**Pre-script** (dated 26 June 2020): Our ideas have evolved into a full-blown realistic (or *classical*) interpretation of all things quantum-mechanical. In addition, I note the dark force has amused himself by removing some material. So no use to read this. Read my recent papers instead. ðŸ™‚

**Original post**:

We’ve come aÂ *veryÂ *long way. Now we’re ready for the Big Stuff. We’ll look at the rules forÂ *transformingÂ *amplitudes from one ‘base’ to ‘another’. [In quantum mechanics, however, we’ll talk about a ‘representation’, rather than a ‘base’, as we’ll reserve the latter term for a ‘base’Â *state*.] In addition, we’ll look at how physicists model how amplitudes evolve over time using the so-calledÂ *HamiltonianÂ *matrix. So let’s go for it.

**Transformations:Â how should we think about them?**

In my previous post, I presented the following hypothetical set-up: we have an S-filter and a T-filter in series, but the T-filter at the angle Î± with respect to the first. In case you forgot: these ‘filters’ are modified Stern-Gerlach apparatuses, designed to split a particle beam according to the angular momentum in the direction of the gradient of the magnetic field, in which we may place masks to filter out one or more states.

The idea is illustrated in the hypothetical *exampleÂ *below. The unpolarized beam goes through S, but we have masks blocking all particles with zero or negative spin in the z-direction, i.e. *with respect to S*. Hence, all particles entering the T-filter are in the +S state. Now, we assume the set-up of the T-filter is such that it filters out all particles with positive or negative spin. Hence, only particles with zero spin goÂ through. So we’ve got something like this:

However, we need to be careful as what we are saying here. The T-apparatus is tilted, so the gradient of the magnetic field is different. To be precise, it’s got the same tilt as the T-filter itself (Î±). Hence, it will be filtering out all particles with positive or negative spinÂ *with respect to T*.Â So, unlike what you might think at first, some fraction of the particles in the +S state will get through the T-filter, and come out in the 0T state. In fact, we know how many, because we have formulas for situations like this. To be precise, in this case, we should apply the following formula:

âŒ© 0T | +S âŒª = Â âˆ’(1/âˆš2)Â·sinÎ±

This is a real-valued amplitude. As usual, we get the following probability by taking the absolute square, so P = |âˆ’(1/âˆš2)Â·sinÎ±|^{2Â }=Â (1/2)Â·sin^{2}Î±, which gives us the following graph of P:

The probability varies between 0 Â (for Î± = 0 or Ï€) and 1/2 = 0.5 (for Î± = Ï€/2 or 3Ï€/2). Now, this graph may or may not make sense to you, so you should think about it. You’ll admit it makes sense to find P = 0 for Î± = 0, but what about the non-zero values?

Think about what this would mean in classical terms: we’ve got a beam of particles whose angular momentum is ‘up’ in the *z*-direction. To be precise, this means that J_{z}Â =Â +Ä§. [Angular momentum and the quantum of action have the same dimension: the jouleÂ·second.] So that’s the *maximumÂ *value out of the three permitted values, which are +Ä§, 0 and â€“Ä§. Note that the particles here must beÂ *bosons*. So you may think we’re talking *photons*, in *practice* but… Well… No. As I’ll explain in a later post, the photon is a spin-one particle but it’s quite particular, because it has noÂ â€˜zero spinâ€™-state. Don’t worry about it here – but it’s really quite remarkable. So, instead of thinking of a photon, you should think ofÂ some composite matter particle obeying Bose-Einstein statistics. These are not so rare as you may think: all matter-particles that contain an *even* number of fermions –Â like elementary particles – haveÂ integer spin – but… Well… Their spin number is usually *zero *– not one. So… Well… Feynman’s particle here is somewhat theoretical – but it doesn’t matter. Let’s move on.Â ðŸ™‚

Let’s look at another transformation formula. More in particular, let’s look at the formula we (should)Â get for âŒ© 0T | âˆ’S âŒª as a function of Î±. So we change theÂ set-up of the S-filter to ensureÂ all particles entering T have negative spin. The formula is:

âŒ© 0T | âˆ’S âŒª = Â +(1/âˆš2)Â·sinÎ±

That gives the same probabilities: |+(1/âˆš2)Â·sinÎ±|^{2Â }=Â (1/2)Â·sin^{2}Î±. Adding |âŒ© 0T | +S âŒª|^{2}Â and |âŒ© 0T | âˆ’S âŒª|^{2Â }gives us aÂ *totalÂ *probability equal toÂ sin^{2}Î±, which is equal to 1 if Î± = Ï€/2 or 3Ï€/2. We may be tempted toÂ interpret this as follows: if a particle is in the +S *or* âˆ’S state before entering the T-apparatus, and the T-apparatus is tilted at an angle Î± = Ï€/2 or 3Ï€/2 with respect to the S-apparatus, then this particle will come out of the T-apparatus in the 0T-state. No ambiguity here: P = 1.

Is this strange? Well… Let’s think about what it means to tilt the T-apparatus. You’ll have to admit that, if the apparatus is tilted at the angle Ï€/2 or 3Ï€/2, it’s going to measure the angular momentum in the *x*-direction. [The *y*-direction is the common axis of both apparatuses here.] So… Well… It’s pretty plausible, isn’t it? If all of the angular momentum is in the positive or negative *z*-direction, then it’s *not *going to*Â *have any angular momentum in the *x*-direction, right? AndÂ not having any angular momentum in the *x*-direction effectively corresponds to being in the 0T-state, right?

**Oh ! Is itÂ thatÂ easy?**

Well… **No! Not at all!Â **TheÂ reasoningÂ above shows how easy it is to be led astray.Â We forgot to normalize. Remember, if we integrate the probability density function over its domain, i.e. Î± âˆˆ [0, 2Ï€], then we *have toÂ *get one, as all probabilities have to add up to one. The definite integral ofÂ (1/2)Â·sin^{2}Î± over [0, 2Ï€] is equal to Ï€/2 (the definite integral of the sine or cosine squared over a full cycle is equal toÂ Ï€), so we need to multiply this function by 2/Ï€ to get the actual probability density function, i.e. (1/Ï€)Â·sin^{2}Î±. It’s got the same shape, obviously, but it gives us maximum probabilities equal toÂ 1/Ï€ â‰ˆ 0.32 for Î± = Ï€/2 or 3Ï€/2, instead of 1/2 = 0.5.

Likewise, theÂ sin^{2}Î± function we got when adding |âŒ© 0T | +S âŒª|^{2}Â and |âŒ© 0T | âˆ’S âŒª|^{2Â }should also beÂ normalized. One really needs to keep one’s wits about oneself here. What we’re saying here is that we have a particle that is *eitherÂ *in the +S *or* theÂ âˆ’S state, so let’s say that the chance is 50/50 to be inÂ *eitherÂ *of the two states. We then have these probabilities |âŒ© 0T | +S âŒª|^{2}Â andÂ |âŒ© 0T | âˆ’S âŒª|^{2}, which we calculated as (1/Ï€)Â·sin^{2}Î±. So the totalÂ *combinedÂ *probability is equal toÂ 0.5Â·(1/Ï€)Â·sin^{2}Î± +Â 0.5Â·(1/Ï€)Â·sin^{2}Î± =Â (1/Ï€)Â·sin^{2}Î±.Â So we’re now *weighing *the two (1/Ï€)Â·sin^{2}Î± functions – and it doesn’t matter ifÂ the weights are 50/50 or 75/25 or whatever, as long asÂ the two weights add up to one. The bottom line is:Â we get the same (1/Ï€)Â·sin^{2}Î±Â function for P, and the same *maximum* probability 1/Ï€ â‰ˆ 0.32 for Î± = Ï€/2 or 3Ï€/2.

**So we don’t get unity: P â‰ 1 for Î± = Ï€/2 or 3Ï€/2.** Why not? Think about it. The classical analysis made sense, didn’t it? If the angular momentum is *all* inÂ the *z*-direction (or in *one *of the two *z*-directions, I should say), then we can*not*Â have *any *of it in the *x*-direction, can it? Well… The surprising answer is: yes, we can.Â TheÂ remarkable thingÂ is that, in quantum physics, we actually *never* have *all*Â of the angular momentum in one direction. As I explained in my post on spin and angular momentum, theÂ classical concepts of angular momentum, and the related magnetic moment, have their limits in quantum mechanics. In quantum physics, we find that the *magnitudeÂ *of a vector quantity, like angular momentum, or the related magnetic moment, is generallyÂ *not *equal to*Â *the maximum value of the component of that quantity *in any direction*. The general rule is that theÂ *maximum value* of any *component* of **J**Â in whatever direction â€“ i.e. +Ä§ in the example we’re discussing here â€“Â is *smaller* than theÂ *magnitudeÂ *of **J**Â â€“ which I calculated in the mentioned post as |**J**| = J = +âˆš2Â·Ä§ â‰ˆ 1.414Â·Ä§, so that’s almost 1.5 times Ä§! So it’s *quite a bit *smaller!Â The upshot is that we can*not* associate any *preciseÂ *and unambiguousÂ direction with quantities like the angular momentumÂ **J**Â or the magnetic momentÂ **Î¼**. So the answer is:Â the angular momentum canÂ *neverÂ *beÂ *all* in the *z*-direction, so we can *alwaysÂ *haveÂ *some of itÂ *in theÂ *x*-direction, and so that explains the amplitudes and probabilities we’re having here.

*Huh?*

Yep. I know. We never seem to get out of this ‘weirdness’, but then that’s how quantum physics is like. Feynman warned us upfront:

“Because atomic behavior is so unlike ordinary experience, it is very difficult to get used to, and it appears peculiar and mysterious to everyoneâ€”both to the novice and to the experienced physicist. Even the experts do not understand it the way they would like to, and it is perfectly reasonable that they should not, because all of direct, human experience and of human intuition applies to large objects. We know how large objects will act, but things on a small scale just do not act that way. So we have to learn about them in a sort of abstract or imaginative fashion and not by connection with our direct experience.”

As I see it, quantum physics is about explaining all sorts of weird stuff, like electron interference and tunneling and what have you, so it shouldn’t surprise us that the framework is as weird as the stuff it’s trying to explain. ðŸ™‚ So… Well… All we can do is to try to go along with it, isn’t it? And so that’s what we’ll do here. ðŸ™‚

**Transformations: the formulas**

We need to distinguish various cases here. The first case is the case explained above: the T-apparatus shares the same *y*-axisÂ â€“ along which the particles moveÂ â€“ but it’s tilted. To be precise, we should say that it’s *rotatedÂ *about the common y-axis by the angleÂ Î±. That implies we can relate the x’, y’, z’ coordinate system of T to the x, y, z coordinate system of S through the following equations:Â zâ€²Â =Â zÂ·cosÎ±Â +Â xÂ·sinÎ±, xâ€²Â =Â xÂ·cosÎ±Â âˆ’Â zÂ·sinÎ±, andÂ yâ€²Â =Â y. Then the transformation amplitudes are:

We used the formula forÂ âŒ© 0T | +S âŒªÂ and âŒ© 0T | âˆ’S âŒª above, and you can play with the formulas above by imagining the related set-up of the S and T filters, such as the one below:

If you do your homework (just check what formula and what set-upÂ this corresponds to), you should find the following graph for the amplitude and the probability as a function ofÂ Î±: the graph is zero for Î± = Ï€, but is non-zero everywhere else. As with the other example, you should think about this. It makes senseâ€”sort of, that is. ðŸ™‚

OK. Next case. Now we’re going to rotate the T-apparatus around the *z*-axis by some angleÂ Î². To illustrate what we’re doing here, we need to take a ‘top view’ of our apparatus, as shown below, which shows a rotation over 90Â°. More in general, for any angleÂ Î², the coordinate transformation is given byÂ zâ€²Â =Â z, xâ€²Â =Â xÂ·cosÎ²Â +Â yÂ·sinÎ², yâ€²Â =Â yÂ·cosÎ²Â âˆ’Â xÂ·sinÎ². [So it’s quite similar to case 1: we’re only rotating the thing in a different plane.]

The transformation amplitudes are now given by:

As you can see, we getÂ *complex-valuedÂ *transformation amplitudes, unlike our first case, which yieldedÂ *real-valuedÂ *transformation amplitudes. That’s just the way it is. Nobody says transformation amplitudes have to be real-valued. On the contrary, one wouldÂ *expectÂ *them to be complex numbers. ðŸ™‚ Having said that, the combined *set *of transformation formulas is, obviously, rather remarkable. The amplitude to go from theÂ +S stateÂ to, say, theÂ 0T state is zero. Also, when our particleÂ has zero spin when coming out of S, it will always haveÂ zero spinÂ when and if it goes through T. In fact, the *absolute value *of those *e*^{Â±iÎ²} functions is also equal to one, so theyÂ are also associated with probabilities that are equal to one: |*e*^{Â±iÎ²}|^{2} = 1^{2} = Â 1. So… Well… Those formulas are simple and weird at the same time, aren’t they? They sure give usÂ plenty of stuff to think about, I’d say.

So what’s next? Well… Not all that much. We’re sort of done, really. Indeed,Â it’s just a property of space that we can getÂ *anyÂ *rotation of T by *combining* the two rotations above. As I only want to introduce the basic concepts here, I’ll refer you to Feynman for the details ofÂ *howÂ *exactly that’s being done. [He illustrates it for spin-1/2 particles in particular.] I’ll just wrap up here by generalizing our results fromÂ *baseÂ *states toÂ *anyÂ *state.

**Transformations: generalization**

We mentioned a couple of times already that the base states are like a particular coordinate system: we will usually describe a state in terms ofÂ *baseÂ *states indeed. More in particular, choosing S as our *representation*,Â we’ll say:

The stateÂ Ï† is defined by the three numbers:

C_{+}Â =Â âŒ© +S | Ï† âŒª,

C_{0}Â =Â âŒ© 0S | Ï† âŒª,

C_{âˆ’}Â =Â âŒ© âˆ’S | Ï† âŒª.

Now, the veryÂ *sameÂ *state can, of course, also be described in the ‘T system’, so then our numbers â€“ i.e. the ‘components’ of Ï† â€“Â would be equal to:

C’_{+}Â =Â âŒ© +T | Ï† âŒª, C’_{0}Â =Â âŒ© 0T | Ï† âŒª, and C’_{âˆ’}Â =Â âŒ© âˆ’T | Ï† âŒª.

So how can we go from the unprimed ‘coordinates’ to the primed ones? The trick is to use the second of the three quantum math ‘Laws’ which I introduced in my previous post:

Just replaceÂ Ï‡ in [II] by +T, 0T and/orÂ â€“T. More in general, if we denoteÂ +T, 0T orÂ â€“T byÂ *j*T, we can re-write this ‘Law’ as:

So the âŒ© *j*T |* i*S âŒª amplitudes are those nine transformation amplitudes. Now, we can represent those nine amplitudes in a nice three-by-three matrix and, yes, we’ll call that matrix the *transformation matrix*. So now you know what that is.

To conclude, I should note that it’s only because we’re talking spin-one particles here that we have three base states here and, hence, three ‘components’, which we denoted byÂ C_{+}, C_{âˆ’}Â and C_{0}, which transform the way they do when going from one representation to another, and so that is very much like what vectors do when we move to a different coordinate system, which is why *spin-one particles are often referred to as ‘ vector particles‘*. [I am just mentioning this in case you’d come across the term and wonder why they’re being called that way. Now you know.] In fact, if we have three base states, in respect to whatever representation, and we define some state Ï† in terms of them, then we can always re-define that state in terms of the following ‘special’ set of components:

The set is ‘special’ because one can show (you can do that yourself that by using those transformation laws) that these components transformÂ *exactlyÂ *the way asÂ x, y, z transform to xâ€², yâ€², zâ€². But so I’ll leave at this.

[…]

Oh… What about the Hamiltonian? Well… I’ll save that for my next posts, as my posts have become longer and longer, and so it’s probably a good idea to separate them out. ðŸ™‚

**Post scriptum: transformations for spin-1/2 particles**

You should actually really check out that chapter of Feynman. The transformation matrices for spin-1/2 particles look different because… Well… Because there’s only two base states for spin-1/2 particles. It’s a pretty technical chapter, but then spin-1/2 particles are the ones that make up the world. ðŸ™‚

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