This post should be the last in the series on this topic. It’s intended to wrap up a few loose ends. First, something on symmetries, perhaps. Why do we have *two* possible values for that phase factor that we need to add when we’re exchanging the role of two particles (or two detectors or whatever), *two* ways of calculating probabilities, but then *three* types of statistics only? Where’s the symmetry here? Let me recapitulate the situation.

**Degrees of freedom**

In the previous post, we noted that *e ^{i}*

^{ωθ},

*e*

^{–}

^{i}^{ωθ}, and any linear combination A

*e*

^{i}^{ωθ }+ B

*e*

^{–}

^{i}^{ωθ}(where A and B can be any complex number), are acceptable functional forms for the wave function

*f*(θ) representing the situation below: particle a goes into detector 1 and particle b goes into detector 2, with the angle of incidence equal to θ.

We also noted that the relevant wave function for the situation below, i.e. particle a going into detector 2 and particle b going into detector 1, was *e ^{i}*

^{δ}

*f*(π–θ), with δ = 0 (for Bose-Einstein interaction)

*or*δ = ± π (for Fermi-Dirac interaction). Finally, we noted that this phase factor might or might not be there when analyzing scattering of non-identical particles, but that it didn’t matter anyway, because there’s no interference (so we have classical (i.e. Maxwell-Boltzmann) statistics) and the phase factor just disappears when calculating the individual probabilities for the two possibilities (a) and (b).

*The table below summarizes the results for **f*(θ) = *e ^{i}*

^{ωθ }and for

*f*(θ) =

*e*

^{–i}

^{ωθ}. The results are symmetrical, i.e. the probability density function for the three situations (i.e. the green, red and blue function in the table below) is

*even*: P(θ) = P(–θ).

The math involved is simple and complicated at the same time. We know that *e ^{i}*

^{ωθ }= cos(ωθ)+

*i*sin(ωθ) and

*e*

^{–i}

^{ωθ }= cos(–ωθ)+

*i*sin(–ωθ) = cos(ωθ)–

*i*sin(ωθ) have the same real part (i.e. the cosine term) but an opposite imaginary part (i.e. the sine term). When combining them, it is also the imaginary part that makes the difference, as shown below: the real part of 2.5

*e*

^{i}^{ωθ }+ 1.5

*e*

^{–i}

^{ωθ }is equal to the real part of 1.5

*e*

^{i}^{ωθ }+ 2.5

*e*

^{–i}

^{ωθ }but the imaginary part of these two functions (i.e. 2.5

*e*

^{i}^{ωθ }+ 1.5

*e*

^{–i}

^{ωθ }and 1.5

*e*

^{i}^{ωθ }+ 2.5

*e*

^{–i}

^{ωθ }respectively), shown in green and red respectively) differs and, I might add, they differ

*not in amplitude but in phase*.

Hence, from a mathematical point of view, we have basically three things to play with, which is strange, because, in light of the symmetry of the two situations, we would expect four. Indeed, we have (i) a phase factor δ which may take one of two possible values (0 or ±π), and (ii) two formulas for calculating the probabilities (one that assumes no interference, and one that does). Combining these two should yield four different outcomes. Yet, we have only three. The reason is obvious, of course: when there’s no interference, the phase factor doesn’t matter, so we have the following overview indeed (with three types of statistics only):

**Combining statistics**

The next loose end concerns combinations of two or more types of statistics. Electrons are, in principle, fermions and, hence, Fermi-Dirac statistics should apply. However, they have an additional property: spin, which comes in two varieties (up or down). If the spin of the ‘bombarding’ electron (particle a) is the same as the spin of the ‘target’ (particle b), then the particles can’t be distinguished and, hence, are identical indeed (i.e. figure (a) and (b) on the left-hand side). However, when the spins are antiparallel (up versus down), then the electrons *are *distinguishable, in principle (and in practice) and, hence, we should treat them as non-identical particles. Hence, it’s the probabilities that add up in that case, not the amplitudes.

If our source produces electrons with random spin, i.e. if it’s a so-called *unpolarized *beam, then the odds are fifty-fifty that the spin will be up *or *down. Combined with the other beam, that yields four possible cases, with a like probability of occurring (25%), as shown in the table below.

So how does that look like? Well… While the formula for the total probability may look daunting, the graph is simple and easy to understand. Let us, once again, assume that *f*(θ) = *e ^{i}*

^{ωθ}. Based on the formulas in the other table above, we see that the second and third term are two constants, and it’s only the first term that models the interference. Hence, we have a graph on top of a baseline, as shown below. The blue graph assumes ω = 1, so we have two maxima (and minima) only, while the green graph is based on the assumption that ω = 4 (so we have 8 maxima and minima).

Combining the same simple laws for adding probabilities and/or amplitudes can explain any situation. A slightly more complex problem is illustrated below. The process involved is the scattering of *neutrons* from a crystal, so the graph shows the neutron intensity as a function of the angle. The process allows for an *exchange of spin *between a nucleus in the crystal and the neutron. For more detail, however, I’ll refer you to the Grandmaster himself (Feynman, *Lectures*, Vol. III-3-8).

**The wave equation revisited**

Finally, one last loose end concerns the math behind the wave *equation*, i.e. the second-order differential equation that defines the functional form of our wave function *f*(θ). We gave that equation (the *Schrödinger *equation) already:

Now, because we’re not going into the detail of the underlying spatial and time variables as everything is merged into one variable only (θ) in this analysis, the shape of ‘our’ Schrödinger equation is something like this:

d^{2}*f*(θ)/dθ^{2} = –ω^{2}*f*(θ)

Now, *that *equation basically says that any function whose *second derivative *is proportional to the function itself (with the factor of proportionality equal to –ω^{2}) will do. We know the general solution for this equation. It’s *f*(θ) = A*e*^{iωθ} + B*e*^{–iωθ}, with A and B representing complex numbers too, although we should immediately say that we will get *real-valued *solutions if our initial conditions are expressed in real numbers. I won’t dwell on that. The only thing I would like to point out is the not-so-subtle difference between a real exponential (e.g. *e*^{θ}) and a complex exponential (e.g. *e*^{iθ}).

Indeed, if, instead of some *negative *number –ω^{2}, we would have some positive factor λ^{2 }in that differential equation above (so we’d write d^{2}*f*(θ)/dθ^{2} = λ^{2}*f*(θ) instead), then our general solution would be *f*(θ) = A*e*^{λθ} + B*e*^{–λθ}, and that solution would be *real-valued*, **always** (so A and B are also *real *numbers here, as opposed to the A and B in the *f*(θ) = A*e*^{iωθ} + B*e*^{–iωθ} expression). How do these *real-valued *general solution for d^{2}*f*(θ)/dθ^{2} = λ^{2}*f*(θ) look like? The green graph below shows *e*^{θ} + *e*^{–θ} for θ ranging between –π and +π. Equating λ with 2 or 4, instead of 1 (as we’re doing below), would not change the general *shape *of the graph: it just makes it ‘narrower’ (you can experiment yourself with some web-based plotter). **The point to note is that we do not have a periodic function here.**

So what’s going on? Well… Nothing much. It’s the magic of complex numbers indeed. It’s easy to see that any *f*(θ) = *e*^{λθ }function solves d^{2}*f*(θ)/dθ^{2} = λ^{2}*f*(θ), because *f*”(θ) = d^{2}*e*^{λθ}/dθ^{2} = λ^{2}*e*^{λθ }= λ^{2}*f*(θ) indeed. Likewise, *f*(θ) = *e*^{−λθ }is a solution too, because *f*”(θ) also equals d^{2}*e*^{–λθ}/dθ^{2} = (–λ)(–λ)*e*^{–λθ }= λ^{2}*f*(θ). But so we get what we get: real exponential functions, which tend to infinity.

Now, suddenly we jot down d^{2}*f*(θ)/dθ^{2} = –λ^{2}*f*(θ) and we allow for *complex *roots, because negative numbers do not have any real root. More specifically, we write (±*i*λ)^{2} = (*i*λ)(*i*λ) = (−*i*λ)(−*i*λ) = –λ^{2 }and so we’re done: our general solution is now *f*(θ) = A*e*^{iλθ} + B*e*^{–iλθ}. The only difference (which, indeed, is not-so-subtle) is that little *i* in the exponentials (and, most likely, some *i*‘s in the coefficients A and B as well): the imaginary unit, which gives our numbers a *direction* also (besides their magnitude, which real numbers also have). And so ** that radically alters the nature of our solution**: unlike a real exponential, a

*complex*exponential

*is*a

*periodic*function. It goes round and round and round, and please do note that that remark applies to the real-valued solutions for the differential equation as well!

What to say? Complex numbers are magical indeed. I’ve dwelled on them a couple of times already (see, for instance, my June 2014 post on them), and so I won’t do it *again*. We can *explain* them as many times as we want, sort of, but, as far as I am concerned, they never *really* lose their magic. :-)

**Post scriptum**: The table above, which was based on the simple assumption that *f*(θ) was equal to *f*(θ) = *e ^{i}*

^{ωθ}, suggests that the Bose-Einstein probability function and the Fermi-Dirac probability function are essentially the same, except for a phase difference of π/2. Is that a general conclusion? Frankly, I thought so… Initially, that is. However, it depends very much on the functional form that we take for

*f*(θ). If we take a more complicated functional form, such as

*f*(θ) = 2.5

*e*

^{i}^{ω}

^{θ }+ 1.5

*e*

^{–iω}

^{θ}, for example, we get in trouble. In that case, we’d write

*f*(θ) as:

f(θ) = 2.5cos(ωθ) + *i*2.5sin(ωθ) + 1.5cos(ωθ) – *i*1.5sin(ωθ) = 4cos(ωθ) + *i*sin(ωθ)

Likewise, we’d write *f*(π–θ) as:

*f*(π–θ) = 2.5cos[ω(π–θ)] + *i*2.5sin[ω(π–θ)] + 1.5cos[ω(π–θ)] – *i*1.5sin[ω(π–θ)]

= 4cos[ω(π–θ)] + *i*sin(π–ωθ) = –4cos[ω(π–θ)] + *i*sin[ω(π–θ)].

Hence, the probability |*f*(θ)|^{2 }is 4cos^{2}[ω(π–θ)] **+ **sin^{2}[ω(π–θ)] = 2 + 0.5 = 2.5. The probability |*f*(π–θ)|^{2 }will be the same, and so the total probability is 5.

Now, *f*(θ) + *f*(π–θ) will be equal to 2*i*sin[ω(π–θ)] (the cosines fall away), while *f*(θ) – *f*(π–θ) = 8cos[ω(π–θ)] (the sines fall away). Hence, we will be getting nonsensical values for the *total *probability here. Hence, I’ve probably done something stupid.

In fact, now that I look at my table again, I note I made the mistake of writing *f*(π–θ) as *f*(π–θ) = cos(π–ωθ) + *i*sin(π–ωθ), while I should have written *f*(π–θ) = cos[ω(π–θ)] + *i*sin[ω(π–θ)] = cos(ωπ–ωθ) + *i*sin(ωπ–ωθ). So there were substitutions there that were *not *entirely *kosher*. However, because I assumed ω is some natural number (cf. the need for modes, i.e. a number of oscillations that ‘fits’ the space) and, hence, we’re talking whole *multiples *of π here, I would think that most of the *fundamental *results still stand, notably those in regard to the general shape of these functions (and also those in regard to the phenomenon of Bose-Einstein and Fermi-Dirac statistics approaching Maxwell-Boltzmann statistics when ω is large) still stand. Strictly speaking, however, the formulas in the table would be valid only for ω = 1, 3, 5 etcetera. I am pretty sure the solution for ω = 2, 4, 6 etcetera is similar: either the real or the imaginary parts of the complex numbers involved will cancel each other out, I would think. However, I’ll leave it to my son to correct my mistakes, a few years from now. :-) And if he doesn’t, *you* are, of course, also free to have a go at it. :-) Please do let me know what *you* find.