Photons are weird. In fact, I already wrote some fairly confused posts on them. This post is probably even more confusing. If anything, it shows how easy it is to get lost when thinking things through. In any case, it did help me to make sense of it all and, hence, perhaps it will help you too.
Electrons and photons: similarities and differences
All elementary particles are weird. As Feynman puts it, in the very first paragraph of his Lectures on Quantum Mechanics : “Historically, the electron, for example, was thought to behave like a particle, and then it was found that in many respects it behaved like a wave. So it really behaves like neither. Now we have given up. We say: “It is like neither. There is one lucky break, however—electrons behave just like light. The quantum behavior of atomic objects (electrons, protons, neutrons, photons, and so on) is the same for all, they are all “particle waves,” or whatever you want to call them. So what we learn about the properties of electrons will apply also to all “particles,” including photons of light.” (Feynman’s Lectures, Vol. III, Chapter 1, Section 1)
I wouldn’t dare to argue with Feynman, of course… But… Photons are like electrons, and then they are not. For starters, photons do not have mass, and they are bosons, force-carrying ‘particles’ obeying very different quantum-mechanical rules, referred to as Bose-Einstein statistics. I’ve written about that in the past, so I won’t do that again here. It’s probably sufficient to remind the reader that these rules imply that the so-called Pauli exclusion principle does not apply to them: bosons like to crowd together, thereby occupying the same quantum state–unlike their counterparts, the so-called fermions or matter-particles: quarks (which make up protons and neutrons) and leptons (including electrons and neutrinos), which can’t do that: two electrons can only sit on top of each other if their spins are opposite (so that makes their quantum state different), and there’s no place whatsoever to add a third one–because there are only two possible values for the spin: up or down.
From all that I’ve been writing so far, I am sure you have some kind of picture of matter-particles now, and notably of the electron: when everything is said and done, it’s a point-like particle defined by some weird wave function–the so-called ‘probability wave’. But what about the photon? They are point-like particles too, aren’t they? Hence, why wouldn’t we associate them with a probability wave too? Do they have a de Broglie wavelength?
Before answering that question, let me present that ‘picture’ of the electron once again.
The wave function for electrons
The electron ‘picture’ can be represented in a number of ways but one of the more scientifically correct ones – whatever that means – is that of a spatially confined wave function representing a complex quantity referred to as the probability amplitude. The animation below (which I took from Wikipedia) visualizes such wave functions. As you know by now, the wave function is usually represented by the Greek letter psi (ψ), and it is often referred to, as mentioned above, as a ‘probability wave’, although that term is quite misleading. Why? You surely know that by now: the wave function represents a probability amplitude, not a probability.
That being said, probability amplitude and probability are obviously related: if we square the psi function (so we square all these amplitudes), then we get the actual probability of finding that electron at point x. That’s the so-called probability density function on the right of each function. [I should be fully correct now and note that we are talking the absolute square here, or the squared norm: remember that the square of a complex number can be negative, as evidenced by the definition of i: i2 = –1. In fact, if there's only an imaginary part, then its square is always negative.]
Below, I’ve inserted another image, which gives a static picture (i.e. one that is not varying in time) of the wave function of an electron. To be precise: it’s the wave function for an electron on the 5d orbital of a hydrogen orbital. I also took it from Wikipedia and so I’ll just copy the explanation here: “The solid body shows the places where the electron’s probability density is above a certain value (0.02), as calculated from the probability amplitude.” As for the colors, this image uses the so-called HSL color system to represent complex numbers: each complex number is represented by a unique color, with a different hue (H), saturation (S) and lightness (L). [Just google if you want to know how that works exactly.]
The Uncertainty Principle revisited
The wave function is usually given as a function in space and time: ψ = ψ(x, t). However, I should also remind you that we have a similar function in the ‘momentum space’. Indeed, the position-space and momentum-space wave functions are related through the Uncertainty Principle. To be precise: they are Fourier transforms of each other – but don’t be put off by that statement. I’ll just quickly jot down the Uncertainty Principle once again:
σx·σp ≥ ħ/2
This is the so-called Kennard formulation of the Principle: it measures the uncertainty (usually written as Δ), of both the position (Δx) as well as momentum (Δp), in terms of the standard deviation–that’s the σ (sigma) symbol–around the mean. Hence, the assumption is that both x and p follow some kind of distribution–and that’s usually a nice “bell curve” in the textbooks. Finally, let me also remind you how tiny that physical constant ħ actually is: about 6.58×10−16 eV·s.
At this point, you may wonder about the units. A position is expressed in distance units, and momentum… Euh… [...] Momentum is mass times velocity, so it’s kg·m/s. Hence, the dimension of the product on the left-hand side of the inequality is m·kg·m/s = kg·m2/s. So what about this eV·s dimension on the right-hand side? Well… The electronvolt is a unit of energy, and so we can convert it to joules. Now, a joule is a newton-meter (N·m), which is the unit for both energy and work. So how do we relate the two sides? Simple: a newton can also be expressed in SI units: 1 N = 1 kg·m/s2: one newton is the force needed to give a mass of 1 kg an acceleration of 1 m/s per second. So just substitute and you’ll see the dimension on the right-hand side is kg·(m/s2)·m·s = kg·m2/s, so it comes out alright. Why this digression? Not sure. Perhaps just to remind you also that the Uncertainty Principle can also be expressed in terms of energy and time:
ΔE·Δt ≥ ħ/2
That expression makes it clear the units on both sides of the inequality are, indeed, the same, but it’s not so obvious to relate the two expressions of the same Uncertainty Principle. I’ll just note that energy and time, just like position and momentum, are also so-called complementary variables in quantum mechanics. We have the same duality for the de Broglie relation, which I’ll also jot down here:
λ = h/p and f = E/h
In these two complementary equations, λ is the wavelength of the (complex-valued) de Broglie wave, and f is its frequency. A stupid question perhaps: what’s the velocity of the de Broglie wave? Well… As you should know from previous posts, the mathematically correct answer involves distinguishing the group and phase velocity of a wave, but the easy answer is: the de Broglie wave of a particle moves with the particle :-) and, hence, its velocity is, obviously, the speed of the particle which, for electrons, is usually non-relativistic (i.e. rather slow as compared to the speed of light).
Before proceeding, I need to make some more introductory remarks. The first is that the Uncertainty Principle implies that we cannot assign a precise wavelength (or a equally precise frequency) to a de Broglie wave: if there is a spread in p (and, hence, in E), then there will be a spread in λ (and in f). That’s good, because a regular wave with an exact frequency would not give us any information about the location. Frankly, I always had a lot of trouble understanding this, so I’ll just quote the expert teacher (Feynman) on this:
“The amplitude to find a particle at a place can, in some circumstances, vary in space and time, let us say in one dimension, in this manner: ψ = Aei(ωt−kx) , where ω is the frequency, which is related to the classical idea of the energy through E = ħω, and k is the wave number, which is related to the momentum through p = ħk. [These are equivalent formulations of the de Broglie relations using the angular frequency and the wave number instead of wavelength and frequency.] We would say the particle had a definite momentum p if the wave number were exactly k, that is, a perfect wave which goes on with the same amplitude everywhere. The ψ = Aei(ωt−kx) equation [then] gives the [complex-valued probability] amplitude, and if we take the absolute square, we get the relative probability for finding the particle as a function of position and time. This is a constant, which means that the probability to find a [this] particle is the same anywhere.” (Feynman’s Lectures, I-48-5)
Of course, that’s a problem: if the probability to find a particle is the same anywhere, then the particle can be anywhere, and, hence, that means it’s actually nowhere. Hence, that wave function doesn’t serve the purpose. In short, that nice ψ = Aei(ωt−kx) function is completely useless in terms of representing an electron, or any other actual particle moving through space. So what to do?
The Wikipedia article on the Uncertainty Principle has this wonderful animation that shows how we can superimpose several waves to form a wave packet. And so that’s what we want: a wave packet traveling through (and limited in) space. I’ve copied the animation below. You should note that it shows only one part of the complex probability amplitude: just visualize the other part (imaginary if the wave below is the real part, and vice versa if the wave below would happen to represent the imaginary part of the probability amplitude). The illustration basically illustrates a mathematical operation (a Fourier analysis–and that’s not the same as that Fourier transform I mentioned above, although these two mathematical concepts obviously have a few things in common) that separates a wave packet into a finite or (potentially) infinite number of component waves. Indeed, note how, in the illustration below, the frequency increases gradually (or, what amounts to the same, the wavelength gets smaller) and, with every wave we add to the packet, it becomes increasingly localized. So that illustrates how the ‘uncertainty’ about (or the spread in) the frequency is inversely related to the uncertainty in position:
Δx = 1/[Δ(1/λ)] = h/Δp.
Frankly, I must admit both Feynman’s explanation and the animation above don’t quite convince me, because I can perfectly imagine a localized wave train with a fixed frequency and wavelength, like the one below, which I’ll re-use later. I’ve made this wave train myself: it’s a standard sine or cosine function multiplied with another easy function generating the envelope. You can easily make one like this yourself. This thing is localized in space and, as mentioned above, it has a fixed frequency and wavelength.
In any case, I need to move on. If you, my reader, would have any suggestion here (I obviously don’t quite get it here), please let me know. As for now, however, I’ll just continue wandering. Let me proceed with two more remarks:
- What about the amplitude of a de Broglie wave? Well… It’s just like this perceived problem with fixed wavelengths: frankly, I couldn’t find much about that in textbooks either. However, there’s an obvious constraint: when everything is said and done, all probabilities must take a value between 0 and 1, and they must also all add up to exactly 1. So that’s a so-called normalization condition that obviously imposes some constraints on the (complex-valued) probability amplitudes of our wave function.
- The complex waves above are so-called standing waves: their frequencies only take discrete values. You can, in fact, see that from the first animation. Likewise, there is no continuous distribution of energies. Well… Let me be precise and qualify that statement: in fact, when the electron is free, it can have any energy. It’s only when it’s bound that it must take one or another out of a set of allowed values, as illustrated below.
Now, you will also remember that an electron absorbs or emits a photon when it goes from one energy level to the other, so it absorbs or emits radiation. And, of course, you will also remember that the frequency of the absorbed or emitted light is related to those energy levels. More specifically, the frequency of the light emitted in a transition from, let’s say, energy level E3 to E1 will be written as ν31 = (E3 – E1)/h. This frequency will be one of the so-called characteristic frequencies of the atom and will define a specific so-called spectral emission line.
Now that’s where the confusion starts, because that ν31 = (E3 – E1)/h equation is – from a mathematical point of view – identical to the de Broglie equation, which assigns a de Broglie wave to a particle: f = E/h. While mathematically similar, the formulas represent very different things. The most obvious remark to make is that a de Broglie wave is a matter-wave and, as such, quite obviously, that it has nothing to do with an electromagnetic wave. Let me be even more explicit. A de Broglie wave is not a ‘real’ wave, in a sense (but, of course, that’s a very unscientific statement to make); it’s a psi function, so it represents these weird mathematical quantities–complex probability amplitudes–which allow us to calculate the probability of finding the particle at position x or, if it’s a wave function for the momentum-space, to find a value p for its momentum. In contrast, a photon that’s emitted or absorbed represents a ‘real’ disturbance of the electromagnetic field propagating through space. Hence, that frequency ν is something very different than f, which is why we another symbol for it (ν is the Greek letter nu, not to be confused with the v we use for velocity). [Of course, let's not get into a philosophical discussion on how 'real' an electromagnetic field is here.]
That being said, we also know light is emitted in discrete energy packets: in fact, that’s how photons were defined originally, first by Planck and then by Einstein. Now, when an electron falls from one energy level in an atom to another (lower) energy level, it emits one – and only one – photon with that particular wavelength and energy. The question then is: how should we picture that photon? Does it also have some more or less defined position and momentum? Can we also associate some wave function – i.e. a de Broglie wave – with it?
When you google for an answer to this simple question, you will get very complicated and often contradictory answers. The very short answer I got from a nuclear scientist – you can imagine that these guys don’t have the time to give a more nuanced answer to idiots like me – was simple: No. One does not associate photons with a de Broglie wave.
When he gave me that answer, I first thought: of course not. A de Broglie wave is a ‘matter wave’, and photons aren’t matter. Period. That being said, the answer is not so simple. Photons do behave like electrons, don’t they? There’s diffraction (when you send a photon through one slit) and interference (when photons go through two slits, and there’s interference even when they go through one by one, just like electrons), and so on and so on. Most importantly, the Uncertainty Principle obviously also applies to them and, hence, one would expect to be able to associate some kind of wave function with them, wouldn’t one?
Who knows? That’s why I wrote this post. I don’t know, so I thought it would be nice to just freewheel a bit on this question. So be warned: nothing of what I write below has been researched really, so critical comments and corrections from actual specialists are more than welcome.
The shape of a photon wave
The answer in regard to the definition of a photon’s position and momentum is, obviously, unambiguous: photons are, per definition, little lumps of energy indeed, and so we detect them as such and, hence, they do occupy some physical space: we detect them somewhere, and we usually do so at a rather precise point in time.
They obviously also have some momentum. In fact, I calculated that momentum in one of my previous posts (Light: Relating Waves to Photons). It was related to the magnetic field vector, which we usually never mention when discussing light – because it’s so tiny as compared to the electric field vector – but so it’s there and a photon’s momentum (in the direction of propagation) is as tiny as the magnetic field vector for an electromagnetic wave traveling through space: p = E/c, with E = νh and c = 3×108. Hence, the momentum of a photon is only a tiny fraction of the photon’s energy. In fact, because it’s so tiny (remember that the energy of a photon is only one or two eV, and so that’s a very tiny unit, and so we have to divide that by c, which is a huge number obviously…), I’ll just forget about it here for a while and focus on that electric field vector only.
So… A photon is, in essence, a electromagnetic disturbance and so, when trying to picture a photon, we can think of some oscillating electric field vector traveling through–and also limited in–space. In short, in the classical world – and in the classical world only of course – a photon must be some electromagnetic wave train, like the one below–perhaps.
Hmm… Why would it have that shape? I don’t know. Your guess is as good as mine. But you’re right to be skeptical. In fact, the wave train above has the same shape as Feynman’s representation of a particle (see below) as a ‘probability wave’ traveling through–and limited in–space.
So, what about it? Let me first remind you once again (I just can’t stress this point enough it seems) that Feynman’s representation – and most are based on his, it seems – is, once again, misleading because it suggests that ψ(x) is some real number. It’s not. In the image above, the vertical axis should not represent some real number (and it surely should not represent a probability, i.e. some real positive number between 0 and 1) but a probability amplitude, i.e. a complex number in which both the real and imaginary part are important. As mentioned above, this wave function will, of course, give you all the probabilities you need when you take its (absolute) square, but so it’s not the same: the image above gives you only one part of the complex-valued wave function (it could be either the real or the imaginary part, in fact), which is why I find it misleading.
But let me go back to the first illustration: the vertical axis of the first illustration is not ψ but E – the electric field vector. So there’s no imaginary part here: just a real number, representing the strength–or magnitude I should say– of the electric field E as a function of the space coordinate x. [Can magnitudes be negative? Of course ! In that case, the field vector points the other way !] Does this suggestion for how a photon could look like make sense? In quantum mechanics, the answer is obviously: no. But in the classical worldview? Well… Maybe. [...] You should be skeptical, however. Even in the classical world, the answer is: most probably not. I know you won’t like it (because the formula doesn’t look easy), but let me remind you of the formula for the (vertical) component of E as a function of the acceleration of some charge q:
The charge q (i.e. the source of the radiation) is, of course, our electron that’s emitting the photon as it jumps from a higher to a lower energy level (or, vice versa, absorbing it). This formula basically states that the magnitude of the electric field (E) is proportional to the acceleration (a) of the charge (with t–r/c the retarded argument). Hence, the suggested shape of E as a function of x implies that the acceleration of the electron is initially quite small, that it swings between positive and negative, that these swings then become larger and larger to reach some maximum, and then they become smaller and smaller again to then die down completely. In short, it’s like a transient wave: “a short-lived burst of energy in a system caused by a sudden change of state”, as Wikipedia defines it.
[...] Well… No. An actual transient looks more like what’s depicted below: no gradual increase in amplitude but big swings initially which then dampen to zero. In other words, if our photon is a transient electromagnetic disturbance caused by a ‘sudden burst of energy’ (which is what that electron jump is, I would think), then its representation will, much more likely, resemble a damped wave, like the one below, rather than Feynman’s picture of a moving matter-particle.
In fact, we’d have to flip the image, both vertically and horizontally, because the acceleration of the source and the field are related as shown below. The vertical flip is because of the minus sign in the formula for E(t). The horizontal flip is because of the minus sign in the (t – r/c) term, the retarded argument: if we add a little time (Δt), we get the same value for a(t−r/c) as we would have if we had subtracted a little distance: Δr=−cΔt. So that’s why E as a function of r (or of x), i.e. as a function in space, is a ‘reversed’ plot of the acceleration as a function of time.
So we’d have something like below.
What does this resemble? It’s not a vibrating string (although I do start to understand the attractiveness of string theory now: vibrating strings are great as energy storage systems, so the idea of a photon being some kind of vibrating string sounds great, doesn’t it?). It’s not resembling a bullwhip effect either, because the oscillation of a whip is confined by a different envelope (see below). And, no, it’s also definitely not a trumpet.
It’s just what it is: an electromagnetic transient traveling through space. Would this be realistic as a ‘picture’ of a photon? Frankly, I don’t know. I’ve looked at a lot of stuff but didn’t find anything on this really. The easy answer, of course, is quite straightforward: we’re not interested in the shape of a photon because we know it is not an electromagnetic wave. It’s a ‘wavicle’, just like an electron.
Sure. I know that too. Feynman told me. :-) But then why wouldn’t we associate some wave function with it? Please tell me, because I really can’t find much of an answer to that question in the literature, and so that’s why I am freewheeling here. So just go along with me for a while, and come up with another suggestion. As I said above, your bet is as good as mine. All that I know is that there’s one thing we need to explain when considering the various possibilities: a photon has a very well-defined frequency (which defines its color in the visible light spectrum) and so our wave train should – in my humble opinion – also have that frequency. At least for quite a while–and then I mean most of the time, or on average at least. Otherwise the concept of a frequency – or a wavelength – doesn’t make sense. Indeed, if the photon has no defined wavelength or frequency, then it has no color, and a photon should have a color: that’s what the Planck relation is all about.
What would be your alternative? I mean… Doesn’t it make sense to think that, when jumping from one energy level to the other, the electron would initially sort of overshoot its new equilibrium position, to then overshoot it again on the other side, and so on and so on, but with an amplitude that becomes smaller and smaller as the oscillation dies out? In short, if we look at radiation as being caused by atomic oscillators, why would we not go all the way and think of them as oscillators subject to some damping force? Just think about it. :-)
The size of a photon wave
Let’s forget about the shape for a while and think about size. We’ve got an electromagnetic train here. So how long would it be? Well… Feynman calculated the Q of these atomic oscillators: it’s of the order of 108 (see his Lectures, I-34-3: it’s a wonderfully simple exercise, and one that really shows his greatness as a physics teacher) and, hence, this wave train will last about 10–8 seconds (that’s the time it takes for the radiation to die out by a factor 1/e). That’s not very long but, taking into account the rather spectacular speed of light (3×108 m/s), that still makes for a wave train with a length of 3 meter. Three meter !? Holy sh**! That’s like infinity on an atomic scale! Such length surely doesn’t match the picture of a photon as a fundamental particle which cannot be broken up, does it? This surely cannot be right and, if it is, then there surely must be some way to break this thing up. It can’t be ‘elementary’, can it?
Well… You’re right, of course. I shouldn’t be doing these classical analyses of a photon, but then I think it actually is kind of instructive. So please do double-check but that’s what it is, it seems. For sodium light (I am just continuing Feynman’s example) here, which has a frequency of 500 THz (500×1012 oscillations per second) and a wavelength of 600 nm (600×10–9 meter), that length corresponds to some five million oscillations. All packed into one photon? One photon with a length of three meters? You must be joking, right?
Sure. I am joking here–but, as far as jokes go, this one is fairly robust from a scientific point of view, isn’t it? :-) Again, please do double-check and correct me, but all what I’ve written so far is not all that speculative. It corresponds to all what I’ve read about it: only one photon is produced per electron in any de-excitation, and its energy is determined by the number of energy levels it drops, as illustrated (for a simple hydrogen atom) below. For those who continue to be skeptical about my sanity here, I’ll quote Feynman once again:
“What happens in a light source is that first one atom radiates, then another atom radiates, and so forth, and we have just seen that atoms radiate a train of waves only for about 10–8 sec; after 10–8 sec, some atom has probably taken over, then another atom takes over, and so on. So the phases can really only stay the same for about 10–8 sec. Therefore, if we average for very much more than 10–8 sec, we do not see an interference from two different sources, because they cannot hold their phases steady for longer than 10–8 sec. With photocells, very high-speed detection is possible, and one can show that there is an interference which varies with time, up and down, in about 10–8 sec.” (Feynman’s Lectures, I-34-4)
So… Well… Now it’s up to you. I am going along here with the assumption that a photon, from a classical world perspective, should indeed be something that’s several meters long and something that packs like five million oscillations. So, while we usually measure stuff in seconds, or hours, or years, and, hence, while we would that think 10–8 seconds is short, a photon would actually be a very stretched-out transient that occupies quite a lot of space. I should also add that, in light of that scale (3 meter), the dampening seems to happen rather slowly!
I can see you shaking your head, for various reasons. First because this type of analysis is not appropriate. [Yes. I know. A photon should not be viewed as an electromagnetic wave. It's a discrete packet of energy. Period.] Second, I guess you may find the math involved in this post not to your liking, even if it’s quite simple and I am not doing anything spectacular here. [...] Whatever. I don’t care. I’ll just bulldozer on.
What about the ‘vertical’ dimension, the y and the z coordinates in space? We’ve got this long snaky thing: how thick-bodied is it?
Here, we need to watch our language. It’s not very obvious to associate a photon with some kind of cross-section normal to its direction of propagation. Not at all actually. Indeed, as mentioned above, the vertical axis of that graph showing the wave train does not indicate some spatial position: it’s not a y- (or z-)coordinate but the magnitude of an electric field vector. [Just to underline the fact that this magnitude has nothing to do with spatial coordinates: note that the value of that magnitude depends on our unit, so it's really got nothing to do with an actual position in space-time.]
However, that being said, perhaps we can do something with that idea of a cross-section. In nuclear physics, the term ‘cross-section’ would usually refer to the so-called Thompson scattering cross-section, which can be defined rather loosely as the target area for the incident wave (i.e. the photon): it is, in fact, a surface which can be calculated from what is referred to as the classical electron radius, which is about 2.82×10–15 m. Just to compare: you may or may not remember the so-called Bohr radius of an atom, which is about 5.29×10–11 m, so that’s a length that’s about 20,000 times longer. To be fully complete, let me give you the exact value for the Thompson scattering cross-section: 6.62×10–29 m2 (note that this is a surface indeed, so we have m squared as a unit, not m).
Now, let me remind you – once again – that we should not associate the oscillation of the electric field vector with something actually happening in space: an electromagnetic field does not move in a medium and, hence, it’s not like a water or sound wave, which makes molecules go up and down as it propagates through its medium. To put it simply: there’s nothing that’s wriggling in space as that photon is flashing through space. However, when it does hit an electron, that electron will effectively ‘move’ (or vibrate or wriggle or whatever you can imagine) as a result of the incident electromagnetic field.
That’s what’s depicted and labeled below: there is a so-called ‘radial component’ of the electric field, and I would say: that’s our photon! [What else would it be?] The illustration below shows that this ‘radial’ component is just E for the incident beam and that, for the scattered beam, it is, in fact, determined by the electron motion caused by the incident beam through that relation described above, in which a is the normal component (i.e. normal to the direction of propagation of the outgoing beam) of the electron’s acceleration.
Now, before I proceed, let me remind you once again that the above illustration is, once again, one of those illustrations that only wants to convey an idea, and so we should not attach too much importance to it: the world at the smallest scale is best not represented by a billiard ball model. In addition, I should also note that the illustration above was taken from the Wikipedia article on elastic scattering (i.e. Thomson scattering), which is only a special case of the more general Compton scattering that actually takes place. It is, in fact, the low-energy limit. Photons with higher energy will usually be absorbed, and then there will be a re-emission, but, in the process, there will be a loss of energy in this ‘collision’ and, hence, the scattered light will have lower energy (and, hence, lower frequency and longer wavelength). But – Hey! – now that I think of it: that’s quite compatible with my idea of damping, isn’t it? :-) [If you think I've gone crazy, I am really joking here: when it's Compton scattering, there's no 'lost' energy: the electron will recoil and, hence, its momentum will increase. That's what's shown below (credit goes to the HyperPhysics site).]
In any case, I don’t want to make this post too long. I do think we’re getting something here in terms of our objective of picturing a photon–using classical concepts that is! A photon should be a long wave train – a very long wave train actually – but its effective ‘radius’ should be of the same order as the classical electron radius, one would think. Or, much more likely, much smaller. If it’s more or less the same radius, then it would be in the order of femtometers (1 fm = 1 fermi = 1×10–15 m). That’s good because that’s a typical length-scale in nuclear physics. For example, it would be comparable with the radius of a proton. So we look at a photon here as something very different – because it’s so incredibly long (as mentioned above, three meter is not an atomic scale at all!) – but as something which does have some kind of ‘radius’ that is normal to its direction of propagation and equal or, more likely, much smaller than the classical electron radius. [Why smaller? First, an electron is obviously fairly massive as compared to a photon (if only because an electron has a rest mass and a photon hasn't). Second, it's the electron that absorbs a photon–not the other way around.]
Now, that radius determines the area in which it may produce some effect, like hitting an electron, for example, or like being detected in a photon detector, which is just what this so-called radius of an atom or an electron is all about: the area which is susceptible of being hit by some particle (including a photon), or which is likely to emit some particle (including a photon). What is exactly, we don’t know: it’s still as spooky as an electron and, therefore, it also does not make all that much sense to talk about its exact position in space. However, if we’d talk about its position, then we should obviously also invoke the Uncertainty Principle, which will give us some upper and lower bounds for its actual position, just like it does for any other particle: the uncertainty about its position will be related to the uncertainty about its momentum, and more knowledge about the former, will implies less knowledge about the latter, and vice versa. Therefore, we can also associate some complex wave function with this photon which is – for all practical purposes – a de Broglie wave. Now how should we visualize that wave?
The shape and size of a photon’s probability wave
I am actually not going to offer anything specific here. First, it’s all speculation. Second, I think I’ve written too much rubbish already. However, if you’re still reading, and you like this kind of unorthodox application of electromagnetics, then the following remarks may stimulate your imagination.
First we should note that if we’re going to have a wave function for the photon in position-space (as opposed to momentum-space), its argument will not only be x and t, but also y and z. In fact, when trying to visualize this wave function, we should probably first think of keeping x and t constant and then how a little complex-valued wave train normal to the direction of propagation would look like.
What about its frequency? You may think that, if we know the frequency of this photon, and its energy, and its momentum (we know all about this photon, don’t we), then we can also associate some de Broglie frequency with this photon. Well… Yes and no. The simplicity of these de Broglie relations (λ = h/p and f = E/h) suggests we can, indeed, assign some frequency (f) or wavelength (λ) to it, all within the limits imposed by the Uncertainty Principle. But we know that we should not end up with a wave function that, when squared, gives us probabilities for each and every point in space. No. The wave function needs to be confined in space and, hence, we’re also talking a wave train here, and a very short one in this case. Indeed, while, in our reasoning here, we look at the photon as being somewhere, we know it should be somewhere within one or two femtometer of our line of sight.
Now, what’s the typical energy of a photon again? Well… Let’s calculate it for that sodium light. E = hν, so we have to multiply Planck’s constant (h = 4.135×10−15 eV·s) with the photon frequency (ν = 500×1012 oscillations/s), so that’s about 2 eV. I haven’t checked this but it should be about right: photons in the visible light spectrum have energies ranging from 1.5 to 3.5 eV. Not a lot but something. Now, what’s the de Broglie frequency and wavelength associated with that energy level? Hmm… Well… It’s the same formula, so we actually get the same frequency and wavelength: 500×1012 Hz and 600 nm (nanometer) for the wavelength. So how do we pack that into our one or two femtometer space? Hmm… Let’s think: one nanometer is a million femtometer, isn’t it? And so we’ve got a de Broglie wavelength of 600 nanometer?
Oh-oh… We must be doing something wrong here, isn’t it?
Yeah. I guess so. Here I’ll quote Feynman again: “We cannot define a unique wavelength for a short wave train. Such a wave train does not have a definite wavelength; there is an indefiniteness in the wave number that is related to the finite length of the train.”
I had equally much trouble with this as with that other statement of Feynman–and then I mean on the ‘impossibility’ of a wave train with a fixed frequency. But now I think it’s very simple actually: a very very short burst is just not long enough to define a wavelength or a frequency: there’s a few up and downs, which are more likely than not to be very irregular, and that’s it. No nice sinusoidal shape. It’s as simple as that… I think. :-)
In fact–now that I am here–there’s something else I didn’t quite understand when reading physics: everyone who writes about light or matter waves seems to be focused on the frequency of these waves only. There’s little or nothing on the amplitude. Now, the energy of a physical wave, and of a light wave, does not only depend on its frequency, but also on the amplitude. In fact, we all know that doubling, tripling or quadrupling the frequency of a wave will double, triple or quadruple its energy (that’s obvious from the E = hν relation), but we tend to forget that the energy of a wave is also proportional to the square of its amplitude, for which I’ll use the symbol A, so we can write: E ∝ A2. Hence, if we double, triple or quadruple the amplitude of a wave, its energy will be multiplied by four, nine and sixteen respectively!
The same relationship obviously holds between probability amplitudes and probability densities: if we double, triple or quadruple the probability amplitudes, then the associated probabilities obviously also get multiplied by four, nine and sixteen respectively! This obviously establishes some kind of relation between the shape of the electromagnetic wave train and the probability wave: if the electromagnetic wave train (i.e. the photon itself) packs a lot of energy upfront (cf. the initial overshooting and the gradual dying out), then we should expect the probability amplitudes to be ‘bigger’ there as well. [Note that we can't directly compare two complex numbers in terms of one being 'bigger' or 'smaller' than the other, but you know what I mean: their absolute square will be bigger.]
Well… Nothing. I can’t say anything more about this. However, to compensate for the fact that I didn’t get anywhere with my concept of a de Broglie wave for a photon – and, hence, I let you down, I guess – I’ll explore the relationship between amplitude, frequency and size of a wave train somewhat more in detail. It may inspire you when thinking yourself about a ‘probability wave’ for a photon. And then… Well… I will write some kind of conclusion, which may or may not give the answer(s) that you are looking for.
The relation between amplitude, frequency and energy
From what I wrote above, it’s obvious that there are two ways of packing more energy in a (real) wave, or a (sufficiently long) wave train:
- We can increase the frequency, and so that results in a linear increase in energy (twice the frequency is twice the energy).
- We increase the amplitude, and that results in an exponential (quadratic) increase in energy (double all amplitudes, and you pack four times more energy in that wave).
With a ‘real’ wave, I obviously mean either a wave that’s traveling in a medium or, in this case, an electromagnetic wave. OK. So what? Well… It’s probably quite reasonable to assume that both factors come into play when an electron emits a photon. Indeed, if the difference between the two energy levels is larger, then the photon will not only have a higher frequency (i.e. we’re talking light (or electromagnetic radiation) in the upper ranges of the spectrum then) but one should also expect that the initial overshooting – and, hence, the initial oscillation – will also be larger. In short, we’ll have larger amplitudes. Hence, higher-energy photons will pack even more energy upfront. They will also have higher frequency, because of the Planck relation. So, yes, both factors would come into play.
What about the length of these wave trains? Would it make them shorter? Yes. I’ll refer you to Feynman’s Lectures to verify that the wavelength appears in the numerator of the formula for Q. Hence, higher frequency means shorter wavelength and, hence, lower Q. Now, I am not quite sure (I am not sure about anything I am writing here it seems) but this may or may not be the reason for yet another statement I never quite understood: photons with higher and higher energy are said to become smaller and smaller, and when they reach the Planck scale, they are said to become black holes.
What’s the conclusion? Well… I’ll leave it to you to think about this. Let me make a bold statement here: that transient above actually is the wave function.
You’ll say: What !? What about normalization? All probabilities have to add up to one and, surely, those magnitudes of the electric field vector wouldn’t add up to one, would they?
My answer to that is simple: that’s just a question of units, i.e. of normalization indeed. So just measure the field strength in some other unit and it will come all right.
[...] But… Yes? What? Well… Those magnitudes are real numbers, not complex numbers.
I am not sure how to answer that one but there’s two things I could say:
- Real numbers are complex numbers too: it’s just that their imaginary part is zero)
- When working with waves, and especially with transients, we’ve always represented them using the complex exponential function. For example, we would write a wave function whose amplitude varies sinusoidally in space and time as Aei(ωt−k·r), with ω the (angular) frequency and k the wave number (so that’s the wavelength expressed in radians per unit distance).
Frankly, think about it: where is the photon? It’s that three-meter long transient, isn’t it? And the probability to find it somewhere is the (absolute) square of some complex number, right? And then we have a wave function already, representing an electromagnetic wave, for which we know that the energy which it packs is the square of its amplitude, as well as being proportional to its frequency. We also know we’re more likely to detect something with high energy than something with low energy, don’t we? So why would we hesitate?
But then what about these probability amplitudes being a function of the y and z coordinates?
Well… Frankly, I’ve started to wonder if a photon actually has a radius. If it doesn’t have a mass, it’s probably the only real point-like particle (i.e. a particle not occupying any space) – as opposed to all other matter-particles, which do have mass.
I don’t know. Your guess is as good as mine. Maybe our concepts of amplitude and frequency of a photon are not very relevant. Perhaps it’s only energy that counts. We know that a photon has a more or less well-defined energy level (within the limits of the Uncertainty Principle) and, hence, our ideas about how that energy actually gets distributed over the frequency, the amplitude and the length of that ‘transient’ have no relation with reality. Perhaps we like to think of a photon as a transient electromagnetic wave, because we’re used to thinking in terms of waves and fields, but perhaps a photon is just a point-like thing indeed, with a wave function that’s got the same shape as that transient. :-)
Post scriptum: I should apologize to you, my dear reader. It’s obvious that, in quantum mechanics, we don’t think of a photon as having some frequency and some wavelength and some dimension in space: it’s just an elementary particle with energy interacting with other elementary particles with energy, and we use these coupling constants and what have you to work with them. We don’t think of photons as a three-meter long snake swirling through space. So, when I write that “our concepts of amplitude and frequency of a photon are maybe not very relevant” when trying to picture a photon, and that “perhaps, it’s only energy that counts”, I actually don’t mean “maybe” or “perhaps”. I mean: Of course !!! In the quantum-mechanical world view, that is.
So I apologize for posting nonsense. However, as all of this nonsense helps me to make sense of these things myself, I’ll just continue. :-) I seem to move very slowly on this Road to Reality, but the good thing about moving slowly, is that it will – hopefully – give me the kind of ‘deeper’ understanding I want, i.e. an understanding beyond the formulas and mathematical and physical models. In the end, that’s all that I am striving for when pursuing this ‘hobby’ of mine. Nothing more, nothing less. :-)