**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 *a*Â to point *b*. If we identify point *aÂ *by the position vector ** r_{1}**Â and point

*bÂ*by the position vectorÂ

**, and using Dirac’s fancyÂ**

*r*_{2}*bra-ketÂ*notation, then it’s written as:

So we have a vector dot product here: **p**âˆ™*r*_{12}Â = |**p**|âˆ™|*r*_{12}|Â·Â *cos*Î¸ = pâˆ™*r*_{12}Â·*cos*Î±. The angle here (Î±) is the angle between theÂ **pÂ **andÂ *r*_{12}Â vector. All good. Well… No. We’ve got a problem. When it comes to calculating *probabilities*, the Î± angle doesn’t matter: |*e*^{iÂ·Î¸}/*r*|^{2}Â = 1/*r*^{2}. Hence, for the probability, we get: P = |Â âŒ©**r**_{2}|**r**_{1}âŒª |^{2}Â =Â 1/*r*_{12}^{2}. ** Always !** Now that’s strange. The Î¸ =Â

**p**âˆ™

*r*_{12}/

*Ä§*Â argument gives us a different phase depending on the angle (Î±) between

**p**and

*r*_{12}. But… Well… Think of it:Â

*cos*Î± goes from 1 to 0 when Î± goes from 0 to Â±90Â° and, of course, is

*negative*when

**p**and

*r*_{12}Â 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·e*

^{âˆ’i}^{âˆ™Î¸}Â =Â

*aÂ·e*

^{âˆ’i}^{âˆ™}

^{(Eâˆ™t âˆ’Â pâˆ™x)/Ä§}=

*a*Â·

*e*

^{âˆ’i}

^{âˆ™}

^{(Eâˆ™t)/Ä§}Â·

*e*

^{i}^{âˆ™}

^{(pâˆ™x)/Ä§}

The only difference is that the âŒ©**r**_{2}|**r**_{1}âŒª 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—10^{âˆ’19}Â J = 9Ã—10^{âˆ’19}Â J. Hence, their momentum is equal to p = E/*c* = (9Ã—10^{âˆ’19}Â NÂ·m)/(3Ã—10^{5}Â m/s) = 3Ã—10^{âˆ’24}Â 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 *nanometer*Â scale (10^{âˆ’9}Â m) or, even better, theÂ *angstroms *scale ((10^{âˆ’9}Â m).*Â *

So… Well… Again: what can we do with Feynman’s formula? Perhaps he didn’t give us a *propagator*Â 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·

*r*Â (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 *a*Â·*e*^{iÂ·Î¸}Â = (1/*r*)Â·*e*^{iÂ·S/Ä§}Â =Â *e*^{iÂ·pâˆ™r/Ä§}/*r*

We’ll use an index to denote the various paths: *r*_{0}Â is the straight-line path and *r*_{i}Â 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 *= *t*_{0Â }=Â *r*_{0}/*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 time*Â 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 *j*Â is given by:

Î´*SÂ *= pÂ·*r*_{j}Â âˆ’Â pÂ·*r*_{i}Â = pÂ·(*r*_{j}Â âˆ’ *r*_{i}) = 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 *S*Â 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 *m*Â·*c*^{2}Â = 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”*S*Â = 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”*S*Â = 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”*S*Â = 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 *arrows*Â 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 *S*_{0}Â is the action for *r*_{0}, then *S*_{1}Â = *S*_{0}Â + *Ä§Â *and *S*_{2}Â = *S*_{0}Â + 2Â·*Ä§ *are still good, but *S*_{3}Â = *S*_{0}Â + 3Â·*Ä§*Â isÂ *notÂ *good. Why? Because the difference in the phase angles is Î”Î¸Â =Â *S*_{1}/*Ä§*Â âˆ’Â *S*_{0}/*Ä§*Â = (*S*_{0}Â + *Ä§*)/*Ä§*Â âˆ’Â *S*_{0}/*Ä§* = 1 andÂ Î”Î¸ =Â *S*_{2}/*Ä§*Â âˆ’Â *S*_{0}/*Ä§*Â = (*S*_{0}Â + 2Â·*Ä§*)/*Ä§*Â âˆ’Â *S*_{0}/*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 *S*_{3}/*Ä§*Â âˆ’Â *S*_{0}/*Ä§*Â = (*S*_{0}Â + 3Â·*Ä§*)/*Ä§*Â âˆ’Â *S*_{0}/*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 *S _{n}*Â =

*S*

_{0}Â +

*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 (

*) 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.*

**p**We will also assume that the phase angle forÂ *S*_{0}Â 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”*S*_{n}Â =Â *S _{n}*Â âˆ’

*S*

_{0}Â =

*n*, and the associated phase angle Î¸

_{n}Â = Î”Î¸

_{n}Â is the same. In short, the amplitude for each path reduces to Ïˆ

_{n}Â =

*e*

^{iÂ·n}/

*r*

_{0}. 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

*r*Â·

*e*

^{iÂ·Î¸}Â =

*r*Â·(

*cos*Î¸ +

*i*Â·

*sin*Î¸) =

*r*Â·

*cos*Î¸ +

*i*Â·

*r*Â·

*sin*Î¸ formula. Needless to say, |

*r*Â·

*e*

^{iÂ·Î¸}|

^{2}Â = |

*r|*

^{2}Â·

*|*

*e*

^{iÂ·Î¸}|

^{2}

*Â =*|

*r*|

^{2}Â·(

*cos*

^{2}Î¸ +

*sin*

^{2}Î¸) =

*r*. Finally, when adding complex numbers, we add the real and imaginary parts respectively, and we’ll denote the Ïˆ

_{0}Â + Ïˆ

_{1}Â +Ïˆ

_{2}Â + … sum as Î¨.

Now, we also need to see how our Î”*S*Â = Î”pÂ·Î”*r*Â 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”*S*_{n}Â = Î”p* _{n}*Â·Î”

*r*Â =Â

_{n}*n*Â·Î”p

_{1}Â·Î”

*r*

_{1}. It also makes sense that you may want Î”p

*Â and Î”*

_{n}*r*Â to be proportional to Î”p

_{n}_{1}Â and Î”

*r*

_{1}Â respectively. Combining both, the assumption would be this:

Î”p* _{n}*Â =Â âˆš

*n*Â·Î”p

_{1Â }andÂ Î”

*r*Â =Â âˆš

_{n}*n*Â·Î”

*r*

_{1}

So now we just need to decide how we will distribute Î”*S*_{1}Â =Â *Ä§Â *= 1 over Î”p_{1}Â and Î”*r*_{1}Â respectively. For example, if we’d assume Î”p_{1}Â = 1, then Î”*r*_{1}Â = *Ä§*/Î”p_{1}Â = 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 wrong*Â *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 *increase*Â 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 (*r*_{0}), thenÂ *acÂ *could be one of the crooked paths (*r _{n}*). To simplify, we’ll assume

*isosceles*triangles, soÂ

*aÂ*equalsÂ

*c*Â and, hence,

*r*Â = 2Â·

_{n}*a*= 2Â·

*c*. We will also assume theÂ successive paths are separated by the same vertical distance (

*h =Â h*

_{1}) right in the middle, so

*h*Â =Â

_{b}*h*

_{n}Â =

*n*Â·

*h*

_{1}.Â It is then easy to show the following:This gives the following graph for

*r*Â = 10 and

_{n}*h*

_{1Â }= 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”*r*_{1}, the velocity *c*_{1}Â must be equal to (*r*_{0}Â + Î”*r*_{1})/*tÂ *= *r*_{0}/*tÂ *+ Î”*r*_{1}/*t =*Â *cÂ *+ Î”*r*_{1}/*tÂ *= *c*_{0}*Â *+ Î”*r*_{1}/*t*. We can write *c*_{1}Â as *c*_{1}Â =Â *c*_{0}*Â *+ Î”*c*_{1}, so Î”*c*_{1}Â = Î”*r*_{1}/*t**.*Â Now, theÂ *ratioÂ *of p_{1}Â and p_{0}Â will be equal to theÂ *ratioÂ *of *c*_{1}Â andÂ *c*_{0}Â because p_{1}/p_{0Â }= (m*c*_{1})/m*c*_{0}) = *c*_{1}/*c*_{0}. Hence, we have the following formula for p_{1}:

p_{1}Â = p_{0}Â·*c*_{1}/*c*_{0}Â = p_{0}Â·(*c*_{0}*Â *+ Î”*c*_{1})/*c*_{0}Â = p_{0}Â·[1 + Î”*r*_{1}/(*c*_{0}*Â·t*) = p_{0}Â·(1 + Î”*r*_{1}/*r*_{0})

ForÂ p* _{n}*, the logic is the same, so we write:

p* _{n}*Â = p

_{0}Â·

*c*/

_{n}*c*

_{0}Â = p

_{0}Â·(

*c*

_{0}

*Â*+ Î”

*c*)/

_{n}*c*

_{0}Â = p

_{0}Â·[1 + Î”

*r*/(

_{n}*c*

_{0}

*Â·t*) = p

_{0}Â·(1 + Î”

*r*/

_{n}*r*

_{0})

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 *very*Â 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 *h*_{1}/*r*_{0}Â 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 r_{1}Â = ,Â *r*_{2}, *r*_{2},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 *one*Â 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·*h*_{n}Â·*r*_{1}, 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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