Conditions for Boundedness of Spectral Measures of Perturbations of Self-Adjoint Operators?
Suppose $A$ is an unbounded self-adjoint operator in a Hilbert space $H$ with discrete spectrum $$lambda_0 < lambda_1 < cdots$$ bounded below with lowest eigenvalue $lambda_0$, lowest eigenstate $psi_0$, and spectral gap $lambda_1 - lambda_0 > 0$.
Let $B$ be a bounded self-adjoint operator in $H$ with absolutely continuous spectrum.
Are there any conditions on $A, B$ which imply that the spectral measure $dmu$ of $O=A+B$ at $psi_0$ is bounded, i.e. $dmu$ compact support?
Recall that the spectral measure of a self-adjoint operator $O$ in a Hilbert space $(H, langle cdot, cdot rangle)$ at a normalized state $psi in H$, $||psi||=1$ is the probability measure $dmu$ on $mathbb{R}$ determined by $O$ and $psi$ implicitly by
$$int_{- infty}^{+infty} e^{textbf{i} t lambda} d mu(lambda) = langle psi , e^{textbf{i} t O} psi rangle $$ for all $t in mathbb{R}$.
operator-theory spectral-theory perturbation-theory self-adjoint-operators
asked Dec 10 '18 at 20:41
Swallow Tail
512
add a comment |
Suppose $A$ is an unbounded self-adjoint operator in a Hilbert space $H$ with discrete spectrum $$lambda_0 < lambda_1 < cdots$$ bounded below with lowest eigenvalue $lambda_0$, lowest eigenstate $psi_0$, and spectral gap $lambda_1 - lambda_0 > 0$.
Let $B$ be a bounded self-adjoint operator in $H$ with absolutely continuous spectrum.
Are there any conditions on $A, B$ which imply that the spectral measure $dmu$ of $O=A+B$ at $psi_0$ is bounded, i.e. $dmu$ compact support?
Recall that the spectral measure of a self-adjoint operator $O$ in a Hilbert space $(H, langle cdot, cdot rangle)$ at a normalized state $psi in H$, $||psi||=1$ is the probability measure $dmu$ on $mathbb{R}$ determined by $O$ and $psi$ implicitly by
$$int_{- infty}^{+infty} e^{textbf{i} t lambda} d mu(lambda) = langle psi , e^{textbf{i} t O} psi rangle $$ for all $t in mathbb{R}$.
operator-theory spectral-theory perturbation-theory self-adjoint-operators
asked Dec 10 '18 at 20:41
Swallow Tail
512
"the spectral measure $dmu$ of $O=A+B$ at $psi_0$ is bounded, i.e. $dmu$ compact support" Do you mean that $O$ does not have a (discrete) lowest eigenvector?
– Keith McClary
Dec 11 '18 at 4:01
No. My question very well could apply to O=A+B with discrete spectrum and lowest eigenvalue with a gap. Suppose this case applies, so O=A+B has discrete spectrum. A priori the spectral measure of O at $psi_0$ is supported on the discrete spectrum of O which is usually infinite. Then in this case I’m asking what assumptions guarantee this support is on finitely many eigenvalues of O.
– Swallow Tail
Dec 11 '18 at 19:21
I think I understand the question. I can't think of an example. Do you have any?
– Keith McClary
Dec 13 '18 at 17:59
add a comment |
Suppose $A$ is an unbounded self-adjoint operator in a Hilbert space $H$ with discrete spectrum $$lambda_0 < lambda_1 < cdots$$ bounded below with lowest eigenvalue $lambda_0$, lowest eigenstate $psi_0$, and spectral gap $lambda_1 - lambda_0 > 0$.
Let $B$ be a bounded self-adjoint operator in $H$ with absolutely continuous spectrum.
Are there any conditions on $A, B$ which imply that the spectral measure $dmu$ of $O=A+B$ at $psi_0$ is bounded, i.e. $dmu$ compact support?
Recall that the spectral measure of a self-adjoint operator $O$ in a Hilbert space $(H, langle cdot, cdot rangle)$ at a normalized state $psi in H$, $||psi||=1$ is the probability measure $dmu$ on $mathbb{R}$ determined by $O$ and $psi$ implicitly by
$$int_{- infty}^{+infty} e^{textbf{i} t lambda} d mu(lambda) = langle psi , e^{textbf{i} t O} psi rangle $$ for all $t in mathbb{R}$.
operator-theory spectral-theory perturbation-theory self-adjoint-operators
asked Dec 10 '18 at 20:41
Swallow Tail
512
Suppose $A$ is an unbounded self-adjoint operator in a Hilbert space $H$ with discrete spectrum $$lambda_0 < lambda_1 < cdots$$ bounded below with lowest eigenvalue $lambda_0$, lowest eigenstate $psi_0$, and spectral gap $lambda_1 - lambda_0 > 0$.
Let $B$ be a bounded self-adjoint operator in $H$ with absolutely continuous spectrum.
Are there any conditions on $A, B$ which imply that the spectral measure $dmu$ of $O=A+B$ at $psi_0$ is bounded, i.e. $dmu$ compact support?
Recall that the spectral measure of a self-adjoint operator $O$ in a Hilbert space $(H, langle cdot, cdot rangle)$ at a normalized state $psi in H$, $||psi||=1$ is the probability measure $dmu$ on $mathbb{R}$ determined by $O$ and $psi$ implicitly by
$$int_{- infty}^{+infty} e^{textbf{i} t lambda} d mu(lambda) = langle psi , e^{textbf{i} t O} psi rangle $$ for all $t in mathbb{R}$.
operator-theory spectral-theory perturbation-theory self-adjoint-operators
operator-theory spectral-theory perturbation-theory self-adjoint-operators
asked Dec 10 '18 at 20:41
Swallow Tail
512
asked Dec 10 '18 at 20:41
Swallow Tail
512
asked Dec 10 '18 at 20:41
Swallow Tail
512
asked Dec 10 '18 at 20:41
Swallow Tail
512
asked Dec 10 '18 at 20:41
Swallow Tail
512
512
"the spectral measure $dmu$ of $O=A+B$ at $psi_0$ is bounded, i.e. $dmu$ compact support" Do you mean that $O$ does not have a (discrete) lowest eigenvector?
– Keith McClary
Dec 11 '18 at 4:01
No. My question very well could apply to O=A+B with discrete spectrum and lowest eigenvalue with a gap. Suppose this case applies, so O=A+B has discrete spectrum. A priori the spectral measure of O at $psi_0$ is supported on the discrete spectrum of O which is usually infinite. Then in this case I’m asking what assumptions guarantee this support is on finitely many eigenvalues of O.
– Swallow Tail
Dec 11 '18 at 19:21
I think I understand the question. I can't think of an example. Do you have any?
– Keith McClary
Dec 13 '18 at 17:59
add a comment |
"the spectral measure $dmu$ of $O=A+B$ at $psi_0$ is bounded, i.e. $dmu$ compact support" Do you mean that $O$ does not have a (discrete) lowest eigenvector?
– Keith McClary
Dec 11 '18 at 4:01
No. My question very well could apply to O=A+B with discrete spectrum and lowest eigenvalue with a gap. Suppose this case applies, so O=A+B has discrete spectrum. A priori the spectral measure of O at $psi_0$ is supported on the discrete spectrum of O which is usually infinite. Then in this case I’m asking what assumptions guarantee this support is on finitely many eigenvalues of O.
– Swallow Tail
Dec 11 '18 at 19:21
I think I understand the question. I can't think of an example. Do you have any?
– Keith McClary
Dec 13 '18 at 17:59
"the spectral measure $dmu$ of $O=A+B$ at $psi_0$ is bounded, i.e. $dmu$ compact support" Do you mean that $O$ does not have a (discrete) lowest eigenvector?
– Keith McClary
Dec 11 '18 at 4:01
"the spectral measure $dmu$ of $O=A+B$ at $psi_0$ is bounded, i.e. $dmu$ compact support" Do you mean that $O$ does not have a (discrete) lowest eigenvector?
– Keith McClary
Dec 11 '18 at 4:01
No. My question very well could apply to O=A+B with discrete spectrum and lowest eigenvalue with a gap. Suppose this case applies, so O=A+B has discrete spectrum. A priori the spectral measure of O at $psi_0$ is supported on the discrete spectrum of O which is usually infinite. Then in this case I’m asking what assumptions guarantee this support is on finitely many eigenvalues of O.
– Swallow Tail
Dec 11 '18 at 19:21
No. My question very well could apply to O=A+B with discrete spectrum and lowest eigenvalue with a gap. Suppose this case applies, so O=A+B has discrete spectrum. A priori the spectral measure of O at $psi_0$ is supported on the discrete spectrum of O which is usually infinite. Then in this case I’m asking what assumptions guarantee this support is on finitely many eigenvalues of O.
– Swallow Tail
Dec 11 '18 at 19:21
I think I understand the question. I can't think of an example. Do you have any?
– Keith McClary
Dec 13 '18 at 17:59
I think I understand the question. I can't think of an example. Do you have any?
– Keith McClary
Dec 13 '18 at 17:59
add a comment |
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"the spectral measure $dmu$ of $O=A+B$ at $psi_0$ is bounded, i.e. $dmu$ compact support" Do you mean that $O$ does not have a (discrete) lowest eigenvector?
– Keith McClary
Dec 11 '18 at 4:01
No. My question very well could apply to O=A+B with discrete spectrum and lowest eigenvalue with a gap. Suppose this case applies, so O=A+B has discrete spectrum. A priori the spectral measure of O at $psi_0$ is supported on the discrete spectrum of O which is usually infinite. Then in this case I’m asking what assumptions guarantee this support is on finitely many eigenvalues of O.
– Swallow Tail
Dec 11 '18 at 19:21
I think I understand the question. I can't think of an example. Do you have any?
– Keith McClary
Dec 13 '18 at 17:59