Mixing
Is $operatorname{tr}|A|=operatorname{tr}|A^dagger|$?
Is it true that $operatorname{tr}|A|=operatorname{tr}|A^dagger|$ for any operator in a Hilbert space? I can prove this statement for normal operators such that $[A,A^dagger]=0$. I want to know is there any proof or counter example for general case?
hilbert-spaces
edited Nov 21 '18 at 19:53
Adrian Keister
4,79851933
asked Nov 21 '18 at 19:47
mathvc_mathvc_
678
add a comment |
Is it true that $operatorname{tr}|A|=operatorname{tr}|A^dagger|$ for any operator in a Hilbert space? I can prove this statement for normal operators such that $[A,A^dagger]=0$. I want to know is there any proof or counter example for general case?
hilbert-spaces
edited Nov 21 '18 at 19:53
Adrian Keister
4,79851933
asked Nov 21 '18 at 19:47
mathvc_mathvc_
678
What does the vertical line mean? Do they mean just the trace or something else?
– Dog_69
Nov 21 '18 at 20:27
@Dog_69 $|A|=sqrt{A A^{dagger}}$
– mathvc_
Nov 21 '18 at 20:28
Then, if you space is not finite dimensional I would say not, because in general $A^{daggerdagger}neq A$.
– Dog_69
Nov 21 '18 at 21:55
1
@Dog_69 Well at least for bounded operators, the ${}^dagger$ map is an involution, i.e. $A^{daggerdagger}=A$. For unbounded operators, this is a whole different thing involving possible domain problems etc. - then again, I personally would rather not tackle the above question for unbounded operators for various reasons.
– Frederik vom Ende
Nov 22 '18 at 14:28
@FrederikvomEnde Got it. Thanks.
– Dog_69
Nov 22 '18 at 16:31
add a comment |
Is it true that $operatorname{tr}|A|=operatorname{tr}|A^dagger|$ for any operator in a Hilbert space? I can prove this statement for normal operators such that $[A,A^dagger]=0$. I want to know is there any proof or counter example for general case?
hilbert-spaces
edited Nov 21 '18 at 19:53
Adrian Keister
4,79851933
asked Nov 21 '18 at 19:47
mathvc_mathvc_
678
Is it true that $operatorname{tr}|A|=operatorname{tr}|A^dagger|$ for any operator in a Hilbert space? I can prove this statement for normal operators such that $[A,A^dagger]=0$. I want to know is there any proof or counter example for general case?
hilbert-spaces
hilbert-spaces
edited Nov 21 '18 at 19:53
Adrian Keister
4,79851933
asked Nov 21 '18 at 19:47
mathvc_mathvc_
678
edited Nov 21 '18 at 19:53
Adrian Keister
4,79851933
asked Nov 21 '18 at 19:47
mathvc_mathvc_
678
edited Nov 21 '18 at 19:53
Adrian Keister
4,79851933
edited Nov 21 '18 at 19:53
Adrian Keister
4,79851933
edited Nov 21 '18 at 19:53
Adrian Keister
4,79851933
4,79851933
asked Nov 21 '18 at 19:47
mathvc_mathvc_
678
asked Nov 21 '18 at 19:47
mathvc_mathvc_
678
asked Nov 21 '18 at 19:47
mathvc_mathvc_
678
678
What does the vertical line mean? Do they mean just the trace or something else?
– Dog_69
Nov 21 '18 at 20:27
@Dog_69 $|A|=sqrt{A A^{dagger}}$
– mathvc_
Nov 21 '18 at 20:28
Then, if you space is not finite dimensional I would say not, because in general $A^{daggerdagger}neq A$.
– Dog_69
Nov 21 '18 at 21:55
1
@Dog_69 Well at least for bounded operators, the ${}^dagger$ map is an involution, i.e. $A^{daggerdagger}=A$. For unbounded operators, this is a whole different thing involving possible domain problems etc. - then again, I personally would rather not tackle the above question for unbounded operators for various reasons.
– Frederik vom Ende
Nov 22 '18 at 14:28
@FrederikvomEnde Got it. Thanks.
– Dog_69
Nov 22 '18 at 16:31
add a comment |
What does the vertical line mean? Do they mean just the trace or something else?
– Dog_69
Nov 21 '18 at 20:27
@Dog_69 $|A|=sqrt{A A^{dagger}}$
– mathvc_
Nov 21 '18 at 20:28
Then, if you space is not finite dimensional I would say not, because in general $A^{daggerdagger}neq A$.
– Dog_69
Nov 21 '18 at 21:55
1
@Dog_69 Well at least for bounded operators, the ${}^dagger$ map is an involution, i.e. $A^{daggerdagger}=A$. For unbounded operators, this is a whole different thing involving possible domain problems etc. - then again, I personally would rather not tackle the above question for unbounded operators for various reasons.
– Frederik vom Ende
Nov 22 '18 at 14:28
@FrederikvomEnde Got it. Thanks.
– Dog_69
Nov 22 '18 at 16:31
What does the vertical line mean? Do they mean just the trace or something else?
– Dog_69
Nov 21 '18 at 20:27
What does the vertical line mean? Do they mean just the trace or something else?
– Dog_69
Nov 21 '18 at 20:27
@Dog_69 $|A|=sqrt{A A^{dagger}}$
– mathvc_
Nov 21 '18 at 20:28
@Dog_69 $|A|=sqrt{A A^{dagger}}$
– mathvc_
Nov 21 '18 at 20:28
Then, if you space is not finite dimensional I would say not, because in general $A^{daggerdagger}neq A$.
– Dog_69
Nov 21 '18 at 21:55
Then, if you space is not finite dimensional I would say not, because in general $A^{daggerdagger}neq A$.
– Dog_69
Nov 21 '18 at 21:55
1
1
@Dog_69 Well at least for bounded operators, the ${}^dagger$ map is an involution, i.e. $A^{daggerdagger}=A$. For unbounded operators, this is a whole different thing involving possible domain problems etc. - then again, I personally would rather not tackle the above question for unbounded operators for various reasons.
– Frederik vom Ende
Nov 22 '18 at 14:28
@Dog_69 Well at least for bounded operators, the ${}^dagger$ map is an involution, i.e. $A^{daggerdagger}=A$. For unbounded operators, this is a whole different thing involving possible domain problems etc. - then again, I personally would rather not tackle the above question for unbounded operators for various reasons.
– Frederik vom Ende
Nov 22 '18 at 14:28
@FrederikvomEnde Got it. Thanks.
– Dog_69
Nov 22 '18 at 16:31
@FrederikvomEnde Got it. Thanks.
– Dog_69
Nov 22 '18 at 16:31
add a comment |
1 Answer
1
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Note: Any references to theorems etc. I will make in the following are with respect to the book "Methods of Modern Mathematical Physics. I: Functional Analysis" by Reed & Simon (1980).
If your Hilbert space $mathcal H$ is separable and $Ainmathcal B(mathcal H)$, i.e. $A$ is a bounded operator on $mathcal H$, your assertion is indeed true for the following reason. As you might know, on separable Hilbert spaces one defines the trace class to be
$$
mathcal B^1(mathcal H):=lbrace Ainmathcal B(mathcal H),|,operatorname{tr}(sqrt{A^dagger A})<inftyrbrace,.
$$
Here one uses that on positive semi-definite operators, the trace is well-defined (independent of the chosen orthonormal basis), linear and takes values in $[0,infty]$. Now if $Ainmathcal B^1(mathcal H)$ then $A$ is compact (Theorem VI.21) which is equivalent to admitting a singular value decomposition
$$
A=sum_{ninmathbb N}sigma_n(A)langle psi_n,cdotranglephi_ntag{1}
$$
with unique non-negative numbers $(sigma_n(A))_{ninmathbb N}$ and orthonormal systems $(psi_n)_{ninmathbb N}$, $(phi_n)_{ninmathbb N}$ in $mathcal H$ where the sum converges in the operator norm (Theorem VI.17). Due to the above singular value decompoition, it is evident that every compact operator $A$ on $mathcal H$ satisfies
$$
operatorname{tr}(sqrt{A^dagger A})=sum_{ninmathbb N}sigma_n(A),,tag{2}
$$
where the right-hand side (and thus both sides) might take the value $infty$. This is enough preparation to prove the statement in question.
Proposition. Let $Ainmathcal B(mathcal H)$. Then $operatorname{tr}(sqrt{A^dagger A})=operatorname{tr}(sqrt{A A^dagger})$, where one side (and thus both sides) might take the value $infty$.
Proof. We may assume that at either $operatorname{tr}(sqrt{A^dagger A})$ or $operatorname{tr}(sqrt{A A^dagger})$ (or both) are finite - otherwise we would obviously be done due to $operatorname{tr}(sqrt{A^dagger A})=infty=operatorname{tr}(sqrt{A A^dagger})$. W.l.o.g. $operatorname{tr}(sqrt{A^dagger A})<infty$ so $Ainmathcal B^1(mathcal H)$ and $operatorname{tr}(sqrt{A^dagger A})=sum_{ninmathbb N}sigma_n(A)$. By (1), obviously $A^dagger$ is compact as well with $sigma_n(A^dagger)=(sigma_n(A))^*=sigma_n(A)$ and thus
$$
operatorname{tr}(sqrt{A^dagger A})=sum_{ninmathbb N}sigma_n(A)=sum_{ninmathbb N}sigma_n(A^dagger)=operatorname{tr}(sqrt{A A^dagger})<infty
$$
which concludes the proof. $quadsquare$
answered Nov 22 '18 at 12:52
Frederik vom EndeFrederik vom Ende
6371321
add a comment |
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Note: Any references to theorems etc. I will make in the following are with respect to the book "Methods of Modern Mathematical Physics. I: Functional Analysis" by Reed & Simon (1980).
If your Hilbert space $mathcal H$ is separable and $Ainmathcal B(mathcal H)$, i.e. $A$ is a bounded operator on $mathcal H$, your assertion is indeed true for the following reason. As you might know, on separable Hilbert spaces one defines the trace class to be
$$
mathcal B^1(mathcal H):=lbrace Ainmathcal B(mathcal H),|,operatorname{tr}(sqrt{A^dagger A})<inftyrbrace,.
$$
Here one uses that on positive semi-definite operators, the trace is well-defined (independent of the chosen orthonormal basis), linear and takes values in $[0,infty]$. Now if $Ainmathcal B^1(mathcal H)$ then $A$ is compact (Theorem VI.21) which is equivalent to admitting a singular value decomposition
$$
A=sum_{ninmathbb N}sigma_n(A)langle psi_n,cdotranglephi_ntag{1}
$$
with unique non-negative numbers $(sigma_n(A))_{ninmathbb N}$ and orthonormal systems $(psi_n)_{ninmathbb N}$, $(phi_n)_{ninmathbb N}$ in $mathcal H$ where the sum converges in the operator norm (Theorem VI.17). Due to the above singular value decompoition, it is evident that every compact operator $A$ on $mathcal H$ satisfies
$$
operatorname{tr}(sqrt{A^dagger A})=sum_{ninmathbb N}sigma_n(A),,tag{2}
$$
where the right-hand side (and thus both sides) might take the value $infty$. This is enough preparation to prove the statement in question.
Proposition. Let $Ainmathcal B(mathcal H)$. Then $operatorname{tr}(sqrt{A^dagger A})=operatorname{tr}(sqrt{A A^dagger})$, where one side (and thus both sides) might take the value $infty$.
Proof. We may assume that at either $operatorname{tr}(sqrt{A^dagger A})$ or $operatorname{tr}(sqrt{A A^dagger})$ (or both) are finite - otherwise we would obviously be done due to $operatorname{tr}(sqrt{A^dagger A})=infty=operatorname{tr}(sqrt{A A^dagger})$. W.l.o.g. $operatorname{tr}(sqrt{A^dagger A})<infty$ so $Ainmathcal B^1(mathcal H)$ and $operatorname{tr}(sqrt{A^dagger A})=sum_{ninmathbb N}sigma_n(A)$. By (1), obviously $A^dagger$ is compact as well with $sigma_n(A^dagger)=(sigma_n(A))^*=sigma_n(A)$ and thus
$$
operatorname{tr}(sqrt{A^dagger A})=sum_{ninmathbb N}sigma_n(A)=sum_{ninmathbb N}sigma_n(A^dagger)=operatorname{tr}(sqrt{A A^dagger})<infty
$$
which concludes the proof. $quadsquare$
answered Nov 22 '18 at 12:52
Frederik vom EndeFrederik vom Ende
6371321
add a comment |
Note: Any references to theorems etc. I will make in the following are with respect to the book "Methods of Modern Mathematical Physics. I: Functional Analysis" by Reed & Simon (1980).
If your Hilbert space $mathcal H$ is separable and $Ainmathcal B(mathcal H)$, i.e. $A$ is a bounded operator on $mathcal H$, your assertion is indeed true for the following reason. As you might know, on separable Hilbert spaces one defines the trace class to be
$$
mathcal B^1(mathcal H):=lbrace Ainmathcal B(mathcal H),|,operatorname{tr}(sqrt{A^dagger A})<inftyrbrace,.
$$
Here one uses that on positive semi-definite operators, the trace is well-defined (independent of the chosen orthonormal basis), linear and takes values in $[0,infty]$. Now if $Ainmathcal B^1(mathcal H)$ then $A$ is compact (Theorem VI.21) which is equivalent to admitting a singular value decomposition
$$
A=sum_{ninmathbb N}sigma_n(A)langle psi_n,cdotranglephi_ntag{1}
$$
with unique non-negative numbers $(sigma_n(A))_{ninmathbb N}$ and orthonormal systems $(psi_n)_{ninmathbb N}$, $(phi_n)_{ninmathbb N}$ in $mathcal H$ where the sum converges in the operator norm (Theorem VI.17). Due to the above singular value decompoition, it is evident that every compact operator $A$ on $mathcal H$ satisfies
$$
operatorname{tr}(sqrt{A^dagger A})=sum_{ninmathbb N}sigma_n(A),,tag{2}
$$
where the right-hand side (and thus both sides) might take the value $infty$. This is enough preparation to prove the statement in question.
Proposition. Let $Ainmathcal B(mathcal H)$. Then $operatorname{tr}(sqrt{A^dagger A})=operatorname{tr}(sqrt{A A^dagger})$, where one side (and thus both sides) might take the value $infty$.
Proof. We may assume that at either $operatorname{tr}(sqrt{A^dagger A})$ or $operatorname{tr}(sqrt{A A^dagger})$ (or both) are finite - otherwise we would obviously be done due to $operatorname{tr}(sqrt{A^dagger A})=infty=operatorname{tr}(sqrt{A A^dagger})$. W.l.o.g. $operatorname{tr}(sqrt{A^dagger A})<infty$ so $Ainmathcal B^1(mathcal H)$ and $operatorname{tr}(sqrt{A^dagger A})=sum_{ninmathbb N}sigma_n(A)$. By (1), obviously $A^dagger$ is compact as well with $sigma_n(A^dagger)=(sigma_n(A))^*=sigma_n(A)$ and thus
$$
operatorname{tr}(sqrt{A^dagger A})=sum_{ninmathbb N}sigma_n(A)=sum_{ninmathbb N}sigma_n(A^dagger)=operatorname{tr}(sqrt{A A^dagger})<infty
$$
which concludes the proof. $quadsquare$
answered Nov 22 '18 at 12:52
Frederik vom EndeFrederik vom Ende
6371321
add a comment |
Note: Any references to theorems etc. I will make in the following are with respect to the book "Methods of Modern Mathematical Physics. I: Functional Analysis" by Reed & Simon (1980).
If your Hilbert space $mathcal H$ is separable and $Ainmathcal B(mathcal H)$, i.e. $A$ is a bounded operator on $mathcal H$, your assertion is indeed true for the following reason. As you might know, on separable Hilbert spaces one defines the trace class to be
$$
mathcal B^1(mathcal H):=lbrace Ainmathcal B(mathcal H),|,operatorname{tr}(sqrt{A^dagger A})<inftyrbrace,.
$$
Here one uses that on positive semi-definite operators, the trace is well-defined (independent of the chosen orthonormal basis), linear and takes values in $[0,infty]$. Now if $Ainmathcal B^1(mathcal H)$ then $A$ is compact (Theorem VI.21) which is equivalent to admitting a singular value decomposition
$$
A=sum_{ninmathbb N}sigma_n(A)langle psi_n,cdotranglephi_ntag{1}
$$
with unique non-negative numbers $(sigma_n(A))_{ninmathbb N}$ and orthonormal systems $(psi_n)_{ninmathbb N}$, $(phi_n)_{ninmathbb N}$ in $mathcal H$ where the sum converges in the operator norm (Theorem VI.17). Due to the above singular value decompoition, it is evident that every compact operator $A$ on $mathcal H$ satisfies
$$
operatorname{tr}(sqrt{A^dagger A})=sum_{ninmathbb N}sigma_n(A),,tag{2}
$$
where the right-hand side (and thus both sides) might take the value $infty$. This is enough preparation to prove the statement in question.
Proposition. Let $Ainmathcal B(mathcal H)$. Then $operatorname{tr}(sqrt{A^dagger A})=operatorname{tr}(sqrt{A A^dagger})$, where one side (and thus both sides) might take the value $infty$.
Proof. We may assume that at either $operatorname{tr}(sqrt{A^dagger A})$ or $operatorname{tr}(sqrt{A A^dagger})$ (or both) are finite - otherwise we would obviously be done due to $operatorname{tr}(sqrt{A^dagger A})=infty=operatorname{tr}(sqrt{A A^dagger})$. W.l.o.g. $operatorname{tr}(sqrt{A^dagger A})<infty$ so $Ainmathcal B^1(mathcal H)$ and $operatorname{tr}(sqrt{A^dagger A})=sum_{ninmathbb N}sigma_n(A)$. By (1), obviously $A^dagger$ is compact as well with $sigma_n(A^dagger)=(sigma_n(A))^*=sigma_n(A)$ and thus
$$
operatorname{tr}(sqrt{A^dagger A})=sum_{ninmathbb N}sigma_n(A)=sum_{ninmathbb N}sigma_n(A^dagger)=operatorname{tr}(sqrt{A A^dagger})<infty
$$
which concludes the proof. $quadsquare$
answered Nov 22 '18 at 12:52
Frederik vom EndeFrederik vom Ende
6371321
Note: Any references to theorems etc. I will make in the following are with respect to the book "Methods of Modern Mathematical Physics. I: Functional Analysis" by Reed & Simon (1980).
If your Hilbert space $mathcal H$ is separable and $Ainmathcal B(mathcal H)$, i.e. $A$ is a bounded operator on $mathcal H$, your assertion is indeed true for the following reason. As you might know, on separable Hilbert spaces one defines the trace class to be
$$
mathcal B^1(mathcal H):=lbrace Ainmathcal B(mathcal H),|,operatorname{tr}(sqrt{A^dagger A})<inftyrbrace,.
$$
Here one uses that on positive semi-definite operators, the trace is well-defined (independent of the chosen orthonormal basis), linear and takes values in $[0,infty]$. Now if $Ainmathcal B^1(mathcal H)$ then $A$ is compact (Theorem VI.21) which is equivalent to admitting a singular value decomposition
$$
A=sum_{ninmathbb N}sigma_n(A)langle psi_n,cdotranglephi_ntag{1}
$$
with unique non-negative numbers $(sigma_n(A))_{ninmathbb N}$ and orthonormal systems $(psi_n)_{ninmathbb N}$, $(phi_n)_{ninmathbb N}$ in $mathcal H$ where the sum converges in the operator norm (Theorem VI.17). Due to the above singular value decompoition, it is evident that every compact operator $A$ on $mathcal H$ satisfies
$$
operatorname{tr}(sqrt{A^dagger A})=sum_{ninmathbb N}sigma_n(A),,tag{2}
$$
where the right-hand side (and thus both sides) might take the value $infty$. This is enough preparation to prove the statement in question.
Proposition. Let $Ainmathcal B(mathcal H)$. Then $operatorname{tr}(sqrt{A^dagger A})=operatorname{tr}(sqrt{A A^dagger})$, where one side (and thus both sides) might take the value $infty$.
Proof. We may assume that at either $operatorname{tr}(sqrt{A^dagger A})$ or $operatorname{tr}(sqrt{A A^dagger})$ (or both) are finite - otherwise we would obviously be done due to $operatorname{tr}(sqrt{A^dagger A})=infty=operatorname{tr}(sqrt{A A^dagger})$. W.l.o.g. $operatorname{tr}(sqrt{A^dagger A})<infty$ so $Ainmathcal B^1(mathcal H)$ and $operatorname{tr}(sqrt{A^dagger A})=sum_{ninmathbb N}sigma_n(A)$. By (1), obviously $A^dagger$ is compact as well with $sigma_n(A^dagger)=(sigma_n(A))^*=sigma_n(A)$ and thus
$$
operatorname{tr}(sqrt{A^dagger A})=sum_{ninmathbb N}sigma_n(A)=sum_{ninmathbb N}sigma_n(A^dagger)=operatorname{tr}(sqrt{A A^dagger})<infty
$$
which concludes the proof. $quadsquare$
answered Nov 22 '18 at 12:52
Frederik vom EndeFrederik vom Ende
6371321
answered Nov 22 '18 at 12:52
Frederik vom EndeFrederik vom Ende
6371321
answered Nov 22 '18 at 12:52
Frederik vom EndeFrederik vom Ende
6371321
answered Nov 22 '18 at 12:52
Frederik vom EndeFrederik vom Ende
6371321
6371321
add a comment |
add a comment |
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What does the vertical line mean? Do they mean just the trace or something else?
– Dog_69
Nov 21 '18 at 20:27
@Dog_69 $|A|=sqrt{A A^{dagger}}$
– mathvc_
Nov 21 '18 at 20:28
Then, if you space is not finite dimensional I would say not, because in general $A^{daggerdagger}neq A$.
– Dog_69
Nov 21 '18 at 21:55
1
@Dog_69 Well at least for bounded operators, the ${}^dagger$ map is an involution, i.e. $A^{daggerdagger}=A$. For unbounded operators, this is a whole different thing involving possible domain problems etc. - then again, I personally would rather not tackle the above question for unbounded operators for various reasons.
– Frederik vom Ende
Nov 22 '18 at 14:28
@FrederikvomEnde Got it. Thanks.
– Dog_69
Nov 22 '18 at 16:31