Mixing
Von Neumann ergodic theorem for purely continuous spectrum
$begingroup$
$newcommand{1}{1negthickspace{mathrm{I}}}$
Von Neumann ergodic theorem states that, if $U(t)$ is a one-parameter group of unitaries acting on a Hilbert space $mathcal{H}$, we have
$$
lim_{Tto infty} parallel X(T)(psi) - P_0(psi) parallel = 0
$$
for all $psiinmathcal{H}$ where $parallel bullet parallel $ is the norm in $mathcal{H}$,
$$
X(T) := frac{1}{T}int_0^T dt U(t) , ,
$$
and $P_0$ is the orthogonal projector onto the invariant subspace $V_0$ of $U(t)$, $V_0={ phiin mathcal{H} |, U(t)phi=phi forall t}$.
Although reminiscent of quantum mechanics it is hard to apply the above theorem in a quantum mechanical setting.
In quantum mechanics we evolve states or operators with the adjoint action (respectively Schroedinger and Heisenberg evolution). Let us pick the Heisenberg evolution:
$$
mathcal{U}(t)[A] :=U(t)^dagger A U(t) , ,
$$
where $U(t)=e^{-i t H}$, where $H$ is a self-adjoint (Hamiltonian) operator.
Define, analogously as before,
$$
mathcal{X}(T) := frac{1}{T}int_0^T dt mathcal{U}(t) , .
$$
Let us consider the case where $H$ has purely continuous spectrum (for example the free article on the line).
Question Assume $H$ has purely continuous spectrum, what is
$$
lim_{Tto infty} mathcal{X}(T) = ?$$
Remarks
1) Clearly we have, for all $T$, and trace class $xi$
$$
mathrm{Tr} Big ( mathcal{X}(T)[xi] Big )= mathrm{Tr} xi
$$
2) It seems that the only invariant operator of $mathcal{U}(t)$ is the identity $1$. Unfortunately this operator is not trace class. Hence is not in $B(mathcal{H})$ seen as an Hilbert space with Hilbert-Schmidt scalar product (which is the setting one would consider when trying to apply the Von Neumann ergodic theorem).
3) In a way the question is: Is there a way to give a meaning to the limit such that it is non-zero?
It seems that this problem should have been well studied but I could not find any reference.
spectral-theory quantum-mechanics ergodic-theory
asked Jan 31 at 2:06
lcvlcv
742410
$endgroup$
add a comment |
$begingroup$
$newcommand{1}{1negthickspace{mathrm{I}}}$
Von Neumann ergodic theorem states that, if $U(t)$ is a one-parameter group of unitaries acting on a Hilbert space $mathcal{H}$, we have
$$
lim_{Tto infty} parallel X(T)(psi) - P_0(psi) parallel = 0
$$
for all $psiinmathcal{H}$ where $parallel bullet parallel $ is the norm in $mathcal{H}$,
$$
X(T) := frac{1}{T}int_0^T dt U(t) , ,
$$
and $P_0$ is the orthogonal projector onto the invariant subspace $V_0$ of $U(t)$, $V_0={ phiin mathcal{H} |, U(t)phi=phi forall t}$.
Although reminiscent of quantum mechanics it is hard to apply the above theorem in a quantum mechanical setting.
In quantum mechanics we evolve states or operators with the adjoint action (respectively Schroedinger and Heisenberg evolution). Let us pick the Heisenberg evolution:
$$
mathcal{U}(t)[A] :=U(t)^dagger A U(t) , ,
$$
where $U(t)=e^{-i t H}$, where $H$ is a self-adjoint (Hamiltonian) operator.
Define, analogously as before,
$$
mathcal{X}(T) := frac{1}{T}int_0^T dt mathcal{U}(t) , .
$$
Let us consider the case where $H$ has purely continuous spectrum (for example the free article on the line).
Question Assume $H$ has purely continuous spectrum, what is
$$
lim_{Tto infty} mathcal{X}(T) = ?$$
Remarks
1) Clearly we have, for all $T$, and trace class $xi$
$$
mathrm{Tr} Big ( mathcal{X}(T)[xi] Big )= mathrm{Tr} xi
$$
2) It seems that the only invariant operator of $mathcal{U}(t)$ is the identity $1$. Unfortunately this operator is not trace class. Hence is not in $B(mathcal{H})$ seen as an Hilbert space with Hilbert-Schmidt scalar product (which is the setting one would consider when trying to apply the Von Neumann ergodic theorem).
3) In a way the question is: Is there a way to give a meaning to the limit such that it is non-zero?
It seems that this problem should have been well studied but I could not find any reference.
spectral-theory quantum-mechanics ergodic-theory
asked Jan 31 at 2:06
lcvlcv
742410
$endgroup$
$begingroup$
This is literally the situation the theorem you quoted deals with: you have strong convergence to the projection $P_0$ on the eigenspace of $H$ for eigenvalue $0$ (of course, $P_0=0$ is possible, for example in your situation when there is only continuous spectrum).
$endgroup$
– user138530
Feb 1 at 1:06
$begingroup$
@ChristianRemling yes I understand, but I would also say that $mathcal{X}(T)[1]=1$ no?
$endgroup$
– lcv
Feb 1 at 1:14
$begingroup$
@ChristianRemling In a way $mathcal{X}(T)$ "converges" to $P_0 = (I/d)mathrm{Tr}(bullet)$ where the dimension of the Hilbert space $d=infty$. So $P_0=0$ as you say. However If could project $mathcal{X}(T)$ onto a finite dimensional space $V$ I could get a finite result. However $V$ should be invariant under $H$... which again is impossible.
$endgroup$
– lcv
Feb 1 at 1:34
add a comment |
$begingroup$
$newcommand{1}{1negthickspace{mathrm{I}}}$
Von Neumann ergodic theorem states that, if $U(t)$ is a one-parameter group of unitaries acting on a Hilbert space $mathcal{H}$, we have
$$
lim_{Tto infty} parallel X(T)(psi) - P_0(psi) parallel = 0
$$
for all $psiinmathcal{H}$ where $parallel bullet parallel $ is the norm in $mathcal{H}$,
$$
X(T) := frac{1}{T}int_0^T dt U(t) , ,
$$
and $P_0$ is the orthogonal projector onto the invariant subspace $V_0$ of $U(t)$, $V_0={ phiin mathcal{H} |, U(t)phi=phi forall t}$.
Although reminiscent of quantum mechanics it is hard to apply the above theorem in a quantum mechanical setting.
In quantum mechanics we evolve states or operators with the adjoint action (respectively Schroedinger and Heisenberg evolution). Let us pick the Heisenberg evolution:
$$
mathcal{U}(t)[A] :=U(t)^dagger A U(t) , ,
$$
where $U(t)=e^{-i t H}$, where $H$ is a self-adjoint (Hamiltonian) operator.
Define, analogously as before,
$$
mathcal{X}(T) := frac{1}{T}int_0^T dt mathcal{U}(t) , .
$$
Let us consider the case where $H$ has purely continuous spectrum (for example the free article on the line).
Question Assume $H$ has purely continuous spectrum, what is
$$
lim_{Tto infty} mathcal{X}(T) = ?$$
Remarks
1) Clearly we have, for all $T$, and trace class $xi$
$$
mathrm{Tr} Big ( mathcal{X}(T)[xi] Big )= mathrm{Tr} xi
$$
2) It seems that the only invariant operator of $mathcal{U}(t)$ is the identity $1$. Unfortunately this operator is not trace class. Hence is not in $B(mathcal{H})$ seen as an Hilbert space with Hilbert-Schmidt scalar product (which is the setting one would consider when trying to apply the Von Neumann ergodic theorem).
3) In a way the question is: Is there a way to give a meaning to the limit such that it is non-zero?
It seems that this problem should have been well studied but I could not find any reference.
spectral-theory quantum-mechanics ergodic-theory
asked Jan 31 at 2:06
lcvlcv
742410
$endgroup$
$newcommand{1}{1negthickspace{mathrm{I}}}$
Von Neumann ergodic theorem states that, if $U(t)$ is a one-parameter group of unitaries acting on a Hilbert space $mathcal{H}$, we have
$$
lim_{Tto infty} parallel X(T)(psi) - P_0(psi) parallel = 0
$$
for all $psiinmathcal{H}$ where $parallel bullet parallel $ is the norm in $mathcal{H}$,
$$
X(T) := frac{1}{T}int_0^T dt U(t) , ,
$$
and $P_0$ is the orthogonal projector onto the invariant subspace $V_0$ of $U(t)$, $V_0={ phiin mathcal{H} |, U(t)phi=phi forall t}$.
Although reminiscent of quantum mechanics it is hard to apply the above theorem in a quantum mechanical setting.
In quantum mechanics we evolve states or operators with the adjoint action (respectively Schroedinger and Heisenberg evolution). Let us pick the Heisenberg evolution:
$$
mathcal{U}(t)[A] :=U(t)^dagger A U(t) , ,
$$
where $U(t)=e^{-i t H}$, where $H$ is a self-adjoint (Hamiltonian) operator.
Define, analogously as before,
$$
mathcal{X}(T) := frac{1}{T}int_0^T dt mathcal{U}(t) , .
$$
Let us consider the case where $H$ has purely continuous spectrum (for example the free article on the line).
Question Assume $H$ has purely continuous spectrum, what is
$$
lim_{Tto infty} mathcal{X}(T) = ?$$
Remarks
1) Clearly we have, for all $T$, and trace class $xi$
$$
mathrm{Tr} Big ( mathcal{X}(T)[xi] Big )= mathrm{Tr} xi
$$
2) It seems that the only invariant operator of $mathcal{U}(t)$ is the identity $1$. Unfortunately this operator is not trace class. Hence is not in $B(mathcal{H})$ seen as an Hilbert space with Hilbert-Schmidt scalar product (which is the setting one would consider when trying to apply the Von Neumann ergodic theorem).
3) In a way the question is: Is there a way to give a meaning to the limit such that it is non-zero?
It seems that this problem should have been well studied but I could not find any reference.
spectral-theory quantum-mechanics ergodic-theory
spectral-theory quantum-mechanics ergodic-theory
asked Jan 31 at 2:06
lcvlcv
742410
asked Jan 31 at 2:06
lcvlcv
742410
edited Feb 1 at 0:38
lcv
asked Jan 31 at 2:06
lcvlcv
742410
asked Jan 31 at 2:06
lcvlcv
742410
asked Jan 31 at 2:06
lcvlcv
742410
742410
$begingroup$
This is literally the situation the theorem you quoted deals with: you have strong convergence to the projection $P_0$ on the eigenspace of $H$ for eigenvalue $0$ (of course, $P_0=0$ is possible, for example in your situation when there is only continuous spectrum).
$endgroup$
– user138530
Feb 1 at 1:06
$begingroup$
@ChristianRemling yes I understand, but I would also say that $mathcal{X}(T)[1]=1$ no?
$endgroup$
– lcv
Feb 1 at 1:14
$begingroup$
@ChristianRemling In a way $mathcal{X}(T)$ "converges" to $P_0 = (I/d)mathrm{Tr}(bullet)$ where the dimension of the Hilbert space $d=infty$. So $P_0=0$ as you say. However If could project $mathcal{X}(T)$ onto a finite dimensional space $V$ I could get a finite result. However $V$ should be invariant under $H$... which again is impossible.
$endgroup$
– lcv
Feb 1 at 1:34
add a comment |
$begingroup$
This is literally the situation the theorem you quoted deals with: you have strong convergence to the projection $P_0$ on the eigenspace of $H$ for eigenvalue $0$ (of course, $P_0=0$ is possible, for example in your situation when there is only continuous spectrum).
$endgroup$
– user138530
Feb 1 at 1:06
$begingroup$
@ChristianRemling yes I understand, but I would also say that $mathcal{X}(T)[1]=1$ no?
$endgroup$
– lcv
Feb 1 at 1:14
$begingroup$
@ChristianRemling In a way $mathcal{X}(T)$ "converges" to $P_0 = (I/d)mathrm{Tr}(bullet)$ where the dimension of the Hilbert space $d=infty$. So $P_0=0$ as you say. However If could project $mathcal{X}(T)$ onto a finite dimensional space $V$ I could get a finite result. However $V$ should be invariant under $H$... which again is impossible.
$endgroup$
– lcv
Feb 1 at 1:34
$begingroup$
This is literally the situation the theorem you quoted deals with: you have strong convergence to the projection $P_0$ on the eigenspace of $H$ for eigenvalue $0$ (of course, $P_0=0$ is possible, for example in your situation when there is only continuous spectrum).
$endgroup$
– user138530
Feb 1 at 1:06
$begingroup$
This is literally the situation the theorem you quoted deals with: you have strong convergence to the projection $P_0$ on the eigenspace of $H$ for eigenvalue $0$ (of course, $P_0=0$ is possible, for example in your situation when there is only continuous spectrum).
$endgroup$
– user138530
Feb 1 at 1:06
$begingroup$
@ChristianRemling yes I understand, but I would also say that $mathcal{X}(T)[1]=1$ no?
$endgroup$
– lcv
Feb 1 at 1:14
$begingroup$
@ChristianRemling yes I understand, but I would also say that $mathcal{X}(T)[1]=1$ no?
$endgroup$
– lcv
Feb 1 at 1:14
$begingroup$
@ChristianRemling In a way $mathcal{X}(T)$ "converges" to $P_0 = (I/d)mathrm{Tr}(bullet)$ where the dimension of the Hilbert space $d=infty$. So $P_0=0$ as you say. However If could project $mathcal{X}(T)$ onto a finite dimensional space $V$ I could get a finite result. However $V$ should be invariant under $H$... which again is impossible.
$endgroup$
– lcv
Feb 1 at 1:34
$begingroup$
@ChristianRemling In a way $mathcal{X}(T)$ "converges" to $P_0 = (I/d)mathrm{Tr}(bullet)$ where the dimension of the Hilbert space $d=infty$. So $P_0=0$ as you say. However If could project $mathcal{X}(T)$ onto a finite dimensional space $V$ I could get a finite result. However $V$ should be invariant under $H$... which again is impossible.
$endgroup$
– lcv
Feb 1 at 1:34
add a comment |
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$begingroup$
This is literally the situation the theorem you quoted deals with: you have strong convergence to the projection $P_0$ on the eigenspace of $H$ for eigenvalue $0$ (of course, $P_0=0$ is possible, for example in your situation when there is only continuous spectrum).
$endgroup$
– user138530
Feb 1 at 1:06
$begingroup$
@ChristianRemling yes I understand, but I would also say that $mathcal{X}(T)[1]=1$ no?
$endgroup$
– lcv
Feb 1 at 1:14
$begingroup$
@ChristianRemling In a way $mathcal{X}(T)$ "converges" to $P_0 = (I/d)mathrm{Tr}(bullet)$ where the dimension of the Hilbert space $d=infty$. So $P_0=0$ as you say. However If could project $mathcal{X}(T)$ onto a finite dimensional space $V$ I could get a finite result. However $V$ should be invariant under $H$... which again is impossible.
$endgroup$
– lcv
Feb 1 at 1:34