photon 'detectors' - how reliable?
Bill Wadge
where they are describing one of the stock quantum mechanics
experiments, and noticed something along the lines of
.. towards a detector which records every photon ..
Is this possible? Can one really build a device so sensitive that it will
detect 100% reliably every photon that enters it, and never
go off by accident?
Seems unlikely, but then I'm not a physicist ...
Bill Wadge, Computer Science, University of Victoria.
wwa...@csr.uvic.ca
Sean Merritt
> I was browsing through one of the many 'pop' physics books,
> where they are describing one of the stock quantum mechanics
> experiments, and noticed something along the lines of
>
> .. towards a detector which records every photon ..
>
> Is this possible? Can one really build a device so sensitive that it will
> detect 100% reliably every photon that enters it, and never
> go off by accident?
>
> Seems unlikely, but then I'm not a physicist ...
First, Bill if you are really interested in this stuff, get yourself an
Electro-Optics Handbook, put out by RCA(Solid State division, Electro
Optics and Devices division. This and your last question are fully
answered there.
What your asking about is the "quantum efficiency" of the detector.
If this is a device that uses a method of converting photon
flux into an electrical current(photo-electric device) if we
ignore noise:
qe = number of charge carriers generated divided by number of photons
generated.
let i be the current
" e " " charge
Popt the omptimal power output of the source
h*nu = energy of the photons(h is Planck's constant)
qe =(i/e)/ (Popt/h*nu)
the parameters here are i and Popt. It is easy to see that you
can't hope to get 100% qe, infact real devices seldom do better
than 25%. I think the best photomultipliers are something less
than this. If you consider quantum noise or S/N it just lowers
the qe more. Although it is much more mathematically complicated.
Usually the qe is specified interms of wavelength, and a function
called "responsivity" R = R (lambda).
qe(lambda) = [R(lambda)*h*c]/(e*lambda)
the units of R are [area/power] power is in watts or lumens.
somettimes itis given in [volume/power].
-sjm
--
Sean J. Merritt |"Road-kill has it's seasons just like
Dept of Physics Boston University|anything, there's possums in the autumn
mer...@macro.bu.edu |and farm cats in the spring." T. Waits
Daniel Seeman
>where they are describing one of the stock quantum mechanics
>experiments, and noticed something along the lines of
>
> .. towards a detector which records every photon ..
>
>Is this possible? Can one really build a device so sensitive that it will
>detect 100% reliably every photon that enters it, and never
>go off by accident?
>
>Seems unlikely, but then I'm not a physicist ...
>
Enclose yourself in a room with a light that is controlled by a dim switch. Dim
the light until you can barely see your hand. Then, look at the bulb. At this
point, your eye is detecting every photon that is comming your way (from the
bulb). True, the eye is pretty unique in that its design is ---well pretty
good. Apparently our designer (nature,evolution, god---arguably all are one in
the same) was quite "skilled." However, there are other devices (natural and
man-made) that have the same (and better) efficiency.
One thing that may help you consider this is that a light bulb emits photons in
all directions (except down into the socket---the metal circuit casing stops the
photons). But you only see the ones that are directly along your line of sight.
All other photons in the room are detected or felt by something else (like the
walls of the room heat slightly etc...). So, you do not (nor does any other
singular detector) see ALL the photons emitted by a light source unless that
source is directing all its photons toward the detector. When you look at it
in that "light---pun intended" these detectors do not seem so awesome.
Hope this helps...
dks.
r...@enh.nist.gov
>>In article <wwadge.727584610@csr> wwa...@csr.UVic.CA (Bill Wadge) writes:
>>> Is this possible? Can one really build a device so sensitive that it will
>>> detect 100% reliably every photon that enters it, and never
>>> go off by accident?
>>>
>>> Seems unlikely, but then I'm not a physicist ...
>>
>>If this is a device that uses a method of converting photon
>>flux into an electrical current(photo-electric device) if we
>>ignore noise:
[description of detector responsivity deleted]
>>the parameters here are i and Popt. It is easy to see that you
>>can't hope to get 100% qe, infact real devices seldom do better
>>than 25%. I think the best photomultipliers are something less
>>than this. If you consider quantum noise or S/N it just lowers
>>the qe more. Although it is much more mathematically complicated.
>>
Depending on the photon energy this is not always the case. Detectors that
have quantum efficiencies greater than 1 are available, but you need to work in
the extreme ultraviolet or soft x-ray region. These are not photoemmitters,
but semicondutor devices where the incident radiation creates electron-hole
pairs in a diode. For example, silcon has an electron-hole pair energy of
about 3.63eV (I think). If a photon with 7.26eV is absorbed, you can get two
electron-hole pairs, and the quantum efficiency is 2, or 200% depending on how
you want to put it.
>>Usually the qe is specified interms of wavelength, and a function
>>called "responsivity" R = R (lambda).
>>
>>
>>qe(lambda) = [R(lambda)*h*c]/(e*lambda)
>>
>>the units of R are [area/power] power is in watts or lumens.
>>somettimes itis given in [volume/power].
>>
Well, we just leave it as QE(lamda) in units of electron/photon.
But returning to the original question, even though you can get more than one
electron out of your detector for every photon going in, you don't actually,
detect every photon and no others. As was mentioned there is noise, usually
from thermal excitation of electrons. Also there are loss mechanism that come
into play so not every photon gets detected. There are reflections off the
front surface (and any other boundaries inside) of the detector, there are dead
regions in the detector (SiO2 absorbs photons but does not result in any signal
in a photodiode), and there is always the possibilty that the photon will just
go through the detector without interacting with it at all.
To my mind, the "perfect" detector can't be made, but it is possible to
calibrate detectors by knowing their quantum efficiency and that allows you to
do science as if you did see every photon coming your way, even when you might
only "really" see one in every ten.
+-----------------------------------------------------------------------------+
+ From the desk of: | r...@enh.nist.gov OR rev@nbsenh +
+ Rob Vest |---------------------------------------------+
+ National Institute of | Why are all the good quotes taken by +
+ Standards & Technology | some other .sig file?? +
+-----------------------------------------------------------------------------+
Aephraim M. Steinberg
>where they are describing one of the stock quantum mechanics
>experiments, and noticed something along the lines of
>
> .. towards a detector which records every photon ..
>
>Is this possible? Can one really build a device so sensitive that it will
>detect 100% reliably every photon that enters it, and never
>go off by accident?
>
>Seems unlikely, but then I'm not a physicist ...
Does seem unlikely, doesn't it? But it will happen one day (just about).
Some photodiodes currently quote quantum efficiencies very close to
100%, but not when operating in single-photon detection mode. That is, they
yield a continuous current proportional to the incoming intensity and
supposedly nearly every incoming photon yields an electron-hole pair. (I
don't know how-- or whether-- this last fact is confirmed.)
The SINGLE-photon detectors of preference until recently were photomultiplier
tubes, with efficiencies ranging from less than 1% to perhaps 5%. This is
one of the reasons (along with the geometric nature of the decay process
originally used) that the "detection loophole" was not closed by experiments
like those of Clauser and of Aspect. In recent years, pmt's have been
superceded by avalanche photodiodes, frequently with quantum efficiencies
of 10 or 20%. We have one photodiode from EG&G which is about 40% efficient.
VERY recently, however, there has been some progress towards nearly-100%
efficient detectors. EG&G offers rather expensive "single-photon counting
modules" which we have measured to have efficiencies of approximately 75%
when they are sufficiently overbiased-- the active element in these modules
is still an APD. On a different tack, Rockwell has been working on
"solid state photomultipliers" which theoretically should have quantum
efficiencies as high as 90 or 95%, although unlike the APD's, they need to
be cooled to near-liquid Helium temperatures. We have measured efficiencies
close to 70% for these devices as well, and it is quite likely that there
were other losses present during that measurement, so the actual efficiency
could well be higher. Needless to say, both companies (and others) are
working towards further improvements.
As far as I know, most people only care in a "linear" fashion about
such efficiencies. For tests of Bell inequalities, on the other hand,
there is a sharp cutoff of 83% (ignoring all other experimental imperfections)
for eliminating the detection loophole. (Although some recent papers
propose ways of reducing this cutoff to around 70%, it is nearly certain
that there will always be some sharp cutoff, and that these schemes will
be less feasible than those which require 83%.)
Anyway, don't trust everything you read. The same physics book that
mentioned 100% efficient photodetectors could well have mentioned frictionless
pulleys and massless ropes: With many of the former and one of the latter,
one could construct a 100% efficient photodetector simply by amplifying
the "light pressure" acting on a ball attached to one end of the rope!
--
Aephraim M. Steinberg | "WHY must I treat the measuring
UCB Physics | device classically?? What will
aeph...@physics.berkeley.edu | happen to me if I don't??"
| -- Eugene Wigner
Aephraim M. Steinberg
>In article <wwadge.727584610@csr> wwa...@csr.UVic.CA (Bill Wadge) writes:
>
>>
>> .. towards a detector which records every photon ..
>>
>> Is this possible? Can one really build a device so sensitive that it will
>> detect 100% reliably every photon that enters it, and never
>> go off by accident?
>
>Electro-Optics Handbook, put out by RCA(Solid State division, Electro
>Optics and Devices division. This and your last question are fully
>answered there.
>
>the parameters here are i and Popt. It is easy to see that you
>can't hope to get 100% qe, infact real devices seldom do better
>than 25%. I think the best photomultipliers are something less
Sorry, in my earlier post I forgot to answer the "by accident" question.
These detectors DO periodically go off by accident, primarily due to
thermal fluctuations (this is why the devices are run cold, typically
around -25C) and impurities in the semiconductors. The resulting
pulses are called "dark counts"-- by the way, your eye has much the same
problem (it too acts as a single-photon detector in sufficiently low light,
albeit of quite low efficiency. I don't think anyone knows for certain
how much of the efficiency problem is due to the intrinsic "quantum
efficiency" of the cells in your retina and how much is due to geometric
effects). Typical APD's have dark count rates of 500 or 1000 counts per
second, but can register 100s of thousands of real counts per second
before they start to saturate (and drop in efficiency). The saturation
is usually due to the speed of the electronics, and can therefore be
pushed to higher count rates by building faster circuitry.
The SPCM's I mentioned are "super-low" impurity-level devices, so their
dark count rates are as low as 50 counts per second. (They too begin
to saturate around 100,000 cps, but they currently have comparitively
slow electronics, which should change soon. The APD operates in a
"metastable" state like a supercooled liquid-- it is held above the
diode's breakdown voltage, but cold enough that it takes about 20ms for
the breakdown to occur. A single photon is usually sufficient to trigger
an avalanche of electron-hole pair-production, which yields a macroscopically
detectable current. After the avalanche, the device needs to be "reset,"
or "quenched," and the speed with which this is done determines the
saturation level. The time of detection, however, is accurate to
something on the order of 100ps.)
Also, a lot of the losses in many of these detectors have to do with
reflections off the devices. The effective efficiency can be improved
by using the same trick that makes cats' eyes appear to glow: a mirror
which gives the photons a "second chance" to be detected.
Benjamin Weiner
>What your asking about is the "quantum efficiency" of the detector.
> ... It is easy to see that you
>can't hope to get 100% qe, infact real devices seldom do better
>than 25%. I think the best photomultipliers are something less
>than this. If you consider quantum noise or S/N it just lowers
>the qe more. Although it is much more mathematically complicated.
Quantum efficiency of any detector is strongly dependent on the
incident waveband - but it can get pretty high, at certain wavelengths:
P. Lena's _Observational Astrophysics_ has a graph (p.172) which
shows a thinned CCD achieving 80% quantum efficiency in the range
450-800 nm, which is why people sometimes say CCD's detect practically
every photon (in the waveband!). Thinning the CCD has its own problems ...
Tsuchiya Masahiro
>>
>>>
>>> .. towards a detector which records every photon ..
>>>
accurate number of photon, you may use photochemical reaction which
quantum efficiency is 1.
masahiro@depertment of chemistry, university of tsukuba
michael kagalenko
>In article <wwadge.727584610@csr> wwa...@csr.UVic.CA (Bill Wadge) writes:
>>I was browsing through one of the many 'pop' physics books,
>>where they are describing one of the stock quantum mechanics
>>experiments, and noticed something along the lines of
>>
>> .. towards a detector which records every photon ..
>>
>>Is this possible? Can one really build a device so sensitive that it will
>>detect 100% reliably every photon that enters it, and never
>>go off by accident?
>>
>>Seems unlikely, but then I'm not a physicist ...
>
>Does seem unlikely, doesn't it? But it will happen one day (just about).
>
scattering (photon-detector interaction is inelastic, of course), there
always exists non-zero elastic part in the cross-section (see Landau's
QM I, chapter about non-elastic scattering).
>
>Anyway, don't trust everything you read. The same physics book that
>mentioned 100% efficient photodetectors could well have mentioned frictionless
>pulleys and massless ropes: With many of the former and one of the latter,
>one could construct a 100% efficient photodetector simply by amplifying
>the "light pressure" acting on a ball attached to one end of the rope!
The cross-section of photon-ball int. includes elastic part, so your
detector is not 100% efficient.
michael kagalenko
>dks.
Vain hope, since what you're saying is wrong.
John C. Baez
>where they are describing one of the stock quantum mechanics
>experiments, and noticed something along the lines of
>
> .. towards a detector which records every photon ..
>
>Is this possible? Can one really build a device so sensitive that it will
>detect 100% reliably every photon that enters it, and never
>go off by accident?
Certainly the answer is "no" unless one puts limits on the frequencies
of the photons one is trying to detect. In fact, there are going to be
more and more photons of lower and lower frequency (or energy), no
matter where you look. This is the "infrared divergence" of QED.
More practically speaking, there is always the chance that something
will foul up and you won't detect a photon you should. The notion of a
perfect detector is an idealization.
thomas.j.roberts
The qe of a detector is HIGHLY energy dependent.
Once you get above the e+e- pair-production threshold (about
1 MeV - that's a gamma ray, MUCH more energetic than visible
light), there are numerous detectors which approach 100% qe.
Tom Roberts att!ihlpl!tjrob TJ...@IHLPL.ATT.COM
Sean Merritt
mer...@macro.bu.edu (Sean Merritt) writes:
>What your asking about is the "quantum efficiency" of the detector.
> ... It is easy to see that you
>can't hope to get 100% qe, infact real devices seldom do better
>than 25%. I think the best photomultipliers are something less
>than this. If you consider quantum noise or S/N it just lowers
>the qe more. Although it is much more mathematically complicated.
# Quantum efficiency of any detector is strongly dependent on the
# incident waveband - but it can get pretty high, at certain wavelengths:
# P. Lena's _Observational Astrophysics_ has a graph (p.172) which
# shows a thinned CCD achieving 80% quantum efficiency in the range
# 450-800 nm, which is why people sometimes say CCD's detect practically
# every photon (in the waveband!). Thinning the CCD has its own problems ...
You erased, as others who poited this out, where I specified detectors
that exploit the the photo-electric effect. I don't know enough
about Charged Coupled Devices, do they use the photo-electric effect?
Does anyone know a manufacturer of CCD's, I should like to get a
databook. Also what are the costs as compared to a photo-voltaic
cell?
Daniel Seeman
Is this description *ABSOLUTELY* correct to the last photon? You are correct, no
it is not. Your eye suffers from inefficiencies too. But I put this out to
show that detectors are in use today that are sufficiently efficient. And to
that extent, the exercise is not wrong.
I discussed this matter with
Scott Chase and he feels that I (and my optometrist) should be a bit more
precise with these statements. Only *PERFECT* eyes under *IDEAL* conditions
would be *CAPABLE* of reacting to significantly high percentage (we discussed
efficiencies over %80 or so) of incident photons.
Why don't you ask your optometrist? Be sure to set up the experiment for her so
she knows what your are "looking" for. I would be interested in what she says.
But at least one reputable optometrist thinks the efficiency we discuss is
at least plausible (maybe not probable though). Also the person at school who
taught me this simple exercise is a pretty good reference too. Certainly he
was not counting beans at the time but rather was trying to illustrate a point.
That is what I was trying to do too (eg. there are some very efficient detectors
in use today and they are not necessarily "high-tech" mechanical solutions).
dks.
r...@enh.nist.gov
>that exploit the the photo-electric effect. I don't know enough
>about Charged Coupled Devices, do they use the photo-electric effect?
CCDs are semiconductor based technology based on photons being absorbed in the
creation of electron-hole pairs in a diode. One of the carriers (say
electrons) are shipped to the back side of the diode by the internal
potentials and charge up that region. The charge is measured after some time
period and that (with the known quantum efficiency of that element) allows you
to measure the accumulated photon flux since the last measurement. The charge
is dumped after the measurement. CCDs are usually (maybe always?) multi-
element arrays used for producing spacially resolved images. If you're not
interested in an image, but just in a measure of photon flux you can get a
simple photodiode for less money. (CCDs have some electronics included in the
package that photodiodes don't as well as being multi-element and therefore a
more complicated to make.) Photodiodes use the same physical principle, except
that you measure the photocurrent directly instead of accumulating charge.
I work with diodes, not CCDs, so this description may not be entirly correct,
but I think it's the right general idea.
SCOTT I CHASE
>
>I discussed this matter with
>Scott Chase and he feels that I (and my optometrist) should be a bit more
>precise with these statements. Only *PERFECT* eyes under *IDEAL* conditions
>would be *CAPABLE* of reacting to significantly high percentage (we discussed
>efficiencies over %80 or so) of incident photons.
I guess I should explain my point of view. No matter what type of
detector you choose (eyes, CCD, avalanche diode, etc.), you can make
the claim that perfect {insert detector name} under ideal conditions
can be 100% efficient. But this is not meaningful in practice. No
practical detector is 100% efficient, and eyes don't even come close.
Probably the worst problem with eyes is that they are connected to a
brain that can easily be fooled. Put a person in a very dark room and
allow them to acclimatize - then ask them to respond vocally whenever
they see a single photon . My best guess is that the person will
respond many more times than you actually illuminate them with your
one-photon source. I don't know whether the problem is with false signals
generated in the eye itself, or with the brain.
Without knowing the detailed physiology of the eye, I also imagine that
the effective detector area is not equal to the area of the retina - even
neglecting the entirely dead region (fovea, right?) near the optic nerve.
Rods and cones must have finite dead time and efficiency, etc. All these
factors fold together to give you a net efficiency less than 100%.
For an optometrist to blithly make the off-the-cuff claim that
your eyes are 100% efficient detectors, is probably just sloppy talk. I
presume that optometrists know better.
To the original poster's question about whether 100% efficient detectors
exist, I think that the best answer will probably take the form "No, but
for high enough energy photons at low enough rates, it is possible to
construct detectors of type {insert detector name here} with efficiency
blah-blah %." As a non-expert on photon detection, I'm not sure what
the best detectors are. Others have mentioned several varieties already.
-Scott
--------------------
Scott I. Chase "It is not a simple life to be a single cell,
SIC...@CSA2.LBL.GOV although I have no right to say so, having
been a single cell so long ago myself that I
have no memory at all of that stage of my
life." - Lewis Thomas
Aephraim M. Steinberg
>>In article <wwadge.727584610@csr> wwa...@csr.UVic.CA (Bill Wadge) writes:
>>>I was browsing through one of the many 'pop' physics books,
>>>
>>>
>>>Is this possible? Can one really build a device so sensitive that it will
>>>detect 100% reliably every photon that enters it, and never
>>>go off by accident?
>>>
>>>Seems unlikely, but then I'm not a physicist ...
>>
>>Does seem unlikely, doesn't it? But it will happen one day (just about).
>>
>No, it won't. It's general property of QM that if you have inelastic
>scattering (photon-detector interaction is inelastic, of course), there
>always exists non-zero elastic part in the cross-section (see Landau's
>QM I, chapter about non-elastic scattering).
Thank you for the correction. I haven't looked up the reference yet,
but I can believe the result; by adding "(just about)" I mean to acknowledge
that no process is 100% efficient, but that one could get arbitrarily
close in principle.
Nonetheless, I am curious about a particular inelastic process which seems
to me to be 100% efficient (in principle, of course), and I wonder whether
(i) Some idealization I assume is not even possible in principle, OR
(ii) Your statement holds not for cross-sections but rather for scattering
amplitudes, implying that interference effects could be arranged to make
the total elastic cross-section go to zero.
The instance I have in mind is a superconducting micromaser cavity. If
an atom in an excited state, whose excitation is resonant with the cavity,
is sent through the maser, it has some chance of decaying. As the atom
propagates through the cavity, it undergoes Rabi oscillations, emitting
a photon, reabsorbing it, and so on. Walther's group in Munich has
shown that by correctly selecting the atom's velocity, it is possible to
cause it to decay most of the time. Now, if the cavity is lossless,
it seems to me that there exist incident k-vectors for which the probability
is exactly 100%. True, in a real experiment, there is a superposition of
different velocities, and the cavity isn't ENTIRELY lossless, but are these
issues of fundamental importance, or is it perhaps unfair to neglect edge
effects when the atom enters or exits the cavity, or is my picture correct IN
PRINCIPLE, and the theorem somehow evaded in this case?
michael kagalenko
I recall some experiments (by Vavilov) where the intensity of light was
gradually reduced until observer was able to see fluctuations of intensity
due to quantum nature of light. If I'm not mistaken, observer compared
intensities in two points of diffraction picture. Our eyes have fairly
good sencitivity. Not 100%, however.
michael kagalenko
one can really say that resistance of superconductor is 0 ; it's
possible to prove that it's less than $\epsilon $(what is the current
value of it, BTW ?) You can show than terms up to the sertain order in
resistance are zero; but the possibility for nasty high-oreder effect
which will cause dissipation still remains.
Corrections ?
Benjamin Weiner
>You erased, as others who poited this out, where I specified detectors
>that exploit the the photo-electric effect. I don't know enough
>about Charged Coupled Devices, do they use the photo-electric effect?
Whoops, sorry. I wasn't trying to correct you, just adding a little
information on something else. Yes: CCDs consist of semiconductor
layers; an incident photon is absorbed, creating loose charges, which
is confined to its pixel by an ingenious layout of the n and p layers;
further ingenuity allows you to read the pixels out in order at the end
of your exposure.
>Does anyone know a manufacturer of CCD's, I should like to get a
>databook. Also what are the costs as compared to a photo-voltaic
>cell?
TI and Tek are the biggies, it seems to me. They are expensive, though,
and it seems to me that CCDs are most useful in imaging or other highly
position-sensitive devices (they usually come in sizes 256x256 to 800x800
or so pixels) - although I may only think this because of astronomy.
Also they need to be fairly cold, to reduce thermal noise, so the cooling
system adds expense and bulk.