Is the quantum relative entropy a Bregman divergence?
$begingroup$
This is related to a broader question about quantum information geometry - this question is more specific.
A fundamental concept in Amari's treatment of information geometry is that of a Bregman divergence. Given a convex function $psi(mathbf{x})$, a Bregman divergence is defined as
$$
D_psi(mathbf{x}|mathbf{x}_0) = psi(mathbf{x}) - psi(mathbf{x}_0) - nabla psi(mathbf{x}_0)(mathbf{x}-mathbf{x}_0).
$$
Bregman divergences are non-negative and are zero when $mathbf{x}=mathbf{x}_0$, and they behave in some ways like the square of a distance between $mathbf{x}_0$ and $mathbf{x}$. (Though taking the square root does not form a metric.)
An example is the Kullback-Leibler divergence between two probability distributions, $D_{KL}(P|P^0)=sum_{i=1}^N p_ilogfrac{p_i}{p^0_i}$. In the case of finite support can be seen (for example) by parameterising each distribution by its first $N-1$ probabilities (so $p_N=1-sum_{i=1}^{N-1}p_i$) and then taking $psi$ to be the negative entropy, $psi=sum_{i=1}^{N}p_ilog p_i$.
In quantum theory, the natural analog of the Kullback-Leibler divergence is the quantum relative entropy, $$
D_Q(rho|rho_0) = mathrm{Tr},rho(logrho - logrho_0),
$$
where $rho$ and $rho_0$ are density matrices and $log$ is the matrix logarithm.
My question is, in the case of a finite Hilbert space (or indeed in general), can the quantum relative entropy be expressed as a Bregman divergence? If so, what is the right way to parameterise $rho$ so that this becomes clear?
convex-analysis information-theory quantum-mechanics information-geometry
asked Jun 8 '18 at 3:37
NathanielNathaniel
1,6931433
$endgroup$
add a comment |
$begingroup$
This is related to a broader question about quantum information geometry - this question is more specific.
A fundamental concept in Amari's treatment of information geometry is that of a Bregman divergence. Given a convex function $psi(mathbf{x})$, a Bregman divergence is defined as
$$
D_psi(mathbf{x}|mathbf{x}_0) = psi(mathbf{x}) - psi(mathbf{x}_0) - nabla psi(mathbf{x}_0)(mathbf{x}-mathbf{x}_0).
$$
Bregman divergences are non-negative and are zero when $mathbf{x}=mathbf{x}_0$, and they behave in some ways like the square of a distance between $mathbf{x}_0$ and $mathbf{x}$. (Though taking the square root does not form a metric.)
An example is the Kullback-Leibler divergence between two probability distributions, $D_{KL}(P|P^0)=sum_{i=1}^N p_ilogfrac{p_i}{p^0_i}$. In the case of finite support can be seen (for example) by parameterising each distribution by its first $N-1$ probabilities (so $p_N=1-sum_{i=1}^{N-1}p_i$) and then taking $psi$ to be the negative entropy, $psi=sum_{i=1}^{N}p_ilog p_i$.
In quantum theory, the natural analog of the Kullback-Leibler divergence is the quantum relative entropy, $$
D_Q(rho|rho_0) = mathrm{Tr},rho(logrho - logrho_0),
$$
where $rho$ and $rho_0$ are density matrices and $log$ is the matrix logarithm.
My question is, in the case of a finite Hilbert space (or indeed in general), can the quantum relative entropy be expressed as a Bregman divergence? If so, what is the right way to parameterise $rho$ so that this becomes clear?
convex-analysis information-theory quantum-mechanics information-geometry
asked Jun 8 '18 at 3:37
NathanielNathaniel
1,6931433
$endgroup$
$begingroup$
$rho = psi circ lambda$, where $lambda$ is the eigenvalue map and your underlying Hilbert space is the real-symmetric matrices with the trace inner product.
$endgroup$
– max_zorn
Jun 8 '18 at 3:56
add a comment |
$begingroup$
This is related to a broader question about quantum information geometry - this question is more specific.
A fundamental concept in Amari's treatment of information geometry is that of a Bregman divergence. Given a convex function $psi(mathbf{x})$, a Bregman divergence is defined as
$$
D_psi(mathbf{x}|mathbf{x}_0) = psi(mathbf{x}) - psi(mathbf{x}_0) - nabla psi(mathbf{x}_0)(mathbf{x}-mathbf{x}_0).
$$
Bregman divergences are non-negative and are zero when $mathbf{x}=mathbf{x}_0$, and they behave in some ways like the square of a distance between $mathbf{x}_0$ and $mathbf{x}$. (Though taking the square root does not form a metric.)
An example is the Kullback-Leibler divergence between two probability distributions, $D_{KL}(P|P^0)=sum_{i=1}^N p_ilogfrac{p_i}{p^0_i}$. In the case of finite support can be seen (for example) by parameterising each distribution by its first $N-1$ probabilities (so $p_N=1-sum_{i=1}^{N-1}p_i$) and then taking $psi$ to be the negative entropy, $psi=sum_{i=1}^{N}p_ilog p_i$.
In quantum theory, the natural analog of the Kullback-Leibler divergence is the quantum relative entropy, $$
D_Q(rho|rho_0) = mathrm{Tr},rho(logrho - logrho_0),
$$
where $rho$ and $rho_0$ are density matrices and $log$ is the matrix logarithm.
My question is, in the case of a finite Hilbert space (or indeed in general), can the quantum relative entropy be expressed as a Bregman divergence? If so, what is the right way to parameterise $rho$ so that this becomes clear?
convex-analysis information-theory quantum-mechanics information-geometry
asked Jun 8 '18 at 3:37
NathanielNathaniel
1,6931433
$endgroup$
This is related to a broader question about quantum information geometry - this question is more specific.
A fundamental concept in Amari's treatment of information geometry is that of a Bregman divergence. Given a convex function $psi(mathbf{x})$, a Bregman divergence is defined as
$$
D_psi(mathbf{x}|mathbf{x}_0) = psi(mathbf{x}) - psi(mathbf{x}_0) - nabla psi(mathbf{x}_0)(mathbf{x}-mathbf{x}_0).
$$
Bregman divergences are non-negative and are zero when $mathbf{x}=mathbf{x}_0$, and they behave in some ways like the square of a distance between $mathbf{x}_0$ and $mathbf{x}$. (Though taking the square root does not form a metric.)
An example is the Kullback-Leibler divergence between two probability distributions, $D_{KL}(P|P^0)=sum_{i=1}^N p_ilogfrac{p_i}{p^0_i}$. In the case of finite support can be seen (for example) by parameterising each distribution by its first $N-1$ probabilities (so $p_N=1-sum_{i=1}^{N-1}p_i$) and then taking $psi$ to be the negative entropy, $psi=sum_{i=1}^{N}p_ilog p_i$.
In quantum theory, the natural analog of the Kullback-Leibler divergence is the quantum relative entropy, $$
D_Q(rho|rho_0) = mathrm{Tr},rho(logrho - logrho_0),
$$
where $rho$ and $rho_0$ are density matrices and $log$ is the matrix logarithm.
My question is, in the case of a finite Hilbert space (or indeed in general), can the quantum relative entropy be expressed as a Bregman divergence? If so, what is the right way to parameterise $rho$ so that this becomes clear?
convex-analysis information-theory quantum-mechanics information-geometry
convex-analysis information-theory quantum-mechanics information-geometry
asked Jun 8 '18 at 3:37
NathanielNathaniel
1,6931433
asked Jun 8 '18 at 3:37
NathanielNathaniel
1,6931433
asked Jun 8 '18 at 3:37
NathanielNathaniel
1,6931433
asked Jun 8 '18 at 3:37
NathanielNathaniel
1,6931433
asked Jun 8 '18 at 3:37
NathanielNathaniel
1,6931433
1,6931433
$begingroup$
$rho = psi circ lambda$, where $lambda$ is the eigenvalue map and your underlying Hilbert space is the real-symmetric matrices with the trace inner product.
$endgroup$
– max_zorn
Jun 8 '18 at 3:56
add a comment |
$begingroup$
$rho = psi circ lambda$, where $lambda$ is the eigenvalue map and your underlying Hilbert space is the real-symmetric matrices with the trace inner product.
$endgroup$
– max_zorn
Jun 8 '18 at 3:56
$begingroup$
$rho = psi circ lambda$, where $lambda$ is the eigenvalue map and your underlying Hilbert space is the real-symmetric matrices with the trace inner product.
$endgroup$
– max_zorn
Jun 8 '18 at 3:56
$begingroup$
$rho = psi circ lambda$, where $lambda$ is the eigenvalue map and your underlying Hilbert space is the real-symmetric matrices with the trace inner product.
$endgroup$
– max_zorn
Jun 8 '18 at 3:56
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I think the following paper answers your question: https://users.renyi.hu/~petz/pdf/112bregman.pdf
Abstract: In this paper the Bregman operator divergence is introduced
for density matrices by differentiation of the matrix-valued function
$x mapsto x log x$. This quantity is compared with the relative
operator entropy of Fujii and Kamei. It turns out that the trace is
the usual Umegaki's relative entropy which is the only intersection of
the classes of quasi-entropies and Bregman divergences.
edited Dec 27 '18 at 22:24
dantopa
6,53942244
answered Dec 27 '18 at 21:38
MarioMario
113
$endgroup$
$begingroup$
Thank you, that does indeed look like an answer - it shows that the quantum relative entropy is a Bregman divergence with $psi=operatorname{Tr}rhologrho$ (in my notation). I'll accept this after reading it more thoroughly.
$endgroup$
– Nathaniel
Dec 28 '18 at 0:18
add a comment |
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1 Answer
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active
oldest
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1 Answer
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oldest
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$begingroup$
I think the following paper answers your question: https://users.renyi.hu/~petz/pdf/112bregman.pdf
Abstract: In this paper the Bregman operator divergence is introduced
for density matrices by differentiation of the matrix-valued function
$x mapsto x log x$. This quantity is compared with the relative
operator entropy of Fujii and Kamei. It turns out that the trace is
the usual Umegaki's relative entropy which is the only intersection of
the classes of quasi-entropies and Bregman divergences.
edited Dec 27 '18 at 22:24
dantopa
6,53942244
answered Dec 27 '18 at 21:38
MarioMario
113
$endgroup$
$begingroup$
Thank you, that does indeed look like an answer - it shows that the quantum relative entropy is a Bregman divergence with $psi=operatorname{Tr}rhologrho$ (in my notation). I'll accept this after reading it more thoroughly.
$endgroup$
– Nathaniel
Dec 28 '18 at 0:18
add a comment |
$begingroup$
I think the following paper answers your question: https://users.renyi.hu/~petz/pdf/112bregman.pdf
Abstract: In this paper the Bregman operator divergence is introduced
for density matrices by differentiation of the matrix-valued function
$x mapsto x log x$. This quantity is compared with the relative
operator entropy of Fujii and Kamei. It turns out that the trace is
the usual Umegaki's relative entropy which is the only intersection of
the classes of quasi-entropies and Bregman divergences.
edited Dec 27 '18 at 22:24
dantopa
6,53942244
answered Dec 27 '18 at 21:38
MarioMario
113
$endgroup$
$begingroup$
Thank you, that does indeed look like an answer - it shows that the quantum relative entropy is a Bregman divergence with $psi=operatorname{Tr}rhologrho$ (in my notation). I'll accept this after reading it more thoroughly.
$endgroup$
– Nathaniel
Dec 28 '18 at 0:18
add a comment |
$begingroup$
I think the following paper answers your question: https://users.renyi.hu/~petz/pdf/112bregman.pdf
Abstract: In this paper the Bregman operator divergence is introduced
for density matrices by differentiation of the matrix-valued function
$x mapsto x log x$. This quantity is compared with the relative
operator entropy of Fujii and Kamei. It turns out that the trace is
the usual Umegaki's relative entropy which is the only intersection of
the classes of quasi-entropies and Bregman divergences.
edited Dec 27 '18 at 22:24
dantopa
6,53942244
answered Dec 27 '18 at 21:38
MarioMario
113
$endgroup$
I think the following paper answers your question: https://users.renyi.hu/~petz/pdf/112bregman.pdf
Abstract: In this paper the Bregman operator divergence is introduced
for density matrices by differentiation of the matrix-valued function
$x mapsto x log x$. This quantity is compared with the relative
operator entropy of Fujii and Kamei. It turns out that the trace is
the usual Umegaki's relative entropy which is the only intersection of
the classes of quasi-entropies and Bregman divergences.
edited Dec 27 '18 at 22:24
dantopa
6,53942244
answered Dec 27 '18 at 21:38
MarioMario
113
edited Dec 27 '18 at 22:24
dantopa
6,53942244
edited Dec 27 '18 at 22:24
dantopa
6,53942244
edited Dec 27 '18 at 22:24
dantopa
6,53942244
6,53942244
answered Dec 27 '18 at 21:38
MarioMario
113
answered Dec 27 '18 at 21:38
MarioMario
113
answered Dec 27 '18 at 21:38
MarioMario
113
113
$begingroup$
Thank you, that does indeed look like an answer - it shows that the quantum relative entropy is a Bregman divergence with $psi=operatorname{Tr}rhologrho$ (in my notation). I'll accept this after reading it more thoroughly.
$endgroup$
– Nathaniel
Dec 28 '18 at 0:18
add a comment |
$begingroup$
Thank you, that does indeed look like an answer - it shows that the quantum relative entropy is a Bregman divergence with $psi=operatorname{Tr}rhologrho$ (in my notation). I'll accept this after reading it more thoroughly.
$endgroup$
– Nathaniel
Dec 28 '18 at 0:18
$begingroup$
Thank you, that does indeed look like an answer - it shows that the quantum relative entropy is a Bregman divergence with $psi=operatorname{Tr}rhologrho$ (in my notation). I'll accept this after reading it more thoroughly.
$endgroup$
– Nathaniel
Dec 28 '18 at 0:18
$begingroup$
Thank you, that does indeed look like an answer - it shows that the quantum relative entropy is a Bregman divergence with $psi=operatorname{Tr}rhologrho$ (in my notation). I'll accept this after reading it more thoroughly.
$endgroup$
– Nathaniel
Dec 28 '18 at 0:18
add a comment |
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$begingroup$
$rho = psi circ lambda$, where $lambda$ is the eigenvalue map and your underlying Hilbert space is the real-symmetric matrices with the trace inner product.
$endgroup$
– max_zorn
Jun 8 '18 at 3:56