Mixing
Integral representation of log of operators
$begingroup$
$def1{mathbb{1}}$
Suppose we're in a "good enough" (finite for example) space, and we have positive (semi)-definite operators $P$ and $Q$.
Let $log{(P)}$ and $log{(P)}$ be logarithms of $P$ and $Q$, defined either as the power series, the logarithm of the Jordan decomposition with the change of basis or the inverse of the exponential function. I'm not sure if these are all equivalent, but let's suppose they are, for now at least.
Then we have the following integral representation
$$
log{(P)} - log{(Q)} = int_0^infty Big(frac{1}{Q + x1} - frac{1}{P + x1}Big) dx
$$
where $1$ is the identity, and $frac{1}{P + x1} = (P+x1)^{-1}$.
Is there a simple proof of this property?
I wouldn't know where to start since I'm not sure about the correct/universal definition of the logarithm of an operator.
Any help is appreciated !
Also I'd be thankful if anyone can direct me to a universal definition of $e^P$ and $log{P}$. I believe that for $e^P$ the power series is enough and the equivalence to Jordan decomposition + $operatorname{exp}$ is straightforward, but then for $log{(P)}$ it doesn't always converge, so maybe there's a different way to define it.
operator-theory hilbert-spaces matrix-calculus operator-algebras quantum-mechanics
asked Jan 18 at 15:27
KoljaKolja
590310
$endgroup$
add a comment |
$begingroup$
$def1{mathbb{1}}$
Suppose we're in a "good enough" (finite for example) space, and we have positive (semi)-definite operators $P$ and $Q$.
Let $log{(P)}$ and $log{(P)}$ be logarithms of $P$ and $Q$, defined either as the power series, the logarithm of the Jordan decomposition with the change of basis or the inverse of the exponential function. I'm not sure if these are all equivalent, but let's suppose they are, for now at least.
Then we have the following integral representation
$$
log{(P)} - log{(Q)} = int_0^infty Big(frac{1}{Q + x1} - frac{1}{P + x1}Big) dx
$$
where $1$ is the identity, and $frac{1}{P + x1} = (P+x1)^{-1}$.
Is there a simple proof of this property?
I wouldn't know where to start since I'm not sure about the correct/universal definition of the logarithm of an operator.
Any help is appreciated !
Also I'd be thankful if anyone can direct me to a universal definition of $e^P$ and $log{P}$. I believe that for $e^P$ the power series is enough and the equivalence to Jordan decomposition + $operatorname{exp}$ is straightforward, but then for $log{(P)}$ it doesn't always converge, so maybe there's a different way to define it.
operator-theory hilbert-spaces matrix-calculus operator-algebras quantum-mechanics
asked Jan 18 at 15:27
KoljaKolja
590310
$endgroup$
2
$begingroup$
The sign in the displayed equation is wrong.
$endgroup$
– MaoWao
Jan 18 at 15:58
$begingroup$
Indeed. Thanks !
$endgroup$
– Kolja
Jan 18 at 16:24
add a comment |
$begingroup$
$def1{mathbb{1}}$
Suppose we're in a "good enough" (finite for example) space, and we have positive (semi)-definite operators $P$ and $Q$.
Let $log{(P)}$ and $log{(P)}$ be logarithms of $P$ and $Q$, defined either as the power series, the logarithm of the Jordan decomposition with the change of basis or the inverse of the exponential function. I'm not sure if these are all equivalent, but let's suppose they are, for now at least.
Then we have the following integral representation
$$
log{(P)} - log{(Q)} = int_0^infty Big(frac{1}{Q + x1} - frac{1}{P + x1}Big) dx
$$
where $1$ is the identity, and $frac{1}{P + x1} = (P+x1)^{-1}$.
Is there a simple proof of this property?
I wouldn't know where to start since I'm not sure about the correct/universal definition of the logarithm of an operator.
Any help is appreciated !
Also I'd be thankful if anyone can direct me to a universal definition of $e^P$ and $log{P}$. I believe that for $e^P$ the power series is enough and the equivalence to Jordan decomposition + $operatorname{exp}$ is straightforward, but then for $log{(P)}$ it doesn't always converge, so maybe there's a different way to define it.
operator-theory hilbert-spaces matrix-calculus operator-algebras quantum-mechanics
asked Jan 18 at 15:27
KoljaKolja
590310
$endgroup$
$def1{mathbb{1}}$
Suppose we're in a "good enough" (finite for example) space, and we have positive (semi)-definite operators $P$ and $Q$.
Let $log{(P)}$ and $log{(P)}$ be logarithms of $P$ and $Q$, defined either as the power series, the logarithm of the Jordan decomposition with the change of basis or the inverse of the exponential function. I'm not sure if these are all equivalent, but let's suppose they are, for now at least.
Then we have the following integral representation
$$
log{(P)} - log{(Q)} = int_0^infty Big(frac{1}{Q + x1} - frac{1}{P + x1}Big) dx
$$
where $1$ is the identity, and $frac{1}{P + x1} = (P+x1)^{-1}$.
Is there a simple proof of this property?
I wouldn't know where to start since I'm not sure about the correct/universal definition of the logarithm of an operator.
Any help is appreciated !
Also I'd be thankful if anyone can direct me to a universal definition of $e^P$ and $log{P}$. I believe that for $e^P$ the power series is enough and the equivalence to Jordan decomposition + $operatorname{exp}$ is straightforward, but then for $log{(P)}$ it doesn't always converge, so maybe there's a different way to define it.
operator-theory hilbert-spaces matrix-calculus operator-algebras quantum-mechanics
operator-theory hilbert-spaces matrix-calculus operator-algebras quantum-mechanics
asked Jan 18 at 15:27
KoljaKolja
590310
asked Jan 18 at 15:27
KoljaKolja
590310
edited Jan 18 at 16:24
Kolja
asked Jan 18 at 15:27
KoljaKolja
590310
asked Jan 18 at 15:27
KoljaKolja
590310
asked Jan 18 at 15:27
KoljaKolja
590310
590310
2
$begingroup$
The sign in the displayed equation is wrong.
$endgroup$
– MaoWao
Jan 18 at 15:58
$begingroup$
Indeed. Thanks !
$endgroup$
– Kolja
Jan 18 at 16:24
add a comment |
2
$begingroup$
The sign in the displayed equation is wrong.
$endgroup$
– MaoWao
Jan 18 at 15:58
$begingroup$
Indeed. Thanks !
$endgroup$
– Kolja
Jan 18 at 16:24
2
2
$begingroup$
The sign in the displayed equation is wrong.
$endgroup$
– MaoWao
Jan 18 at 15:58
$begingroup$
The sign in the displayed equation is wrong.
$endgroup$
– MaoWao
Jan 18 at 15:58
$begingroup$
Indeed. Thanks !
$endgroup$
– Kolja
Jan 18 at 16:24
$begingroup$
Indeed. Thanks !
$endgroup$
– Kolja
Jan 18 at 16:24
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
We should assume that $P,Q>0$ in order that $log P$ and $log Q$ are well-defined. Suppose we are in a finite dimensional inner product space. Then by the spectral decomposition theorem, there exist unitary matrices $U,V$ such that
$$
P=UAU^*,quad Q=VBV^*
$$for some diagonal matrices $A,B>0$. Note that $log P= Uleft(log Aright) U^*$ and $log Q =Vleft(log Bright) V^*$. Now, we have
$$begin{eqnarray}
int_0^N (P+x1)^{-1}dx = Uleft(int_0^N (A+x1)^{-1}dxright)U^*&=&Uleft(log (A+N1)-log Aright)U^*\&=&Uleft(log left(1+frac{A}{N}right) right)U^*-log P+log Ncdot 1
end{eqnarray}$$ and similarly
$$
int_0^N (Q+x1)^{-1}dx=Vleft(log left(1+frac{B}{N}right)right)V^*-log Q+log Ncdot 1.
$$ Here, $1$ denotes the identity operator. Hence,
$$begin{eqnarray}
int_0^N (P+x1)^{-1}-(Q+x1)^{-1}dx&=&-log P+log Q+\&&+Uleft(log left(1+frac{A}{N}right)right)U^*-Vleft(log left(1+frac{B}{N}right)right)V^*\
&to&-log P+log Q
end{eqnarray}$$ as $Ntoinfty$. This proves
$$
int_0^infty (P+x1)^{-1}-(Q+x1)^{-1}dx=-log P+log Q.
$$ If we are in a Hilbert space, then similar proof is possible by the spectral representation $$
P=int_{(0,infty)}lambda dE_1(lambda), quad Q=int_{(0,infty)}lambda dE_2(lambda).
$$
answered Jan 18 at 15:56
SongSong
15.7k1737
$endgroup$
$begingroup$
Thank you, this was very helpful
$endgroup$
– Kolja
Jan 18 at 16:40
add a comment |
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$begingroup$
We should assume that $P,Q>0$ in order that $log P$ and $log Q$ are well-defined. Suppose we are in a finite dimensional inner product space. Then by the spectral decomposition theorem, there exist unitary matrices $U,V$ such that
$$
P=UAU^*,quad Q=VBV^*
$$for some diagonal matrices $A,B>0$. Note that $log P= Uleft(log Aright) U^*$ and $log Q =Vleft(log Bright) V^*$. Now, we have
$$begin{eqnarray}
int_0^N (P+x1)^{-1}dx = Uleft(int_0^N (A+x1)^{-1}dxright)U^*&=&Uleft(log (A+N1)-log Aright)U^*\&=&Uleft(log left(1+frac{A}{N}right) right)U^*-log P+log Ncdot 1
end{eqnarray}$$ and similarly
$$
int_0^N (Q+x1)^{-1}dx=Vleft(log left(1+frac{B}{N}right)right)V^*-log Q+log Ncdot 1.
$$ Here, $1$ denotes the identity operator. Hence,
$$begin{eqnarray}
int_0^N (P+x1)^{-1}-(Q+x1)^{-1}dx&=&-log P+log Q+\&&+Uleft(log left(1+frac{A}{N}right)right)U^*-Vleft(log left(1+frac{B}{N}right)right)V^*\
&to&-log P+log Q
end{eqnarray}$$ as $Ntoinfty$. This proves
$$
int_0^infty (P+x1)^{-1}-(Q+x1)^{-1}dx=-log P+log Q.
$$ If we are in a Hilbert space, then similar proof is possible by the spectral representation $$
P=int_{(0,infty)}lambda dE_1(lambda), quad Q=int_{(0,infty)}lambda dE_2(lambda).
$$
answered Jan 18 at 15:56
SongSong
15.7k1737
$endgroup$
$begingroup$
Thank you, this was very helpful
$endgroup$
– Kolja
Jan 18 at 16:40
add a comment |
$begingroup$
We should assume that $P,Q>0$ in order that $log P$ and $log Q$ are well-defined. Suppose we are in a finite dimensional inner product space. Then by the spectral decomposition theorem, there exist unitary matrices $U,V$ such that
$$
P=UAU^*,quad Q=VBV^*
$$for some diagonal matrices $A,B>0$. Note that $log P= Uleft(log Aright) U^*$ and $log Q =Vleft(log Bright) V^*$. Now, we have
$$begin{eqnarray}
int_0^N (P+x1)^{-1}dx = Uleft(int_0^N (A+x1)^{-1}dxright)U^*&=&Uleft(log (A+N1)-log Aright)U^*\&=&Uleft(log left(1+frac{A}{N}right) right)U^*-log P+log Ncdot 1
end{eqnarray}$$ and similarly
$$
int_0^N (Q+x1)^{-1}dx=Vleft(log left(1+frac{B}{N}right)right)V^*-log Q+log Ncdot 1.
$$ Here, $1$ denotes the identity operator. Hence,
$$begin{eqnarray}
int_0^N (P+x1)^{-1}-(Q+x1)^{-1}dx&=&-log P+log Q+\&&+Uleft(log left(1+frac{A}{N}right)right)U^*-Vleft(log left(1+frac{B}{N}right)right)V^*\
&to&-log P+log Q
end{eqnarray}$$ as $Ntoinfty$. This proves
$$
int_0^infty (P+x1)^{-1}-(Q+x1)^{-1}dx=-log P+log Q.
$$ If we are in a Hilbert space, then similar proof is possible by the spectral representation $$
P=int_{(0,infty)}lambda dE_1(lambda), quad Q=int_{(0,infty)}lambda dE_2(lambda).
$$
answered Jan 18 at 15:56
SongSong
15.7k1737
$endgroup$
$begingroup$
Thank you, this was very helpful
$endgroup$
– Kolja
Jan 18 at 16:40
add a comment |
$begingroup$
We should assume that $P,Q>0$ in order that $log P$ and $log Q$ are well-defined. Suppose we are in a finite dimensional inner product space. Then by the spectral decomposition theorem, there exist unitary matrices $U,V$ such that
$$
P=UAU^*,quad Q=VBV^*
$$for some diagonal matrices $A,B>0$. Note that $log P= Uleft(log Aright) U^*$ and $log Q =Vleft(log Bright) V^*$. Now, we have
$$begin{eqnarray}
int_0^N (P+x1)^{-1}dx = Uleft(int_0^N (A+x1)^{-1}dxright)U^*&=&Uleft(log (A+N1)-log Aright)U^*\&=&Uleft(log left(1+frac{A}{N}right) right)U^*-log P+log Ncdot 1
end{eqnarray}$$ and similarly
$$
int_0^N (Q+x1)^{-1}dx=Vleft(log left(1+frac{B}{N}right)right)V^*-log Q+log Ncdot 1.
$$ Here, $1$ denotes the identity operator. Hence,
$$begin{eqnarray}
int_0^N (P+x1)^{-1}-(Q+x1)^{-1}dx&=&-log P+log Q+\&&+Uleft(log left(1+frac{A}{N}right)right)U^*-Vleft(log left(1+frac{B}{N}right)right)V^*\
&to&-log P+log Q
end{eqnarray}$$ as $Ntoinfty$. This proves
$$
int_0^infty (P+x1)^{-1}-(Q+x1)^{-1}dx=-log P+log Q.
$$ If we are in a Hilbert space, then similar proof is possible by the spectral representation $$
P=int_{(0,infty)}lambda dE_1(lambda), quad Q=int_{(0,infty)}lambda dE_2(lambda).
$$
answered Jan 18 at 15:56
SongSong
15.7k1737
$endgroup$
We should assume that $P,Q>0$ in order that $log P$ and $log Q$ are well-defined. Suppose we are in a finite dimensional inner product space. Then by the spectral decomposition theorem, there exist unitary matrices $U,V$ such that
$$
P=UAU^*,quad Q=VBV^*
$$for some diagonal matrices $A,B>0$. Note that $log P= Uleft(log Aright) U^*$ and $log Q =Vleft(log Bright) V^*$. Now, we have
$$begin{eqnarray}
int_0^N (P+x1)^{-1}dx = Uleft(int_0^N (A+x1)^{-1}dxright)U^*&=&Uleft(log (A+N1)-log Aright)U^*\&=&Uleft(log left(1+frac{A}{N}right) right)U^*-log P+log Ncdot 1
end{eqnarray}$$ and similarly
$$
int_0^N (Q+x1)^{-1}dx=Vleft(log left(1+frac{B}{N}right)right)V^*-log Q+log Ncdot 1.
$$ Here, $1$ denotes the identity operator. Hence,
$$begin{eqnarray}
int_0^N (P+x1)^{-1}-(Q+x1)^{-1}dx&=&-log P+log Q+\&&+Uleft(log left(1+frac{A}{N}right)right)U^*-Vleft(log left(1+frac{B}{N}right)right)V^*\
&to&-log P+log Q
end{eqnarray}$$ as $Ntoinfty$. This proves
$$
int_0^infty (P+x1)^{-1}-(Q+x1)^{-1}dx=-log P+log Q.
$$ If we are in a Hilbert space, then similar proof is possible by the spectral representation $$
P=int_{(0,infty)}lambda dE_1(lambda), quad Q=int_{(0,infty)}lambda dE_2(lambda).
$$
answered Jan 18 at 15:56
SongSong
15.7k1737
answered Jan 18 at 15:56
SongSong
15.7k1737
answered Jan 18 at 15:56
SongSong
15.7k1737
answered Jan 18 at 15:56
SongSong
15.7k1737
15.7k1737
$begingroup$
Thank you, this was very helpful
$endgroup$
– Kolja
Jan 18 at 16:40
add a comment |
$begingroup$
Thank you, this was very helpful
$endgroup$
– Kolja
Jan 18 at 16:40
$begingroup$
Thank you, this was very helpful
$endgroup$
– Kolja
Jan 18 at 16:40
$begingroup$
Thank you, this was very helpful
$endgroup$
– Kolja
Jan 18 at 16:40
add a comment |
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2
$begingroup$
The sign in the displayed equation is wrong.
$endgroup$
– MaoWao
Jan 18 at 15:58
$begingroup$
Indeed. Thanks !
$endgroup$
– Kolja
Jan 18 at 16:24