# 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. ðŸ™‚ Let’s go.

Let’s first look at Feynman’s (simplified) formula for the amplitude of a photon to go from point aÂ to point b. If we identify point aÂ by the position vector r1Â and point bÂ by the position vectorÂ r2, and using Dirac’s fancyÂ bra-ketÂ notation, then it’s written as:

So we have a vector dot product here: pâˆ™r12Â = |p|âˆ™|r12|Â·Â cosÎ¸ = pâˆ™r12Â·cosÎ±. The angle here (Î±) is the angle between theÂ pÂ andÂ r12Â vector. All good. Well… No. We’ve got a problem. When it comes to calculating probabilities, the Î± angle doesn’t matter: |eiÂ·Î¸/r|2Â = 1/r2. Hence, for the probability, we get: P = |Â âŒ©r2|r1âŒª |2Â =Â 1/r122. Always ! Now that’s strange. The Î¸ =Â pâˆ™r12/Ä§Â argument gives us a different phase depending on the angle (Î±) between p and r12. But… Well… Think of it:Â cosÎ± goes from 1 to 0 when Î± goes from 0 to Â±90Â° and, of course, is negative when p and r12Â have opposite directions but… Well… According to this formula, the probabilitiesÂ doÂ not depend on the direction of the momentum. That’s just weird, I think. Did Feynman, in his iconicÂ 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:

Ïˆ(x, t) = aÂ·eâˆ’iâˆ™Î¸Â =Â aÂ·eâˆ’iâˆ™(Eâˆ™t âˆ’Â pâˆ™x)/Ä§= aÂ·eâˆ’iâˆ™(Eâˆ™t)/Ä§Â·eiâˆ™(pâˆ™x)/Ä§

The only difference is that the âŒ©r2|r1âŒª sort of abstracts away from time, so… Well… Let’s get a feel for the quantities. Let’s think of a photon carryingÂ some typical amount of energy. Hence, let’s talk visible light and, therefore, photons of a few eV only – say 5.625 eV = 5.625Ã—1.6Ã—10âˆ’19Â J = 9Ã—10âˆ’19Â J. Hence, their momentum is equal to p = E/c = (9Ã—10âˆ’19Â NÂ·m)/(3Ã—105Â m/s) = 3Ã—10âˆ’24Â NÂ·s. That’s tiny but that’s only becauseÂ newtonsÂ andÂ secondsÂ 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 experimentalÂ 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 ofÂ Î¸ that is equal to 13.6 million. Hence, theÂ densityÂ 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:Â Î¸ is a phase angle) when we go down to the nanometerÂ scale (10âˆ’9Â m) or, even better, theÂ angstroms scale ((10âˆ’9Â m).Â

So… Well… Again: what can we do with Feynman’s formula? Perhaps he didn’t give us a propagatorÂ 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 ðŸ™‚ – I think… Well… Yes. That’s it. Feynman wants us to think about it. ðŸ™‚ Are you joking again, Mr. Feynman?Â ðŸ™‚Â So let’s assume the reasonable thing: let’s assume it gives us the amplitude to go from point a toÂ point bÂ by the position vectorÂ along some path r.Â So, then, in line with what we wrote in our previous post, let’s say pÂ·rÂ (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 writeÂ the formula like this:

ÏˆÂ =Â aÂ·eiÂ·Î¸Â = (1/r)Â·eiÂ·S/Ä§Â =Â eiÂ·pâˆ™r/Ä§/r

We’ll use an index to denote the various paths: r0Â is the straight-line path and riÂ is any (other) path.Â Now, quantum mechanics tells us we should calculate this amplitudeÂ 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 PlanckÂ units:Â Î¸ =Â S/Ä§.Â

The time interval is given by tÂ = t0Â =Â r0/c, for all paths. Why is the time interval the same for all paths? Because we think of a photon going from some specificÂ point in space and in timeÂ to some otherÂ specificÂ point in space and in time. Indeed, when everything is said and done, we do think of light as traveling from pointÂ a to pointÂ bÂ at the speed of light (c). In fact, all of the weird stuff here is all about trying to explain howÂ it does that. ðŸ™‚

Now, if we would think of the photon actually traveling along this or that path, then this implies its velocityÂ along any of the nonlinear paths will be largerÂ thanÂ c, which is OK. That’s just the weirdness of quantum mechanics, and you should actuallyÂ notÂ 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. ðŸ™‚

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

Î´SÂ = pÂ·rjÂ âˆ’Â pÂ·riÂ = pÂ·(rjÂ âˆ’ ri) = pÂ·Î”r

I’ll explain theÂ Î´S <Â 2Ï€Ä§/3 thing in a moment. Let’s first pause and think about theÂ uncertainty and how we’re modeling it. We can effectively think of the variation in SÂ as some uncertaintyÂ in the action: Î´SÂ = Î”S = pÂ·Î”r. However, if SÂ is also equal to energy times time (SÂ = EÂ·t), and we insist tÂ is the same for all paths, then we must have some uncertainty in the energy, right? Hence, we can write Î´SÂ as Î”S = Î”EÂ·t. But, of course, E =Â E =Â mÂ·c2Â = pÂ·c, so we will have an uncertainty in the momentum as well. Hence, the variation inÂ SÂ should be written as:

Î´SÂ = Î”SÂ = Î”pÂ·Î”r

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

Î´SÂ = Î”SÂ = Î”pÂ·Î”r =Â Î”EÂ·Î”r/c = Î”EÂ·Î”t with Î”t =Â Î”r/c

So we have the two expressions for the Uncertainty Principle here: Î”SÂ = Î”pÂ·Î”r =Â Î”EÂ·Î”t. Just be careful with the interpretation of Î”t: it’s just the equivalent of Î”r. We just express the uncertainty in distance in secondsÂ using the (absolute) speed of light. We are notÂ changing our spacetime interval: we’re still looking at a photon going fromÂ aÂ toÂ bÂ inÂ tÂ seconds,Â exactly. Let’s now look at theÂ Î´S <Â 2Ï€Ä§/3 thing. If we’re addingÂ twoÂ amplitudes (twoÂ arrowsÂ 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 2Ï€/3 rad. The illustration below shows how you can figure that out geometrically.Hence, if S0Â is the action for r0, then S1Â = S0Â + Ä§Â and S2Â = S0Â + 2Â·Ä§ are still good, but S3Â = S0Â + 3Â·Ä§Â isÂ notÂ good. Why? Because the difference in the phase angles is Î”Î¸Â =Â S1/Ä§Â âˆ’Â S0/Ä§Â = (S0Â + Ä§)/Ä§Â âˆ’Â S0/Ä§ = 1 andÂ Î”Î¸ =Â S2/Ä§Â âˆ’Â S0/Ä§Â = (S0Â + 2Â·Ä§)/Ä§Â âˆ’Â S0/Ä§ = 2 respectively, so that’s 57.3Â°Â and 114.6Â°Â respectively and that’s, effectively,Â lessÂ than 120Â°. In contrast,Â for the next path, we find that Î”Î¸Â =Â S3/Ä§Â âˆ’Â S0/Ä§Â = (S0Â + 3Â·Ä§)/Ä§Â âˆ’Â S0/Ä§ = 3, so that’s 171.9Â°. 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:Â SnÂ = S0Â + n. Of course, nÂ = 1, 2,… etcetera, right? Well… Maybe not. We areÂ measuringÂ action in units ofÂ Ä§, butÂ do we actually think actionÂ comesÂ in units ofÂ Ä§?Â I am not sure. It would make sense, intuitively, butâ€¦ Wellâ€¦ Thereâ€™s uncertainty on the energy (E) and the momentum (p) of our photon, right? And how accurately can we measure the distance? So thereâ€™s some randomness everywhere. ðŸ˜¦ So let’s leave that question open as for now.

We will also assume that the phase angle forÂ S0Â is equal to 0 (or some multiple of 2Ï€, if you want). That’s just a matter of choosing the origin of time. This makes it really easy: Î”SnÂ =Â SnÂ âˆ’ S0Â = n, and the associated phase angle Î¸nÂ = Î”Î¸nÂ is the same. In short, the amplitude for each path reduces to ÏˆnÂ = eiÂ·n/r0. So we need to add these firstÂ andÂ thenÂ 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 rÂ·eiÂ·Î¸Â = rÂ·(cosÎ¸ + iÂ·sinÎ¸) = rÂ·cosÎ¸ + iÂ·rÂ·sinÎ¸ formula. Needless to say, |rÂ·eiÂ·Î¸|2Â = |r|2Â·|eiÂ·Î¸|2Â = |r|2Â·(cos2Î¸ + sin2Î¸) = r. Finally, when adding complex numbers, we add the real and imaginary parts respectively, and we’ll denote the Ïˆ0Â + Ïˆ1Â +Ïˆ2Â + … sum as Î¨.

Now, we also need to see how our Î”SÂ = Î”pÂ·Î”rÂ 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:Â Î”SnÂ = Î”pnÂ·Î”rnÂ =Â nÂ·Î”p1Â·Î”r1. It also makes sense that you may want Î”pnÂ and Î”rnÂ to be proportional to Î”p1Â and Î”r1Â respectively. Combining both, the assumption would be this:

Î”pnÂ =Â âˆšnÂ·Î”p1Â andÂ Î”rnÂ =Â âˆšnÂ·Î”r1

So now we just need to decide how we will distribute Î”S1Â =Â Ä§Â = 1 over Î”p1Â and Î”r1Â respectively. For example, if we’d assume Î”p1Â = 1, then Î”r1Â = Ä§/Î”p1Â = 1/1 = 1. These are the calculations. I will let you analyze them. ðŸ™‚Well… We get a weird result. It reminds me ofÂ Feynman’s explanation of the partial reflection of light, shown below, but… Well… That doesn’t make much sense, does it?

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

Does it? Are we doingÂ something wrongÂ here? These amplitudes should reflect the ones that are reflected in those nice animations (like this one, for example, which is part of thatâ€™s part of the Wikipedia article on Feynmanâ€™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 distanceÂ rÂ varies as a function ofÂ n.Â

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

Well… Maybe. Let’s try it.Â Hmm… We need to think about the geometry here. Look at the triangle below.Â IfÂ bÂ is the straight-line path (r0), thenÂ acÂ could be one of the crooked paths (rn). To simplify, we’ll assume isosceles triangles, soÂ aÂ equalsÂ cÂ and, hence, rnÂ = 2Â·a = 2Â·c. We will also assume theÂ successive paths are separated by the same vertical distance (h =Â h1) right in the middle, so hbÂ =Â hnÂ = nÂ·h1.Â It is then easy to show the following:This gives the following graph for rnÂ = 10 and h1Â = 0.01.

Is this the right step increase? Not sure. We can vary the values in our spreadsheet. Let’s first build it. TheÂ 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 (nÂ = 1). In order to cover the extraÂ distance Î”r1, the velocity c1Â must be equal to (r0Â + Î”r1)/tÂ = r0/tÂ + Î”r1/t =Â cÂ + Î”r1/tÂ = c0Â + Î”r1/t. We can write c1Â as c1Â =Â c0Â + Î”c1, so Î”c1Â = Î”r1/t.Â Now, theÂ ratioÂ of p1Â  and p0Â will be equal to theÂ ratioÂ of c1Â andÂ c0Â because p1/p0Â = (mc1)/mc0) = c1/c0. Hence, we have the following formula for p1:

p1Â = p0Â·c1/c0Â = p0Â·(c0Â + Î”c1)/c0Â = p0Â·[1 + Î”r1/(c0Â·t) = p0Â·(1 + Î”r1/r0)

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

pnÂ = p0Â·cn/c0Â = p0Â·(c0Â + Î”cn)/c0Â = p0Â·[1 + Î”rn/(c0Â·t) = p0Â·(1 + Î”rn/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.Â

Pretty interesting. In fact, this looksÂ reallyÂ good. TheÂ probabilityÂ 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 veryÂ meaningful result with this model. Sweet ! ðŸ™‚ I’m lovin’ it ! ðŸ™‚ And, here you go, this is (part of) the calculation table, so you can see what I am doing. ðŸ™‚

The graphs below look even better: I just changed the h1/r0Â ratio from 1/100 to 1/10. The probability stabilizes almost immediately. ðŸ™‚ 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 ! ðŸ™‚

ðŸ™‚ This is good stuff… ðŸ™‚

Post scriptum (19 September 2017): There is an obvious inconsistency in the model above, and in the calculations. We assume there is a path r1Â = ,Â 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.

In three-dimensional space, these lines become surfaces. Hence, rather than adding oneÂ arrow for everyÂ Î´Â Â having oneÂ 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 Ï€Â·hnÂ·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.That’s because we increase theÂ weightÂ of the paths that are further removed from the center. So… Well… We shouldn’t be doing that, I guess. ðŸ™‚ I’ll you look for the right formula, OK? Let me know when you found it. ðŸ™‚

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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’sÂ Lectures,Â 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 iconicÂ exposÃ©Â 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 actualÂ trajectory – which Feynman writes as x(t) =Â x(t)Â +Â Î·(t) so as to distinguish it from some other nearby path x(t) – willÂ minimizeÂ the value of the following 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 ðŸ™‚ – 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 theÂ sumÂ 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.

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

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

Note the integrand: KEÂ âˆ’ PEÂ = mÂ·v2. Strange, isn’t it? It’sÂ likeÂ EÂ =Â mÂ·c2, right? We get aÂ weird cubic function, which I plotted below (blue). I added the function for theÂ heightÂ (but inÂ millimeter) because of the different scales.

So what’s going on? The action concept is interesting. As theÂ productÂ 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 timeÂ that is needed should matter as well, right? Yes. But the question is:Â how, exactly? Let’s analyze what happens fromÂ tÂ = 0 toÂ tÂ = 3.2 seconds, so that’s the trajectory fromÂ hÂ = 0 to the highest point (hÂ = 50 m). The actionÂ that is required toÂ bring our 5 kg object there would be equal to FÂ·hÂ·t = mÂ·gÂ·hÂ·tÂ =Â 5Ã—9.8Ã—50Ã—3.2 = 7828.9Â JÂ·s. [I use non-rounded values in my calculations.]Â However, our action integral tells us it’s only 5219.6Â JÂ·s. The difference (2609.3 JÂ·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 asÂ action. So our action integral gives us a netÂ value, so to speak.

To be precise, we can calculate the time rate of change of the kinetic energy as d(KE)/dtÂ = âˆ’1533.7 + 480.2Â·t, so that’s a linear function of time. The graph below shows how it works. The time rate of change is initially negative, asÂ 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:Â vÂ =Â v0Â âˆ’ gÂ·t, remember? Hence, atÂ tÂ = v0/g = 31.3/9.8 = 3.2 s, the velocity becomesÂ negativeÂ so our cannon ball is, effectively, falling down. Of course, as it falls down and gains speed, it covers more and more distance per secondÂ and, therefore, the associated actionÂ also goes up exponentially. Just re-define our starting point atÂ tÂ = 3.2 s. The mÂ·v0tÂ·(v0Â âˆ’Â gt) term is zero at that point, and so then it’s only the mÂ·g2Â·t3/3 term that counts.

So… Yes. That’s clear enough. But it still doesn’t answer the fundamental question: how does that minimization of SÂ (or the maximization ofÂ âˆ’S) work,Â exactly? Well… It’s not likeÂ NatureÂ knows it wants to go from pointÂ aÂ to pointÂ b, and then sort of works out some least actionÂ algorithm. No. The true path is given by the force law which,Â at every point in spacetime, will accelerate, or decelerate, our object at a rateÂ aÂ that is equal to the ratio of the force and the mass of our object. In this case, we write:Â aÂ = F/mÂ = mÂ·g/m = g, so that’s the acceleration of gravity. That’s the onlyÂ realÂ thing: all of the above is just math, someÂ 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 choosingÂ 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 isÂ noÂ optimization. No teleology.Â 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 minimumÂ 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 talkingÂ velocities, we’ll usually want to talkÂ momentum. Photons have no rest mass (m0Â = 0), but they do haveÂ momentumÂ because of their energy: for a photon, the E = mÂ·c2Â equation can be rewritten as E = pÂ·c, and the Einstein-Planck relation for photons tells us the photon energy (E) is related to the frequency (f): E = hÂ·f. Now, for a photon, the wavelength is given by fÂ = c/Î».Â Hence, p = E/c = hÂ·f/c= h/Î» = Ä§Â·k.

OK. What’s the action integral? What’s the kinetic and potential energy? Let’s just try the energy: E = mÂ·c2. It reflects theÂ KEÂ âˆ’ PEÂ = mÂ·v2Â formula we used above. Of course, the energy of a photon doesÂ notÂ 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 vectorsÂ r1Â andÂ r2Â for point aÂ andÂ b respectively. We can then define a vector pointing fromÂ r1Â toÂ r2, which we will write as r12. The distance between the two points is then, obviously, equal to|r12| = âˆšr122Â =Â r12. Our photon travels at the speed of light, so theÂ timeÂ interval will be equal toÂ tÂ = r12/c. So we get a very simple formula for the action:Â SÂ = EÂ·t = pÂ·cÂ·tÂ = pÂ·cÂ·r12/cÂ = pÂ·r12. Now, it may or may not make sense to assume that the directionÂ of the momentum of our photon and the direction of r12Â are somewhat different, so we’ll want to re-write this as a vector dot product: S =Â pÂ·r12. [Of course, you know theÂ pâˆ™r12Â dot product equals |p|âˆ™|r12|Â· cosÎ¸ = pâˆ™r12Â·cosÎ¸, with Î¸ the angle betweenÂ pÂ andÂ r12. If the angle is the same, then cosÎ¸ is equal to 1. If the angle is Â± Ï€/2, then itâ€™s 0.]

So now we minimize the action so as to determine the actualÂ path? No. We have this weird stopwatchÂ stuff in quantum mechanics.Â We’ll use this S =Â pÂ·r12Â 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 Ä§, because… Well… Planck’s constant is a pretty fundamental unit of action, right? ðŸ™‚ So we write Î¸ = S/Ä§Â =Â pÂ·r12/Ä§.
2. Î¸ usually denotes an angle, right? Right. Î¸ = pÂ·r12/Ä§Â is the so-called phase of… Well… A proper wavefunction:

Ïˆ(p,Â r12) = aÂ·eiÂ·Î¸Â = (1/r12)Â·eiÂ·pâˆ™r12/Ä§Â Â Â

Wow !Â I realize you may never have seen this… Well… It’s myÂ derivation of what physicists refer to as theÂ propagator functionÂ 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):This formulation looks slightly better because it uses Diracs bra-ketÂ notation:Â the initialÂ state of our photon is written as âŒ©Â r1|Â and its final state is, accordingly, |r2âŒª. But it’s the same: it’s the amplitude for our photon to go from point aÂ to pointÂ b. In case you wonder, the 1/r12Â 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 (orÂ intensity, if you want): they get diluted as the distance increases. [Note that we get the inverse square (1/r122)Â when calculating a probability, which we do byÂ taking the absolute square of our amplitude:Â |(1/r12)Â·eiÂ·pâˆ™r12/Ä§|2Â = |1/r122)|2Â·|eiÂ·pâˆ™r12/Ä§|2Â = 1/r122.]

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

â€œHere is how it works: Suppose that for all paths, SÂ is very large compared to Ä§.Â One path contributes a certain amplitude. For a nearby path, the phase is quite different, because with an enormous SÂ even a small change in SÂ means a completely different phaseâ€”because Ä§Â is so tiny. So nearby paths will normally cancel their effects out in taking the sumâ€”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 Ä§). 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, butÂ a guy who identifies himself asÂ Juan David) I think it’s great ! ðŸ™‚

## 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 forÂ theÂ amplitudeÂ of a photon to go from one pointÂ to another:The 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 cÂ (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:

Ïˆ(p,Â r12) =Â aÂ·eiÂ·Î¸Â = (1/r12)Â·eiÂ·S/Ä§Â = eiÂ·pâˆ™r12/Ä§/r12

Again, S =Â pÂ·r12Â is just the amount ofÂ actionÂ we calculate for the path. Action is energy over some time (1 NÂ·mÂ·s = 1 JÂ·s), or momentum over some distance (1 kgÂ·(m/s)Â·m = 1 NÂ·(s2/m)Â·(m/s)Â·m) = 1 NÂ·mÂ·s). For a photon traveling at the speed of light, we have E = pÂ·c, and tÂ =Â r12/c, so we get a very simple formula for the action:Â SÂ = EÂ·tÂ = pÂ·r12. Now, we know that, in quantum mechanics, we have to add the amplitudes for the various paths between r1Â and r2Â so we get a ‘final arrow’ whose absolute square gives us the probability of… Well… Our photon going from r1Â 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.Our 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 biggestÂ contributionÂ 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 doesnâ€™t matter what direction â€“ as long as itâ€™s more or lessÂ 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 otherÂ possibleÂ 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 nearbyÂ paths, are associated with amplitudes that have the same phase, more or less.

However, when the slit is very narrow, there is a problem. AsÂ 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 (Î¸) of our amplitudes aÂ·eiÂ·Î¸Â = (1/r12)Â·eiÂ·S/Ä§Â is measured in units of Ä§:Â Î¸ = S/Ä§. Hence, we should measure the variation in SÂ in units of Ä§. Consider two paths, for example: one for which the action is equal to S, and one for which the action is equal toÂ SÂ +Â Î´SÂ =Â SÂ +Â Ï€Â·Ä§, so Î´SÂ = Ï€Â·Ä§.Â They will cancel each other out:

eiÂ·S/Ä§/r12Â + eiÂ·(SÂ +Â Î´S)/Ä§/r12Â = (1/r12)Â·(eiÂ·S/Ä§/r12Â + eiÂ·(S+Ï€Â·Ä§)/Ä§/r12Â )

= (1/r12)Â·(eiÂ·S/Ä§Â + eiÂ·S/Ä§Â·eiÂ·Ï€) = (1/r12)Â·(eiÂ·S/Ä§Â âˆ’ eiÂ·S/Ä§) = 0

So nearby paths will interfere constructively, so to speak, by making the final arrow larger. In order for that to happen,Â Î´S should be smaller thanÂ 2Ï€Ä§/3 â‰ˆ 2Ä§, as shown below.

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 thanÂ 2Ï€Ä§/3 â‰ˆ 2Ä§Â 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!]
We need to note a few things here. First, unlike what you might think, the amplitudes of theÂ higher and lower path in the drawing do notÂ cancel. On the contrary, the action SÂ 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.

Interesting geometry. How relevant are these zones as we move out from the center, steadily increasing Î´S? 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?Â What’s that?Â Can’t we add more paths? It’s a tricky question. We are measuringÂ action in units ofÂ Ä§, butÂ do we actually think action comesÂ in units ofÂ Ä§?Â 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 effectivelyÂ comesÂ in units of Ä§:Â Ä§Â is, effectively, theÂ scaling factorÂ here.

So how can we have more paths? More arrows? I don’t think so. We measure SÂ as energy over some time, or as momentum over some distance, and we express all these quantities in old-fashioned SI units: newtonÂ for the force,Â meterÂ for the distance, andÂ secondÂ for the time. If we want smaller arrows, we’ll have to use other units, but then the numericalÂ value forÂ Ä§Â 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 isÂ Ä§. 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 Planckâ€™s constant Ä§Â goes to zero, the correct quantum-mechanical laws can be summarized by simply saying: â€˜Forget about all these probability amplitudes. The particle does go on a special path, namely, that one for which SÂ does not vary in the first approximation.â€™”

So thatâ€™s how the Principle of Least Action sort of unifies quantum mechanics as well as classical mechanics. ðŸ™‚

Post scriptumÂ 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 exampleÂ which shows the 1/r12Â factor does its job. ðŸ™‚ Just have a look at it. ðŸ™‚

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