Spectral theorem in QM [Archive]

Kasper J. Larsen

Oct4-04, 12:22 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Hello,\n\n\nCould someone please explain how the following version of the spectral\ntheorem for unbounded operators\n\n"If A is self-adjoint on H then there exists a spectral measure E on\nsigma(A) such that A = int z dE"\n\nimplies the statement below\n\n"If A is self-adjoint on H then there exists an orthonormal basis of H\nconsisting of eigenvectors for A"\n\nwhich is assumed in Sakurai. Even though the case of the\nquantum-mechanical position operator is an explicit counterexample to\nthe latter statement Sakurai considers distributions as eigenvectors\nwhich I find somewhat unsatisfactory.\n\n\nMy questions are as below:\n\n1) How does the first statement of the spectral theorem imply the\nsecond? Can one, e.g., prove that under certain assumptions on the\nspectrum of A ("discreteness"?), A will satisfy the second version of\nthe spectral theorem?\n\n2) Can one combine some of the theory of distributions with the first\nversion of the spectral theorem to make rigorous sense out of the\nnotion of distributions as a kind of eigenvectors?\n\nOr are there other ways of remedying the horrible mathematical MESS in\nSakurai?\n\n\nAny answer or reference to bibliography much appreciated!\n\n\nBest regards,\nKasper J. Larsen\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Hello,

Could someone please explain how the following version of the spectral
theorem for unbounded operators

"If A is self-adjoint on H then there exists a spectral measure E on
\sigma(A) such that A = \int z dE"

implies the statement below

"If A is self-adjoint on H then there exists an orthonormal basis of H
consisting of eigenvectors for A"

which is assumed in Sakurai. Even though the case of the
quantum-mechanical position operator is an explicit counterexample to
the latter statement Sakurai considers distributions as eigenvectors
which I find somewhat unsatisfactory.

My questions are as below:

1) How does the first statement of the spectral theorem imply the
second? Can one, e.g., prove that under certain assumptions on the
spectrum of A ("discreteness"?), A will satisfy the second version of
the spectral theorem?

2) Can one combine some of the theory of distributions with the first
version of the spectral theorem to make rigorous sense out of the
notion of distributions as a kind of eigenvectors?

Or are there other ways of remedying the horrible mathematical MESS in
Sakurai?

Any answer or reference to bibliography much appreciated!

Best regards,
Kasper J. Larsen

Arnold Neumaier

Oct5-04, 06:44 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Kasper J. Larsen wrote:\n>\n> Could someone please explain how the following version of the spectral\n> theorem for unbounded operators\n>\n> "If A is self-adjoint on H then there exists a spectral measure E on\n> sigma(A) such that A = int z dE"\n>\n> implies the statement below\n>\n> "If A is self-adjoint on H then there exists an orthonormal basis of H\n> consisting of eigenvectors for A"\n\nThis is true iff, in addition, A has a discrete spectrum.\nIn this case, the projection valued measure is discrete, and the\nprojections resulting just project to the eigenspaces.\n\n\n> 2) Can one combine some of the theory of distributions with the first\n> version of the spectral theorem to make rigorous sense out of the\n> notion of distributions as a kind of eigenvectors?\n\nYes, using Gelfand triples. The eigenvectors corresponding to the\ncontinuous spectrum belong to the dual of the nuclear space in the triple.\nAn introduction to Gelfand triples from a physcicists point of view can be\nfound, e.g., in\nV.I. Kukulin, V.M. Krasnopol\'sky and I. Hor\\\'a\\v cek,\nTheory of Resonances. Principles and Applications,\nKluwer, Dordrecht 1989.\n\n\nArnold Neumaier\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Kasper J. Larsen wrote:
>
> Could someone please explain how the following version of the spectral
> theorem for unbounded operators
>
> "If A is self-adjoint on H then there exists a spectral measure E on
> \sigma(A) such that A = \int z dE"
>
> implies the statement below
>
> "If A is self-adjoint on H then there exists an orthonormal basis of H
> consisting of eigenvectors for A"

This is true iff, in addition, A has a discrete spectrum.
In this case, the projection valued measure is discrete, and the
projections resulting just project to the eigenspaces.

> 2) Can one combine some of the theory of distributions with the first
> version of the spectral theorem to make rigorous sense out of the
> notion of distributions as a kind of eigenvectors?

Yes, using Gelfand triples. The eigenvectors corresponding to the
continuous spectrum belong to the dual of the nuclear space in the triple.
An introduction to Gelfand triples from a physcicists point of view can be
found, e.g., in
V.I. Kukulin, V.M. Krasnopol'sky and I. Hor\'a\v cek,
Theory of Resonances. Principles and Applications,
Kluwer, Dordrecht 1989.

Arnold Neumaier

Ilja Schmelzer

Oct5-04, 06:59 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>"Kasper J. Larsen" <kjlarsen@hotmail.com> schrieb\n> Could someone please explain how the following version of the spectral\n> theorem for unbounded operators\n>\n> "If A is self-adjoint on H then there exists a spectral measure E on\n> sigma(A) such that A = int z dE"\n>\n> implies the statement below\n>\n> "If A is self-adjoint on H then there exists an orthonormal basis of H\n> consisting of eigenvectors for A"\n>\n> which is assumed in Sakurai. Even though the case of the\n> quantum-mechanical position operator is an explicit counterexample to\n> the latter statement Sakurai considers distributions as eigenvectors\n> which I find somewhat unsatisfactory.\n>\n> My questions are as below:\n>\n> 1) How does the first statement of the spectral theorem imply the\n> second? Can one, e.g., prove that under certain assumptions on the\n> spectrum of A ("discreteness"?), A will satisfy the second version of\n> the spectral theorem?\n\nYes. The connection is the following: Assume the spectrum is discrete,\nthat means you have for an eigenvalue z_0 some environment\n[z_0-e, z_0+e] which does not contain any other parts of the\nspectrum, then\n\nint_{z_0-e}^{z_0+e} dE\n\ndefines the projector on the eigenspace.\n\n> 2) Can one combine some of the theory of distributions with the first\n> version of the spectral theorem to make rigorous sense out of the\n> notion of distributions as a kind of eigenvectors?\n\nThis spectral theorem is the way to make rigorous sense out of\nthe notion of distributions as a kind of eigenvectors.\n\nUsing it, you can define arbitrary functions of the operator A\n\nf(A) = int f(z) dE\n\n(with the special case A = int z dE) which is what you can do\nalso with a decomposition into discrete eigenspaces:\n\nf(A) = sum f(z) |psi_z><psi_z|\n\nwhich works for A with a discrete spectrum\n\nA |psi_z> = z |psi_z>\n\nA distribution dmu = delta(x-x_0)dx is a linear functional:\n\nf(.) -> int f(x) dmu in IR\n\nIts generalization is the operator-values functional\n\nf(.) -> int f(z) dE in L(H,H)\n\nAs in distribution theory you have to do all the things you\nwant to do in terms of such integrals, you have to do all\nthings you want to do with decompositions into eigenspaces\nin terms of such integrals int f(z) dE.\n\nIlja\n\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>"Kasper J. Larsen" <kjlarsen@hotmail.com> schrieb
> Could someone please explain how the following version of the spectral
> theorem for unbounded operators
>
> "If A is self-adjoint on H then there exists a spectral measure E on
> \sigma(A) such that A = \int z dE"
>
> implies the statement below
>
> "If A is self-adjoint on H then there exists an orthonormal basis of H
> consisting of eigenvectors for A"
>
> which is assumed in Sakurai. Even though the case of the
> quantum-mechanical position operator is an explicit counterexample to
> the latter statement Sakurai considers distributions as eigenvectors
> which I find somewhat unsatisfactory.
>
> My questions are as below:
>
> 1) How does the first statement of the spectral theorem imply the
> second? Can one, e.g., prove that under certain assumptions on the
> spectrum of A ("discreteness"?), A will satisfy the second version of
> the spectral theorem?

Yes. The connection is the following: Assume the spectrum is discrete,
that means you have for an eigenvalue z_0 some environment
[z_0-e, z_0+e] which does not contain any other parts of the
spectrum, then

\int_{z_0-e}^{z_0+e} dE

defines the projector on the eigenspace.

> 2) Can one combine some of the theory of distributions with the first
> version of the spectral theorem to make rigorous sense out of the
> notion of distributions as a kind of eigenvectors?

This spectral theorem is the way to make rigorous sense out of
the notion of distributions as a kind of eigenvectors.

Using it, you can define arbitrary functions of the operator A

f(A) = \int f(z) dE

(with the special case A = \int z dE) which is what you can do
also with a decomposition into discrete eigenspaces:

f(A) = sum f(z) |\psi_z><\psi_z|

which works for A with a discrete spectrum

A |\psi_z> = z |\psi_z>

A distribution dmu = \delta(x-x_0)dx is a linear functional:

f(.) -> \int f(x) dmu in IR

Its generalization is the operator-values functional

f(.) -> \int f(z) dE in L(H,H)

As in distribution theory you have to do all the things you
want to do in terms of such integrals, you have to do all
things you want to do with decompositions into eigenspaces
in terms of such integrals \int f(z) dE.

Ilja

grubb@math.niu.edu

Oct6-04, 08:03 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Kasper J. Larsen <kjlarsen@hotmail.com> wrote in message news:<cjs0s2\\$svi\\$1@lfa222122.richmond.edu>...\ n> Hello,\n>\n>\n> Could someone please explain how the following version of the spectral\n> theorem for unbounded operators\n>\n> "If A is self-adjoint on H then there exists a spectral measure E on\n> sigma(A) such that A = int z dE"\n>\n> implies the statement below\n>\n> "If A is self-adjoint on H then there exists an orthonormal basis of H\n> consisting of eigenvectors for A"\n>\n> which is assumed in Sakurai. Even though the case of the\n> quantum-mechanical position operator is an explicit counterexample to\n> the latter statement Sakurai considers distributions as eigenvectors\n> which I find somewhat unsatisfactory.\n>\n>\n> My questions are as below:\n>\n> 1) How does the first statement of the spectral theorem imply the\n> second? Can one, e.g., prove that under certain assumptions on the\n> spectrum of A ("discreteness"?), A will satisfy the second version of\n> the spectral theorem?\n\nWell, if sigma(A) is discrete, we can consider E({x}) for x in sigma(A).\nThis is a closed subspace of H since E is a spectral measure. If you take\nan orthonormal basis in this subspace for each x, then, since E is\na resolution of the identity, i.e. E(R)=I, H is a direct sum of these\nsubspaces, so we get an orthonormal basis for H. Clearly, the\nvectors in E({x}) are eigenvectors of A with eigenvalue x.\n\nThis argument works even if sigma(A) is not exactly discrete, but has,\nsay, only finitely many limit points. Just deal with the limit points\nseparately.\n\n\n> 2) Can one combine some of the theory of distributions with the first\n> version of the spectral theorem to make rigorous sense out of the\n> notion of distributions as a kind of eigenvectors?\n\nThere are difficulties here even in the case of bounded operators as you\nhave pointed out (the position operator on L_2 ([0,1]) is bounded). At\nthe very least, you have to know that the test functions are in your\nHilbert space and that your operator is defined on them. Then you would\nneed to get the distributions as weak limits of elements of the Hilbert\nspace. By looking at the possible limits from vectors in E([x-eps,x+eps]),\nas eps->0, you should be able to select the \'eigenvectors which are\ndistributions with eigenvalue x\'. I have not seen details, though. Sorry.\n\n> Or are there other ways of remedying the horrible mathematical MESS in\n> Sakurai?\n>\n>\n> Any answer or reference to bibliography much appreciated!\n\nI\'d like to see them also :).\n\n>\n>\n> Best regards,\n> Kasper J. Larsen\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Kasper J. Larsen <kjlarsen@hotmail.com> wrote in message news:<cjs0s2$svi$1@lfa222122.richmond.edu>...
> Hello,
>
>
> Could someone please explain how the following version of the spectral
> theorem for unbounded operators
>
> "If A is self-adjoint on H then there exists a spectral measure E on
> \sigma(A) such that A = \int z dE"
>
> implies the statement below
>
> "If A is self-adjoint on H then there exists an orthonormal basis of H
> consisting of eigenvectors for A"
>
> which is assumed in Sakurai. Even though the case of the
> quantum-mechanical position operator is an explicit counterexample to
> the latter statement Sakurai considers distributions as eigenvectors
> which I find somewhat unsatisfactory.
>
>
> My questions are as below:
>
> 1) How does the first statement of the spectral theorem imply the
> second? Can one, e.g., prove that under certain assumptions on the
> spectrum of A ("discreteness"?), A will satisfy the second version of
> the spectral theorem?

Well, if \sigma(A) is discrete, we can consider E({x}) for x in \sigma(A).
This is a closed subspace of H since E is a spectral measure. If you take
an orthonormal basis in this subspace for each x, then, since E is
a resolution of the identity, i.e. E(R)=I, H is a direct sum of these
subspaces, so we get an orthonormal basis for H. Clearly, the
vectors in E({x}) are eigenvectors of A with eigenvalue x.

This argument works even if \sigma(A) is not exactly discrete, but has,
say, only finitely many limit points. Just deal with the limit points
separately.

> 2) Can one combine some of the theory of distributions with the first
> version of the spectral theorem to make rigorous sense out of the
> notion of distributions as a kind of eigenvectors?

There are difficulties here even in the case of bounded operators as you
have pointed out (the position operator on L_2 ([0,1]) is bounded). At
the very least, you have to know that the test functions are in your
Hilbert space and that your operator is defined on them. Then you would
need to get the distributions as weak limits of elements of the Hilbert
space. By looking at the possible limits from vectors in E([x-eps,x+eps]),
as eps->0, you should be able to select the 'eigenvectors which are
distributions with eigenvalue x'. I have not seen details, though. Sorry.

> Or are there other ways of remedying the horrible mathematical MESS in
> Sakurai?
>
>
> Any answer or reference to bibliography much appreciated!

I'd like to see them also :).

>
>
> Best regards,
> Kasper J. Larsen