Wavetrains, Photons, Coherence, and Small Blind Men Describing a Mouse as an Elephant [Archive]

Steve McDaniel

May7-04, 06:40 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>I am not a scientist. I am an engineer and patent agent who has\nwritten a couple of patents for inventors in the photonics field and\nread enough to know that I know very little. I just reviewed a very\nconfusing thread from 1999 on the "Length of wavetrain of a single\nphoton" which makes me very leary of asking simple questions and a bit\ndisappointed in the scientific community generally and of discussion\nthreads specifically. Nevertheless I proceed (ad nauseum even).\n\nHowever, before I pester the reader with a series of rapid fire\nquestions let me provide a bit of background that stimulated my\nresearch into this topic and provide a common foundation for my\nquestions (and the answers). In "Fundamentals of Physics" by Halliday\nand Resnick 1974 edition [my old physics textbook from college] page\n710 states "If we turn from radio sources to common sources of visible\nlight, such as incandescent wires ... we become aware of a fundamental\ndifference [when compared to antennas]. In ... these sources the\nfundamental light emission processes occur in individual atoms and\nthese atoms do not act together in a cooperative (i.e. coherent way).\nThe act of light emission by a single atom [which I thought was a\nphoton] takes, in a typical case, about 10^-8 seconds and the emitted\nlight is properly described as a wavetrain rather than as a wave. For\nemission times such as these the wavetrains are a few meters long."\n\n"Interference effects from ordinary light sources may be produced\n....[The text then proceeds to describe a single slit follow by a\ndouble slit] The diffracted beams [from the double slit] thus\nrepresent the same population of wavetrains and are coherent with\nrespect to each other."\n\nIs the above text refering to a photon as a wavetrain? I had assumed\nthat a wavetrain is a concatenation of photons (such as through\nstimulated emission or antenna excitation) that are locally coherent\nbut may drift as more photons are concatenated resulting in\nincoherence in the same wavetrain. Is that view correct?\n\nHow does one split a wavetrain (a concatenated photon stream?) into\ntwo paths such as the double slits and provide cloned wavetrains to\nboth paths? How does splitting or cloning effect the photons (if\nthere is such a thing) involved in the wavetrains?\n\nAre the streams of photons split in two? Do multiple quanta of\nphotons occupy the same space (i.e. the variable n in plancks equation\nE=nhv). Is a wavetrain then actually a train of concatenated\noverlapping photon sets? Or does n describe the harmonic number or\nenergy level of an oscillating self supporting electromagetic field\ni.e. a photon that moves at the speed of light? Can I subtract quanta\nfrom the photon and essentially clone the photon at lower energy\nlevels?\n\nIs each photon or quanta thereof directed to one slit or the other\nwith a certain probability according to spatial orientation?\n\nIs there really a quanta (i.e. the variable n in plancks equation\nE=nhv) involved with photons or is that simply an atomic effect and\nall bets are off once the photon is emitted? Is there really such a\nthing as photons? See for example THE PHOTON FACT OR FICTION? By BERT\nSCHREIBER.\n\nLast of all. How in the world can the emission of light take so long\nin an atom (several meters worth of light as suggest in the above\nreference) and how can the photons or wavetrains be so much larger\nthat the atoms from which they are emitted or the wavelengths they\ncorrespond to? What is so strange about that is that the small energy\ndrops would seem to take longer to occur/emit than the larger energy\ndrops.\n\nI seek clarification not confusion so please respond accordingly.\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>I am not a scientist. I am an engineer and patent agent who has
written a couple of patents for inventors in the photonics field and
read enough to know that I know very little. I just reviewed a very
confusing thread from 1999 on the "Length of wavetrain of a single
photon" which makes me very leary of asking simple questions and a bit
disappointed in the scientific community generally and of discussion
threads specifically. Nevertheless I proceed (ad nauseum even).

However, before I pester the reader with a series of rapid fire
questions let me provide a bit of background that stimulated my
research into this topic and provide a common foundation for my
questions (and the answers). In "Fundamentals of Physics" by Halliday
and Resnick 1974 edition [my old physics textbook from college] page
710 states "If we turn from radio sources to common sources of visible
light, such as incandescent wires ... we become aware of a fundamental
difference [when compared to antennas]. In ... these sources the
fundamental light emission processes occur in individual atoms and
these atoms do not act together in a cooperative (i.e. coherent way).
The act of light emission by a single atom [which I thought was a
photon] takes, in a typical case, about 10^-8 seconds and the emitted
light is properly described as a wavetrain rather than as a wave. For
emission times such as these the wavetrains are a few meters long."

"Interference effects from ordinary light sources may be produced
....[The text then proceeds to describe a single slit follow by a
double slit] The diffracted beams [from the double slit] thus
represent the same population of wavetrains and are coherent with
respect to each other."

Is the above text refering to a photon as a wavetrain? I had assumed
that a wavetrain is a concatenation of photons (such as through
stimulated emission or antenna excitation) that are locally coherent
but may drift as more photons are concatenated resulting in
incoherence in the same wavetrain. Is that view correct?

How does one split a wavetrain (a concatenated photon stream?) into
two paths such as the double slits and provide cloned wavetrains to
both paths? How does splitting or cloning effect the photons (if
there is such a thing) involved in the wavetrains?

Are the streams of photons split in two? Do multiple quanta of
photons occupy the same space (i.e. the variable n in plancks equation
E=nhv). Is a wavetrain then actually a train of concatenated
overlapping photon sets? Or does n describe the harmonic number or
energy level of an oscillating self supporting electromagetic field
i.e. a photon that moves at the speed of light? Can I subtract quanta
from the photon and essentially clone the photon at lower energy
levels?

Is each photon or quanta thereof directed to one slit or the other
with a certain probability according to spatial orientation?

Is there really a quanta (i.e. the variable n in plancks equation
E=nhv) involved with photons or is that simply an atomic effect and
all bets are off once the photon is emitted? Is there really such a
thing as photons? See for example THE PHOTON FACT OR FICTION? By BERT
SCHREIBER.

Last of all. How in the world can the emission of light take so long
in an atom (several meters worth of light as suggest in the above
reference) and how can the photons or wavetrains be so much larger
that the atoms from which they are emitted or the wavelengths they
correspond to? What is so strange about that is that the small energy
drops would seem to take longer to occur/emit than the larger energy
drops.

I seek clarification not confusion so please respond accordingly.

=?ISO-8859-15?Q?J=FCrgen?= Appel

May11-04, 05:10 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>Steve McDaniel schrieb:\n\n> Is the above text refering to a photon as a wavetrain? I had assumed\n> that a wavetrain is a concatenation of photons (such as through\n> stimulated emission or antenna excitation) that are locally coherent\n> but may drift as more photons are concatenated resulting in\n> incoherence in the same wavetrain. Is that view correct?\n\nStrictly speaking, a photon is a unit of excitation of a particular field\nmode. As such, it has a sharply defined wavevector and energy. That means,\nthat its momentum and frequency are specified to one particular value and\nthat a photon lasts for an infinite time and has infinite length (as is\nobvious for a signal with a delta-frequancy-spectrum).\n\nIn practice such photons cannot be generated nor is it useful to use this\nidealistic picture.\n\nTo understand a more realistic description, one first has to understand,\nthat in quantum mechanics a particle can be in a coherent superposition of\nmany states at the same time and only by measuring, it is projected to a\nparticular one.\nSo observing a single field mode with at most one photon, there can be no\nphoton |psi>=|0>, there can be one photon |psi>=|1> or there can be any\nsuperposition of these states |psi>=alpha |0> + beta |1> with |alpha|^2+\nbeta|^2=1 (This is depictured by Schrödinger\'s cat being in a\nsuperposition of being alive or dead often in literature).\nWhen you actually measure the photon number you will find no photon with\nprobability |alpha|^2 and one photon with probability |beta|^2.\n\n\nWhen one talks about a single photon, what is usually referred to is a\ncoherent superposition of exactly one single photon being in one of a set\nof modes:\n\n|1> := Integral_k psi(k) |1_k> dk with psi(k) a weight function with\nIntegral_k psi(k) dk=1 and |1_k> being the state with exactly 1 photon in\nthe field mode corresponding to the wave vector k (strictly speaking, we\nshould have includes polarization also, but let\'s leave this out for\nclarity by now).\n\nThe state |1> then does occupy a certain spectral width and as such may\nhave a limited duration and length.\n\n> How does one split a wavetrain (a concatenated photon stream?) into\n> two paths such as the double slits and provide cloned wavetrains to\n> both paths?\n\nIf you do not have vacuum any more, you have to solve maxwells equations\nwith the according boundary conditions and then decompose the solutions\ninto their modes. Photons then again are defined as units of excitation of\nthese modes.\nChanging the boundary conditions in this way modifies, what a particular\nphoton is or what photons may exist after all. For example putting to\nhigly reflecting mirrors very close to each other reduces the number of\navailable modes. This decrease of the vacuum mode density has been\nmeasured (casimir effect) and it has been found, that excited atoms decay\nmore slowly if there are less vacuum modes available where they could send\ntheir energy into.\n\n> How does splitting or cloning effect the photons (if\n> there is such a thing) involved in the wavetrains?\nAfter a slit the view of the photon being in one or the other arm leads to\nwrong predictions. You rather get a superposition of the states with the\nphoton propagating in each angle according the the Intensity distribution\nof the diffraction pattern. Only after you actually detect a photon\nspecifying the photon\'s location makes sense.\n\n> Are the streams of photons split in two?\nIn this sense a mach-Zehnder-Interferometer splits photons:\nD2\n|\n/--b---/--- D1\n| |\n->-/---a--/\n\nAfter a photon hits the first 50% beam splitter, it is in a coherent\nsuperposition of being in channel a and b. After they recombine at the\nsecond 50:50 splitter all photons hit detector D1 and none hit D2 (as can\nalso be calculated via classical electrodynamics).\n\nIf you block path b you will see, that now the probability of getting a\nclick at D1 od D2 is equal. So a single photon hitting D2 must somehow\nhave "seen" that something is blocking the path that it did _not_ take =\n>\nImagining a photon as a localized particle after a beamsplitter leads to\nstrange paradoxa. A photon is no "local" concept.\n\n> Do multiple quanta of\n> photons occupy the same space (i.e. the variable n in plancks equation\n> E=nhv).\nYes. A two-photon-state |2> just means that you have two units of\nexcitation in the according mode.\n\n> Is a wavetrain then actually a train of concatenated\n> overlapping photon sets?\nA wavetrain corresponds to a superposition of photons in different modes\n(s.a). They don\'t follow each other like a in a queue, they are just\nexcitations of the field modes making up that pulse giving rise to bigger\namplitudes of the field.\n\n> Is there really a quanta (i.e. the variable n in plancks equation\n> E=nhv) involved with photons or is that simply an atomic effect and\n> all bets are off once the photon is emitted? Is there really such a\n> thing as photons? See for example THE PHOTON FACT OR FICTION? By BERT\n> SCHREIBER.\n\nIn my optinion there is such a thing as the photon number, but it is not\ncorrect to imagine that a light pulse always actually _has_ a defined\nnumber of quanta. In fact, if you attenuate a laser beam to the power of a\nsingle photon, you will find, that you won\'t get a |1>-State but a\ncoherent superposition of the vacuum |0>, the single photon state |1> and\nslight contributions of higher photon number states |n> wich only _on\naverage_ give you a photon number of 1.\n\nThe photon number, the electric field strength and the magnetic field\nstrength for example are noncommuting obervables. They cannot be measured\nsharply at the same time (like momentum and location). If you prepare a\nlight states field strength precicely (what a laser does to a good\napproximation), you cannot predict what a photon-number measurement will\nresult to, you can only predict the statistics.\n\nSo producing a real |1>-state is a major experimental effort and subject of\nrecent research. It\'s very hard to really send a single photon into a well\ndefined spatiotemporal mode without having too much |0> contribution.\nSee http://www.uni-konstanz.de/quantum-optics/quantech/qlight.html\n\n> Last of all. How in the world can the emission of light take so long\n> in an atom (several meters worth of light as suggest in the above\n> reference) and how can the photons or wavetrains be so much larger\n> that the atoms from which they are emitted or the wavelengths they\n> correspond to?\n\nAn excited atom has a certain livetime. After we excite an atom, we cannot\nsay exactly when it will decay, so there is a certain time slot in which\nthe photon will be emitted. Thus after a time the light field is in a\ncoherent superposition of the photon having been emitted at different\ntimes (with contributions according tho the probability that the atom\ndecays just then). This gives the "length" of that photon (which is\nsomewhat pointless, since there is no way to actually measure this\nlength).\n\n> What is so strange about that is that the small energy\n> drops would seem to take longer to occur/emit than the larger energy\n> drops.\n\nWhy so?\n\n\nBest regards,\nJürgen\n--\nGPG key:\nhttp://pgp.mit.edu:11371/pks/lookup?search=3DJ%FCrgen+Appel&op=3Dget\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>Steve McDaniel schrieb:

> Is the above text refering to a photon as a wavetrain? I had assumed
> that a wavetrain is a concatenation of photons (such as through
> stimulated emission or antenna excitation) that are locally coherent
> but may drift as more photons are concatenated resulting in
> incoherence in the same wavetrain. Is that view correct?

Strictly speaking, a photon is a unit of excitation of a particular field
mode. As such, it has a sharply defined wavevector and energy. That means,
that its momentum and frequency are specified to one particular value and
that a photon lasts for an infinite time and has infinite length (as is
obvious for a signal with a \delta-frequancy-spectrum).

In practice such photons cannot be generated nor is it useful to use this
idealistic picture.

To understand a more realistic description, one first has to understand,
that in quantum mechanics a particle can be in a coherent superposition of
many states at the same time and only by measuring, it is projected to a
particular one.
So observing a single field mode with at most one photon, there can be no
photon |\psi>=|0>, there can be one photon |\psi>=|1> or there can be any
superposition of these states |\psi>=\alpha |0> + \beta |1> with |\alpha|^2+\beta|^2=1 (This is depictured by Schrödinger's cat being in a
superposition of being alive or dead often in literature).
When you actually measure the photon number you will find no photon with
probability |\alpha|^2 and one photon with probability |\beta|^2.

When one talks about a single photon, what is usually referred to is a
coherent superposition of exactly one single photon being in one of a set
of modes:

|1> := Integral_k \psi(k) |1_k> dk with \psi(k) a weight function with
Integral_k \psi(k) dk=1 and |1_k> being the state with exactly 1 photon in
the field mode corresponding to the wave vector k (strictly speaking, we
should have includes polarization also, but let's leave this out for
clarity by now).

The state |1> then does occupy a certain spectral width and as such may
have a limited duration and length.

> How does one split a wavetrain (a concatenated photon stream?) into
> two paths such as the double slits and provide cloned wavetrains to
> both paths?

If you do not have vacuum any more, you have to solve maxwells equations
with the according boundary conditions and then decompose the solutions
into their modes. Photons then again are defined as units of excitation of
these modes.
Changing the boundary conditions in this way modifies, what a particular
photon is or what photons may exist after all. For example putting to
higly reflecting mirrors very close to each other reduces the number of
available modes. This decrease of the vacuum mode density has been
measured (casimir effect) and it has been found, that excited atoms decay
more slowly if there are less vacuum modes available where they could send
their energy into.

> How does splitting or cloning effect the photons (if
> there is such a thing) involved in the wavetrains?
After a slit the view of the photon being in one or the other arm leads to
wrong predictions. You rather get a superposition of the states with the
photon propagating in each angle according the the Intensity distribution
of the diffraction pattern. Only after you actually detect a photon
specifying the photon's location makes sense.

> Are the streams of photons split in two?
In this sense a mach-Zehnder-Interferometer splits photons:
D2
|
/--b---/--- D1
| |
->-/---a--/

After a photon hits the first 50% beam splitter, it is in a coherent
superposition of being in channel a and b. After they recombine at the
second 50:50 splitter all photons hit detector D1 and none hit D2 (as can
also be calculated via classical electrodynamics).

If you block path b you will see, that now the probability of getting a
click at D1 od D2 is equal. So a single photon hitting D2 must somehow
have "seen" that something is blocking the path that it did _not_ take =
>
Imagining a photon as a localized particle after a beamsplitter leads to
strange paradoxa. A photon is no "local" concept.

> Do multiple quanta of
> photons occupy the same space (i.e. the variable n in plancks equation
> E=nhv).
Yes. A two-photon-state |2> just means that you have two units of
excitation in the according mode.

> Is a wavetrain then actually a train of concatenated
> overlapping photon sets?
A wavetrain corresponds to a superposition of photons in different modes
(s.a). They don't follow each other like a in a queue, they are just
excitations of the field modes making up that pulse giving rise to bigger
amplitudes of the field.

> Is there really a quanta (i.e. the variable n in plancks equation
> E=nhv) involved with photons or is that simply an atomic effect and
> all bets are off once the photon is emitted? Is there really such a
> thing as photons? See for example THE PHOTON FACT OR FICTION? By BERT
> SCHREIBER.

In my optinion there is such a thing as the photon number, but it is not
correct to imagine that a light pulse always actually _has_ a defined
number of quanta. In fact, if you attenuate a laser beam to the power of a
single photon, you will find, that you won't get a |1>-State but a
coherent superposition of the vacuum |0>, the single photon state |1> and
slight contributions of higher photon number states |n> wich only _on
average_ give you a photon number of 1.

The photon number, the electric field strength and the magnetic field
strength for example are noncommuting obervables. They cannot be measured
sharply at the same time (like momentum and location). If you prepare a
light states field strength precicely (what a laser does to a good
approximation), you cannot predict what a photon-number measurement will
result to, you can only predict the statistics.

So producing a real |1>-state is a major experimental effort and subject of
recent research. It's very hard to really send a single photon into a well
defined spatiotemporal mode without having too much |0> contribution.
See http://www.uni-konstanz.de/quantum-optics/quantech/qlight.html

> Last of all. How in the world can the emission of light take so long
> in an atom (several meters worth of light as suggest in the above
> reference) and how can the photons or wavetrains be so much larger
> that the atoms from which they are emitted or the wavelengths they
> correspond to?

An excited atom has a certain livetime. After we excite an atom, we cannot
say exactly when it will decay, so there is a certain time slot in which
the photon will be emitted. Thus after a time the light field is in a
coherent superposition of the photon having been emitted at different
times (with contributions according tho the probability that the atom
decays just then). This gives the "length" of that photon (which is
somewhat pointless, since there is no way to actually measure this
length).

> What is so strange about that is that the small energy
> drops would seem to take longer to occur/emit than the larger energy
> drops.

Why so?

Best regards,
Jürgen
--
GPG key:
http://pgp.mit.edu:11371/pks/lookup?search=3DJ%FCrgen+Appel&op=3Dget

Charles Francis

May12-04, 01:42 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>In message <2a7bb4d3.0405062307.779143bf@posting.google.com >, Steve\nMcDaniel <stevemcdaniel@connect2.com> writes\n>Last of all. How in the world can the emission of light take so long\n>in an atom (several meters worth of light as suggest in the above\n>reference) and how can the photons or wavetrains be so much larger that\n>the atoms from which they are emitted or the wavelengths they\n>correspond to? What is so strange about that is that the small energy\n>drops would seem to take longer to occur/emit than the larger energy drops.\n\nThe interpretation of quantum effects is a controversial field, and most\nphysicists duck out of it. I hold that the wave represents only our\nknowledge of what would happen in a measurement, not the actual\nparticles motion. The particle is always emitted at a point, and in the\nabsence of measurement we model the probability of where it might be\nfound using wave mechanics. We know that wave mechanics gives the right\nanswers. What we do not know is why.\n\n--\nCharles Francis\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>In message <2a7bb4d3.0405062307.779143bf@posting.google.com>, Steve
McDaniel <stevemcdaniel@connect2.com> writes
>Last of all. How in the world can the emission of light take so long
>in an atom (several meters worth of light as suggest in the above
>reference) and how can the photons or wavetrains be so much larger that
>the atoms from which they are emitted or the wavelengths they
>correspond to? What is so strange about that is that the small energy
>drops would seem to take longer to occur/emit than the larger energy drops.

The interpretation of quantum effects is a controversial field, and most
physicists duck out of it. I hold that the wave represents only our
knowledge of what would happen in a measurement, not the actual
particles motion. The particle is always emitted at a point, and in the
absence of measurement we model the probability of where it might be
found using wave mechanics. We know that wave mechanics gives the right
answers. What we do not know is why.

--
Charles Francis

Oz

May12-04, 05:21 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>Jürgen Appel <jappel@linux01.gwdg.de> writes\n\n>Strictly speaking, a photon is a unit of excitation of a particular field\n>mode. As such, it has a sharply defined wavevector and energy. That means,\n>that its momentum and frequency are specified to one particular value and\n>that a photon lasts for an infinite time and has infinite length (as is\n>obvious for a signal with a delta-frequancy-spectrum).\n\nYes. Clearly impossible.\n\n>In practice such photons cannot be generated nor is it useful to use this\n>idealistic picture.\n\nYes, but ....\n\n>To understand a more realistic description, one first has to understand,\n>that in quantum mechanics a particle can be in a coherent superposition of\n>many states at the same time and only by measuring, it is projected to a\n>particular one.\n\nFine.\n\n>So observing a single field mode with at most one photon, there can be no\n>photon |psi>=|0>, there can be one photon |psi>=|1> or there can be any\n>superposition of these states |psi>=alpha |0> + beta |1> with |alpha|^2+\n>beta|^2=1 (This is depictured by Schrödinger\'s cat being in a\n>superposition of being alive or dead often in literature).\n\nOK, now I\'m an ignorant. I have no real grasp of the dirac\nrepresentation. I would read that as a superposition of some state |0>\nand some state |1>, but wouldn\'t it more typically be a superposition of\na large number of states: sum |n>?\n\nNB Reading on I note that you use |0> as the vacuum state (|>).\nI don\'t really grok a \'vacuum state\', unfortunately, and have no idea of\nits properties.\n\n>When you actually measure the photon number you will find no photon with\n>probability |alpha|^2 and one photon with probability |beta|^2.\n\nYup. You only find one photon of defined energy, this is by definition\nif its a one-photon state.\n\n>When one talks about a single photon, what is usually referred to is a\n>coherent superposition of exactly one single photon being in one of a set\n>of modes:\n>\n>|1> := Integral_k psi(k) |1_k> dk with psi(k) a weight function with\n>Integral_k psi(k) dk=1 and |1_k> being the state with exactly 1 photon in\n>the field mode corresponding to the wave vector k (strictly speaking, we\n>should have includes polarization also, but let\'s leave this out for\n>clarity by now).\n\n<Gulp> Is this the proper way to say what I said above?\n\nIn effect are you not synthesising a waveform by expressing it as its\nfourier series? I have no problem with this whatsoever as a mathematical\ntechnique. As a physical one I have a real problem because each of these\nelements suffers from the problem you expressed above, that is they are\nsinewaves that exist for an infinite time. I would be somewhat happier\nif the waveforms were at least ones that nominally tended to zero\noutside the time of interest. Alternatively, and rather more shocking,\nsome evidence that they don\'t rapidly become zero, although h may be\none\'s friend here.\n\n>The state |1> then does occupy a certain spectral width and as such may\n>have a limited duration and length.\n\nOK.\n\n>This decrease of the vacuum mode density has been\n>measured (casimir effect) and it has been found, that excited atoms decay\n>more slowly if there are less vacuum modes available where they could send\n>their energy into.\n\nLovely stuff.\n\n>> Are the streams of photons split in two?\n>In this sense a mach-Zehnder-Interferometer splits photons:\n\n>If you block path b you will see, that now the probability of getting a\n>click at D1 od D2 is equal. So a single photon hitting D2 must somehow\n>have "seen" that something is blocking the path that it did _not_ take =\n\nI\'m not sure this is a good way to see it. Its easier to see the\nlightbeam as a wave, whereupon there is no need to take this view. One\nthen transfers the \'quantumness\' to the detector (and the emitter in\nfact). The detector can only detect a whole photon, usually because it\nrequires a whole quantum of energy to make the step required to make a\ndetection, and equally the emitter can only lose energy in steps of one\nquantums worth (determined by its own ability to change state). Since\none can never know the precise conditions of either the emitting or\ndetecting atom (probably even in principle) both the emission and\ndetection are apparently random.\n\n>Imagining a photon as a localized particle after a beamsplitter leads to\n>strange paradoxa. A photon is no "local" concept.\n\nI agree. Completely.\n\n>A wavetrain corresponds to a superposition of photons in different modes\n>(s.a). They don\'t follow each other like a in a queue, they are just\n>excitations of the field modes making up that pulse giving rise to bigger\n>amplitudes of the field.\n\nOr you can see it simply as a field, no \'photons\' at all. The\nquantumness applies to the internal energy banding inherent in the\nemitter and/or the detector.\n\n>In my optinion there is such a thing as the photon number, but it is not\n>correct to imagine that a light pulse always actually _has_ a defined\n>number of quanta. In fact, if you attenuate a laser beam to the power of a\n>single photon, you will find, that you won\'t get a |1>-State but a\n>coherent superposition of the vacuum |0>, the single photon state |1> and\n>slight contributions of higher photon number states |n> wich only _on\n>average_ give you a photon number of 1.\n\nOk.\n\n>The photon number, the electric field strength and the magnetic field\n>strength for example are noncommuting obervables. They cannot be measured\n>sharply at the same time (like momentum and location). If you prepare a\n>light states field strength precicely (what a laser does to a good\n>approximation), you cannot predict what a photon-number measurement will\n>result to, you can only predict the statistics.\n\nOK. Without fully understanding the consequences, I\'m happy with that.\n\n>So producing a real |1>-state is a major experimental effort and subject of\n>recent research. It\'s very hard to really send a single photon into a well\n>defined spatiotemporal mode without having too much |0> contribution.\n>See http://www.uni-konstanz.de/quantum-optics/quantech/qlight.html\n\nThat site is really *wild*!\nMindblowing!\n\n>An excited atom has a certain livetime. After we excite an atom, we cannot\n>say exactly when it will decay, so there is a certain time slot in which\n>the photon will be emitted. Thus after a time the light field is in a\n>coherent superposition of the photon having been emitted at different\n>times (with contributions according tho the probability that the atom\n>decays just then). This gives the "length" of that photon (which is\n>somewhat pointless, since there is no way to actually measure this\n>length).\n\nAfter much argument here, I have come to agree.\nSo \'in flight\' a \'photon\' has no defined length, typically no defined\nnumber and behaves precisely like a wave. If it looks like a wave, spaks\nlike a wave then why not call it a wave, even if interactions are more\nconveniently described in a quantumlike manner.\n\n\n--\nOz\nThis post is worth absolutely nothing and is probably fallacious.\n\nBTOPENWORLD address about to cease. DEMON address no longer in use.\n>>Use oz@farmeroz.port995.com (whitelist check on first posting)<<\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>Jürgen Appel <jappel@linux01.gwdg.de> writes

>Strictly speaking, a photon is a unit of excitation of a particular field
>mode. As such, it has a sharply defined wavevector and energy. That means,
>that its momentum and frequency are specified to one particular value and
>that a photon lasts for an infinite time and has infinite length (as is
>obvious for a signal with a \delta-frequancy-spectrum).

Yes. Clearly impossible.

>In practice such photons cannot be generated nor is it useful to use this
>idealistic picture.

Yes, but ....

>To understand a more realistic description, one first has to understand,
>that in quantum mechanics a particle can be in a coherent superposition of
>many states at the same time and only by measuring, it is projected to a
>particular one.

Fine.

>So observing a single field mode with at most one photon, there can be no
>photon |\psi>=|0>, there can be one photon |\psi>=|1> or there can be any
>superposition of these states |\psi>=\alpha |0> + \beta |1> with |\alpha|^2+>\beta|^2=1 (This is depictured by Schrödinger's cat being in a
>superposition of being alive or dead often in literature).

OK, now I'm an ignorant. I have no real grasp of the dirac
representation. I would read that as a superposition of some state |0>
and some state |1>, but wouldn't it more typically be a superposition of
a large number of states: sum |n>?

NB Reading on I note that you use |0> as the vacuum state (|>).
I don't really grok a 'vacuum state', unfortunately, and have no idea of
its properties.

>When you actually measure the photon number you will find no photon with
>probability |\alpha|^2 and one photon with probability |\beta|^2.

Yup. You only find one photon of defined energy, this is by definition
if its a one-photon state.

>When one talks about a single photon, what is usually referred to is a
>coherent superposition of exactly one single photon being in one of a set
>of modes:
>
>|1> := Integral_k \psi(k) |1_k> dk with \psi(k) a weight function with
>Integral_k \psi(k) dk=1 and |1_k> being the state with exactly 1 photon in
>the field mode corresponding to the wave vector k (strictly speaking, we
>should have includes polarization also, but let's leave this out for
>clarity by now).

<Gulp> Is this the proper way to say what I said above?

In effect are you not synthesising a waveform by expressing it as its
fourier series? I have no problem with this whatsoever as a mathematical
technique. As a physical one I have a real problem because each of these
elements suffers from the problem you expressed above, that is they are
sinewaves that exist for an infinite time. I would be somewhat happier
if the waveforms were at least ones that nominally tended to zero
outside the time of interest. Alternatively, and rather more shocking,
some evidence that they don't rapidly become zero, although h may be
one's friend here.

>The state |1> then does occupy a certain spectral width and as such may
>have a limited duration and length.

OK.

>This decrease of the vacuum mode density has been
>measured (casimir effect) and it has been found, that excited atoms decay
>more slowly if there are less vacuum modes available where they could send
>their energy into.

Lovely stuff.

>> Are the streams of photons split in two?
>In this sense a mach-Zehnder-Interferometer splits photons:

>If you block path b you will see, that now the probability of getting a
>click at D1 od D2 is equal. So a single photon hitting D2 must somehow
>have "seen" that something is blocking the path that it did _not_ take =

I'm not sure this is a good way to see it. Its easier to see the
lightbeam as a wave, whereupon there is no need to take this view. One
then transfers the 'quantumness' to the detector (and the emitter in
fact). The detector can only detect a whole photon, usually because it
requires a whole quantum of energy to make the step required to make a
detection, and equally the emitter can only lose energy in steps of one
quantums worth (determined by its own ability to change state). Since
one can never know the precise conditions of either the emitting or
detecting atom (probably even in principle) both the emission and
detection are apparently random.

>Imagining a photon as a localized particle after a beamsplitter leads to
>strange paradoxa. A photon is no "local" concept.

I agree. Completely.

>A wavetrain corresponds to a superposition of photons in different modes
>(s.a). They don't follow each other like a in a queue, they are just
>excitations of the field modes making up that pulse giving rise to bigger
>amplitudes of the field.

Or you can see it simply as a field, no 'photons' at all. The
quantumness applies to the internal energy banding inherent in the
emitter and/or the detector.

>In my optinion there is such a thing as the photon number, but it is not
>correct to imagine that a light pulse always actually _has_ a defined
>number of quanta. In fact, if you attenuate a laser beam to the power of a
>single photon, you will find, that you won't get a |1>-State but a
>coherent superposition of the vacuum |0>, the single photon state |1> and
>slight contributions of higher photon number states |n> wich only _on
>average_ give you a photon number of 1.

Ok.

>The photon number, the electric field strength and the magnetic field
>strength for example are noncommuting obervables. They cannot be measured
>sharply at the same time (like momentum and location). If you prepare a
>light states field strength precicely (what a laser does to a good
>approximation), you cannot predict what a photon-number measurement will
>result to, you can only predict the statistics.

OK. Without fully understanding the consequences, I'm happy with that.

>So producing a real |1>-state is a major experimental effort and subject of
>recent research. It's very hard to really send a single photon into a well
>defined spatiotemporal mode without having too much |0> contribution.
>See http://www.uni-konstanz.de/quantum-optics/quantech/qlight.html

That site is really *wild*!
Mindblowing!

>An excited atom has a certain livetime. After we excite an atom, we cannot
>say exactly when it will decay, so there is a certain time slot in which
>the photon will be emitted. Thus after a time the light field is in a
>coherent superposition of the photon having been emitted at different
>times (with contributions according tho the probability that the atom
>decays just then). This gives the "length" of that photon (which is
>somewhat pointless, since there is no way to actually measure this
>length).

After much argument here, I have come to agree.
So 'in flight' a 'photon' has no defined length, typically no defined
number and behaves precisely like a wave. If it looks like a wave, spaks
like a wave then why not call it a wave, even if interactions are more
conveniently described in a quantumlike manner.

--
Oz
This post is worth absolutely nothing and is probably fallacious.

BTOPENWORLD address about to cease. DEMON address no longer in use.
>>Use oz@farmeroz.port995.com (whitelist check on first posting)<<

Oz

May17-04, 06:16 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>Charles Francis <charles@lluestfarmpoultry.co.uk> writes\n\n>The interpretation of quantum effects is a controversial field, and most\n>physicists duck out of it. I hold that the wave represents only our knowledge of\n>what would happen in a measurement, not the actual particles motion. The\n>particle is always emitted at a point, and in the absence of measurement we\n>model the probability of where it might be found using wave mechanics. We know\n>that wave mechanics gives the right answers. What we do not know is why.\n\nNB: Remember everyone else is expert, I am not.\n\nWhilst I hold the diametrically opposed position, that all particles are\nwaves and (simplistically) cannot really be localised to more than a few\nwavelengths. When we \'detect\' something at a position, typically the\ndetector is something with a very small wavelength (eg an atom), and it\nis the position of this atom that we consider as some measure of the\n\'position\' of the particle we are detecting. Typically (say) a photon on\na photographic film is \'detected\' by a change in a single silver atom.\n\nIts not the position of the photon (IMHO), that you are detecting, but a\nsingle silver atom that has managed to interact with the field enough to\nchange state. That\'s why, if you see the results of a detector that\nshows individual photon absorption, the pattern comes initially as an\napparently random collection of points and only shows the final picture\nafter many points have been made.\n\nThis is the only reasonable way I can rationalise experiments, eg that\nshow TWO highly attenuated lasers, each being shone through a SINGLE\nslit such that there is never more than one \'photon\' in the apparatus at\nany one time, yet still give a diffraction pattern.\n\nUnfortunately using the wave model has serious disadvantages, as far as\nI can see. We are missing, often its unknowable even in principle, the\nprecise states and behaviour of the detector atom (usually atoms). It is\nthus impossible to produce a useful mathematical model to say which\nsilver atom will change state when exposed to (say) a specified EM wave.\nEven worse, as I understand it, the maths would be \'challenging\'.\n\nFortunately mathematics comes to our aid. We can represent a real wave,\nto any level of accuracy required, by a series of sinewaves by using a\nfourier transformation. This allows us to handle many aspects of a\ncomplex waveform using easy to handle sinewaves. Equally we can\nrepresent any waveform by a series of delta functions (ie as point\nparticles), which nicely handle a different set of problems (and very\nuseful ones). Both are accurate descriptions of some waveform but only\nwhen ALL the components are taken as a whole.\n\nNow many particles (I use this description colloquially), particularly\nphotons, are huge compared to the detecting atom. Light is typically\nabout 10^-6m wavelength whilst an atom is around 10^-10m. Hunting about\nfor some actual figures I find (watch line wrap)\n\nhttp://george.ph.utexas.edu/~dsteck/alkalidata/sodiumnumbers.pdf\n[Table 5]\n\nGives sodium at 589nm (508x10^12Hz) having a transition time of 16ns\nThat is (if I have it right) the transition time is equivalent to some\n8M oscillations of the emitted photon. In reverse it takes some 8M\nwiggles to excite a sodium atom or a wavetrain some 5m long.\n\nI think I will stop there.\n\n--\nOz\nThis post is worth absolutely nothing and is probably fallacious.\n\nBTOPENWORLD address about to cease. DEMON address no longer in use.\n>>Use oz@farmeroz.port995.com (whitelist check on first posting)<<\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>Charles Francis <charles@lluestfarmpoultry.co.uk> writes

>The interpretation of quantum effects is a controversial field, and most
>physicists duck out of it. I hold that the wave represents only our knowledge of
>what would happen in a measurement, not the actual particles motion. The
>particle is always emitted at a point, and in the absence of measurement we
>model the probability of where it might be found using wave mechanics. We know
>that wave mechanics gives the right answers. What we do not know is why.

NB: Remember everyone else is expert, I am not.

Whilst I hold the diametrically opposed position, that all particles are
waves and (simplistically) cannot really be localised to more than a few
wavelengths. When we 'detect' something at a position, typically the
detector is something with a very small wavelength (eg an atom), and it
is the position of this atom that we consider as some measure of the
'position' of the particle we are detecting. Typically (say) a photon on
a photographic film is 'detected' by a change in a single silver atom.

Its not the position of the photon (IMHO), that you are detecting, but a
single silver atom that has managed to interact with the field enough to
change state. That's why, if you see the results of a detector that
shows individual photon absorption, the pattern comes initially as an
apparently random collection of points and only shows the final picture
after many points have been made.

This is the only reasonable way I can rationalise experiments, eg that
show TWO highly attenuated lasers, each being shone through a SINGLE
slit such that there is never more than one 'photon' in the apparatus at
any one time, yet still give a diffraction pattern.

Unfortunately using the wave model has serious disadvantages, as far as
I can see. We are missing, often its unknowable even in principle, the
precise states and behaviour of the detector atom (usually atoms). It is
thus impossible to produce a useful mathematical model to say which
silver atom will change state when exposed to (say) a specified EM wave.
Even worse, as I understand it, the maths would be 'challenging'.

Fortunately mathematics comes to our aid. We can represent a real wave,
to any level of accuracy required, by a series of sinewaves by using a
fourier transformation. This allows us to handle many aspects of a
complex waveform using easy to handle sinewaves. Equally we can
represent any waveform by a series of \delta functions (ie as point
particles), which nicely handle a different set of problems (and very
useful ones). Both are accurate descriptions of some waveform but only
when ALL the components are taken as a whole.

Now many particles (I use this description colloquially), particularly
photons, are huge compared to the detecting atom. Light is typically
about 10^-6m wavelength whilst an atom is around 10^-10m. Hunting about
for some actual figures I find (watch line wrap)

http://george.ph.utexas.edu/~dsteck/alkalidata/sodiumnumbers.pdf
[Table 5]

Gives sodium at 589nm (508x10^12Hz) having a transition time of 16ns
That is (if I have it right) the transition time is equivalent to some
8M oscillations of the emitted photon. In reverse it takes some 8M
wiggles to excite a sodium atom or a wavetrain some 5m long.

I think I will stop there.

--
Oz
This post is worth absolutely nothing and is probably fallacious.

BTOPENWORLD address about to cease. DEMON address no longer in use.
>>Use oz@farmeroz.port995.com (whitelist check on first posting)<<

=?ISO-8859-15?Q?J=FCrgen?= Appel

May17-04, 08:04 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>Hi,\n\nOz wrote:\n\n>>So observing a single field mode with at most one photon, there can be no\n>>photon |psi>=|0>, there can be one photon |psi>=|1> or there can be any\n>>superposition of these states |psi>=alpha |0> + beta |1> with |alpha|^2+\n>>beta|^2=1 (This is depictured by Schrödinger\'s cat being in a\n>>superposition of being alive or dead often in literature).\n>\n> OK, now I\'m an ignorant. I have no real grasp of the dirac\n> representation. I would read that as a superposition of some state |0>\n> and some state |1>, but wouldn\'t it more typically be a superposition of\n> a large number of states: sum |n>?\n>\n> NB Reading on I note that you use |0> as the vacuum state (|>).\n> I don\'t really grok a \'vacuum state\', unfortunately, and have no idea of\n> its properties.\n\n|n> describes a state with exactly n photons. So |0> is the vacuum state.\nThe vacuum is the state, which, when you measure its photon number, always\ngives the value 0. (As noted before, the photon number and the electric\nfield strength cannot be measures sharply simultaneously. So even in the\nvaccum state the electric field strength is zero only on average. It\nfluctuates.)\n\nThis notation just refers to one mode of a light field. A complete\ndescription of the vaccum would be |0,0,0,0,...,0>, the photon number in\neach mode being 0.\n\n>>When you actually measure the photon number you will find no photon with\n>>probability |alpha|^2 and one photon with probability |beta|^2.\n>\n> Yup. You only find one photon of defined energy, this is by definition\n> if its a one-photon state.\nNot necessarily. If you are in a superposition of |psi>=alpha |0> + beta\n1>, and you measure this state\'s photon number, the possible results are 0\nand 1. Only after you measured the number to be 1 (0), the state is |1> (\n0>).\n\nThere is a difference between having a light state |psi>=1/sqrt(2) * (|0> +\n|1>) (a coherent superposition) and having either |0> or |1> with 50%\nprobability each (a so called mixture).\n\n>>> Are the streams of photons split in two?\n>>In this sense a mach-Zehnder-Interferometer splits photons:\n>\n>>If you block path b you will see, that now the probability of getting a\n>>click at D1 od D2 is equal. So a single photon hitting D2 must somehow\n>>have "seen" that something is blocking the path that it did _not_ take =\n>\n> I\'m not sure this is a good way to see it. Its easier to see the\n> lightbeam as a wave, whereupon there is no need to take this view. One\n> then transfers the \'quantumness\' to the detector (and the emitter in\n> fact). The detector can only detect a whole photon, usually because it\n> requires a whole quantum of energy to make the step required to make a\n> detection, and equally the emitter can only lose energy in steps of one\n> quantums worth (determined by its own ability to change state). Since\n> one can never know the precise conditions of either the emitting or\n> detecting atom (probably even in principle) both the emission and\n> detection are apparently random.\n\nIn a way, yes. But this picture suffers from the fact, that if you would\nplace many detectors all around your single excited atom, only one may\nclick. If light were a classical wave, nothing would prevent two detectors\nto click synchronously.\n\n\nIt all boils down to the following:\nLight is neither particle nor wave. But if you look at it as a wave, it\ntends to behave as one, if you investigate its particle properties, it\nbehaves as a particle. Trouble arises, if you try to do both at the same\ntime, then both simplified views fail and only quantum mechanics gives the\nright answer.\n\n\n> After much argument here, I have come to agree.\n> So \'in flight\' a \'photon\' has no defined length, typically no defined\n> number\nThat is wrong: If you have "one photon", the photon number is 1, of course.\nWhat is not defined, is the electric field strength then.\n\n\n\n> and behaves precisely like a wave.\nWhy do you think so? It would be equally justified to say, that \'in flight\'\na photon behaves as a particle.\n\n> If it looks like a wave, spaks\n> like a wave then why not call it a wave,\n> even if interactions are more\n> conveniently described in a quantumlike manner.\n\nThat depends on the type of interaction. If it\'s a diffraction-like\ninteraction, it reacts as a wave. If it\'s a quantized interaction, it\nbehaves as a particle.\n\nIt is like with all simplified pictures: geometrical optics for example\nalso works good for the most part, but if you do the right experiments\n(diffraction for example), you will notice, that in fact light just does\nnot consist of rays but is of a much more involved structure...\n\nBest regards,\nJürgen\n--\nGPG key:\nhttp://pgp.mit.edu:11371/pks/lookup?search=J%FCrgen+Appel&op=get\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>Hi,

Oz wrote:

>>So observing a single field mode with at most one photon, there can be no
>>photon |\psi>=|0>, there can be one photon |\psi>=|1> or there can be any
>>superposition of these states |\psi>=\alpha |0> + \beta |1> with |\alpha|^2+>>\beta|^2=1 (This is depictured by Schrödinger's cat being in a
>>superposition of being alive or dead often in literature).
>
> OK, now I'm an ignorant. I have no real grasp of the dirac
> representation. I would read that as a superposition of some state |0>
> and some state |1>, but wouldn't it more typically be a superposition of
> a large number of states: sum |n>?
>
> NB Reading on I note that you use |0> as the vacuum state (|>).
> I don't really grok a 'vacuum state', unfortunately, and have no idea of
> its properties.

|n> describes a state with exactly n photons. So |0> is the vacuum state.
The vacuum is the state, which, when you measure its photon number, always
gives the value . (As noted before, the photon number and the electric
field strength cannot be measures sharply simultaneously. So even in the
vaccum state the electric field strength is zero only on average. It
fluctuates.)

This notation just refers to one mode of a light field. A complete
description of the vaccum would be |0,0,0,0,...,0>, the photon number in
each mode being .

>>When you actually measure the photon number you will find no photon with
>>probability |\alpha|^2 and one photon with probability |\beta|^2.
>
> Yup. You only find one photon of defined energy, this is by definition
> if its a one-photon state.
Not necessarily. If you are in a superposition of |\psi>=\alpha |0> + \beta
1>, and you measure this state's photon number, the possible results are
and 1. Only after you measured the number to be 1 (0), the state is |1> (
0>).

There is a difference between having a light state |\psi>=1/\sqrt(2) * (|0> +|1>) (a coherent superposition) and having either |0> or |1> with 50%
probability each (a so called mixture).

>>> Are the streams of photons split in two?
>>In this sense a mach-Zehnder-Interferometer splits photons:
>
>>If you block path b you will see, that now the probability of getting a
>>click at D1 od D2 is equal. So a single photon hitting D2 must somehow
>>have "seen" that something is blocking the path that it did _not_ take =
>
> I'm not sure this is a good way to see it. Its easier to see the
> lightbeam as a wave, whereupon there is no need to take this view. One
> then transfers the 'quantumness' to the detector (and the emitter in
> fact). The detector can only detect a whole photon, usually because it
> requires a whole quantum of energy to make the step required to make a
> detection, and equally the emitter can only lose energy in steps of one
> quantums worth (determined by its own ability to change state). Since
> one can never know the precise conditions of either the emitting or
> detecting atom (probably even in principle) both the emission and
> detection are apparently random.

In a way, yes. But this picture suffers from the fact, that if you would
place many detectors all around your single excited atom, only one may
click. If light were a classical wave, nothing would prevent two detectors
to click synchronously.

It all boils down to the following:
Light is neither particle nor wave. But if you look at it as a wave, it
tends to behave as one, if you investigate its particle properties, it
behaves as a particle. Trouble arises, if you try to do both at the same
time, then both simplified views fail and only quantum mechanics gives the
right answer.

> After much argument here, I have come to agree.
> So 'in flight' a 'photon' has no defined length, typically no defined
> number
That is wrong: If you have "one photon", the photon number is 1, of course.
What is not defined, is the electric field strength then.

> and behaves precisely like a wave.
Why do you think so? It would be equally justified to say, that 'in flight'
a photon behaves as a particle.

> If it looks like a wave, spaks
> like a wave then why not call it a wave,
> even if interactions are more
> conveniently described in a quantumlike manner.

That depends on the type of interaction. If it's a diffraction-like
interaction, it reacts as a wave. If it's a quantized interaction, it
behaves as a particle.

It is like with all simplified pictures: geometrical optics for example
also works good for the most part, but if you do the right experiments
(diffraction for example), you will notice, that in fact light just does
not consist of rays but is of a much more involved structure...

Best regards,
Jürgen
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