## Simulating interaction of two systems

<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>Igor Khavkine wrote in message\nnews:&lt;f1ac2e6e.0405250810.2bf939be@posting.google.com&gt;.. .\n&gt; "M. Stay" &lt;staym@clear.net.nz&gt; wrote in message\n&gt; news:&lt;1224802492.20040525130116@clear.net.nz&gt;...\n&gt; &gt; I\'m trying to figure out how to simulate the evolution of two systems\n&gt; &gt; of qubits, originally separate, but now brought together. I can get\n&gt; &gt; the Hamiltonian for the combined system without interactions:\n&gt; &gt; H_1 (X) I + I (X) H_2 (where I is the suitable identity)\n&gt; &gt; but any interactions between them need to conserve energy, and I\'m not\n&gt; &gt; sure how to do that. It seems like the only allowed things would be\n&gt; &gt; to do something like rotate the basis between energy states with the\n&gt; &gt; same energy, but that seems to ignore the difficulty of distinguishing\n&gt; &gt; states that are nearly degenerate.\n&gt;\n&gt; As long as the Hamiltonian is time independent, energy conservation takes\n&gt; care of itself.\n\nYes; I wasn\'t very clear. I meant that I wanted the eigenvalues to\nremain the same between H and H+H_int. It\'s straightforward to pick\ninteraction terms that preserve the eigenvalues, but I couldn\'t think\nof any reason to prefer those to interaction terms that don\'t.\n\n&gt; Adding an interaction is not very complicated. You just\n&gt; have to think of two states |a&gt; and |b&gt; between which a transition makes\n&gt; sense, then add c(|a&gt;&lt;b| + |b&gt;&lt;a|) to the Hamiltonian. For example,\n&gt; |a&gt; could be an excited hydrogen atom in an electromagnetic vacuum, and\n&gt; |b&gt; could be the hydrogen atom in its ground state and one emitted photon.\n&gt; Or if you have two spins interacting, |a&gt; could be |+-&gt; and be could be |-+&gt;,\n&gt; this a spin flip.\n\nSo I could add a third subsystem that represents an intermediate state\nwith the photon in the field rather than in either of the original\nsubsystems, and couple anything that wants to emit or absorb photons\nto that third one. That makes sense.\n\n&gt;\n&gt; In general, the form of interaction depends on the physical nature of\n&gt; the system. For an abstract system of qubits, it is rather hard to\n&gt; restrict the possibilities.\n&gt;\n&gt; Hope this helps.\n\nYes, thanks.\n\n&gt;\n&gt; Igor\n\nMike\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">&nbsp;&nbsp;View this Usenet post in original ASCII form </a></div><P></jabberwocky>Igor Khavkine wrote in message
> "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.

> 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 be 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.
>
> Hope this helps.

Yes, thanks.

>
> Igor

Mike

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On Mon, 31 May 2004 21:22:47 $+0000,$ Mike Stay wrote: > Igor Khavkine wrote in message > news:... >> "M. Stay" 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 $c(|a>> |a>$ 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