View Full Version : Interference even with long interferometers
Michael Hobbs
Oct11-04, 03:52 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>\n\nWhen a [Mach-Zehnder] interferometer is set up such that there is only\none photon in the apparatus at a time, we still see an interference\npattern evolve on the detector. This would imply that the beam\nsplitter does not direct half of the photons one way and half of the\nphotons the other way, but actually splits a photon\'s wave in half and\nsends each half down a separate path. Correct me if this is an\nincorrect interpretation.\n\nI also understand that with a laser, it is possible to have a short\npath A through the interferometer and a long path B that can be\nseveral miles longer, which still results in an interference pattern\neven with a single photon. Is this correct?\n\nIf the above assumptions are correct, it is puzzling that the wave\nfront traveling through path A reaches the detector while the wave\nfront traveling through path B is several miles behind it, yet there\nis still an interference pattern. How can the wave traveling through\npath B possibly interfere with the wave traveling through path A,\nunless the waves are several miles long? Is it possible that the wave\nof a single photon is many miles long?\n\nEven with a multiple worlds interpretation, it still doesn\'t work out\nin my mind. Even if the waves travel an infinite number of paths\nsimultaneously through parallel universes, the waves traveling the\npaths through A will always reach the detector long before the waves\ntraveling the paths through B.\n\nCould someone please point out what part of my understanding is in\nerror?\n\nThanks,\n- Mike Hobbs\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>When a [Mach-Zehnder] interferometer is set up such that there is only
one photon in the apparatus at a time, we still see an interference
pattern evolve on the detector. This would imply that the beam
splitter does not direct half of the photons one way and half of the
photons the other way, but actually splits a photon's wave in half and
sends each half down a separate path. Correct me if this is an
incorrect interpretation.
I also understand that with a laser, it is possible to have a short
path A through the interferometer and a long path B that can be
several miles longer, which still results in an interference pattern
even with a single photon. Is this correct?
If the above assumptions are correct, it is puzzling that the wave
front traveling through path A reaches the detector while the wave
front traveling through path B is several miles behind it, yet there
is still an interference pattern. How can the wave traveling through
path B possibly interfere with the wave traveling through path A,
unless the waves are several miles long? Is it possible that the wave
of a single photon is many miles long?
Even with a multiple worlds interpretation, it still doesn't work out
in my mind. Even if the waves travel an infinite number of paths
simultaneously through parallel universes, the waves traveling the
paths through A will always reach the detector long before the waves
traveling the paths through B.
Could someone please point out what part of my understanding is in
error?
Thanks,
- Mike Hobbs
Pierre Asselin
Oct11-04, 01:06 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>\nMichael Hobbs <mike@hobbshouse.org> wrote:\n\n\n> [ ... interferometer with a single photon ... ]\n\n> I also understand that with a laser, it is possible to have a short\n> path A through the interferometer and a long path B that can be\n> several miles longer, which still results in an interference pattern\n> even with a single photon. Is this correct?\n\nI\'m not sure you can get away with "several miles", but yes you can\nget interferences with a very asymmetric interferometer, up to the\nlaser\'s coherence length. Let\'s assume "several miles" anyway.\n\n> If the above assumptions are correct, it is puzzling that the wave\n> front traveling through path A reaches the detector while the wave\n> front traveling through path B is several miles behind it, yet there\n> is still an interference pattern. How can the wave traveling through\n> path B possibly interfere with the wave traveling through path A,\n> unless the waves are several miles long? Is it possible that the wave\n> of a single photon is many miles long?\n\nIt won\'t work with a pulsed laser, it has to be a CW laser --or at\nleast, the pulses have to be several miles long so the waves still\noverlap. Even if you reduce the power level so that there is only\none photon at a time in the interferometer you still get interference\nbecause you don\'t know *when* the photon passed through, any more\nthan you know *which arm* it took.\n\n--\npa at panix dot com\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>Michael Hobbs <mike@hobbshouse.org> wrote:
> [ ... interferometer with a single photon ... ]
> I also understand that with a laser, it is possible to have a short
> path A through the interferometer and a long path B that can be
> several miles longer, which still results in an interference pattern
> even with a single photon. Is this correct?
I'm not sure you can get away with "several miles", but yes you can
get interferences with a very asymmetric interferometer, up to the
laser's coherence length. Let's assume "several miles" anyway.
> If the above assumptions are correct, it is puzzling that the wave
> front traveling through path A reaches the detector while the wave
> front traveling through path B is several miles behind it, yet there
> is still an interference pattern. How can the wave traveling through
> path B possibly interfere with the wave traveling through path A,
> unless the waves are several miles long? Is it possible that the wave
> of a single photon is many miles long?
It won't work with a pulsed laser, it has to be a CW laser --or at
least, the pulses have to be several miles long so the waves still
overlap. Even if you reduce the power level so that there is only
one photon at a time in the interferometer you still get interference
because you don't know *when* the photon passed through, any more
than you know *which arm* it took.
--
pa at panix dot com
Pierre Asselin
Oct14-04, 11:07 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>\nMichael Hobbs <mike@hobbshouse.org> wrote:\n\n> [ ... ]\n> If we assume that a photon\'s wave is only one wavelength long, [ ... ]\n\nThat assumption is wrong.\n\n> If I understand your statement correctly, you\'re saying that even though\n> we may only have one photon in the interferometer per second, *on\n> average*, we may still sometimes have multiple photons in the\n> interferometer at once, which creates the interference?\n\nNo, it doesn\'t work that way. The interference still happens with\na single photon. This fact has nothing to do with the size of the\npath difference between the two legs of the interferometer, i.e.\neven with equal legs your reasoning predicts a loss of interference\nwhen there aren\'t enough photons to fill both legs --and that\ndoesn\'t happen.\n\n--\npa at panix dot com\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>Michael Hobbs <mike@hobbshouse.org> wrote:
> [ ... ]
> If we assume that a photon's wave is only one wavelength long, [ ... ]
That assumption is wrong.
> If I understand your statement correctly, you're saying that even though
> we may only have one photon in the interferometer per second, *on
> average*, we may still sometimes have multiple photons in the
> interferometer at once, which creates the interference?
No, it doesn't work that way. The interference still happens with
a single photon. This fact has nothing to do with the size of the
path difference between the two legs of the interferometer, i.e.
even with equal legs your reasoning predicts a loss of interference
when there aren't enough photons to fill both legs --and that
doesn't happen.
--
pa at panix dot com
<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>\nMichael Hobbs <mike@hobbshouse.org> writes\n>If we assume that a photon\'s wave is only one wavelength long, how is\n>it that we see interference patterns emerge when only one photon is in\n>the apparatus at a time? The waves should only intersect in space at the\n>middle point between the two slits and nowhere else. If we assume that\n>a photon\'s wave is many wavelengths long, then how long is it, exactly?\n\nDon\'t go that way.....\nI did, once, and I regret to say at considerable and boring length.\n\nConsider this experiment, which has been done and showed interference:\n\nTwo *separate* stable lasers.\nTwo slits, each illuminated by *one* laser.\nAttenuated so "never more than one photon in the apparatus at one time".\n[NB Unsurprising, one can use two radio transmitters]\n\nThe conclusion is that a photon is a wave, essentially maxwellian in\nform.\n\nWhen you interrogate those experts who use pointlike photons to analyse\nlight behaviour you find that the mathematical structures they use are\nso put together (clever people, all) that it perfectly emulates a simple\nwavelike explanation. So for interpretational purposes its simpler to\nconsider a photon as a wave.\n\nYou will then ask me how I rationalise away the quantumlike properties\nof photons, which indisputably exist.\n\nQuantumlike properties arise from quantumlike systems. All detectors are\nquantumlike, they *require* an energy gap to function. Its thus\nunsurprising that when photons (ie light) interact with detectors they\ninevitably show quantumlike effects, these are due to the detector.\n\nThat is in almost all circumstances we have an incident wave of *far*\ntoo low an intensity to cause the jumping of the energy gap, but if the\nresonant frequency \'of the gap\' and the frequency of the incident light\nis *nearly right* then energy gets stored in the system until it either\ncauses a jump in the energy gap, or the light/energy is re-radiated\naway.\n\nIt is NOT true that light of the wrong frequency (ie below the energy of\nthe light quantum) will not trigger a jump of the bandgap no matter what\nits intensity. It is, for example, perfectly possible to ionise atoms by\napplying and adequately high *DC* potential. That\'s how sparks form. The\nstress is however many many orders higher than is required if the light\nis at the resonant frequency of the detector atoms.\n\nNB In this I am generally regarded as a crank.\n\n--\nOz\nThis post is worth absolutely nothing and is probably fallacious.\n\nUse oz@farmeroz.port995.com [ozacoohdb@despammed.com functions].\nBTOPENWORLD address has ceased. DEMON address has ceased.\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>Michael Hobbs <mike@hobbshouse.org> writes
>If we assume that a photon's wave is only one wavelength long, how is
>it that we see interference patterns emerge when only one photon is in
>the apparatus at a time? The waves should only intersect in space at the
>middle point between the two slits and nowhere else. If we assume that
>a photon's wave is many wavelengths long, then how long is it, exactly?
Don't go that way.....
I did, once, and I regret to say at considerable and boring length.
Consider this experiment, which has been done and showed interference:
Two *separate* stable lasers.
Two slits, each illuminated by *one* laser.
Attenuated so "never more than one photon in the apparatus at one time".
[NB Unsurprising, one can use two radio transmitters]
The conclusion is that a photon is a wave, essentially maxwellian in
form.
When you interrogate those experts who use pointlike photons to analyse
light behaviour you find that the mathematical structures they use are
so put together (clever people, all) that it perfectly emulates a simple
wavelike explanation. So for interpretational purposes its simpler to
consider a photon as a wave.
You will then ask me how I rationalise away the quantumlike properties
of photons, which indisputably exist.
Quantumlike properties arise from quantumlike systems. All detectors are
quantumlike, they *require* an energy gap to function. Its thus
unsurprising that when photons (ie light) interact with detectors they
inevitably show quantumlike effects, these are due to the detector.
That is in almost all circumstances we have an incident wave of *far*
too low an intensity to cause the jumping of the energy gap, but if the
resonant frequency 'of the gap' and the frequency of the incident light
is *nearly right* then energy gets stored in the system until it either
causes a jump in the energy gap, or the light/energy is re-radiated
away.
It is NOT true that light of the wrong frequency (ie below the energy of
the light quantum) will not trigger a jump of the bandgap no matter what
its intensity. It is, for example, perfectly possible to ionise atoms by
applying and adequately high *DC* potential. That's how sparks form. The
stress is however many many orders higher than is required if the light
is at the resonant frequency of the detector atoms.
NB In this I am generally regarded as a crank.
--
Oz
This post is worth absolutely nothing and is probably fallacious.
Use oz@farmeroz.port995.com [ozacoohdb@despammed.com functions].
BTOPENWORLD address has ceased. DEMON address has ceased.
Maurice Barnhill
Oct17-04, 01:46 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>\nOz wrote:\n\n[snip]\n> Quantumlike properties arise from quantumlike systems. All detectors are\n> quantumlike, they *require* an energy gap to function. Its thus\n> unsurprising that when photons (ie light) interact with detectors they\n> inevitably show quantumlike effects, these are due to the detector.\n>\n> That is in almost all circumstances we have an incident wave of *far*\n> too low an intensity to cause the jumping of the energy gap, but if the\n> resonant frequency \'of the gap\' and the frequency of the incident light\n> is *nearly right* then energy gets stored in the system until it either\n> causes a jump in the energy gap, or the light/energy is re-radiated\n> away.\n\nThen how do you explain the fact that when low-intensity light\nstarts shining on a photocell, electrons sometimes appear before\nthe wave has time classically to transfer a photons-worth of\nenergy to the photocell? The standard explanation works well:\nthe photon is as likely to appear in the front of the wave pulse\nas at the back.\n\n--\nMaurice Barnhill\nmvb@udel.edu [Use ReplyTo, not From]\n[bellatlantic.net is reserved for spam only]\nDepartment of Physics and Astronomy\nUniversity of Delaware\nNewark, DE 19716\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>Oz wrote:
[snip]
> Quantumlike properties arise from quantumlike systems. All detectors are
> quantumlike, they *require* an energy gap to function. Its thus
> unsurprising that when photons (ie light) interact with detectors they
> inevitably show quantumlike effects, these are due to the detector.
>
> That is in almost all circumstances we have an incident wave of *far*
> too low an intensity to cause the jumping of the energy gap, but if the
> resonant frequency 'of the gap' and the frequency of the incident light
> is *nearly right* then energy gets stored in the system until it either
> causes a jump in the energy gap, or the light/energy is re-radiated
> away.
Then how do you explain the fact that when low-intensity light
starts shining on a photocell, electrons sometimes appear before
the wave has time classically to transfer a photons-worth of
energy to the photocell? The standard explanation works well:
the photon is as likely to appear in the front of the wave pulse
as at the back.
--
Maurice Barnhill
mvb@udel.edu [Use ReplyTo, not From]
[bellatlantic.net is reserved for spam only]
Department of Physics and Astronomy
University of Delaware
Newark, DE 19716
Michael Hobbs
Oct17-04, 01:46 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>\nOz <oz@farmeroz.port995.com> wrote:\n>\n> Michael Hobbs <mike@hobbshouse.org> writes\n>>If we assume that a photon\'s wave is only one wavelength long, how is\n>>it that we see interference patterns emerge when only one photon is in\n>>the apparatus at a time? The waves should only intersect in space at the\n>>middle point between the two slits and nowhere else. If we assume that\n>>a photon\'s wave is many wavelengths long, then how long is it, exactly?\n>\n> The conclusion is that a photon is a wave, essentially maxwellian in\n> form.\n>\n\nI wasn\'t trying to get into the wave vs. particle mess, I\'m just wondering\nhow we see the interference we do, even if we just stick with waves.\nLet\'s forget about photons for a moment, if we can, and just consider the\nMaxwellian wave generated by a single electron that drops its orbital.\nSince the electron doesn\'t jitter back and forth when it makes its\ntransition, can we assume that the resulting Maxwellian wave is no more\nthan one wavelength long? (Maybe it does jitter, I don\'t know.)\n\nNow, *if* the Maxwellian wave is just one wavelength long, then consider\nwhat happens with the two waves that are created when it passes through\nthe double slits. The left wave will come in contact with the left side\nof the detector and be absorbed long before the right wave will reach\nthe left side of the detector. Likewise for the right wave with the\nright side of the detector. In fact, the only place where the detector\nwill come in contact with the two waves as they are interfering with\neach other is somewhere between the two slits.\n\nSo how is it that we see an interference pattern along the entire width\nof the detector? Shouldn\'t we only see such an interference pattern when\nthere is a continuous standing wave?\n\n\n> That is in almost all circumstances we have an incident wave of *far*\n> too low an intensity to cause the jumping of the energy gap, but if the\n> resonant frequency \'of the gap\' and the frequency of the incident light\n> is *nearly right* then energy gets stored in the system until it either\n> causes a jump in the energy gap, or the light/energy is re-radiated\n> away.\n\nThat reminds me of another question that I probably shouldn\'t get into\nhere. If an electron requires a photon with energy = fh to jump one\norbital and a photon with energy, say, = 1.7fh to jump two orbitals,\nwhat happens when it encounters a photon with 1.5f? Does it jump one\norbital and somehow bleed off the remaining 0.5fh energy, or does it\njust ignore the photon? Or are electron orbitals stacked in such a way\nthat this would never happen?\n\n\nThanks,\n-- Michael Hobbs\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>Oz <oz@farmeroz.port995.com> wrote:
>
> Michael Hobbs <mike@hobbshouse.org> writes
>>If we assume that a photon's wave is only one wavelength long, how is
>>it that we see interference patterns emerge when only one photon is in
>>the apparatus at a time? The waves should only intersect in space at the
>>middle point between the two slits and nowhere else. If we assume that
>>a photon's wave is many wavelengths long, then how long is it, exactly?
>
> The conclusion is that a photon is a wave, essentially maxwellian in
> form.
>
I wasn't trying to get into the wave vs. particle mess, I'm just wondering
how we see the interference we do, even if we just stick with waves.
Let's forget about photons for a moment, if we can, and just consider the
Maxwellian wave generated by a single electron that drops its orbital.
Since the electron doesn't jitter back and forth when it makes its
transition, can we assume that the resulting Maxwellian wave is no more
than one wavelength long? (Maybe it does jitter, I don't know.)
Now, *if* the Maxwellian wave is just one wavelength long, then consider
what happens with the two waves that are created when it passes through
the double slits. The left wave will come in contact with the left side
of the detector and be absorbed long before the right wave will reach
the left side of the detector. Likewise for the right wave with the
right side of the detector. In fact, the only place where the detector
will come in contact with the two waves as they are interfering with
each other is somewhere between the two slits.
So how is it that we see an interference pattern along the entire width
of the detector? Shouldn't we only see such an interference pattern when
there is a continuous standing wave?
> That is in almost all circumstances we have an incident wave of *far*
> too low an intensity to cause the jumping of the energy gap, but if the
> resonant frequency 'of the gap' and the frequency of the incident light
> is *nearly right* then energy gets stored in the system until it either
> causes a jump in the energy gap, or the light/energy is re-radiated
> away.
That reminds me of another question that I probably shouldn't get into
here. If an electron requires a photon with energy = fh to jump one
orbital and a photon with energy, say, = 1.7fh to jump two orbitals,
what happens when it encounters a photon with 1.5f? Does it jump one
orbital and somehow bleed off the remaining .5fh energy, or does it
just ignore the photon? Or are electron orbitals stacked in such a way
that this would never happen?
Thanks,
-- Michael Hobbs
<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>\nMaurice Barnhill <mvb@udel.edu> writes\n\n>Then how do you explain the fact that when low-intensity light\n>starts shining on a photocell, electrons sometimes appear before\n>the wave has time classically to transfer a photons-worth of\n>energy to the photocell? The standard explanation works well:\n>the photon is as likely to appear in the front of the wave pulse\n>as at the back.\n\nI am discussing this on another thread.\n\nHowever if we take the waveform as a probability curve for seeing a\nphoton then the probability of this happening for a long wavelength (ie\nnarrow linewidth) is very small because the waveform can be millions of\nwavelengths long.\n\nThat\'s before you even consider the effect on the waveform of the\nshutter mechanism.\n\n--\nOz\nThis post is worth absolutely nothing and is probably fallacious.\n\nUse oz@farmeroz.port995.com [ozacoohdb@despammed.com functions].\nBTOPENWORLD address has ceased. DEMON address has ceased.\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>Maurice Barnhill <mvb@udel.edu> writes
>Then how do you explain the fact that when low-intensity light
>starts shining on a photocell, electrons sometimes appear before
>the wave has time classically to transfer a photons-worth of
>energy to the photocell? The standard explanation works well:
>the photon is as likely to appear in the front of the wave pulse
>as at the back.
I am discussing this on another thread.
However if we take the waveform as a probability curve for seeing a
photon then the probability of this happening for a long wavelength (ie
narrow linewidth) is very small because the waveform can be millions of
wavelengths long.
That's before you even consider the effect on the waveform of the
shutter mechanism.
--
Oz
This post is worth absolutely nothing and is probably fallacious.
Use oz@farmeroz.port995.com [ozacoohdb@despammed.com functions].
BTOPENWORLD address has ceased. DEMON address has ceased.
Jon Bell
Oct19-04, 03:29 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>\nOn Sun, 17 Oct 2004, Michael Hobbs wrote:\n\n> Let\'s forget about photons for a moment, if we can, and just consider the\n> Maxwellian wave generated by a single electron that drops its orbital.\n> Since the electron doesn\'t jitter back and forth when it makes its\n> transition, can we assume that the resulting Maxwellian wave is no more\n> than one wavelength long? (Maybe it does jitter, I don\'t know.)\n\nActually, it does, in a certain sense. The probability densities\n(psi_1*)(psi_1) and (psi_2*)(psi_2) of the initial and final states don\'t\nvary with time, and are therefore often called "stationary states".\nDuring the transition, we have a linear combination of the two states,\na.psi_1 + b.psi_2, in which a and b are time-varying coefficients:\ninitially a = 1 and b = 0, and finally a = 0 and b = 1, and in between\nboth have values somewhere between 0 and 1. The probablility density\nproduced by that linear combination, (a*.psi_1* + b*.psi_2*)(a.psi_1 +\nb.psi_2), physically oscillates (sloshes back and forth) with a frequency\ngiven by (E_1 - E_2)/h, which is the frequency of the emittted photon.\n\nI\'ve sometimes wondered what one would get if one treated this sloshing\nprobablity density as a sloshing charge density, and calculated the\nclassical E and B fields that they would produce. Would they actually\ncorrespond somehow to the observed behavior (probabilistic, of course) of\nthe emitted photon?\n\n> If an electron requires a photon with energy = fh to jump one\n> orbital and a photon with energy, say, = 1.7fh to jump two orbitals,\n> what happens when it encounters a photon with 1.5f? Does it jump one\n> orbital and somehow bleed off the remaining 0.5fh energy, or does it\n> just ignore the photon?\n\nThe photon is "ignored." That\'s why gases have an absorption spectrum\nwith sharp dark lines in them ("Fraunhofer lines"). Only those photons\nthat have just the right frequency get absorbed by upward transitions.\n\n--\nJon Bell <jbellm4h@presby.edu> Presbyterian College\nDept. of Physics and Computer Science Clinton, South Carolina USA\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 Sun, 17 Oct 2004, Michael Hobbs wrote:
> Let's forget about photons for a moment, if we can, and just consider the
> Maxwellian wave generated by a single electron that drops its orbital.
> Since the electron doesn't jitter back and forth when it makes its
> transition, can we assume that the resulting Maxwellian wave is no more
> than one wavelength long? (Maybe it does jitter, I don't know.)
Actually, it does, in a certain sense. The probability densities
(\psi_1*)(\psi_1) and (\psi_2*)(\psi_2) of the initial and final states don't
vary with time, and are therefore often called "stationary states".
During the transition, we have a linear combination of the two states,
a.\psi_1 + b.\psi_2, in which a and b are time-varying coefficients:
initially a = 1 and b = 0, and finally a = and b = 1, and in between
both have values somewhere between and 1. The probablility density
produced by that linear combination, (a*.\psi_1* + b*.\psi_2*)(a.\psi_1 +b.\psi_2), physically oscillates (sloshes back and forth) with a frequency
given by (E_1 - E_2)/h, which is the frequency of the emittted photon.
I've sometimes wondered what one would get if one treated this sloshing
probablity density as a sloshing charge density, and calculated the
classical E and B fields that they would produce. Would they actually
correspond somehow to the observed behavior (probabilistic, of course) of
the emitted photon?
> If an electron requires a photon with energy = fh to jump one
> orbital and a photon with energy, say, = 1.7fh to jump two orbitals,
> what happens when it encounters a photon with 1.5f? Does it jump one
> orbital and somehow bleed off the remaining .5fh energy, or does it
> just ignore the photon?
The photon is "ignored." That's why gases have an absorption spectrum
with sharp dark lines in them ("Fraunhofer lines"). Only those photons
that have just the right frequency get absorbed by upward transitions.
--
Jon Bell <jbellm4h@presby.edu> Presbyterian College
Dept. of Physics and Computer Science Clinton, South Carolina USA
<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\nMichael Hobbs <mike@hobbshouse.org> writes\n\n>I wasn\'t trying to get into the wave vs. particle mess, I\'m just wondering\n>how we see the interference we do, even if we just stick with waves.\n>Let\'s forget about photons for a moment, if we can, and just consider the\n>Maxwellian wave generated by a single electron that drops its orbital.\n>Since the electron doesn\'t jitter back and forth when it makes its\n>transition, can we assume that the resulting Maxwellian wave is no more\n>than one wavelength long? (Maybe it does jitter, I don\'t know.)\n\nIt must \'jitter\' because the emission times are known. Some take a very\nlong time, to the extent that they are never seen if the atoms are in\ncollisions more frequently than the emission time. It also gives the\ncharacteristic width of each emission line which is inversely related to\nemission time. If they were one wavelength long then the linewidth would\nbe incredibly wide rather then (typically) very narrow.\n\nCertainly in most cases its very many cycles of the EM wave. Also\nconsider that typically the wavelength of a light photon is many\nthousands of times larger that the emitting atom.\n\n>Now, *if* the Maxwellian wave is just one wavelength long,\n\nIt\'s not. I once did calculate it for a yellow sodium line, IIRC it was\nmillions of wavelengths.\n\n>> That is in almost all circumstances we have an incident wave of *far*\n>> too low an intensity to cause the jumping of the energy gap, but if the\n>> resonant frequency \'of the gap\' and the frequency of the incident light\n>> is *nearly right* then energy gets stored in the system until it either\n>> causes a jump in the energy gap, or the light/energy is re-radiated\n>> away.\n>\n>That reminds me of another question that I probably shouldn\'t get into\n>here. If an electron requires a photon with energy = fh to jump one\n>orbital and a photon with energy, say, = 1.7fh to jump two orbitals,\n>what happens when it encounters a photon with 1.5f? Does it jump one\n>orbital and somehow bleed off the remaining 0.5fh energy, or does it\n>just ignore the photon? Or are electron orbitals stacked in such a way\n>that this would never happen?\n\nExperts will doubtless correct me but IMHO if the atom is completely\nisolated so that there is no way to dump the surplus energy, then the\nlight will go straight through. This is really the usual situation, the\ncollection of atoms will be transparent (usual for gases for example,\nwhere interaction is usually low).\n\n--\nOz\nThis post is worth absolutely nothing and is probably fallacious.\n\nUse oz@farmeroz.port995.com [ozacoohdb@despammed.com functions].\nBTOPENWORLD address has ceased. DEMON address has ceased.\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>Michael Hobbs <mike@hobbshouse.org> writes
>I wasn't trying to get into the wave vs. particle mess, I'm just wondering
>how we see the interference we do, even if we just stick with waves.
>Let's forget about photons for a moment, if we can, and just consider the
>Maxwellian wave generated by a single electron that drops its orbital.
>Since the electron doesn't jitter back and forth when it makes its
>transition, can we assume that the resulting Maxwellian wave is no more
>than one wavelength long? (Maybe it does jitter, I don't know.)
It must 'jitter' because the emission times are known. Some take a very
long time, to the extent that they are never seen if the atoms are in
collisions more frequently than the emission time. It also gives the
characteristic width of each emission line which is inversely related to
emission time. If they were one wavelength long then the linewidth would
be incredibly wide rather then (typically) very narrow.
Certainly in most cases its very many cycles of the EM wave. Also
consider that typically the wavelength of a light photon is many
thousands of times larger that the emitting atom.
>Now, *if* the Maxwellian wave is just one wavelength long,
It's not. I once did calculate it for a yellow sodium line, IIRC it was
millions of wavelengths.
>> That is in almost all circumstances we have an incident wave of *far*
>> too low an intensity to cause the jumping of the energy gap, but if the
>> resonant frequency 'of the gap' and the frequency of the incident light
>> is *nearly right* then energy gets stored in the system until it either
>> causes a jump in the energy gap, or the light/energy is re-radiated
>> away.
>
>That reminds me of another question that I probably shouldn't get into
>here. If an electron requires a photon with energy = fh to jump one
>orbital and a photon with energy, say, = 1.7fh to jump two orbitals,
>what happens when it encounters a photon with 1.5f? Does it jump one
>orbital and somehow bleed off the remaining .5fh energy, or does it
>just ignore the photon? Or are electron orbitals stacked in such a way
>that this would never happen?
Experts will doubtless correct me but IMHO if the atom is completely
isolated so that there is no way to dump the surplus energy, then the
light will go straight through. This is really the usual situation, the
collection of atoms will be transparent (usual for gases for example,
where interaction is usually low).
--
Oz
This post is worth absolutely nothing and is probably fallacious.
Use oz@farmeroz.port995.com [ozacoohdb@despammed.com functions].
BTOPENWORLD address has ceased. DEMON address has ceased.
Michael Hobbs
Oct20-04, 03:18 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>\nOz <oz@farmeroz.port995.com> wrote:\n> Experts will doubtless correct me but IMHO if the atom is completely\n> isolated so that there is no way to dump the surplus energy, then the\n> light will go straight through. This is really the usual situation, the\n> collection of atoms will be transparent (usual for gases for example,\n> where interaction is usually low).\n\nConsidering that the frequency of light can be any arbitrary value, due\nto red shift and blue shift, I find it to be incredible that any electron\ncan absorb the precise energy available in a given beam of light.\nCertainly, there must be *some* rounding going on. Perhaps with some\nconversion to/from thermal kinetic energy?\n\n-- Mike\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>Oz <oz@farmeroz.port995.com> wrote:
> Experts will doubtless correct me but IMHO if the atom is completely
> isolated so that there is no way to dump the surplus energy, then the
> light will go straight through. This is really the usual situation, the
> collection of atoms will be transparent (usual for gases for example,
> where interaction is usually low).
Considering that the frequency of light can be any arbitrary value, due
to red shift and blue shift, I find it to be incredible that any electron
can absorb the precise energy available in a given beam of light.
Certainly, there must be *some* rounding going on. Perhaps with some
conversion to/from thermal kinetic energy?
-- Mike
<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\nMichael Hobbs <mike@hobbshouse.org> writes\n>Oz <oz@farmeroz.port995.com> wrote:\n>> Experts will doubtless correct me but IMHO if the atom is completely\n>> isolated so that there is no way to dump the surplus energy, then the\n>> light will go straight through. This is really the usual situation, the\n>> collection of atoms will be transparent (usual for gases for example,\n>> where interaction is usually low).\n>\n>Considering that the frequency of light can be any arbitrary value, due\n>to red shift and blue shift, I find it to be incredible that any electron\n>can absorb the precise energy available in a given beam of light.\n>Certainly, there must be *some* rounding going on. Perhaps with some\n>conversion to/from thermal kinetic energy?\n\nOf course I simplified. There are some other methods to dump energy\n(\'permanently\') or otherwise accept a range of energy.\n\n1) In a gas (actually any collection of atoms) there will be a range of\nvelocities. The energy of the incident photon will thus not be precisely\nthe same for all the atoms in the collection. Consequently the\ncollection of atoms will absorb a finite range of incident photon energy\nbecause each will have a different velocity. The same effect can be seen\nin reverse with sodium lighting. The conventional sodium lights are\nalmost entirely in two narrow bands in the yellow, and blue/green/red\nobjects look black. High pressure/temp sodium lighting (which is more\nwhite with a tinge of red) widens these bands so they cover most of the\nvisible spectrum. One can see this equivalently as a time-dilation\neffect.\n\n2) In most situations atoms are not isolated, but come into collision\nwith other atoms either \'continually\' (eg a solid) or frequently (eg a\nnon-rareified gas). In these situations its relatively easy for energy\nto be dumped into kinetic energy (eg phonons or translational).\n\n3) Even in an isolated molecule there are ways to dump energy. There\nwill be a whole raft of (quantised) rotational and vibrational modes\nwhich in complex molecules can effectively amount to a continuum.\n\nAfter all if this was not the case the various spectroscopy techniques\nwould have very limited value.\n\n--\nOz\nThis post is worth absolutely nothing and is probably fallacious.\n\nUse oz@farmeroz.port995.com [ozacoohdb@despammed.com functions].\nBTOPENWORLD address has ceased. DEMON address has ceased.\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>Michael Hobbs <mike@hobbshouse.org> writes
>Oz <oz@farmeroz.port995.com> wrote:
>> Experts will doubtless correct me but IMHO if the atom is completely
>> isolated so that there is no way to dump the surplus energy, then the
>> light will go straight through. This is really the usual situation, the
>> collection of atoms will be transparent (usual for gases for example,
>> where interaction is usually low).
>
>Considering that the frequency of light can be any arbitrary value, due
>to red shift and blue shift, I find it to be incredible that any electron
>can absorb the precise energy available in a given beam of light.
>Certainly, there must be *some* rounding going on. Perhaps with some
>conversion to/from thermal kinetic energy?
Of course I simplified. There are some other methods to dump energy
('permanently') or otherwise accept a range of energy.
1) In a gas (actually any collection of atoms) there will be a range of
velocities. The energy of the incident photon will thus not be precisely
the same for all the atoms in the collection. Consequently the
collection of atoms will absorb a finite range of incident photon energy
because each will have a different velocity. The same effect can be seen
in reverse with sodium lighting. The conventional sodium lights are
almost entirely in two narrow bands in the yellow, and blue/green/red
objects look black. High pressure/temp sodium lighting (which is more
white with a tinge of red) widens these bands so they cover most of the
visible spectrum. One can see this equivalently as a time-dilation
effect.
2) In most situations atoms are not isolated, but come into collision
with other atoms either 'continually' (eg a solid) or frequently (eg a
non-rareified gas). In these situations its relatively easy for energy
to be dumped into kinetic energy (eg phonons or translational).
3) Even in an isolated molecule there are ways to dump energy. There
will be a whole raft of (quantised) rotational and vibrational modes
which in complex molecules can effectively amount to a continuum.
After all if this was not the case the various spectroscopy techniques
would have very limited value.
--
Oz
This post is worth absolutely nothing and is probably fallacious.
Use oz@farmeroz.port995.com [ozacoohdb@despammed.com functions].
BTOPENWORLD address has ceased. DEMON address has ceased.
Ralph Hartley
Oct20-04, 10:48 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>\n\nMichael Hobbs wrote:\n\n> Considering that the frequency of light can be any arbitrary value, due\n> to red shift and blue shift, I find it to be incredible that any electron\n> can absorb the precise energy available in a given beam of light.\n\nWhat an atom "cares" about is the frequency in its rest frame. In those\nterms the line may be exceedingly narrow; the frequency must match almost\nexactly.\n\n> Certainly, there must be *some* rounding going on.\n\nYes, the inherent width of the line is inversely proportional to the\nlifetime of the exited state (because the energy of the exited state is\nuncertain). So if the exited state lasts, on average, about one second,\nthen the frequency has to be within about 1Hz (factors of h cancel out,\npossible small factors like Pi ignored).\n\nIn real experiments, the width of a spectral line will generally be\ngreater, for several reasons.\n\nFirst, at nonzero temperature, the atom will have an unknown velocity, and\nthe frequency it absorbs will depend on that.\n\nThat lets astronomers measure the temperature of a gas by observing the\nwidth of a spectral line.\n\nThe same effect can be used to cool atoms in the lab, by bombarding them\nwith light that they can absorb only if they are moving towards the light\nsource. The momentum of the absorbed photon slows the atom down.\n\nEven at zero temperature, the uncertainty between an atom\'s position and\nmomentum may increase the atom\'s line with, depending on how it is prepared.\n\nFinally, it is difficult to expose an atom to perfectly monochromatic\nlight. In fact it is impossible in any experiment carried out in a finite\namount of time. Time and bandwidth are again inversely proportional. To\nmake the light monochromatic to within about 1Hz, it has to last at least\nabout a second.\n\nThat means that to see the inherent line width of an atom, you must expose\nit to light for at least the lifetime of the excited state.\n\nRalph Hartley\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>Michael Hobbs wrote:
> Considering that the frequency of light can be any arbitrary value, due
> to red shift and blue shift, I find it to be incredible that any electron
> can absorb the precise energy available in a given beam of light.
What an atom "cares" about is the frequency in its rest frame. In those
terms the line may be exceedingly narrow; the frequency must match almost
exactly.
> Certainly, there must be *some* rounding going on.
Yes, the inherent width of the line is inversely proportional to the
lifetime of the exited state (because the energy of the exited state is
uncertain). So if the exited state lasts, on average, about one second,
then the frequency has to be within about 1Hz (factors of h cancel out,
possible small factors like \Pi ignored).
In real experiments, the width of a spectral line will generally be
greater, for several reasons.
First, at nonzero temperature, the atom will have an unknown velocity, and
the frequency it absorbs will depend on that.
That lets astronomers measure the temperature of a gas by observing the
width of a spectral line.
The same effect can be used to cool atoms in the lab, by bombarding them
with light that they can absorb only if they are moving towards the light
source. The momentum of the absorbed photon slows the atom down.
Even at zero temperature, the uncertainty between an atom's position and
momentum may increase the atom's line with, depending on how it is prepared.
Finally, it is difficult to expose an atom to perfectly monochromatic
light. In fact it is impossible in any experiment carried out in a finite
amount of time. Time and bandwidth are again inversely proportional. To
make the light monochromatic to within about 1Hz, it has to last at least
about a second.
That means that to see the inherent line width of an atom, you must expose
it to light for at least the lifetime of the excited state.
Ralph Hartley
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