Perturbation Theory: Derivative of a trace.
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The problem that I am looking at is the following perturbation problem from the notes on Trace Inequalities and Quantum Entropy on page 12, the following result is said to follow "by the Spectral Theorem and first order perturbation theory":
Let $f$ be a continuously differentiable function on $(0,infty)$ and $B,C$ self-adjoint (complex) $n times n$ matrices. Then
$$fracddtleft|_t=0right.operatornameTr[f(B + tC)] = operatornameTr[f'(B)C].$$
If someone knows a reference where I can learn why this is true (preferably an article, I could find online or through a University library), then that would be great.
Note: I already received a solution to this using some more powerful perturbation theory $C^1$ unitarily diagonalizing, but I was wondering about less complicated things.
matrices operator-theory c-star-algebras trace perturbation-theory
asked Aug 22 at 2:59
4-ier
5989
add a comment |Â
up vote
2
down vote
favorite
The problem that I am looking at is the following perturbation problem from the notes on Trace Inequalities and Quantum Entropy on page 12, the following result is said to follow "by the Spectral Theorem and first order perturbation theory":
Let $f$ be a continuously differentiable function on $(0,infty)$ and $B,C$ self-adjoint (complex) $n times n$ matrices. Then
$$fracddtleft|_t=0right.operatornameTr[f(B + tC)] = operatornameTr[f'(B)C].$$
If someone knows a reference where I can learn why this is true (preferably an article, I could find online or through a University library), then that would be great.
Note: I already received a solution to this using some more powerful perturbation theory $C^1$ unitarily diagonalizing, but I was wondering about less complicated things.
matrices operator-theory c-star-algebras trace perturbation-theory
asked Aug 22 at 2:59
4-ier
5989
Considering how a trace is simply a sum of terms, the total time derivative evaluated at $t=0$ would just slide into the trace operation. From here, it's simply an extension on the chain rule
– Captain Morgan
Aug 22 at 5:45
Sure... just why does there exist a chain rule for functions defined via the spectral theorem. Maybe, I would need to prove a version of the convergence lemma in Baby Rudin, where I approximate $f'$ by polynomials that have some antiderivatives that converge to $f$ and then I can try to interchange the derivative limit and the limit of polymonials for which the chain rule obviously applies.
– 4-ier
Aug 22 at 6:02
For instance, if $f(x) = x^2$, then we know that $f(B + tC) = B^2 + t(BC + CB) + t^2C^2$ so its derivative in $t$ at zero is $BC + CB$ which does not equal $f'(B)C = 2BC$ in general, but does after applying the trace.
– 4-ier
Aug 22 at 6:10
Yes, but since $rm tr(CB)=rm tr(BC)$ you can combine the terms to obtain $rm tr(2BC)$.
– lynn
Aug 23 at 19:12
Yes, I agree. What I am saying is that for polynomials $f$, I can see that it works because under the trace, everything commutes. However, for a general continuously differentiable function I am not sure if one can make it work.
– 4-ier
Aug 23 at 22:34
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
The problem that I am looking at is the following perturbation problem from the notes on Trace Inequalities and Quantum Entropy on page 12, the following result is said to follow "by the Spectral Theorem and first order perturbation theory":
Let $f$ be a continuously differentiable function on $(0,infty)$ and $B,C$ self-adjoint (complex) $n times n$ matrices. Then
$$fracddtleft|_t=0right.operatornameTr[f(B + tC)] = operatornameTr[f'(B)C].$$
If someone knows a reference where I can learn why this is true (preferably an article, I could find online or through a University library), then that would be great.
Note: I already received a solution to this using some more powerful perturbation theory $C^1$ unitarily diagonalizing, but I was wondering about less complicated things.
matrices operator-theory c-star-algebras trace perturbation-theory
asked Aug 22 at 2:59
4-ier
5989
The problem that I am looking at is the following perturbation problem from the notes on Trace Inequalities and Quantum Entropy on page 12, the following result is said to follow "by the Spectral Theorem and first order perturbation theory":
Let $f$ be a continuously differentiable function on $(0,infty)$ and $B,C$ self-adjoint (complex) $n times n$ matrices. Then
$$fracddtleft|_t=0right.operatornameTr[f(B + tC)] = operatornameTr[f'(B)C].$$
If someone knows a reference where I can learn why this is true (preferably an article, I could find online or through a University library), then that would be great.
Note: I already received a solution to this using some more powerful perturbation theory $C^1$ unitarily diagonalizing, but I was wondering about less complicated things.
matrices operator-theory c-star-algebras trace perturbation-theory
asked Aug 22 at 2:59
4-ier
5989
edited Aug 23 at 22:37
asked Aug 22 at 2:59
4-ier
5989
asked Aug 22 at 2:59
4-ier
5989
asked Aug 22 at 2:59
4-ier
5989
5989
Considering how a trace is simply a sum of terms, the total time derivative evaluated at $t=0$ would just slide into the trace operation. From here, it's simply an extension on the chain rule
– Captain Morgan
Aug 22 at 5:45
Sure... just why does there exist a chain rule for functions defined via the spectral theorem. Maybe, I would need to prove a version of the convergence lemma in Baby Rudin, where I approximate $f'$ by polynomials that have some antiderivatives that converge to $f$ and then I can try to interchange the derivative limit and the limit of polymonials for which the chain rule obviously applies.
– 4-ier
Aug 22 at 6:02
For instance, if $f(x) = x^2$, then we know that $f(B + tC) = B^2 + t(BC + CB) + t^2C^2$ so its derivative in $t$ at zero is $BC + CB$ which does not equal $f'(B)C = 2BC$ in general, but does after applying the trace.
– 4-ier
Aug 22 at 6:10
Yes, but since $rm tr(CB)=rm tr(BC)$ you can combine the terms to obtain $rm tr(2BC)$.
– lynn
Aug 23 at 19:12
Yes, I agree. What I am saying is that for polynomials $f$, I can see that it works because under the trace, everything commutes. However, for a general continuously differentiable function I am not sure if one can make it work.
– 4-ier
Aug 23 at 22:34
add a comment |Â
Considering how a trace is simply a sum of terms, the total time derivative evaluated at $t=0$ would just slide into the trace operation. From here, it's simply an extension on the chain rule
– Captain Morgan
Aug 22 at 5:45
Sure... just why does there exist a chain rule for functions defined via the spectral theorem. Maybe, I would need to prove a version of the convergence lemma in Baby Rudin, where I approximate $f'$ by polynomials that have some antiderivatives that converge to $f$ and then I can try to interchange the derivative limit and the limit of polymonials for which the chain rule obviously applies.
– 4-ier
Aug 22 at 6:02
For instance, if $f(x) = x^2$, then we know that $f(B + tC) = B^2 + t(BC + CB) + t^2C^2$ so its derivative in $t$ at zero is $BC + CB$ which does not equal $f'(B)C = 2BC$ in general, but does after applying the trace.
– 4-ier
Aug 22 at 6:10
Yes, but since $rm tr(CB)=rm tr(BC)$ you can combine the terms to obtain $rm tr(2BC)$.
– lynn
Aug 23 at 19:12
Yes, I agree. What I am saying is that for polynomials $f$, I can see that it works because under the trace, everything commutes. However, for a general continuously differentiable function I am not sure if one can make it work.
– 4-ier
Aug 23 at 22:34
Considering how a trace is simply a sum of terms, the total time derivative evaluated at $t=0$ would just slide into the trace operation. From here, it's simply an extension on the chain rule
– Captain Morgan
Aug 22 at 5:45
Considering how a trace is simply a sum of terms, the total time derivative evaluated at $t=0$ would just slide into the trace operation. From here, it's simply an extension on the chain rule
– Captain Morgan
Aug 22 at 5:45
Sure... just why does there exist a chain rule for functions defined via the spectral theorem. Maybe, I would need to prove a version of the convergence lemma in Baby Rudin, where I approximate $f'$ by polynomials that have some antiderivatives that converge to $f$ and then I can try to interchange the derivative limit and the limit of polymonials for which the chain rule obviously applies.
– 4-ier
Aug 22 at 6:02
Sure... just why does there exist a chain rule for functions defined via the spectral theorem. Maybe, I would need to prove a version of the convergence lemma in Baby Rudin, where I approximate $f'$ by polynomials that have some antiderivatives that converge to $f$ and then I can try to interchange the derivative limit and the limit of polymonials for which the chain rule obviously applies.
– 4-ier
Aug 22 at 6:02
For instance, if $f(x) = x^2$, then we know that $f(B + tC) = B^2 + t(BC + CB) + t^2C^2$ so its derivative in $t$ at zero is $BC + CB$ which does not equal $f'(B)C = 2BC$ in general, but does after applying the trace.
– 4-ier
Aug 22 at 6:10
For instance, if $f(x) = x^2$, then we know that $f(B + tC) = B^2 + t(BC + CB) + t^2C^2$ so its derivative in $t$ at zero is $BC + CB$ which does not equal $f'(B)C = 2BC$ in general, but does after applying the trace.
– 4-ier
Aug 22 at 6:10
Yes, but since $rm tr(CB)=rm tr(BC)$ you can combine the terms to obtain $rm tr(2BC)$.
– lynn
Aug 23 at 19:12
Yes, but since $rm tr(CB)=rm tr(BC)$ you can combine the terms to obtain $rm tr(2BC)$.
– lynn
Aug 23 at 19:12
Yes, I agree. What I am saying is that for polynomials $f$, I can see that it works because under the trace, everything commutes. However, for a general continuously differentiable function I am not sure if one can make it work.
– 4-ier
Aug 23 at 22:34
Yes, I agree. What I am saying is that for polynomials $f$, I can see that it works because under the trace, everything commutes. However, for a general continuously differentiable function I am not sure if one can make it work.
– 4-ier
Aug 23 at 22:34
add a comment |Â
1 Answer
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The result is easy to check if $f(x)$ was a polynomial. For a general continuously differentiable function $f(x)$ and a given $epsilon>0$, choose a polynomial $p(x)$ such that $|f(x)-p(x)|+|f'(x)-p'(x)|<epsilon$ on an interval $I$ containing all of the eigenvalues of $B$ and $C$. One has
$$dfracddtTr(p(B+tC))(0)=Tr(p'(B)C),$$
and so for
$$g(t)=|Tr(p(B+tC))-Tr(p(B))-Tr(p'(B)C)t|$$
we have $g(t)/t rightarrow 0$ as $trightarrow 0$.
First note that $$|Tr(p'(B)C)-Tr(f'(B)C)|=|Tr((p'(B)-f'(B))C)| $$ $$leq |p'(B)-f'(B)| |C|<n epsilon |C|.$$
Moreover,
$$|Tr(p(B+tC))-Tr(f(B+tC))|=|Tr(p(B+tC)-f(B+tC))|<n epsilon,$$
and similarly
$$|Tr(p(B))-Tr(f(B))|<n epsilon.$$
Putting all these inequalities together, we have
$$h(t)=|Tr(f(B+tC))-Tr(f(B))-Tr(f'(B)C)t|leq g(t)+(2+|C|)nepsilon,$$
By letting $epsilon rightarrow 0$, we have $0leq h(t)leq g(t)$ which implies the formula.
answered Aug 24 at 0:13
Marco
1,3287
I'm fine up until getting $h(t) leq g(t)$, because this inequality depended on $|p - f| < epsilon$ and $g = g_p$. Once $epsilon < |p - f|$ the inequality is false. You would need to control $g_p(t)$ as $epsilon to 0$.
– 4-ier
Aug 24 at 3:03
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
The result is easy to check if $f(x)$ was a polynomial. For a general continuously differentiable function $f(x)$ and a given $epsilon>0$, choose a polynomial $p(x)$ such that $|f(x)-p(x)|+|f'(x)-p'(x)|<epsilon$ on an interval $I$ containing all of the eigenvalues of $B$ and $C$. One has
$$dfracddtTr(p(B+tC))(0)=Tr(p'(B)C),$$
and so for
$$g(t)=|Tr(p(B+tC))-Tr(p(B))-Tr(p'(B)C)t|$$
we have $g(t)/t rightarrow 0$ as $trightarrow 0$.
First note that $$|Tr(p'(B)C)-Tr(f'(B)C)|=|Tr((p'(B)-f'(B))C)| $$ $$leq |p'(B)-f'(B)| |C|<n epsilon |C|.$$
Moreover,
$$|Tr(p(B+tC))-Tr(f(B+tC))|=|Tr(p(B+tC)-f(B+tC))|<n epsilon,$$
and similarly
$$|Tr(p(B))-Tr(f(B))|<n epsilon.$$
Putting all these inequalities together, we have
$$h(t)=|Tr(f(B+tC))-Tr(f(B))-Tr(f'(B)C)t|leq g(t)+(2+|C|)nepsilon,$$
By letting $epsilon rightarrow 0$, we have $0leq h(t)leq g(t)$ which implies the formula.
answered Aug 24 at 0:13
Marco
1,3287
I'm fine up until getting $h(t) leq g(t)$, because this inequality depended on $|p - f| < epsilon$ and $g = g_p$. Once $epsilon < |p - f|$ the inequality is false. You would need to control $g_p(t)$ as $epsilon to 0$.
– 4-ier
Aug 24 at 3:03
add a comment |Â
up vote
0
down vote
The result is easy to check if $f(x)$ was a polynomial. For a general continuously differentiable function $f(x)$ and a given $epsilon>0$, choose a polynomial $p(x)$ such that $|f(x)-p(x)|+|f'(x)-p'(x)|<epsilon$ on an interval $I$ containing all of the eigenvalues of $B$ and $C$. One has
$$dfracddtTr(p(B+tC))(0)=Tr(p'(B)C),$$
and so for
$$g(t)=|Tr(p(B+tC))-Tr(p(B))-Tr(p'(B)C)t|$$
we have $g(t)/t rightarrow 0$ as $trightarrow 0$.
First note that $$|Tr(p'(B)C)-Tr(f'(B)C)|=|Tr((p'(B)-f'(B))C)| $$ $$leq |p'(B)-f'(B)| |C|<n epsilon |C|.$$
Moreover,
$$|Tr(p(B+tC))-Tr(f(B+tC))|=|Tr(p(B+tC)-f(B+tC))|<n epsilon,$$
and similarly
$$|Tr(p(B))-Tr(f(B))|<n epsilon.$$
Putting all these inequalities together, we have
$$h(t)=|Tr(f(B+tC))-Tr(f(B))-Tr(f'(B)C)t|leq g(t)+(2+|C|)nepsilon,$$
By letting $epsilon rightarrow 0$, we have $0leq h(t)leq g(t)$ which implies the formula.
answered Aug 24 at 0:13
Marco
1,3287
I'm fine up until getting $h(t) leq g(t)$, because this inequality depended on $|p - f| < epsilon$ and $g = g_p$. Once $epsilon < |p - f|$ the inequality is false. You would need to control $g_p(t)$ as $epsilon to 0$.
– 4-ier
Aug 24 at 3:03
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The result is easy to check if $f(x)$ was a polynomial. For a general continuously differentiable function $f(x)$ and a given $epsilon>0$, choose a polynomial $p(x)$ such that $|f(x)-p(x)|+|f'(x)-p'(x)|<epsilon$ on an interval $I$ containing all of the eigenvalues of $B$ and $C$. One has
$$dfracddtTr(p(B+tC))(0)=Tr(p'(B)C),$$
and so for
$$g(t)=|Tr(p(B+tC))-Tr(p(B))-Tr(p'(B)C)t|$$
we have $g(t)/t rightarrow 0$ as $trightarrow 0$.
First note that $$|Tr(p'(B)C)-Tr(f'(B)C)|=|Tr((p'(B)-f'(B))C)| $$ $$leq |p'(B)-f'(B)| |C|<n epsilon |C|.$$
Moreover,
$$|Tr(p(B+tC))-Tr(f(B+tC))|=|Tr(p(B+tC)-f(B+tC))|<n epsilon,$$
and similarly
$$|Tr(p(B))-Tr(f(B))|<n epsilon.$$
Putting all these inequalities together, we have
$$h(t)=|Tr(f(B+tC))-Tr(f(B))-Tr(f'(B)C)t|leq g(t)+(2+|C|)nepsilon,$$
By letting $epsilon rightarrow 0$, we have $0leq h(t)leq g(t)$ which implies the formula.
answered Aug 24 at 0:13
Marco
1,3287
The result is easy to check if $f(x)$ was a polynomial. For a general continuously differentiable function $f(x)$ and a given $epsilon>0$, choose a polynomial $p(x)$ such that $|f(x)-p(x)|+|f'(x)-p'(x)|<epsilon$ on an interval $I$ containing all of the eigenvalues of $B$ and $C$. One has
$$dfracddtTr(p(B+tC))(0)=Tr(p'(B)C),$$
and so for
$$g(t)=|Tr(p(B+tC))-Tr(p(B))-Tr(p'(B)C)t|$$
we have $g(t)/t rightarrow 0$ as $trightarrow 0$.
First note that $$|Tr(p'(B)C)-Tr(f'(B)C)|=|Tr((p'(B)-f'(B))C)| $$ $$leq |p'(B)-f'(B)| |C|<n epsilon |C|.$$
Moreover,
$$|Tr(p(B+tC))-Tr(f(B+tC))|=|Tr(p(B+tC)-f(B+tC))|<n epsilon,$$
and similarly
$$|Tr(p(B))-Tr(f(B))|<n epsilon.$$
Putting all these inequalities together, we have
$$h(t)=|Tr(f(B+tC))-Tr(f(B))-Tr(f'(B)C)t|leq g(t)+(2+|C|)nepsilon,$$
By letting $epsilon rightarrow 0$, we have $0leq h(t)leq g(t)$ which implies the formula.
answered Aug 24 at 0:13
Marco
1,3287
answered Aug 24 at 0:13
Marco
1,3287
answered Aug 24 at 0:13
Marco
1,3287
answered Aug 24 at 0:13
Marco
1,3287
1,3287
I'm fine up until getting $h(t) leq g(t)$, because this inequality depended on $|p - f| < epsilon$ and $g = g_p$. Once $epsilon < |p - f|$ the inequality is false. You would need to control $g_p(t)$ as $epsilon to 0$.
– 4-ier
Aug 24 at 3:03
add a comment |Â
I'm fine up until getting $h(t) leq g(t)$, because this inequality depended on $|p - f| < epsilon$ and $g = g_p$. Once $epsilon < |p - f|$ the inequality is false. You would need to control $g_p(t)$ as $epsilon to 0$.
– 4-ier
Aug 24 at 3:03
I'm fine up until getting $h(t) leq g(t)$, because this inequality depended on $|p - f| < epsilon$ and $g = g_p$. Once $epsilon < |p - f|$ the inequality is false. You would need to control $g_p(t)$ as $epsilon to 0$.
– 4-ier
Aug 24 at 3:03
I'm fine up until getting $h(t) leq g(t)$, because this inequality depended on $|p - f| < epsilon$ and $g = g_p$. Once $epsilon < |p - f|$ the inequality is false. You would need to control $g_p(t)$ as $epsilon to 0$.
– 4-ier
Aug 24 at 3:03
add a comment |Â
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Considering how a trace is simply a sum of terms, the total time derivative evaluated at $t=0$ would just slide into the trace operation. From here, it's simply an extension on the chain rule
– Captain Morgan
Aug 22 at 5:45
Sure... just why does there exist a chain rule for functions defined via the spectral theorem. Maybe, I would need to prove a version of the convergence lemma in Baby Rudin, where I approximate $f'$ by polynomials that have some antiderivatives that converge to $f$ and then I can try to interchange the derivative limit and the limit of polymonials for which the chain rule obviously applies.
– 4-ier
Aug 22 at 6:02
For instance, if $f(x) = x^2$, then we know that $f(B + tC) = B^2 + t(BC + CB) + t^2C^2$ so its derivative in $t$ at zero is $BC + CB$ which does not equal $f'(B)C = 2BC$ in general, but does after applying the trace.
– 4-ier
Aug 22 at 6:10
Yes, but since $rm tr(CB)=rm tr(BC)$ you can combine the terms to obtain $rm tr(2BC)$.
– lynn
Aug 23 at 19:12
Yes, I agree. What I am saying is that for polynomials $f$, I can see that it works because under the trace, everything commutes. However, for a general continuously differentiable function I am not sure if one can make it work.
– 4-ier
Aug 23 at 22:34