## Re: Intensity of a Photon

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no,location=no, scrollbars=yes,resize=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>\n"Curious" <curious11112001@yahoo.com> wrote in message\nnews:34a4f456.0404011123.63ca6994@posting.google.com...\n> What determines it?\n>\n> What is it measured in?\n>\nWell, if a question is not ideally posed, we should try to steer it into a good\nanswer. Intensity isn\'t a good concept for a single photon. The closest answer\nis to think of energy of a single photon, not "intensity." In MKS units, energy\n"U" [to distinguish from electric field "E" I suppose] is given according to U =\nh*nu, where h is Planck\'s constant and nu is frequency. We can see the\ndifference in photon frequency and energy: higher energy photons look blue, and\nlower energy photons look red, and beyond a certain range we can\'t see them at\nall. Normally, intensity for light means either the total energy per second\n[power] emitted by a source, in watts, or on a surface, it\'s the power received\nper square meter. In full sunlight, the earth\'s surface receives up to about 2\nkW per square meter from photons of different individual energies.\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">&nbsp;&nbsp;View this Usenet post in original ASCII form </a></div><P></jabberwocky>"Curious" <curious11112001@yahoo.com> wrote in message
> What determines it?
>
> What is it measured in?
>

Well, if a question is not ideally posed, we should try to steer it into a good
answer. Intensity isn't a good concept for a single photon. The closest answer
is to think of energy of a single photon, not "intensity." In MKS units, energy
"U" [to distinguish from electric field "E" I suppose] is given according to U =
$h*\nu,$ where h is Planck's constant and \nu is frequency. We can see the
difference in photon frequency and energy: higher energy photons look blue, and
lower energy photons look red, and beyond a certain range we can't see them at
all. Normally, intensity for light means either the total energy per second
[power] emitted by a source, in watts, or on a surface, it's the power received
per square meter. In full sunlight, the earth's surface receives up to about 2
kW per square meter from photons of different individual energies.

 PhysOrg.com physics news on PhysOrg.com >> A quantum simulator for magnetic materials>> Atomic-scale investigations solve key puzzle of LED efficiency>> Error sought & found: State-of-the-art measurement technique optimised

wrote in message\nnews:030420041853345987%fusioneer@directway.com...\n\n| energy/area, power/area are concepts dealing with the idea of\n| continuous surfaces whereas the idea of a photon as a descrete quantum\n| of energy doesn\'t mesh with the notion of an area. In other words we\n| can\'t rationally mix classical concepts with quantum concepts. I know,\n| I know, it is done all of the time but I think rather to the detriment\n| of understanding.\n|\n| The idea that a photon of a given wavelength can have an intensity that\n| varies flies in the face of the notion of discrete quanta also for the\n| photon would necessarily have to be composed of a finite number of\n| subcomponents. Then one might suppose that two photons of the same\n| given wavelength might be composed of unequal quantities of this\n| hypothetical subcomponent for such photons to have a variation in\n| \'intensity\'. But then, again, what do I know?\n\nYes, I have been personally struggling with this for some time now. If say\nyou have a certain number of identical photons passing thru a certain area\nperpendicular to the photon\'s travel direction, this defines an intensity.\nHowever, photons being bosons, can occupy the same "space" as each other.\nThis would seem to indicate that the area could or would be defined by a\nsingle photon of the group of identical photons. Especially for soft\nphotons. IOW, as you take identical photons away, the area stays the same\nand when you get down to the last single photon, you still have the same\namount of area. ???\n\nFrediFizzx\n\n\n');">   View this Usenet post in original ASCII form

"CCRyder" wrote in message news:030420041853345987%fusioneer@directway.com... | energy/area, power/area are concepts dealing with the idea of | continuous surfaces whereas the idea of a photon as a descrete quantum | of energy doesn't mesh with the notion of an area. In other words we | can't rationally mix classical concepts with quantum concepts. I know, | I know, it is done all of the time but I think rather to the detriment | of understanding. | | The idea that a photon of a given wavelength can have an intensity that | varies flies in the face of the notion of discrete quanta also for the | photon would necessarily have to be composed of a finite number of | subcomponents. Then one might suppose that two photons of the same | given wavelength might be composed of unequal quantities of this | hypothetical subcomponent for such photons to have a variation in | 'intensity'. But then, again, what do I know? Yes, I have been personally struggling with this for some time now. If say you have a certain number of identical photons passing thru a certain area perpendicular to the photon's travel direction, this defines an intensity. However, photons being bosons, can occupy the same "space" as each other. This would seem to indicate that the area could or would be defined by a single photon of the group of identical photons. Especially for soft photons. IOW, as you take identical photons away, the area stays the same and when you get down to the last single photon, you still have the same amount of area. ??? FrediFizzx



In <34a4f456.0404011123.63ca6994@posting.google.com> Curious wrote: > What determines it? > > What is it measured in? Intensity, that is flux per steradian (photometrists) or flux per area ( everyone else), is related to the number of photons per second. http://math.ucr.edu/home/baez/photon/schmoton.htm -- Andrew Resnick, Ph. D. National Center for Microgravity Research NASA Glenn Research Center

## Re: Intensity of a Photon

<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>\n"FrediFizzx" &lt;fredifizzx@hotmail.com&gt; wrote in message\nnews:c4pmus\\$2j2r2o\\$1@ID-185976.news.uni-berlin.de...\n&gt; "CCRyder" &lt;fusioneer@directway.com&gt; wrote in message\n&gt; news:030420041853345987%fusioneer@directway.com...\n&gt;\n&gt; | energy/area, power/area are concepts dealing with the idea of\n&gt; | continuous surfaces whereas the idea of a photon as a descrete quantum\n&gt; | of energy doesn\'t mesh with the notion of an area. In other words we\n&gt; | can\'t rationally mix classical concepts with quantum concepts. I know,\n&gt; | I know, it is done all of the time but I think rather to the detriment\n&gt; | of understanding.\n&gt; |\n&gt; | The idea that a photon of a given wavelength can have an intensity that\n&gt; | varies flies in the face of the notion of discrete quanta also for the\n&gt; | photon would necessarily have to be composed of a finite number of\n&gt; | subcomponents. Then one might suppose that two photons of the same\n&gt; | given wavelength might be composed of unequal quantities of this\n&gt; | hypothetical subcomponent for such photons to have a variation in\n&gt; | \'intensity\'. But then, again, what do I know?\n&gt;\n&gt; Yes, I have been personally struggling with this for some time now. If\nsay\n&gt; you have a certain number of identical photons passing thru a certain area\n&gt; perpendicular to the photon\'s travel direction, this defines an intensity.\n&gt; However, photons being bosons, can occupy the same "space" as each other.\n\nThe photons do not occupy the same "space" as each other. The can share the\nsame quantum state, which is a different thing.\n\n\n&gt; This would seem to indicate that the area could or would be defined by a\n&gt; single photon of the group of identical photons.\n\nIt would if your presumption of a boson\'s properties were correct. However,\nit is not.\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">&nbsp;&nbsp;View this Usenet post in original ASCII form </a></div><P></jabberwocky>"FrediFizzx" <fredifizzx@hotmail.com> wrote in message
news:c4pmus$2j2r2o$1@ID-185976.news.uni-berlin.de...
> "CCRyder" <fusioneer@directway.com> wrote in message
> news:030420041853345987%fusioneer@directway.com...
>
> | energy/area, power/area are concepts dealing with the idea of
> | continuous surfaces whereas the idea of a photon as a descrete quantum
> | of energy doesn't mesh with the notion of an area. In other words we
> | can't rationally mix classical concepts with quantum concepts. I know,
> | I know, it is done all of the time but I think rather to the detriment
> | of understanding.
> |
> | The idea that a photon of a given wavelength can have an intensity that
> | varies flies in the face of the notion of discrete quanta also for the
> | photon would necessarily have to be composed of a finite number of
> | subcomponents. Then one might suppose that two photons of the same
> | given wavelength might be composed of unequal quantities of this
> | hypothetical subcomponent for such photons to have a variation in
> | 'intensity'. But then, again, what do I know?
>
> Yes, I have been personally struggling with this for some time now. If

say
> you have a certain number of identical photons passing thru a certain area
> perpendicular to the photon's travel direction, this defines an intensity.
> However, photons being bosons, can occupy the same "space" as each other.

The photons do not occupy the same "space" as each other. The can share the
same quantum state, which is a different thing.

> This would seem to indicate that the area could or would be defined by a
> single photon of the group of identical photons.

It would if your presumption of a boson's properties were correct. However,
it is not.



FrediFizzx writes >Yes, I have been personally struggling with this for some time now. If say >you have a certain number of identical photons passing thru a certain area >perpendicular to the photon's travel direction, this defines an intensity. >However, photons being bosons, can occupy the same "space" as each other. >This would seem to indicate that the area could or would be defined by a >single photon of the group of identical photons. Especially for soft >photons. IOW, as you take identical photons away, the area stays the same >and when you get down to the last single photon, you still have the same >amount of area. ??? I struggled with this too. It seems (at least Others Here Who Know These Things say) that the number of photons in a beam is indeterminate or unknown. I believe the appropriate model exemplifies this. ============== Warning: I am nothing like an expert, read with caution. ============== To me it suggested a different tack, which is that light is a wave (astonishing, I know) and the quantum aspect is primarily due to its interaction with quantised objects (typically atoms). It is a fact that two independent photon sources will interfere with each other and the clincher to me was that two separate heavily attenuated laser beams each sent through one slit of a two slit diffraction apparatus still gave an interference pattern even though both were so heavily attenuated that there was negligible chance of two 'photons' being in the apparatus at any one time. I then consider the quantum aspect, typically line emission/absorption as a resonance effect where a very low level oscillating wave eventually transfers/absorbs enough energy to cause an energy level jump. Note that this does NOT (as far as my limited knowledge can tell) alter the physics of QM in general, I am pretty happy with the idea of superpositions, entanglement and so on which seem to me to fall very naturally into a wave-like explanation. The only potential problem is that I have now dispensed completely with the idea of a particle at a point, after all if everything is a (extended) wave, it makes no sense to ask if its precisely *here*. At best one can say its mostly in a volume, and then only if the volume is of the order of a wavelength, and preferably very many wavelengths (I suppose you could make an argument for saying $- say-$ its 50% in some volume). In general this isn't a problem experimentally as measurements of position are rarely (ever?) smaller than the wave that is being detected. For example detecting a photon on a photographic plate means detecting an *ATOM* that has absorbed a photonsworth of energy, and atoms are pretty small, wavelengthwise. Ie its the atom you are detecting, NOT the beam of photons (which is indeterminate). This begs the question of why we have quantised *particles* with identical and well defined properties (like electrons) as well as rather indeterminate things like photons. If my logic is to 'explain' these then I have to demand an internal mechanism that constrains their properties to energy jumps/levels. That is, in essence, there must be some form of oscillation. Here I have no option but to be vague and speculative. My current, ad-hoc and very tentative concept is that it is energy oscillating between an electric dimension and a time direction. Put crudely oscillating between electric field and mass. Note that just as an electron in an S-orbital has no angular momentum, there need be no reason why an electron should show an actual oscillation of (say) electric field. Well, that's my 2c worth anyway. Until someone comes along and shows me why this cannot be so..... Flying along waiting to be Shot Down In Flames .... -- Oz This post is worth absolutely nothing and is probably fallacious. DEMON address no longer in use.



In article , Michael Varney wrote: > The photons do not occupy the same "space" as each other. The can share the > same quantum state, which is a different thing. Nothing like the air of authority to settle such arguments. Except we must be sure that it is not really the authority of air. :-). The idea that bosons cannot be in the same 'space' is forbidden by what 'law' in physics? What experimental data confirms your beleif in this matter? Fermions, for example, cannot occupy the same quantum state but pairs of electrons, one spin up the other spin down appear to be able to occupy the same 'space'. CCRyder



CCRyder wrote: > In article , Michael Varney > wrote: > > >>The photons do not occupy the same "space" as each other. The can share the >>same quantum state, which is a different thing. > > > Nothing like the air of authority to settle such arguments. Except we > must be sure that it is not really the authority of air. :-). The idea > that bosons cannot be in the same 'space' is forbidden by what 'law' in > physics? What experimental data confirms your beleif in this matter? > Fermions, for example, cannot occupy the same quantum state but pairs > of electrons, one spin up the other spin down appear to be able to > occupy the same 'space'. It is not even legitimate to imagine individual particles. There aren't "two bosons" the way there are say a 7-ball and an 8-ball left on the table. The boson field has occupation number 2. The point is the symmetry of the entire arrangement under interchange of particles and it is totally verified experimentally (lasers, superfluids). This can't be overemphasized - there is not only no way to imagine individual bosons in a collective amalgam of them, there is no point to it, because the very thing that describes them is symmetrical. You might as well try to tell two points on a circle apart. $-drl$





Michael Varney wrote: > "FrediFizzx" wrote in message >> Yes, I have been personally struggling with this for some time now. >> If say you have a certain number of identical photons passing thru a >> certain area perpendicular to the photon's travel direction, this >> defines an intensity. However, photons being bosons, can occupy the >> same "space" as each other. > The photons do not occupy the same "space" as each other. The can > share the same quantum state, which is a different thing. Photon states consist of two parts: (a) the mode function, which is a normalised solution of Maxwell's equations, and (b) the state of the quantum simple harmonic oscillator (QSHO) living inside the mode. Complicated photonic states might consist of some modes+QSHO-state combinations in various superpositions $and/or$ mixtures with other mode+QSHO-state combinations. The intensity of a photon state confined to a single mode will be given by a number derived from the mode function, and then multiplied by the number of quanta in its QSHO state. Any two QSHO quanta from the same mode could quite reasonably be described as occupying the same space. -- ---------------------------------+--------------------------------- Dr. Paul Kinsler Blackett Laboratory (QOLS) (ph) $+44-20-759-47520$ (fax) 47714 Imperial College London, Dr.Paul.Kinsler@physics.org SW7 2BW, United Kingdom. http://www.qols.ph.ic.ac.uk/~kinsle/



In infinite wisdom CCRyder answered: > In article , Michael Varney > wrote: > > >>The photons do not occupy the same "space" as each other. The can share the >>same quantum state, which is a different thing. > > > Nothing like the air of authority to settle such arguments. Except we > must be sure that it is not really the authority of air. :-). The idea > that bosons cannot be in the same 'space' is forbidden by what 'law' in > physics? What experimental data confirms your beleif in this matter? > Fermions, for example, cannot occupy the same quantum state but pairs > of electrons, one spin up the other spin down appear to be able to > occupy the same 'space'. This is curious, I'm given to understand that electrons have no structure, that every new experiment sets a smaller upper bound on the, err, size of an electron. And then there's the small matter of Heisenberg. How can anyone localize the exact position of two electrons to the precision needed to determine that they occupy the same position. And if indeed, one could do this, by what manner would it be determined that there are two electrons rather than one? Then again, perhaps your use of the words 'same space' does not correspond to 'exact same position' as I read it. Rich > CCRyder



In article <407408D7.2010704@somewhere.com>, Rich wrote: > In infinite wisdom CCRyder answered: > > In article , Michael Varney > > wrote: > > > > > >>The photons do not occupy the same "space" as each other. The can share the > >>same quantum state, which is a different thing. > > > > > > Nothing like the air of authority to settle such arguments. Except we > > must be sure that it is not really the authority of air. :-). The idea > > that bosons cannot be in the same 'space' is forbidden by what 'law' in > > physics? What experimental data confirms your belief in this matter? > > Fermions, for example, cannot occupy the same quantum state but pairs > > of electrons, one spin up the other spin down appear to be able to > > occupy the same 'space'. > > This is curious, I'm given to understand that electrons have no > structure, that every new experiment sets a smaller upper bound > on the, err, size of an electron. And then there's the small matter > of Heisenberg. It is *supposed* that electrons have no structure. The failure to discover a structure is not equivalent to the lack of one. If one applies the axiomatic idea that quanta can only have motion with respect to other quanta then position with respect to a coordinate system is an absolutely inappropriate concept that perhaps only serves us well at the classical world and to the classical limit; and, in fact, the application and extrapolation of that single axiom completely reveals the basic reason why the Heisenberg Uncertainty Principle exists. Position is a classical concept that has been shoehorned into a quantum world. For instance, the notion of gas pressure is mathematically represented as a continuous force applied to a continuous surface whereas both concepts and their mathematical representations, we must be brave enough to acknowledge, are inappropriate to the quantum world. > How can anyone localize the exact position of two electrons to > the precision needed to determine that they occupy the same > position. And if indeed, one could do this, by what manner would > it be determined that there are two electrons rather than one? I can think of the perfect instance which I mentioned in my first post, Cooper Pairing, where electrons are not just tightly bound but rather are united in such a fashion so as to produce a null motion gradient structure that rapidly induces other nearby electrons to overlap in the same momentum space (producing a fast cascading effect which rapidly involves the entire superconducting material sample) if not directly in the same place where 'the same place' may be identified not with respect to an ordinary coordinate system but rather with the idea that both electrons are exactly the same distance from every other quantum particle in the universe. 'Position' in such a quantum scheme is only relational. > > Then again, perhaps your use of the words 'same space' does not > correspond to 'exact same position' as I read it. > Rich Probably $not:-}$. CCRyder



CCRyder wrote: > In article <407408D7.2010704@somewhere.com>, Rich > wrote: > > >>In infinite wisdom CCRyder answered: >> >>>In article , Michael Varney >>> wrote: >>> >>> >>> >>>>The photons do not occupy the same "space" as each other. The can share the >>>>same quantum state, which is a different thing. >>> >>> >>>Nothing like the air of authority to settle such arguments. Except we >>>must be sure that it is not really the authority of air. :-). The idea >>>that bosons cannot be in the same 'space' is forbidden by what 'law' in >>>physics? What experimental data confirms your belief in this matter? >>>Fermions, for example, cannot occupy the same quantum state but pairs >>>of electrons, one spin up the other spin down appear to be able to >>>occupy the same 'space'. >> >>This is curious, I'm given to understand that electrons have no >>structure, that every new experiment sets a smaller upper bound >>on the, err, size of an electron. And then there's the small matter >>of Heisenberg. > > > It is *supposed* that electrons have no structure. Physical electros _do_ have structure. it is given by the form factor, computed to some approximation in most QFT books. Only bare electrons are structureless; but they are unobservable fictions of the same kind as virtual photons. Arnold Neumaier



In article <407BE664.1010808@univie.ac.at>, Arnold Neumaier wrote: > CCRyder wrote: > > In article <407408D7.2010704@somewhere.com>, Rich > > wrote: > > > > > >>In infinite wisdom CCRyder answered: > >> > >>>In article , Michael Varney > >>> wrote: > >>> > >>> > >>> > >>>>The photons do not occupy the same "space" as each other. The can share > >>>>the > >>>>same quantum state, which is a different thing. > >>> > >>> > >>>Nothing like the air of authority to settle such arguments. Except we > >>>must be sure that it is not really the authority of air. :-). The idea > >>>that bosons cannot be in the same 'space' is forbidden by what 'law' in > >>>physics? What experimental data confirms your belief in this matter? > >>>Fermions, for example, cannot occupy the same quantum state but pairs > >>>of electrons, one spin up the other spin down appear to be able to > >>>occupy the same 'space'. > >> > >>This is curious, I'm given to understand that electrons have no > >>structure, that every new experiment sets a smaller upper bound > >>on the, err, size of an electron. And then there's the small matter > >>of Heisenberg. > > > > > > It is *supposed* that electrons have no structure. > > Physical electros _do_ have structure. it is given by the form factor, > computed to some approximation in most QFT books. Only bare electrons > are structureless; but they are unobservable fictions of the same kind > as virtual photons. > > > Arnold Neumaier What possibly could you mean by the term 'bare electrons'? Do you envision electrons as singularities? Do you conceive of electrons as unconnected to the rest of the universe? My lack of formal education on these matters has me envisioning electron's structure as consisting only of their relationships to the rest of the particles in the universe. That *is* their structure. Bear with me for a moment or at least humor me. If the unit positive charge is a sink then why not envision that every electron (as a source particle with the unit negative charge) provides a single component of the multicomponent sink structure? If there were a thousand sink particles (protons) in the universe and a thousand source particles (electrons) then each electron would be connected to every proton. The unit charge of a sink type particle would have 1000 subcomponents and the unit charge of a source type particle would also have 1000 subcomponents. What's wrong with such a model? Conservation of charge is preserved. The idea of a subcomponent begs the question of the meaning of such a subcomponent which seems pretty straightforward as a motion relationship which is always binary in nature due to the relativity of motion. But then what do I know? CCRyder antispam measure/lose the 't' for email.



In infinite wisdom CCRyder answered: > In article <407408D7.2010704@somewhere.com>, Rich > wrote: > > >>In infinite wisdom CCRyder answered: >> >>>In article , Michael Varney >>> wrote: >>> >>> >>> >>>>The photons do not occupy the same "space" as each other. The can share the >>>>same quantum state, which is a different thing. >>> >>> >>>Nothing like the air of authority to settle such arguments. Except we >>>must be sure that it is not really the authority of air. :-). The idea >>>that bosons cannot be in the same 'space' is forbidden by what 'law' in >>>physics? What experimental data confirms your belief in this matter? >>>Fermions, for example, cannot occupy the same quantum state but pairs >>>of electrons, one spin up the other spin down appear to be able to >>>occupy the same 'space'. >> >>This is curious, I'm given to understand that electrons have no >>structure, that every new experiment sets a smaller upper bound >>on the, err, size of an electron. And then there's the small matter >>of Heisenberg. > > > It is *supposed* that electrons have no structure. The failure to > discover a structure is not equivalent to the lack of one. Good point. But still, this is the currently available information. > If one > applies the axiomatic idea that quanta can only have motion with > respect to other quanta then position with respect to a coordinate > system is an absolutely inappropriate concept that perhaps only serves > us well at the classical world and to the classical limit; and, in > fact, the application and extrapolation of that single axiom completely > reveals the basic reason why the Heisenberg Uncertainty Principle > exists. Position is a classical concept that has been shoehorned into > a quantum world. For instance, the notion of gas pressure is > mathematically represented as a continuous force applied to a > continuous surface whereas both concepts and their mathematical > representations, we must be brave enough to acknowledge, are > inappropriate to the quantum world. Sounds more like it's the wrong tool for the job than any fault with QM or classical gas pressure laws. >>How can anyone localize the exact position of two electrons to >>the precision needed to determine that they occupy the same >>position. And if indeed, one could do this, by what manner would >>it be determined that there are two electrons rather than one? > > I can think of the perfect instance which I mentioned in my first post, > Cooper Pairing, where electrons are not just tightly bound but rather > are united in such a fashion so as to produce a null motion gradient > structure that rapidly induces other nearby electrons to overlap in the > same momentum space (producing a fast cascading effect which rapidly > involves the entire superconducting material sample) if not directly in > the same place where 'the same place' may be identified not with > respect to an ordinary coordinate system but rather with the idea that > both electrons are exactly the same distance from every other quantum > particle in the universe. 'Position' in such a quantum scheme is only > relational. But cooper pairs are not said to occupy the "same 'space'" to the best of my knowledge. They just act in unison. Somewhat belatedly it occurred to me that there is a much stronger reason to question whether 2 electrons could occupy the "same 'space'", electrostatic repulsion. It's the strongest force shy of the nuclear force, and unlike gravity it's not attractive (in this case). Two electrons may occupy the same quantum state, but there seem to be a few problems with them occupying the same space (irregardless of coordinate systems or lack thereof). Rich >>Then again, perhaps your use of the words 'same space' does not >>correspond to 'exact same position' as I read it. > > >>Rich > > > Probably $not:-}$. > > CCRyder