Generalised eigenvalues in the residual spectrum The Next CEO of Stack OverflowResidual spectrum is emptySpectrum of the unbounded operator $ipartial_x$Spectrum proofsSpectrum of unbounded Operators + Spectral TheoremComplete ONS and pure point spectrumThe point spectrum and residual spectrum of an operator on $l_2$ related to backward shiftMeaning of the continuous spectrum and the residual spectrumResidual Spectrum of $(Tf)(x) = int^x_0 f(y) dy + x f(x)$Meaning of terminologies “regular value”, “point spectrum”, “continous spectrum” and “residual spectrum”Residual spectrum of a Hermitian operator
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Generalised eigenvalues in the residual spectrum
The Next CEO of Stack OverflowResidual spectrum is emptySpectrum of the unbounded operator $ipartial_x$Spectrum proofsSpectrum of unbounded Operators + Spectral TheoremComplete ONS and pure point spectrumThe point spectrum and residual spectrum of an operator on $l_2$ related to backward shiftMeaning of the continuous spectrum and the residual spectrumResidual Spectrum of $(Tf)(x) = int^x_0 f(y) dy + x f(x)$Meaning of terminologies “regular value”, “point spectrum”, “continous spectrum” and “residual spectrum”Residual spectrum of a Hermitian operator
$begingroup$
For an operator $T$ on a Hilbert space the residual spectrum is defined as the set of complex numbers $lambda$ for which $(T-lambda I)^-1$ exists but is not densely defined. The question is: could there be a complex number in the residual spectrum which is a generalised eigenvalue? By generalised eigenvalue I mean that there exists a sequence $x_n$ of unit vectors that satisfy:
$$
(T-lambda I)x_n longrightarrow 0.
$$
I was thinking about this from a physics point of view (discarding the fact that observables are assumed to be self adjoint) and wondering if one wanted to make an "observable" out of something which is hermitian but doesn't have an empty residual spectrum (like momentum on $L^2(0,infty))$ would the problem be that for such operators the system could in principle collapse to some state with a complex spectral value associated to it? This spectral value would have to be in the residual spectrum because the continuous and point spectra of hermitian operators must be real.
functional-analysis hilbert-spaces physics
asked Mar 27 at 18:27
Michele GalliMichele Galli
1759
$endgroup$
add a comment |
$begingroup$
For an operator $T$ on a Hilbert space the residual spectrum is defined as the set of complex numbers $lambda$ for which $(T-lambda I)^-1$ exists but is not densely defined. The question is: could there be a complex number in the residual spectrum which is a generalised eigenvalue? By generalised eigenvalue I mean that there exists a sequence $x_n$ of unit vectors that satisfy:
$$
(T-lambda I)x_n longrightarrow 0.
$$
I was thinking about this from a physics point of view (discarding the fact that observables are assumed to be self adjoint) and wondering if one wanted to make an "observable" out of something which is hermitian but doesn't have an empty residual spectrum (like momentum on $L^2(0,infty))$ would the problem be that for such operators the system could in principle collapse to some state with a complex spectral value associated to it? This spectral value would have to be in the residual spectrum because the continuous and point spectra of hermitian operators must be real.
functional-analysis hilbert-spaces physics
asked Mar 27 at 18:27
Michele GalliMichele Galli
1759
$endgroup$
$begingroup$
Note that self-adjoint operators never have resudual spectrum. And that what you call a generalized eigenvalue is usually known as an approximate eigenvalue.
$endgroup$
– amsmath
Mar 27 at 18:45
$begingroup$
yes indeed, that’s precisely the source of my question, i was just trying to get a picture of why in quantum mechanics you insist that operators are self adjoint and thus have empty residual spectrum, and i was thinking that perhaps this is because if the residual spectrum is not empty, you could have complex spectral values, which is unphysical.
$endgroup$
– Michele Galli
Mar 27 at 18:58
$begingroup$
What you want is an operator $T$ in a Hilbert space $mathcal H$ (may be bounded) such that $ker T = 0$ and $operatornameranT$ is neither closed nor dense in $mathcal H$. I am pretty sure such operators exist, but I cannot give you an example right now.
$endgroup$
– amsmath
Mar 27 at 19:25
$begingroup$
Thanks! I was just trying to figure out why insisting on the empty residual spectrum is physically necessary (which is why you impose self adjoint)
$endgroup$
– Michele Galli
Mar 27 at 19:27
add a comment |
$begingroup$
For an operator $T$ on a Hilbert space the residual spectrum is defined as the set of complex numbers $lambda$ for which $(T-lambda I)^-1$ exists but is not densely defined. The question is: could there be a complex number in the residual spectrum which is a generalised eigenvalue? By generalised eigenvalue I mean that there exists a sequence $x_n$ of unit vectors that satisfy:
$$
(T-lambda I)x_n longrightarrow 0.
$$
I was thinking about this from a physics point of view (discarding the fact that observables are assumed to be self adjoint) and wondering if one wanted to make an "observable" out of something which is hermitian but doesn't have an empty residual spectrum (like momentum on $L^2(0,infty))$ would the problem be that for such operators the system could in principle collapse to some state with a complex spectral value associated to it? This spectral value would have to be in the residual spectrum because the continuous and point spectra of hermitian operators must be real.
functional-analysis hilbert-spaces physics
asked Mar 27 at 18:27
Michele GalliMichele Galli
1759
$endgroup$
For an operator $T$ on a Hilbert space the residual spectrum is defined as the set of complex numbers $lambda$ for which $(T-lambda I)^-1$ exists but is not densely defined. The question is: could there be a complex number in the residual spectrum which is a generalised eigenvalue? By generalised eigenvalue I mean that there exists a sequence $x_n$ of unit vectors that satisfy:
$$
(T-lambda I)x_n longrightarrow 0.
$$
I was thinking about this from a physics point of view (discarding the fact that observables are assumed to be self adjoint) and wondering if one wanted to make an "observable" out of something which is hermitian but doesn't have an empty residual spectrum (like momentum on $L^2(0,infty))$ would the problem be that for such operators the system could in principle collapse to some state with a complex spectral value associated to it? This spectral value would have to be in the residual spectrum because the continuous and point spectra of hermitian operators must be real.
functional-analysis hilbert-spaces physics
functional-analysis hilbert-spaces physics
asked Mar 27 at 18:27
Michele GalliMichele Galli
1759
asked Mar 27 at 18:27
Michele GalliMichele Galli
1759
asked Mar 27 at 18:27
Michele GalliMichele Galli
1759
asked Mar 27 at 18:27
Michele GalliMichele Galli
1759
asked Mar 27 at 18:27
Michele GalliMichele Galli
1759
1759
$begingroup$
Note that self-adjoint operators never have resudual spectrum. And that what you call a generalized eigenvalue is usually known as an approximate eigenvalue.
$endgroup$
– amsmath
Mar 27 at 18:45
$begingroup$
yes indeed, that’s precisely the source of my question, i was just trying to get a picture of why in quantum mechanics you insist that operators are self adjoint and thus have empty residual spectrum, and i was thinking that perhaps this is because if the residual spectrum is not empty, you could have complex spectral values, which is unphysical.
$endgroup$
– Michele Galli
Mar 27 at 18:58
$begingroup$
What you want is an operator $T$ in a Hilbert space $mathcal H$ (may be bounded) such that $ker T = 0$ and $operatornameranT$ is neither closed nor dense in $mathcal H$. I am pretty sure such operators exist, but I cannot give you an example right now.
$endgroup$
– amsmath
Mar 27 at 19:25
$begingroup$
Thanks! I was just trying to figure out why insisting on the empty residual spectrum is physically necessary (which is why you impose self adjoint)
$endgroup$
– Michele Galli
Mar 27 at 19:27
add a comment |
$begingroup$
Note that self-adjoint operators never have resudual spectrum. And that what you call a generalized eigenvalue is usually known as an approximate eigenvalue.
$endgroup$
– amsmath
Mar 27 at 18:45
$begingroup$
yes indeed, that’s precisely the source of my question, i was just trying to get a picture of why in quantum mechanics you insist that operators are self adjoint and thus have empty residual spectrum, and i was thinking that perhaps this is because if the residual spectrum is not empty, you could have complex spectral values, which is unphysical.
$endgroup$
– Michele Galli
Mar 27 at 18:58
$begingroup$
What you want is an operator $T$ in a Hilbert space $mathcal H$ (may be bounded) such that $ker T = 0$ and $operatornameranT$ is neither closed nor dense in $mathcal H$. I am pretty sure such operators exist, but I cannot give you an example right now.
$endgroup$
– amsmath
Mar 27 at 19:25
$begingroup$
Thanks! I was just trying to figure out why insisting on the empty residual spectrum is physically necessary (which is why you impose self adjoint)
$endgroup$
– Michele Galli
Mar 27 at 19:27
$begingroup$
Note that self-adjoint operators never have resudual spectrum. And that what you call a generalized eigenvalue is usually known as an approximate eigenvalue.
$endgroup$
– amsmath
Mar 27 at 18:45
$begingroup$
Note that self-adjoint operators never have resudual spectrum. And that what you call a generalized eigenvalue is usually known as an approximate eigenvalue.
$endgroup$
– amsmath
Mar 27 at 18:45
$begingroup$
yes indeed, that’s precisely the source of my question, i was just trying to get a picture of why in quantum mechanics you insist that operators are self adjoint and thus have empty residual spectrum, and i was thinking that perhaps this is because if the residual spectrum is not empty, you could have complex spectral values, which is unphysical.
$endgroup$
– Michele Galli
Mar 27 at 18:58
$begingroup$
yes indeed, that’s precisely the source of my question, i was just trying to get a picture of why in quantum mechanics you insist that operators are self adjoint and thus have empty residual spectrum, and i was thinking that perhaps this is because if the residual spectrum is not empty, you could have complex spectral values, which is unphysical.
$endgroup$
– Michele Galli
Mar 27 at 18:58
$begingroup$
What you want is an operator $T$ in a Hilbert space $mathcal H$ (may be bounded) such that $ker T = 0$ and $operatornameranT$ is neither closed nor dense in $mathcal H$. I am pretty sure such operators exist, but I cannot give you an example right now.
$endgroup$
– amsmath
Mar 27 at 19:25
$begingroup$
What you want is an operator $T$ in a Hilbert space $mathcal H$ (may be bounded) such that $ker T = 0$ and $operatornameranT$ is neither closed nor dense in $mathcal H$. I am pretty sure such operators exist, but I cannot give you an example right now.
$endgroup$
– amsmath
Mar 27 at 19:25
$begingroup$
Thanks! I was just trying to figure out why insisting on the empty residual spectrum is physically necessary (which is why you impose self adjoint)
$endgroup$
– Michele Galli
Mar 27 at 19:27
$begingroup$
Thanks! I was just trying to figure out why insisting on the empty residual spectrum is physically necessary (which is why you impose self adjoint)
$endgroup$
– Michele Galli
Mar 27 at 19:27
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let $S : mathcal Htomathcal H$ be any linear operator with non-closed range and $ker S = 0$ (for example $Sf = tcdot f$ on $L^2(0,1)$ satisfies this). Now, let $M$ be any closed, proper infinite-dimensional subspace of $mathcal H$ and let $V : overlineoperatornameranSto M$ be unitary. Then $T := VS$ satisfies the requirements.
answered Mar 27 at 19:39
amsmathamsmath
3,278419
$endgroup$
add a comment |
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$begingroup$
Let $S : mathcal Htomathcal H$ be any linear operator with non-closed range and $ker S = 0$ (for example $Sf = tcdot f$ on $L^2(0,1)$ satisfies this). Now, let $M$ be any closed, proper infinite-dimensional subspace of $mathcal H$ and let $V : overlineoperatornameranSto M$ be unitary. Then $T := VS$ satisfies the requirements.
answered Mar 27 at 19:39
amsmathamsmath
3,278419
$endgroup$
add a comment |
$begingroup$
Let $S : mathcal Htomathcal H$ be any linear operator with non-closed range and $ker S = 0$ (for example $Sf = tcdot f$ on $L^2(0,1)$ satisfies this). Now, let $M$ be any closed, proper infinite-dimensional subspace of $mathcal H$ and let $V : overlineoperatornameranSto M$ be unitary. Then $T := VS$ satisfies the requirements.
answered Mar 27 at 19:39
amsmathamsmath
3,278419
$endgroup$
add a comment |
$begingroup$
Let $S : mathcal Htomathcal H$ be any linear operator with non-closed range and $ker S = 0$ (for example $Sf = tcdot f$ on $L^2(0,1)$ satisfies this). Now, let $M$ be any closed, proper infinite-dimensional subspace of $mathcal H$ and let $V : overlineoperatornameranSto M$ be unitary. Then $T := VS$ satisfies the requirements.
answered Mar 27 at 19:39
amsmathamsmath
3,278419
$endgroup$
Let $S : mathcal Htomathcal H$ be any linear operator with non-closed range and $ker S = 0$ (for example $Sf = tcdot f$ on $L^2(0,1)$ satisfies this). Now, let $M$ be any closed, proper infinite-dimensional subspace of $mathcal H$ and let $V : overlineoperatornameranSto M$ be unitary. Then $T := VS$ satisfies the requirements.
answered Mar 27 at 19:39
amsmathamsmath
3,278419
answered Mar 27 at 19:39
amsmathamsmath
3,278419
answered Mar 27 at 19:39
amsmathamsmath
3,278419
answered Mar 27 at 19:39
amsmathamsmath
3,278419
3,278419
add a comment |
add a comment |
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$begingroup$
Note that self-adjoint operators never have resudual spectrum. And that what you call a generalized eigenvalue is usually known as an approximate eigenvalue.
$endgroup$
– amsmath
Mar 27 at 18:45
$begingroup$
yes indeed, that’s precisely the source of my question, i was just trying to get a picture of why in quantum mechanics you insist that operators are self adjoint and thus have empty residual spectrum, and i was thinking that perhaps this is because if the residual spectrum is not empty, you could have complex spectral values, which is unphysical.
$endgroup$
– Michele Galli
Mar 27 at 18:58
$begingroup$
What you want is an operator $T$ in a Hilbert space $mathcal H$ (may be bounded) such that $ker T = 0$ and $operatornameranT$ is neither closed nor dense in $mathcal H$. I am pretty sure such operators exist, but I cannot give you an example right now.
$endgroup$
– amsmath
Mar 27 at 19:25
$begingroup$
Thanks! I was just trying to figure out why insisting on the empty residual spectrum is physically necessary (which is why you impose self adjoint)
$endgroup$
– Michele Galli
Mar 27 at 19:27