Absolutely continuous spectrum invariant under unitary equivalence
I am doing an exercise calculating the absolutely continuous spectrum of some operator $A$ by calculating the spectrum of a different operator $B$ unitarily equivalent to $A$, i.e.
$$A = U B U^*.$$
I know that $sigma (A) = sigma(B)$ as is easily seen by playing with the resolvent. But I cannot as easily conclude $sigma_{ac}(A) = sigma_{ac}(B)$. Am I missing something obvious? How can one show this? I first thought about something along
$$sigma_{ac}(A) = sigma(A mid_{mathcal H_{ac}}) = sigma(UBU^* mid_{mathcal H_{ac}}) stackrel{?}{=} sigma(U B mid_{stackrel{sim}{mathcal H_{ac}}
}U^*) = sigma_{ac}(B),$$
but I don't think the third equality is justifiable, especially we have different Hilbert spaces $mathcal H, stackrel{sim}{mathcal H}$.
functional-analysis analysis operator-theory mathematical-physics spectral-theory
asked Nov 29 '18 at 20:37
bavor42bavor42
30419
add a comment |
I am doing an exercise calculating the absolutely continuous spectrum of some operator $A$ by calculating the spectrum of a different operator $B$ unitarily equivalent to $A$, i.e.
$$A = U B U^*.$$
I know that $sigma (A) = sigma(B)$ as is easily seen by playing with the resolvent. But I cannot as easily conclude $sigma_{ac}(A) = sigma_{ac}(B)$. Am I missing something obvious? How can one show this? I first thought about something along
$$sigma_{ac}(A) = sigma(A mid_{mathcal H_{ac}}) = sigma(UBU^* mid_{mathcal H_{ac}}) stackrel{?}{=} sigma(U B mid_{stackrel{sim}{mathcal H_{ac}}
}U^*) = sigma_{ac}(B),$$
but I don't think the third equality is justifiable, especially we have different Hilbert spaces $mathcal H, stackrel{sim}{mathcal H}$.
functional-analysis analysis operator-theory mathematical-physics spectral-theory
asked Nov 29 '18 at 20:37
bavor42bavor42
30419
add a comment |
I am doing an exercise calculating the absolutely continuous spectrum of some operator $A$ by calculating the spectrum of a different operator $B$ unitarily equivalent to $A$, i.e.
$$A = U B U^*.$$
I know that $sigma (A) = sigma(B)$ as is easily seen by playing with the resolvent. But I cannot as easily conclude $sigma_{ac}(A) = sigma_{ac}(B)$. Am I missing something obvious? How can one show this? I first thought about something along
$$sigma_{ac}(A) = sigma(A mid_{mathcal H_{ac}}) = sigma(UBU^* mid_{mathcal H_{ac}}) stackrel{?}{=} sigma(U B mid_{stackrel{sim}{mathcal H_{ac}}
}U^*) = sigma_{ac}(B),$$
but I don't think the third equality is justifiable, especially we have different Hilbert spaces $mathcal H, stackrel{sim}{mathcal H}$.
functional-analysis analysis operator-theory mathematical-physics spectral-theory
asked Nov 29 '18 at 20:37
bavor42bavor42
30419
I am doing an exercise calculating the absolutely continuous spectrum of some operator $A$ by calculating the spectrum of a different operator $B$ unitarily equivalent to $A$, i.e.
$$A = U B U^*.$$
I know that $sigma (A) = sigma(B)$ as is easily seen by playing with the resolvent. But I cannot as easily conclude $sigma_{ac}(A) = sigma_{ac}(B)$. Am I missing something obvious? How can one show this? I first thought about something along
$$sigma_{ac}(A) = sigma(A mid_{mathcal H_{ac}}) = sigma(UBU^* mid_{mathcal H_{ac}}) stackrel{?}{=} sigma(U B mid_{stackrel{sim}{mathcal H_{ac}}
}U^*) = sigma_{ac}(B),$$
but I don't think the third equality is justifiable, especially we have different Hilbert spaces $mathcal H, stackrel{sim}{mathcal H}$.
functional-analysis analysis operator-theory mathematical-physics spectral-theory
functional-analysis analysis operator-theory mathematical-physics spectral-theory
asked Nov 29 '18 at 20:37
bavor42bavor42
30419
asked Nov 29 '18 at 20:37
bavor42bavor42
30419
asked Nov 29 '18 at 20:37
bavor42bavor42
30419
asked Nov 29 '18 at 20:37
bavor42bavor42
30419
asked Nov 29 '18 at 20:37
bavor42bavor42
30419
30419
add a comment |
add a comment |
1 Answer
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For each $hin H$, you have the corresponding spectral measure $mu_h^A$. Since
$$
langle f(UAU^*)h,hrangle=langle f(A)U^*h,U^*hrangle,
$$
it follows that $$mu_h^{UAU^*}=mu_{U^*h}^A.$$ Replacing $h$ with $Uh$ we get
$$mu_{Uh}^{UAU^*}=mu_{h}^A, hin H.$$ From this you can deduce that $hin H_{ac}^A $ if and only if $Uhin H_{ac}^B$. In other words, $U:H_{ac}^Ato H_{ac}^B$ is an isomorphism. This shows exactly what you want
$$
UAU^*|_{H_{ac}^{UAU^*}}=U(A|_{H_{ac}^A})U^*
$$
answered Nov 29 '18 at 21:50
Martin ArgeramiMartin Argerami
124k1176175
Very clear. Thanks!
– bavor42
Nov 30 '18 at 9:47
A small question left though: How does one see $f(UAU^*) = Uf(A)U^*$? By checking the properties of the measurable functional calculus?
– bavor42
Nov 30 '18 at 10:09
Yes. Works for polynomials, and if two Borel measures on a compact subset of $mathbb C $ agree on polynomials, then they are equal.
– Martin Argerami
Nov 30 '18 at 12:25
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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active
oldest
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active
oldest
votes
For each $hin H$, you have the corresponding spectral measure $mu_h^A$. Since
$$
langle f(UAU^*)h,hrangle=langle f(A)U^*h,U^*hrangle,
$$
it follows that $$mu_h^{UAU^*}=mu_{U^*h}^A.$$ Replacing $h$ with $Uh$ we get
$$mu_{Uh}^{UAU^*}=mu_{h}^A, hin H.$$ From this you can deduce that $hin H_{ac}^A $ if and only if $Uhin H_{ac}^B$. In other words, $U:H_{ac}^Ato H_{ac}^B$ is an isomorphism. This shows exactly what you want
$$
UAU^*|_{H_{ac}^{UAU^*}}=U(A|_{H_{ac}^A})U^*
$$
answered Nov 29 '18 at 21:50
Martin ArgeramiMartin Argerami
124k1176175
Very clear. Thanks!
– bavor42
Nov 30 '18 at 9:47
A small question left though: How does one see $f(UAU^*) = Uf(A)U^*$? By checking the properties of the measurable functional calculus?
– bavor42
Nov 30 '18 at 10:09
Yes. Works for polynomials, and if two Borel measures on a compact subset of $mathbb C $ agree on polynomials, then they are equal.
– Martin Argerami
Nov 30 '18 at 12:25
add a comment |
For each $hin H$, you have the corresponding spectral measure $mu_h^A$. Since
$$
langle f(UAU^*)h,hrangle=langle f(A)U^*h,U^*hrangle,
$$
it follows that $$mu_h^{UAU^*}=mu_{U^*h}^A.$$ Replacing $h$ with $Uh$ we get
$$mu_{Uh}^{UAU^*}=mu_{h}^A, hin H.$$ From this you can deduce that $hin H_{ac}^A $ if and only if $Uhin H_{ac}^B$. In other words, $U:H_{ac}^Ato H_{ac}^B$ is an isomorphism. This shows exactly what you want
$$
UAU^*|_{H_{ac}^{UAU^*}}=U(A|_{H_{ac}^A})U^*
$$
answered Nov 29 '18 at 21:50
Martin ArgeramiMartin Argerami
124k1176175
Very clear. Thanks!
– bavor42
Nov 30 '18 at 9:47
A small question left though: How does one see $f(UAU^*) = Uf(A)U^*$? By checking the properties of the measurable functional calculus?
– bavor42
Nov 30 '18 at 10:09
Yes. Works for polynomials, and if two Borel measures on a compact subset of $mathbb C $ agree on polynomials, then they are equal.
– Martin Argerami
Nov 30 '18 at 12:25
add a comment |
For each $hin H$, you have the corresponding spectral measure $mu_h^A$. Since
$$
langle f(UAU^*)h,hrangle=langle f(A)U^*h,U^*hrangle,
$$
it follows that $$mu_h^{UAU^*}=mu_{U^*h}^A.$$ Replacing $h$ with $Uh$ we get
$$mu_{Uh}^{UAU^*}=mu_{h}^A, hin H.$$ From this you can deduce that $hin H_{ac}^A $ if and only if $Uhin H_{ac}^B$. In other words, $U:H_{ac}^Ato H_{ac}^B$ is an isomorphism. This shows exactly what you want
$$
UAU^*|_{H_{ac}^{UAU^*}}=U(A|_{H_{ac}^A})U^*
$$
answered Nov 29 '18 at 21:50
Martin ArgeramiMartin Argerami
124k1176175
For each $hin H$, you have the corresponding spectral measure $mu_h^A$. Since
$$
langle f(UAU^*)h,hrangle=langle f(A)U^*h,U^*hrangle,
$$
it follows that $$mu_h^{UAU^*}=mu_{U^*h}^A.$$ Replacing $h$ with $Uh$ we get
$$mu_{Uh}^{UAU^*}=mu_{h}^A, hin H.$$ From this you can deduce that $hin H_{ac}^A $ if and only if $Uhin H_{ac}^B$. In other words, $U:H_{ac}^Ato H_{ac}^B$ is an isomorphism. This shows exactly what you want
$$
UAU^*|_{H_{ac}^{UAU^*}}=U(A|_{H_{ac}^A})U^*
$$
answered Nov 29 '18 at 21:50
Martin ArgeramiMartin Argerami
124k1176175
answered Nov 29 '18 at 21:50
Martin ArgeramiMartin Argerami
124k1176175
answered Nov 29 '18 at 21:50
Martin ArgeramiMartin Argerami
124k1176175
answered Nov 29 '18 at 21:50
Martin ArgeramiMartin Argerami
124k1176175
124k1176175
Very clear. Thanks!
– bavor42
Nov 30 '18 at 9:47
A small question left though: How does one see $f(UAU^*) = Uf(A)U^*$? By checking the properties of the measurable functional calculus?
– bavor42
Nov 30 '18 at 10:09
Yes. Works for polynomials, and if two Borel measures on a compact subset of $mathbb C $ agree on polynomials, then they are equal.
– Martin Argerami
Nov 30 '18 at 12:25
add a comment |
Very clear. Thanks!
– bavor42
Nov 30 '18 at 9:47
A small question left though: How does one see $f(UAU^*) = Uf(A)U^*$? By checking the properties of the measurable functional calculus?
– bavor42
Nov 30 '18 at 10:09
Yes. Works for polynomials, and if two Borel measures on a compact subset of $mathbb C $ agree on polynomials, then they are equal.
– Martin Argerami
Nov 30 '18 at 12:25
Very clear. Thanks!
– bavor42
Nov 30 '18 at 9:47
Very clear. Thanks!
– bavor42
Nov 30 '18 at 9:47
A small question left though: How does one see $f(UAU^*) = Uf(A)U^*$? By checking the properties of the measurable functional calculus?
– bavor42
Nov 30 '18 at 10:09
A small question left though: How does one see $f(UAU^*) = Uf(A)U^*$? By checking the properties of the measurable functional calculus?
– bavor42
Nov 30 '18 at 10:09
Yes. Works for polynomials, and if two Borel measures on a compact subset of $mathbb C $ agree on polynomials, then they are equal.
– Martin Argerami
Nov 30 '18 at 12:25
Yes. Works for polynomials, and if two Borel measures on a compact subset of $mathbb C $ agree on polynomials, then they are equal.
– Martin Argerami
Nov 30 '18 at 12:25
add a comment |
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