it is possible to determine from these definition whether the sate I mentioned above is pure or mixed. How can we? Or perhaps there's some method that doesn't use the definitions directly?
Could anyone please clear up this problem for me ?
quantum-state quantum-information mathematics density-matrix matrix-representation
quantum-state quantum-information mathematics density-matrix matrix-representation
edited 30 mins ago
Mithrandir24601♦
2,4352932
2,4352932
asked 4 hours ago
can'tcauchycan'tcauchy
1455
1455
$begingroup$
Is it possible that you instead mean the density matrix $rho=beginpmatrix3/4&1/2\1/2&1/4endpmatrix$?
$endgroup$
– Mithrandir24601♦
29 mins ago
add a comment |
$begingroup$
Is it possible that you instead mean the density matrix $rho=beginpmatrix3/4&1/2\1/2&1/4endpmatrix$?
$endgroup$
– Mithrandir24601♦
29 mins ago
$begingroup$
Is it possible that you instead mean the density matrix $rho=beginpmatrix3/4&1/2\1/2&1/4endpmatrix$?
$endgroup$
– Mithrandir24601♦
29 mins ago
$begingroup$
Is it possible that you instead mean the density matrix $rho=beginpmatrix3/4&1/2\1/2&1/4endpmatrix$?
$endgroup$
– Mithrandir24601♦
29 mins ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
First, the example that you give is not a density matrix (they must have trace 1).
Second, you’re asking how to go from the matrix into an operator representation that is not unique. So, there are many ways of doing this. However, a particularly natural way of decomposing it is using the spectral decomposition. The weights are the eigenvalues and the states are the corresponding eigenvectors.
However, if all you want to do is determine if the state is mixed, there’s a simpler way: calculate the trace of the square of the density matrix. If it’s 1, the state is pure. If it’s less than 1, the state is mixed. If it’s more than 1, you’ve messed up.
$endgroup$
add a comment |
$begingroup$
By spectral theorem density matrices are diagonizable, since they are hermitian (also they are positive semi-definite and have trace 1). That means that there is a set of $n$ non-negative eigenvalues $lambda_i$ with $n$ corresponding mutually orthogonal eigenvectors $|v_irangle$ such that
$$
rho = sum_i=1^nv_irangle langle v_i
$$
This matrix represents pure state only if it has exactly one non-zero eigenvalue (it must be equal to 1 since $tr(rho)=sum_i=1^nlambda_i=1$).
So, to analyze density matrix you just need to find eigenvalues and eigenvectors.
Though, to check if $rho$ is pure it is enough to verify the equality
$$
tr(rho^2)=1
$$
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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active
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$begingroup$
First, the example that you give is not a density matrix (they must have trace 1).
Second, you’re asking how to go from the matrix into an operator representation that is not unique. So, there are many ways of doing this. However, a particularly natural way of decomposing it is using the spectral decomposition. The weights are the eigenvalues and the states are the corresponding eigenvectors.
However, if all you want to do is determine if the state is mixed, there’s a simpler way: calculate the trace of the square of the density matrix. If it’s 1, the state is pure. If it’s less than 1, the state is mixed. If it’s more than 1, you’ve messed up.
$endgroup$
add a comment |
$begingroup$
First, the example that you give is not a density matrix (they must have trace 1).
Second, you’re asking how to go from the matrix into an operator representation that is not unique. So, there are many ways of doing this. However, a particularly natural way of decomposing it is using the spectral decomposition. The weights are the eigenvalues and the states are the corresponding eigenvectors.
However, if all you want to do is determine if the state is mixed, there’s a simpler way: calculate the trace of the square of the density matrix. If it’s 1, the state is pure. If it’s less than 1, the state is mixed. If it’s more than 1, you’ve messed up.
$endgroup$
add a comment |
$begingroup$
First, the example that you give is not a density matrix (they must have trace 1).
Second, you’re asking how to go from the matrix into an operator representation that is not unique. So, there are many ways of doing this. However, a particularly natural way of decomposing it is using the spectral decomposition. The weights are the eigenvalues and the states are the corresponding eigenvectors.
However, if all you want to do is determine if the state is mixed, there’s a simpler way: calculate the trace of the square of the density matrix. If it’s 1, the state is pure. If it’s less than 1, the state is mixed. If it’s more than 1, you’ve messed up.
$endgroup$
First, the example that you give is not a density matrix (they must have trace 1).
Second, you’re asking how to go from the matrix into an operator representation that is not unique. So, there are many ways of doing this. However, a particularly natural way of decomposing it is using the spectral decomposition. The weights are the eigenvalues and the states are the corresponding eigenvectors.
However, if all you want to do is determine if the state is mixed, there’s a simpler way: calculate the trace of the square of the density matrix. If it’s 1, the state is pure. If it’s less than 1, the state is mixed. If it’s more than 1, you’ve messed up.
answered 4 hours ago
DaftWullieDaftWullie
15.5k1642
15.5k1642
add a comment |
add a comment |
$begingroup$
By spectral theorem density matrices are diagonizable, since they are hermitian (also they are positive semi-definite and have trace 1). That means that there is a set of $n$ non-negative eigenvalues $lambda_i$ with $n$ corresponding mutually orthogonal eigenvectors $|v_irangle$ such that
$$
rho = sum_i=1^nv_irangle langle v_i
$$
This matrix represents pure state only if it has exactly one non-zero eigenvalue (it must be equal to 1 since $tr(rho)=sum_i=1^nlambda_i=1$).
So, to analyze density matrix you just need to find eigenvalues and eigenvectors.
Though, to check if $rho$ is pure it is enough to verify the equality
$$
tr(rho^2)=1
$$
$endgroup$
add a comment |
$begingroup$
By spectral theorem density matrices are diagonizable, since they are hermitian (also they are positive semi-definite and have trace 1). That means that there is a set of $n$ non-negative eigenvalues $lambda_i$ with $n$ corresponding mutually orthogonal eigenvectors $|v_irangle$ such that
$$
rho = sum_i=1^nv_irangle langle v_i
$$
This matrix represents pure state only if it has exactly one non-zero eigenvalue (it must be equal to 1 since $tr(rho)=sum_i=1^nlambda_i=1$).
So, to analyze density matrix you just need to find eigenvalues and eigenvectors.
Though, to check if $rho$ is pure it is enough to verify the equality
$$
tr(rho^2)=1
$$
$endgroup$
add a comment |
$begingroup$
By spectral theorem density matrices are diagonizable, since they are hermitian (also they are positive semi-definite and have trace 1). That means that there is a set of $n$ non-negative eigenvalues $lambda_i$ with $n$ corresponding mutually orthogonal eigenvectors $|v_irangle$ such that
$$
rho = sum_i=1^nv_irangle langle v_i
$$
This matrix represents pure state only if it has exactly one non-zero eigenvalue (it must be equal to 1 since $tr(rho)=sum_i=1^nlambda_i=1$).
So, to analyze density matrix you just need to find eigenvalues and eigenvectors.
Though, to check if $rho$ is pure it is enough to verify the equality
$$
tr(rho^2)=1
$$
$endgroup$
By spectral theorem density matrices are diagonizable, since they are hermitian (also they are positive semi-definite and have trace 1). That means that there is a set of $n$ non-negative eigenvalues $lambda_i$ with $n$ corresponding mutually orthogonal eigenvectors $|v_irangle$ such that
$$
rho = sum_i=1^nv_irangle langle v_i
$$
This matrix represents pure state only if it has exactly one non-zero eigenvalue (it must be equal to 1 since $tr(rho)=sum_i=1^nlambda_i=1$).
So, to analyze density matrix you just need to find eigenvalues and eigenvectors.
Though, to check if $rho$ is pure it is enough to verify the equality
$$
tr(rho^2)=1
$$
edited 3 hours ago
answered 3 hours ago
Danylo YDanylo Y
67016
67016
add a comment |
add a comment |
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$begingroup$
Is it possible that you instead mean the density matrix $rho=beginpmatrix3/4&1/2\1/2&1/4endpmatrix$?
$endgroup$
– Mithrandir24601♦
29 mins ago