Playing with amplitudes

Pre-script (dated 26 June 2020): This post got mutilated by the removal of some material by the dark force. You should be able to follow the main story line, however. If anything, the lack of illustrations might actually help you to think things through for yourself. In any case, we now have different views on these concepts as part of our realist interpretation of quantum mechanics, so we recommend you read our recent papers instead of these old blog posts.

Original post:

Let’s play a bit with the stuff we found in our previous post. This is going to be unconventional, or experimental, if you want. The idea is to give you… Well… Some ideas. So you can play yourself. 🙂 Let’s go.

Let’s first look at Feynman’s (simplified) formula for the amplitude of a photon to go from point a to point b. If we identify point by the position vector r1 and point by the position vector r2, and using Dirac’s fancy bra-ket notation, then it’s written as:


So we have a vector dot product here: pr12 = |p|∙|r12|· cosθ = p∙r12·cosα. The angle here (α) is the angle between the and r12 vector. All good. Well… No. We’ve got a problem. When it comes to calculating probabilities, the α angle doesn’t matter: |ei·θ/r|2 = 1/r2. Hence, for the probability, we get: P = | 〈r2|r1〉 |2 = 1/r122. Always ! Now that’s strange. The θ = pr12/ħ argument gives us a different phase depending on the angle (α) between p and r12. But… Well… Think of it: cosα goes from 1 to 0 when α goes from 0 to ±90° and, of course, is negative when p and r12 have opposite directions but… Well… According to this formula, the probabilities do not depend on the direction of the momentum. That’s just weird, I think. Did Feynman, in his iconic Lectures, give us a meaningless formula?

Maybe. We may also note this function looks like the elementary wavefunction for any particle, which we wrote as:

ψ(x, t) = a·e−i∙θ = a·e−i(E∙t − px)/ħ= a·ei(E∙t)/ħ·ei(px)/ħ

The only difference is that the 〈r2|r1〉 sort of abstracts away from time, so… Well… Let’s get a feel for the quantities. Let’s think of a photon carrying some typical amount of energy. Hence, let’s talk visible light and, therefore, photons of a few eV only – say 5.625 eV = 5.625×1.6×10−19 J = 9×10−19 J. Hence, their momentum is equal to p = E/c = (9×10−19 N·m)/(3×105 m/s) = 3×10−24 N·s. That’s tiny but that’s only because newtons and seconds are enormous units at the (sub-)atomic scale. As for the distance, we may want to use the thickness of a playing card as a starter, as that’s what Young used when establishing the experimental fact of light interfering with itself. Now, playing cards in Young’s time were obviously rougher than those today, but let’s take the smaller distance: modern cards are as thin as 0.3 mm. Still, that distance is associated with a value of θ that is equal to 13.6 million. Hence, the density of our wavefunction is enormous at this scale, and it’s a bit of a miracle that Young could see any interference at all ! As shown in the table below, we only get meaningful values (remember: θ is a phase angle) when we go down to the nanometer scale (10−9 m) or, even better, the angstroms scale ((10−9 m). table action

So… Well… Again: what can we do with Feynman’s formula? Perhaps he didn’t give us a propagator function but something that is more general (read: more meaningful) at our (limited) level of knowledge. As I’ve been reading Feynman for quite a while now – like three or four years 🙂 – I think… Well… Yes. That’s it. Feynman wants us to think about it. 🙂 Are you joking again, Mr. Feynman? 🙂 So let’s assume the reasonable thing: let’s assume it gives us the amplitude to go from point a to point by the position vector along some path r. So, then, in line with what we wrote in our previous post, let’s say p·r (momentum over a distance) is the action (S) we’d associate with this particular path (r) and then see where we get. So let’s write the formula like this:

ψ = a·ei·θ = (1/rei·S = ei·p∙r/r

We’ll use an index to denote the various paths: r0 is the straight-line path and ri is any (other) path. Now, quantum mechanics tells us we should calculate this amplitude for every possible path. The illustration below shows the straight-line path and two nearby paths. So each of these paths is associated with some amount of action, which we measure in Planck units: θ = S/ħalternative paths

The time interval is given by = tr0/c, for all paths. Why is the time interval the same for all paths? Because we think of a photon going from some specific point in space and in time to some other specific point in space and in time. Indeed, when everything is said and done, we do think of light as traveling from point a to point at the speed of light (c). In fact, all of the weird stuff here is all about trying to explain how it does that. 🙂

Now, if we would think of the photon actually traveling along this or that path, then this implies its velocity along any of the nonlinear paths will be larger than c, which is OK. That’s just the weirdness of quantum mechanics, and you should actually not think of the photon actually traveling along one of these paths anyway although we’ll often put it that way. Think of something fuzzier, whatever that may be. 🙂

So the action is energy times time, or momentum times distance. Hence, the difference in action between two paths and j is given by:

δ= p·rj − p·ri = p·(rj − ri) = p·Δr

I’ll explain the δS < ħ/3 thing in a moment. Let’s first pause and think about the uncertainty and how we’re modeling it. We can effectively think of the variation in as some uncertainty in the action: δ= ΔS = p·Δr. However, if S is also equal to energy times time (= E·t), and we insist is the same for all paths, then we must have some uncertainty in the energy, right? Hence, we can write δas ΔS = ΔE·t. But, of course, E = E = m·c2 = p·c, so we will have an uncertainty in the momentum as well. Hence, the variation in should be written as:

δ= ΔS = Δp·Δr

That’s just logical thinking: if we, somehow, entertain the idea of a photon going from some specific point in spacetime to some other specific point in spacetime along various paths, then the variation, or uncertainty, in the action will effectively combine some uncertainty in the momentum and the distance. We can calculate Δp as ΔE/c, so we get the following:

δ= ΔS = Δp·Δr = ΔE·Δr/c = ΔE·Δt with ΔtΔr/c

So we have the two expressions for the Uncertainty Principle here: ΔS = Δp·Δr = ΔE·Δt. Just be careful with the interpretation of Δt: it’s just the equivalent of Δr. We just express the uncertainty in distance in seconds using the (absolute) speed of light. We are not changing our spacetime interval: we’re still looking at a photon going from to in seconds, exactly. Let’s now look at the δS < ħ/3 thing. If we’re adding two amplitudes (two arrows or vectors, so to speak) and we want the magnitude of the result to be larger than the magnitude of the two contributions, then the angle between them should be smaller than 120 degrees, so that’s 2π/3 rad. The illustration below shows how you can figure that out geometrically.angles 2Hence, if S0 is the action for r0, then S1 = S0 + ħ and S2 = S0 + 2·ħ are still good, but S3 = S0 + 3·ħ is not good. Why? Because the difference in the phase angles is Δθ = S1/ħ − S0/ħ = (S0 + ħ)/ħ − S0/ħ = 1 and Δθ = S2/ħ − S0/ħ = (S0 + 2·ħ)/ħ − S0/ħ = 2 respectively, so that’s 57.3° and 114.6° respectively and that’s, effectively, less than 120°. In contrast, for the next path, we find that Δθ = S3/ħ − S0/ħ = (S0 + 3·ħ)/ħ − S0/ħ = 3, so that’s 171.9°. So that amplitude gives us a negative contribution.

Let’s do some calculations using a spreadsheet. To simplify things, we will assume we measure everything (time, distance, force, mass, energy, action,…) in Planck units. Hence, we can simply write: Sn = S0 + n. Of course, = 1, 2,… etcetera, right? Well… Maybe not. We are measuring action in units of ħ, but do we actually think action comes in units of ħ? I am not sure. It would make sense, intuitively, but… Well… There’s uncertainty on the energy (E) and the momentum (p) of our photon, right? And how accurately can we measure the distance? So there’s some randomness everywhere. 😦 So let’s leave that question open as for now.

We will also assume that the phase angle for S0 is equal to 0 (or some multiple of 2π, if you want). That’s just a matter of choosing the origin of time. This makes it really easy: ΔSn = Sn − S0 = n, and the associated phase angle θn = Δθn is the same. In short, the amplitude for each path reduces to ψn = ei·n/r0. So we need to add these first and then calculate the magnitude, which we can then square to get a probability. Of course, there is also the issue of normalization (probabilities have to add up to one) but let’s tackle that later. For the calculations, we use Euler’s r·ei·θ = r·(cosθ + i·sinθ) = r·cosθ + i·r·sinθ formula. Needless to say, |r·ei·θ|2 = |r|2·|ei·θ|2 = |r|2·(cos2θ + sin2θ) = r. Finally, when adding complex numbers, we add the real and imaginary parts respectively, and we’ll denote the ψ0 + ψ1 +ψ2 + … sum as Ψ.

Now, we also need to see how our ΔS = Δp·Δr works out. We may want to assume that the uncertainty in p and in r will both be proportional to the overall uncertainty in the action. For example, we could try writing the following: ΔSn = Δpn·Δrn = n·Δp1·Δr1. It also makes sense that you may want Δpn and Δrn to be proportional to Δp1 and Δr1 respectively. Combining both, the assumption would be this:

Δpn = √n·Δpand Δrn = √n·Δr1

So now we just need to decide how we will distribute ΔS1 = ħ = 1 over Δp1 and Δr1 respectively. For example, if we’d assume Δp1 = 1, then Δr1 = ħ/Δp1 = 1/1 = 1. These are the calculations. I will let you analyze them. 🙂newnewWell… We get a weird result. It reminds me of Feynman’s explanation of the partial reflection of light, shown below, but… Well… That doesn’t make much sense, does it?

partial reflection

Hmm… Maybe it does. 🙂 Look at the graph more carefully. The peaks sort of oscillate out so… Well… That might make sense… 🙂

Does it? Are we doing something wrong here? These amplitudes should reflect the ones that are reflected in those nice animations (like this one, for example, which is part of that’s part of the Wikipedia article on Feynman’s path integral formulation of quantum mechanics). So what’s wrong, if anything? Well… Our paths differ by some fixed amount of action, which doesn’t quite reflect the geometric approach that’s used in those animations. The graph below shows how the distance varies as a function of ngeometry

If we’d use a model in which the distance would increase linearly or, preferably, exponentially, then we’d get the result we want to get, right?

Well… Maybe. Let’s try it. Hmm… We need to think about the geometry here. Look at the triangle below. triangle sideIf is the straight-line path (r0), then ac could be one of the crooked paths (rn). To simplify, we’ll assume isosceles triangles, so equals c and, hence, rn = 2·a = 2·c. We will also assume the successive paths are separated by the same vertical distance (h = h1) right in the middle, so hb = hn = n·h1. It is then easy to show the following:r formulaThis gives the following graph for rn = 10 and h= 0.01.r graph

Is this the right step increase? Not sure. We can vary the values in our spreadsheet. Let’s first build it. The photon will have to travel faster in order to cover the extra distance in the same time, so its momentum will be higher. Let’s think about the velocity. Let’s start with the first path (= 1). In order to cover the extra distance Δr1, the velocity c1 must be equal to (r0 + Δr1)/= r0/+ Δr1/t = + Δr1/= c0 + Δr1/t. We can write c1 as c1 = c0 + Δc1, so Δc1 = Δr1/t. Now, the ratio of p1  and p0 will be equal to the ratio of c1 and c0 because p1/p= (mc1)/mc0) = c1/c0. Hence, we have the following formula for p1:

p1 = p0·c1/c0 = p0·(c0 + Δc1)/c0 = p0·[1 + Δr1/(c0·t) = p0·(1 + Δr1/r0)

For pn, the logic is the same, so we write:

pn = p0·cn/c0 = p0·(c0 + Δcn)/c0 = p0·[1 + Δrn/(c0·t) = p0·(1 + Δrn/r0)

Let’s do the calculations, and let’s use meaningful values, so the nanometer scale and actual values for Planck’s constant and the photon momentum. The results are shown below. original

Pretty interesting. In fact, this looks really good. The probability first swings around wildly, because of these zones of constructive and destructive interference, but then stabilizes. [Of course, I would need to normalize the probabilities, but you get the idea, right?] So… Well… I think we get a very meaningful result with this model. Sweet ! 🙂 I’m lovin’ it ! 🙂 And, here you go, this is (part of) the calculation table, so you can see what I am doing. 🙂newnew

The graphs below look even better: I just changed the h1/r0 ratio from 1/100 to 1/10. The probability stabilizes almost immediately. 🙂 So… Well… It’s not as fancy as the referenced animation, but I think the educational value of this thing here is at least as good ! 🙂great

🙂 This is good stuff… 🙂

Post scriptum (19 September 2017): There is an obvious inconsistency in the model above, and in the calculations. We assume there is a path r1 = , r2, r2,etcetera, and then we calculate the action for it, and the amplitude, and then we add the amplitude to the sum. But, surely, we should count these paths twice, in two-dimensional space, that is. Think of the graph: we have positive and negative interference zones that are sort of layered around the straight-line path, as shown below.zones

In three-dimensional space, these lines become surfaces. Hence, rather than adding one arrow for every δ  having one contribution only, we may want to add… Well… In three-dimensional space, the formula for the surface around the straight-line path would probably look like π·hn·r1, right? Hmm… Interesting idea. I changed my spreadsheet to incorporate that idea, and I got the graph below. It’s a nonsensical result, because the probability does swing around, but it gradually spins out of control: it never stabilizes.revisedThat’s because we increase the weight of the paths that are further removed from the center. So… Well… We shouldn’t be doing that, I guess. 🙂 I’ll you look for the right formula, OK? Let me know when you found it. 🙂

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The Principle of Least Action re-visited

Pre-script (dated 26 June 2020): This post got mutilated by the removal of some material by the dark force. You should be able to follow the main story line, however. If anything, the lack of illustrations might actually help you to think things through for yourself. In any case, we now have different views on these concepts as part of our realist interpretation of quantum mechanics, so we recommend you read our recent papers instead of these old blog posts.

Original post:

As I was posting some remarks on the Exercises that come with Feynman’s Lectures, I was thinking I should do another post on the Principle of Least Action, and how it is used in quantum mechanics. It is an interesting matter, because the Principle of Least Action sort of connects classical and quantum mechanics.

Let us first re-visit the Principle in classical mechanics. The illustrations which Feynman uses in his iconic exposé on it are copied below. You know what they depict: some object that goes up in the air, and then comes back down because of… Well… Gravity. Hence, we have a force field and, therefore, some potential which gives our object some potential energy. The illustration is nice because we can apply it any (uniform) force field, so let’s analyze it a bit more in depth.

We know the actual trajectory – which Feynman writes as x(t)x(t) + η(t) so as to distinguish it from some other nearby path x(t) – will minimize the value of the following integral:action integral

In the mentioned post, I try to explain what the formula actually means by breaking it up in two separate integrals: one with the kinetic energy in the integrand and – you guessed it 🙂 – one with the potential energy. We can choose any reference point for our potential energy, of course, but to better reflect the energy conservation principle, we assume PE = 0 at the highest point. This ensures that the sum of the kinetic and the potential energy is zero. For a mass of 5 kg (think of the ubiquitous cannon ball), and a (maximum) height of 50 m, we got the following graph.

new graph 2

Just to make sure, here is how we calculate KE and PE as a function of time:

KE and PE

We can, of course, also calculate the action as a function of time:

action integralNote the integrand: KE − PE = m·v2. Strange, isn’t it? It’s like E = m·c2, right? We get a weird cubic function, which I plotted below (blue). I added the function for the height (but in millimeter) because of the different scales.action graph final

So what’s going on? The action concept is interesting. As the product of force, distance and time, it makes intuitive sense: it’s force over distance over time. To cover some distance in some force field, energy will be used or spent but, clearly, the time that is needed should matter as well, right? Yes. But the question is: how, exactly? Let’s analyze what happens from = 0 to = 3.2 seconds, so that’s the trajectory from = 0 to the highest point (= 50 m). The action that is required to bring our 5 kg object there would be equal to F·h·t = m·g·h·t = 5×9.8×50×3.2 = 7828.9 J·s. [I use non-rounded values in my calculations.] However, our action integral tells us it’s only 5219.6 J·s. The difference (2609.3 J·s) is explained by the initial velocity and, hence, the initial kinetic energy, which we got for free, so to speak, and which, over the time interval, is spent as action. So our action integral gives us a net value, so to speak.

To be precise, we can calculate the time rate of change of the kinetic energy as d(KE)/dt = −1533.7 + 480.2·t, so that’s a linear function of time. The graph below shows how it works. The time rate of change is initially negative, as kinetic energy gets spent and increases the potential energy of our object. At the maximum height, the time of rate of change is zero. The object then starts falling, and the time rate of change becomes positive, as the velocity of our object goes from zero to… Well… The velocity is a linear function of time as well: v0 − g·t, remember? Hence, at = v0/g = 31.3/9.8 = 3.2 s, the velocity becomes negative so our cannon ball is, effectively, falling down. Of course, as it falls down and gains speed, it covers more and more distance per second and, therefore, the associated action also goes up exponentially. Just re-define our starting point at = 3.2 s. The m·v0t·(v0 − gt) term is zero at that point, and so then it’s only the m·g2·t3/3 term that counts.

KE time rate final

So… Yes. That’s clear enough. But it still doesn’t answer the fundamental question: how does that minimization of S (or the maximization of −S) work, exactly? Well… It’s not like Nature knows it wants to go from point to point b, and then sort of works out some least action algorithm. No. The true path is given by the force law which, at every point in spacetime, will accelerate, or decelerate, our object at a rate that is equal to the ratio of the force and the mass of our object. In this case, we write: = F/= m·g/m = g, so that’s the acceleration of gravity. That’s the only real thing: all of the above is just math, some mental construct, so to speak.

Of course, this acceleration, or deceleration, then gives the velocity and the kinetic energy. Hence, once again, it’s not like we’re choosing some average for our kinetic energy: the force (gravity, in this particular case) just give us that average. Likewise, the potential energy depends on the position of our object, which we get from… Well… Where it starts and where it goes, so it also depends on the velocity and, hence, the acceleration or deceleration from the force field. So there is no optimization. No teleology. Newton’s force law gives us the true path. If we drop something down, it will go down in a straight line, because any deviation from it would add to the distance. A more complicated illustration is Fermat’s Principle of Least Time, which combines distance and time. But we won’t go into any further detail here. Just note that, in classical mechanics, the true path can, effectively, be associated with a minimum value for that action integral: any other path will be associated with a higher S. So we’re done with classical mechanics here. What about the Principle of Least Action in quantum mechanics?

The Principle of Least Action in quantum mechanics

We have the uncertainty in quantum mechanics: there is no unique path. However, we can, effectively, associate each possible path with a definite amount of action, which we will also write as S. However, instead of talking velocities, we’ll usually want to talk momentum. Photons have no rest mass (m0 = 0), but they do have momentum because of their energy: for a photon, the E = m·c2 equation can be rewritten as E = p·c, and the Einstein-Planck relation for photons tells us the photon energy (E) is related to the frequency (f): E = h·f. Now, for a photon, the wavelength is given by = c/λ. Hence, p = E/c = h·f/c= h/λ = ħ·k.

OK. What’s the action integral? What’s the kinetic and potential energy? Let’s just try the energy: E = m·c2. It reflects the KE − PE = m·v2 formula we used above. Of course, the energy of a photon does not vary, so the value of our integral is just the energy times the travel time, right? What is the travel time? Let’s do things properly by using vector notations here, so we will have two position vectors rand r2 for point and b respectively. We can then define a vector pointing from r1 to r2, which we will write as r12. The distance between the two points is then, obviously, equal to|r12| = √r122 = r12. Our photon travels at the speed of light, so the time interval will be equal to = r12/c. So we get a very simple formula for the action: = E·t = p·c·= p·c·r12/c = p·r12. Now, it may or may not make sense to assume that the direction of the momentum of our photon and the direction of r12 are somewhat different, so we’ll want to re-write this as a vector dot product: S = p·r12. [Of course, you know the pr12 dot product equals |p|∙|r12cosθ = p∙r12·cosθ, with θ the angle between p and r12. If the angle is the same, then cosθ is equal to 1. If the angle is ± π/2, then it’s 0.]

So now we minimize the action so as to determine the actual path? No. We have this weird stopwatch stuff in quantum mechanics. We’ll use this S = p·r12 value to calculate a probability amplitude. So we’ll associate trajectories with amplitudes, and we just use the action values to do so. This is how it works (don’t ask me why – not now, at least):

  1. We measure action in units of ħ, because… Well… Planck’s constant is a pretty fundamental unit of action, right? 🙂 So we write θ = S/ħ p·r12/ħ.
  2. θ usually denotes an angle, right? Right. θ = p·r12/ħ is the so-called phase of… Well… A proper wavefunction:

ψ(pr12) = a·ei·θ = (1/r12ei·pr12   

Wow ! I realize you may never have seen this… Well… It’s my derivation of what physicists refer to as the propagator function for a photon. If you google it, you may see it written like this (most probably not, however, as it’s usually couched in more abstract math):propagatorThis formulation looks slightly better because it uses Diracs bra-ket notation: the initial state of our photon is written as 〈 r1| and its final state is, accordingly, |r2〉. But it’s the same: it’s the amplitude for our photon to go from point to point b. In case you wonder, the 1/r12 coefficient is there to take care of the inverse square law. I’ll let you think about that for yourself. It’s just like any other physical quantity (or intensity, if you want): they get diluted as the distance increases. [Note that we get the inverse square (1/r122) when calculating a probability, which we do by taking the absolute square of our amplitude: |(1/r12ei·pr12|2 = |1/r122)|2·|ei·pr12|2 = 1/r122.]

So… Well… Now we are ready to understand Feynman’s own summary of his path integral formulation of quantum mechanics:  explanation words:

“Here is how it works: Suppose that for all paths, S is very large compared to ħ. One path contributes a certain amplitude. For a nearby path, the phase is quite different, because with an enormous even a small change in means a completely different phase—because ħ is so tiny. So nearby paths will normally cancel their effects out in taking the sum—except for one region, and that is when a path and a nearby path all give the same phase in the first approximation (more precisely, the same action within ħ). Only those paths will be the important ones.”

You are now, finally, ready to understand that wonderful animation that’s part of the Wikipedia article on Feynman’s path integral formulation of quantum mechanics. Check it out, and let the author (not me, but a guy who identifies himself as Juan David) I think it’s great ! 🙂

Explaining diffraction

All of the above is nice, but how does it work? What’s the geometry? Let me be somewhat more adventurous here. So we have our formula for the amplitude of a photon to go from one point to another:propagatorThe formula is far too simple, if only because it assumes photons always travel at the speed of light. As explained in an older post of mine, a photon also has an amplitude to travel slower or faster than (I know that sounds crazy, but it is what it is) and a more sophisticated propagator function will acknowledge that and, unsurprisingly, ensure the spacetime intervals that are more light-like make greater contributions to the ‘final arrow’, as Feynman (or his student, Ralph Leighton, I should say) put it in his Strange Theory of Light and Matter. However, then we’d need to use four-vector notation and we don’t want to do that here. The simplified formula above serves the purpose. We can re-write it as:

ψ(pr12) = a·ei·θ = (1/r12ei·S = ei·pr12/r12

Again, S = p·r12 is just the amount of action we calculate for the path. Action is energy over some time (1 N·m·s = 1 J·s), or momentum over some distance (1 kg·(m/s)·m = 1 N·(s2/m)·(m/s)·m) = 1 N·m·s). For a photon traveling at the speed of light, we have E = p·c, and r12/c, so we get a very simple formula for the action: = E·t = p·r12. Now, we know that, in quantum mechanics, we have to add the amplitudes for the various paths between r1 and r2 so we get a ‘final arrow’ whose absolute square gives us the probability of… Well… Our photon going from r1 and r2. You also know that we don’t really know what actually happens in-between: we know amplitudes interfere, but that’s what we’re modeling when adding the arrows. Let me copy one of Feynman’s famous drawings so we’re sure we know what we’re talking about.adding-arrowsOur simplified approach (the assumption of light traveling at the speed of light) reduces our least action principle to a least time principle: the arrows associated with the path of least time and the paths immediately left and right of it that make the biggest contribution to the final arrow. Why? Think of the stopwatch metaphor: these stopwatches arrive around the same time and, hence, their hands point more or less in the same direction. It doesn’t matter what direction – as long as it’s more or less the same.

Now let me copy the illustrations he uses to explain diffraction. Look at them carefully, and read the explanation below.

When the slit is large, our photon is likely to travel in a straight line. There are many other possible paths – crooked paths – but the amplitudes that are associated with those other paths cancel each other out. In contrast, the straight-line path and, importantly, the nearby paths, are associated with amplitudes that have the same phase, more or less.

However, when the slit is very narrow, there is a problem. As Feynman puts it, “there are not enough arrows to cancel each other out” and, therefore, the crooked paths are also associated with sizable probabilities. Now how does that work, exactly? Not enough arrows? Why? Let’s have a look at it.

The phase (θ) of our amplitudes a·ei·θ = (1/r12ei·S is measured in units of ħ: θ = S/ħ. Hence, we should measure the variation in in units of ħ. Consider two paths, for example: one for which the action is equal to S, and one for which the action is equal to + δ+ π·ħ, so δ= π·ħ. They will cancel each other out:

ei·S/ħ/r12 + e(S + δS)/ħ/r12 = (1/r12)·(ei·S/ħ/r12 + ei·(S+π·ħ)/ħ/r12 )

= (1/r12)·(ei·S/ħ + ei·S/ħ·ei·π) = (1/r12)·(ei·S/ħ − ei·S/ħ) = 0

So nearby paths will interfere constructively, so to speak, by making the final arrow larger. In order for that to happen, δS should be smaller than 2πħ/3 ≈ 2ħ, as shown below.

alternative paths

Why? That’s just the way the addition of angles work. Look at the illustration below: if the red arrow is the amplitude to which we are adding another, any amplitude whose phase angle is smaller than 2πħ/3 ≈ 2ħ will add something to its length. That’s what the geometry of the situation tells us. [If you have time, you can perhaps find some algebraic proof: let me know the result!]
anglesWe need to note a few things here. First, unlike what you might think, the amplitudes of the higher and lower path in the drawing do not cancel. On the contrary, the action is the same, so their magnitudes just add up. Second, if this logic is correct, we will have alternating zones with paths that interfere positively and negatively, as shown below.interference 2

Interesting geometry. How relevant are these zones as we move out from the center, steadily increasing δS? I am not quite sure. I’d have to get into the math of it all, which I don’t want to do in a blog like this. What I do want to do is re-examine is Feynman’s intuitive explanation of diffraction: when the slit is very narrow, “there are not enough arrows to cancel each other out.”

Huh? What’s that? Can’t we add more paths? It’s a tricky question. We are measuring action in units of ħ, but do we actually think action comes in units of ħ? I am not sure. It would make sense, intuitively, but… Well… There’s uncertainty on the energy (E) and the momentum (p) of our photon, right? And how accurately can we measure the distance? So there’s some randomness everywhere. Having said that, the whole argument does requires us to assume action effectively comes in units of ħħ is, effectively, the scaling factor here.

So how can we have more paths? More arrows? I don’t think so. We measure as energy over some time, or as momentum over some distance, and we express all these quantities in old-fashioned SI units: newton for the force, meter for the distance, and second for the time. If we want smaller arrows, we’ll have to use other units, but then the numerical value for ħ will change too! So… Well… No. I don’t think so. And it’s not because of the normalization rule (all probabilities have to add up to one, so we do some have some re-scaling for that). That doesn’t matter, really. What matters is the physics behind the formula, and the formula tells us the physical reality is ħ. So the geometry of the situation is what it is.

Hmm… I guess that, at this point, we should wrap up our rather intuitive discussion here, and resort to the mathematical formalism of Feynman’s path integral formulation, but you can find that elsewhere.

Post scriptum: I said I would show how the Principle of Least Action is relevant to both classical as well as quantum mechanics. Well… Let me quote the Master once more:

“So in the limiting case in which Planck’s constant ħ goes to zero, the correct quantum-mechanical laws can be summarized by simply saying: ‘Forget about all these probability amplitudes. The particle does go on a special path, namely, that one for which does not vary in the first approximation.’”

So that’s how the Principle of Least Action sort of unifies quantum mechanics as well as classical mechanics. 🙂

Post scriptum 2: In my next post, I’ll be doing some calculations. They will answer the question as to how relevant those zones of positive and negative interference further away from the straight-line path. I’ll give a numerical example which shows the 1/r12 factor does its job. 🙂 Just have a look at it. 🙂

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