Simulating interaction of two systems [Archive]

Mike Stay

May31-04, 04:22 PM

Igor Khavkine

Jun1-04, 02:11 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>On Mon, 31 May 2004 21:22:47 +0000, Mike Stay wrote:\n\n> Igor Khavkine wrote in message\n> news:<f1ac2e6e.0405250810.2bf939be@posting.google. com>...\n>> "M. Stay" <staym@clear.net.nz> wrote in message\n>> news:<1224802492.20040525130116@clear.net.nz>...\n >> > I\'m trying to figure out how to simulate the evolution of two systems\n>> > of qubits, originally separate, but now brought together. I can get\n>> > the Hamiltonian for the combined system without interactions:\n>> > H_1 (X) I + I (X) H_2 (where I is the suitable identity)\n>> > but any interactions between them need to conserve energy, and I\'m not\n>> > sure how to do that. It seems like the only allowed things would be\n>> > to do something like rotate the basis between energy states with the\n>> > same energy, but that seems to ignore the difficulty of distinguishing\n>> > states that are nearly degenerate.\n>>\n>> As long as the Hamiltonian is time independent, energy conservation takes\n>> care of itself.\n>\n> Yes; I wasn\'t very clear. I meant that I wanted the eigenvalues to\n> remain the same between H and H+H_int. It\'s straightforward to pick\n> interaction terms that preserve the eigenvalues, but I couldn\'t think\n> of any reason to prefer those to interaction terms that don\'t.\n\nAs a general rule, if you add a perturbation to the Hamiltonian, your\nenergy levels will shift. I would imagine that it is very difficult to\nfind a perturbation that leaves the energy levels intact. I think so for\nthe very simple reason that the characteristic polynomial of a linear\noperator does not transform in a simple way when the operator is perturbed.\nOr in simpler terms, the determinant det(A+B) is not a simple expression\nin terms of A and B.\n\nThe only sure way to get what you seem to want is to make H_int depend on\nn parameters (where n is the dimension of H_int), diagonalize the\nHamiltonian H+H_int, vary the parameters to fix these eigenvalues to be\nthe same as eigenvalues of H. But I don\'t think you want to do that.\n\nPhysically, one would expect any interaction to have an effect on the\nspacing of energy levels of a quantum system. In fact, much of what we\nknow about the internal interactions in complex quantum systems such as\nmolecules, atoms, nuclei, etc. are deduced from the deviations of observed\nenergy levels from what would be expected from a non-interacting system.\n\n>> Adding an interaction is not very complicated. You just\n>> have to think of two states |a> and |b> between which a transition makes\n>> sense, then add c(|a><b| + |b><a|) to the Hamiltonian. For example,\n>> |a> could be an excited hydrogen atom in an electromagnetic vacuum, and\n>> |b> could be the hydrogen atom in its ground state and one emitted photon.\n>> Or if you have two spins interacting, |a> could be |+-> and |b> could\n>> be |-+>, this a spin flip.\n>\n> So I could add a third subsystem that represents an intermediate state\n> with the photon in the field rather than in either of the original\n> subsystems, and couple anything that wants to emit or absorb photons to\n> that third one. That makes sense.\n>\n>> In general, the form of interaction depends on the physical nature of\n>> the system. For an abstract system of qubits, it is rather hard to\n>> restrict the possibilities.\n\nYes. By introducing photons, you are supplying more information about the\nphysics of your system. As I suggested above, this gives you a physical\nbasis for choosing an interaction.\n\nIgor\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>On Mon, 31 May 2004 21:22:47 +0000, Mike Stay wrote:

> Igor Khavkine wrote in message
> news:<f1ac2e6e.0405250810.2bf939be@posting.google.com>...
>> "M. Stay" <staym@clear.net.nz> wrote in message
>> news:<1224802492.20040525130116@clear.net.nz>...
>> > I'm trying to figure out how to simulate the evolution of two systems
>> > of qubits, originally separate, but now brought together. I can get
>> > the Hamiltonian for the combined system without interactions:
>> > H_1 (X) I + I (X) H_2 (where I is the suitable identity)
>> > but any interactions between them need to conserve energy, and I'm not
>> > sure how to do that. It seems like the only allowed things would be
>> > to do something like rotate the basis between energy states with the
>> > same energy, but that seems to ignore the difficulty of distinguishing
>> > states that are nearly degenerate.
>>
>> As long as the Hamiltonian is time independent, energy conservation takes
>> care of itself.
>
> Yes; I wasn't very clear. I meant that I wanted the eigenvalues to
> remain the same between H and H+H_{int}. It's straightforward to pick
> interaction terms that preserve the eigenvalues, but I couldn't think
> of any reason to prefer those to interaction terms that don't.

As a general rule, if you add a perturbation to the Hamiltonian, your
energy levels will shift. I would imagine that it is very difficult to
find a perturbation that leaves the energy levels intact. I think so for
the very simple reason that the characteristic polynomial of a linear
operator does not transform in a simple way when the operator is perturbed.
Or in simpler terms, the determinant det(A+B) is not a simple expression
in terms of A and B.

The only sure way to get what you seem to want is to make H_{int} depend on
n parameters (where n is the dimension of H_{int}), diagonalize the
Hamiltonian H+H_{int}, vary the parameters to fix these eigenvalues to be
the same as eigenvalues of H. But I don't think you want to do that.

Physically, one would expect any interaction to have an effect on the
spacing of energy levels of a quantum system. In fact, much of what we
know about the internal interactions in complex quantum systems such as
molecules, atoms, nuclei, etc. are deduced from the deviations of observed
energy levels from what would be expected from a non-interacting system.

>> Adding an interaction is not very complicated. You just
>> have to think of two states |a> and |b> between which a transition makes
>> sense, then add c(|a><b| + |b><a|) to the Hamiltonian. For example,
>> |a> could be an excited hydrogen atom in an electromagnetic vacuum, and
>> |b> could be the hydrogen atom in its ground state and one emitted photon.
>> Or if you have two spins interacting, |a> could be |+-> and |b> could
>> be |-+>, this a spin flip.
>
> So I could add a third subsystem that represents an intermediate state
> with the photon in the field rather than in either of the original
> subsystems, and couple anything that wants to emit or absorb photons to
> that third one. That makes sense.
>
>> In general, the form of interaction depends on the physical nature of
>> the system. For an abstract system of qubits, it is rather hard to
>> restrict the possibilities.

Yes. By introducing photons, you are supplying more information about the
physics of your system. As I suggested above, this gives you a physical
basis for choosing an interaction.

Igor