As I was writing about those rotations in my previous post (on electron orbitals), I suddenly felt I should do some more thinking on (1) symmetries and (2) the concept of quantum-mechanical *magnitudes *of vectors. I’ll write about the first topic (symmetries) in some other post. Let’s first tackle the latter concept. Oh… And for those I frightened with my last post… Well… This should really be an *easy* read. More of a short philosophical reflection about quantum mechanics. Not a technical thing. Something intuitive. At least I hope it will come out that way. 🙂

First, you should note that the fundamental idea that quantities like energy, or momentum, may be quantized is a very natural one. In fact, it’s what the early Greek philosophers thought about Nature. Of course, while the idea of quantization comes naturally to us (I think it’s easier to understand than, say, the idea of infinity), it is, perhaps, *not *so easy to deal with it *mathematically*. Indeed, most mathematical ideas – like functions and derivatives – are based on what I’ll loosely refer to as *continuum theory*. So… Yes, quantization does yield some surprising results, like that formula for the magnitude of some vector ** J**:The

*·*

**J****in the**

*J**classical*formula above is, of course, the equally classical vector dot product, and the formula itself is nothing but Pythagoras’ Theorem in three dimensions. Easy. I just put a + sign in front of the square roots so as to remind you we actually always have

*two*square roots and that we should take the

*positive*one. 🙂

I will now show you how we get that quantum-mechanical formula. The logic behind it is fairly straightforward but, at the same time… Well… You’ll see. 🙂 We know that a quantum-mechanical variable – like the spin of an electron, or the angular momentum of an atom – is not continuous but *discrete*: it will have some value *m *= *j*, *j*-1, *j*-2, …, -(*j*-2), -(*j*-1), –*j. *Our *j *here is the *maximum *value of the magnitude of the ** component** of our vector (

**) in the direction of measurement, which – as you know – is usually written as**

*J**J*. Why? Because we will usually choose our coordinate system such that our

_{z}*z*-axis is aligned accordingly. 🙂 Those values

*j*,

*j*-1,

*j*-2, …, -(

*j*-2), -(

*j*-1), –

*j*are separated by one unit. That unit would be Planck’s quantum of action ħ ≈ 1.0545718×10

^{−34}N·m·s – by the way, isn’t it amazing we can actually measure such tiny stuff in some experiment? 🙂 – if

*would happen to be the angular momentum, but the approach here is more general –*

**J***action*can express itself in various ways 🙂 – so the unit doesn’t matter: it’s just the unit, so that’s just

*one*. 🙂 It’s easy to see that this separation implies

*j*must be some integer or half-integer. [Of course, now you might think the values of a series like 2.4, 1.4, 0.4, -0.6, -1.6 are also separated by one unit, but… Well… That would violate the most basic symmetry requirement so… Well… No. Our

*j*has to be an integer or a half-integer. Please also note that the number of possible values for

*m*is equal to 2

*j*+1, as we’ll use that in a moment.]

OK. You’re familiar with this by now and so I should not repeat the obvious. To make things somewhat more real, let’s assume *j *= 3/2, so *m *= 3/2, 1/2, -1/2 or +3/2. Now, we don’t know anything about the system and, therefore, these four values are all equally *likely*. Now, you may *not *agree with this assumption but… Well… You’ll have to agree that, at this point, you can’t come up with anything else that would make sense, right? It’s just like a classical situation: * J *might point in any direction, so we have to give all

*angles*an equal probability. [In fact, I’ll show you – in a minute or so – that you actually have a point here: we should think some more about this assumption – but so that’s for later. I am asking you to just go along with this story as for now.]

So the *expected *value of *J _{z}* is E[

*J*] is equal to E[

_{z}*J*] = (1/4)·(3/2)+(1/4)·(1/2)+(1/4)·(-1/2)+(1/4)·(-3/2) = 0. Nothing new here. We just multiply probabilities with all of the possible values to get an expected value. So we get zero here because our values are distributed symmetrically around the zero point. No surprise. Now, to calculate a

_{z}*magnitude*, we don’t need

*J*but

_{z }*J*

_{z}^{2}. In case you wonder, that’s what this squaring business is all about: we’re abstracting away from the

*direction*and so we’re going to

*square*both positive as well as negative values to then add it all up and take a square

*root*. Now, the

*expected*value of

*J*

_{z}^{2}is equal to E[

*J*] = (1/4)·(3/2)

_{z}^{2}+(1/4)·(1/2)

^{2}+(1/4)·(-1/2)

^{2}+(1/4)·(-3/2)

^{2}= 5/4 = 1.25. Some

*positive*value.

You may note that it’s a bit larger than the average of the *absolute *value of our variable, which is equal to (|3/2|+|1/2|+|-1/2|+|-3/2|)/4 = 1, but that’s just because the squaring favors larger values 🙂 Also note that, of course, we’d also get some positive value if *J _{z }*would be a

*continuous*variable over the [-3/2, +3/2] interval, but I’ll let

*you*think about

*what*positive value we’d get for

*J*

_{z}^{2 }assuming

*J*is uniform distributed over the [-3/2, +3/2] interval, because that calculation is actually

_{z}*not*so straightforward as it may seem at first. In any case, these considerations are not very relevant to our story here, so let’s move on.

Of course, our *z*-direction was random, and so we get the same thing for whatever direction. More in particular, we’ll also get it for the *x*– and *y*-directions: E[*J _{x}*] = E[

*J*] = E[

_{y}*J*] = 5/4. Now, at this point it’s probably good to give you a more generalized formula for these quantities. I think you’ll easily agree to the following one:So now we can apply our

_{z}*classical*

**·**

*J***=**

*J**J*+

_{x}^{2 }*J*+

_{y}^{2 }*J*formula to these quantities by calculating the expected value of

_{z}^{2 }*J*=

**·**

*J***, which is equal to:**

*J*E[** J**·

**] = E[**

*J**J*

_{x}^{2}] + E[

*J*

_{y}^{2}] + E[

*J*

_{z}^{2}] = 3·E[

*J*

_{x}^{2}] = 3·E[

*J*

_{y}^{2}] = 3·E[

*J*

_{z}^{2}]

You should note we’re making use of the E[*X*** **+

*Y*] = E[

*X*]+ E[

*Y*] property here: the expected value of the sum of two variables is equal to the sum of the expected values of the variables, and you should also note this is true even if the individual variables would happen to be correlated – which might or might not be the case. [

**What do you think**]

*is*the case here?For *j *= 3/2, it’s easy to see we get E[** J**·

**] = 3·E[**

*J**J*] = 3·5/4 = (3/2)·(3/2+1) =

_{x}*j*·(

*j*+1). We should now generalize this formula for other values of

*j*, which is

*not*so easy… Hmm… It obviously involves some formula for a series, and I am not good at that… So… Well… I just checked if it was true for

*j*= 1/2 and

*j*= 1 (please check

*that*at least for yourself too!) and then I just believe the authorities on this for all other values of

*j*. 🙂

Now, in a *classical *situation, we know that ** J**·

**product will be the same for whatever direction**

*J**would happen to have, and so its expected value will be equal to its*

**J***constant*value

**·**

*J***. So we can write: E[**

*J***·**

*J***] =**

*J*

**J**·**. So… Well… That’s why we write what we wrote above:**

*J*Makes sense, no? E[** J**·

**] = E[**

*J**J*

_{x}^{2}+

*J*

_{y}^{2}+

*J*

_{z}^{2}] = E[

*J*

_{x}^{2}]+E[

*J*

_{y}^{2}]+E[

*J*

_{z}^{2}] =

*j*·(

*j*+1) =

**J**·**=**

*J**J*

^{2}, so

*J*= +√[

*j*(

*j+*1)], right?

**Hold your horses, man! *** Think! What are we doing here, really? *We didn’t calculate all that much above. We only found that E[

*J*

_{x}^{2}]+E[

*J*

_{y}^{2}]+E[

*J*

_{z}^{2}] = E[

*J*

_{x}^{2}+

*J*

_{y}^{2}+

*J*

_{z}^{2}] =

*j*·(

*j*+1).

**So what?**Well… That’s

*not*a proof that the

*vector actually exists.*

**J****Huh?**

Yes. That * J *vector might just be some theoretical concept. When everything is said and done, all we’ve been doing – or at least, we

*imagined*we did – is those repeated measurements of

*J*,

_{x}*J*and

_{y}*J*here – or whatever subscript you’d want to use, like

_{z}*J*

_{θ,φ}, for example (the example is

*not*random, of course) – and so, of course, it’s only natural that we

*assume*these things are the magnitude of the

*component*(in the direction of measurement) of some

*real*vector that is out there, but then… Well… Who knows? Think of what we wrote about the angular momentum in our previous post on electron orbitals. We

*imagine*– or do like to think – that there’s some angular momentum vector

*out there, which we think of as being “cocked” at some angle, so its projection onto the*

**J***z*-axis gives us those discrete values for

*m*which, for

*j*= 2, for example, are equal to 0, 1 or 2 (and -1 and -2, of course) – like in the illustration below. 🙂But… Well… Note those weird angles: we get something close to 24.1° and then another value close to 54.7°. No symmetry here. 😦 The table below gives some more values for larger

*j*. They’re easy to calculate – it’s, once again, just Pythagoras’ Theorem – but… Well… No symmetries here. Just weird values. [I am

*not*saying the formula for these angles is

*not*straightforward. That formula is easy enough: θ =

*sin*

^{-1}(

*m*/√[

*j*(

*j*+1)]). It’s just… Well… No symmetry. You’ll see why that matters in a moment.]I skipped the half-integer values for

*j*in the table above so you might think they might make it easier to come up with some kind of sensible explanation for the angles. Well… No. They don’t. For example, for

*j*= 1/2 and m = ± 1/2, the angles are ±35.2644° – more or less, that is. 🙂 As you can see, these angles do

*not*nicely cut up our circle in equal pieces, which triggers the obvious question:

**are these angles really**Equal angles do

*equally*likely?*not*correspond to equal distances on the

*z-*axis (in case you don’t appreciate the point, look at the illustration below).

So… Well… Let me summarize the issue on hand as follows: **the idea of the angle of the J vector being randomly distributed is not compatible with the idea of those J_{z} values being equally spaced and equally likely.** The latter idea – equally spaced and equally likely

*J*values – relates to different possible

_{z}*states*of the system being equally likely, so… Well… It’s just a different idea. 😦

Now there is another thing which we should mention here. The maximum value of the *z*-component of our * J *vector is always

*smaller*than that quantum-mechanical magnitude, and quite significantly so for small

*j*, as shown in the table below. It is only for larger values of

*j*that the ratio of the two starts to converge to 1. For example, for

*j*= 25, it is about 1.02, so that’s only 2% off. That’s why physicists tell us that, in quantum mechanics, the angular momentum is never “completely along the

*z*-direction.” It is obvious that this actually challenges the

*idea*of a very precise direction in quantum mechanics, but then that shouldn’t surprise us, does it? After, isn’t this what the Uncertainty Principle is all about?

Different *states*, rather than different *directions*… And then Uncertainty because… Well… Because of discrete variables that won’t split in the middle. Hmm… 😦

Perhaps. Perhaps I should just accept all of this and go along with it… But… Well… I am *really* not satisfied here, despite Feynman’s assurance that that’s OK: “Understanding of these matters comes very slowly, if at all. Of course, one does get better able to know what is going to happen in a quantum-mechanical situation—if that is what understanding means—but one never gets a comfortable feeling that these quantum-mechanical rules are ‘natural’.”

I do want to get that comfortable feeling – on *some *sunny day, at least. 🙂 And so I’ll keep playing with this, until… Well… Until I give up. 🙂 In the meanwhile, if you’d feel you’ve got some better or some more *intuitive* explanation for all of this, please do let me know. I’d be ** very** grateful to you. 🙂

**Post scriptum**: Of course, we would all want to believe that * J somehow exists *because… Well… We want to

*explain*those states somehow, right? I, for one, am

*not*happy with being told to just accept things and shut up. So let me add some remarks here. First, you may think that the narrative above should distinguish between polar and axial vectors. You’ll remember polar vectors are the

*real*vectors, like a radius vector

**, or a force**

*r***, or velocity or (linear) momentum. Axial vectors (also known as pseudo-vectors) are vectors like the angular momentum vector: we sort of**

*F**construct*them from… Well… From

*real*vectors. The angular momentum

**L**, for example, is the vector

*cross*product of the radius vector

*and the linear momentum vector*

**r****: we write**

*p***L**=

**×**

*r***. In that sense, they’re a figment of our imagination. But then… What’s real and unreal? The**

*p**magnitude*of L, for example, does correspond to something real, doesn’t it? And its direction does give us the direction of circulation, right? You’re right. Hence, I think polar and axial vectors are both real – in whatever sense you’d want to define real. Their reality is just different, and that’s reflected in their mathematical behavior: if you change the

*direction*of the axes of your reference frame, polar vectors will change sign too, as opposed to axial vectors: they don’t swap sign. They do something else, which I’ll explain in my next post, where I’ll be talking symmetries.

But let us, for the sake of argument, assume whatever I wrote about those angles applies to *axial *vectors only. Let’s be even more specific, and say it applies to the angular momentum vector only. If that’s the case, we may want to think of a *classical *equivalent for the mentioned lack of a precise direction: free nutation. It’s a complicated thing – even more complicated than the phenomenon of *precession*, which we should be familiar with by now. Look at the illustration below (which I took from an article of a physics professor from Saint Petersburg), which shows both precession as well as nutation. Think of the movement of a spinning top when you release it: its axis will, at first, *nutate *around the axis of precession, before it settles in a more steady precession.The nutation is caused by the gravitational force field, and the nutation movement usually dies out quickly because of *dampening *forces (read: friction). Now, we don’t think of gravitational fields when analyzing angular momentum in quantum mechanics, and we shouldn’t. But there is something else we may want to think of. There is also a phenomenon which is referred to as *free nutation*, i.e. a nutation that is *not *caused by an external force field. The Earth, for example, nutates slowly because of a gravitational pull from the Sun and the other planets – so that’s *not *a free nutation – but, in addition to this, there’s an even smaller wobble – which *is *an example of free nutation – because the Earth is not *exactly* spherical. In fact, the Great Mathematician, Leonhard Euler, had already predicted this, back in 1765, but it took another 125 years or so before an astronomist, Seth Chandler, could finally experimentally confirm and measure it. So they named this wobble the Chandler wobble (Euler already has too many things named after him). 🙂

Now I don’t have much backup here – *none*, actually 🙂 – but why wouldn’t we imagine our electron would also sort of nutate freely because of… Well… Some symmetric asymmetry – something like the slightly elliptical shape of our Earth. 🙂 We may then effectively imagine the angular momentum vector as continually changing direction between a minimum and a maximum angle – something like what’s shown below, perhaps, between 0 and 40 degrees. Think of it as a rotation within a rotation, or an oscillation within an oscillation – or a standing wave within a standing wave. 🙂I am not sure if this approach would solve the problem of our angles and distances – the issue of whether we should think in equally likely *angles* or equally likely *distances *along the z-axis, really – but… Well… I’ll let you play with this. Please do send me some feedback if you think you’ve found something. 🙂

Whatever your solution is, it is likely to involve the equipartition theorem and harmonics, right? Perhaps we can, indeed, imagine standing waves within standing waves, and then standing waves within standing waves. How far can we go? 🙂

**Post scriptum 2**: When re-reading this post, I was thinking I should probably do something with the following idea. If we’ve got a sphere, and we’re thinking of some vector pointing to some point on the *surface *of that sphere, then we’re doing something which is referred to as point picking on the surface of a sphere, and the probability distributions – as a function of the polar and azimuthal angles θ and φ – are quite particular. See the article on the Wolfram site on this, for example. I am not sure if it’s going to lead to some easy explanation of the ‘angle problem’ we’ve laid out here but… Well… It’s surely an element in the explanation. The key idea here is shown in the illustration below: if the direction of our momentum in three-dimensional space is really random, there may still be more of a chance of an orientation towards the equator, rather than towards the pole. So… Well… We need to study the math of this. 🙂 But that’s for later.