Using specific *functional forms *for wave functions to illustrate the difference between classical mechanics and quantum mechanics is enlightening, so let’s see if we can make the example(s) I presented in my previous post even more specific.

**The scattering process revisited**

The situation was simple: we were looking at an elastic scattering process involving two particles, a and b, going into detector 1 or detector 2 (see below). The analysis is simplified because we’re using a (theoretical) center-of-mass reference frame. Hence, we know that particle b will end up in detector 2 if (and only if) particle a goes into detector 1, and vice versa. Hence, there are *two possibilities only*, which are referred to as situation (a) and situation (b) respectively (using the same letters a and b is a bit confusing but just note the brackets), which, if the scattering process involves non-identical particles, correspond to *two probabilities*, which are to be added to find the *combined *probability of finding *some* particle (i.e. particle a *or* b) in *some* detector (i.e. detector 1 *or* 2).

Now, as you know, the key assumption in quantum mechanics (which we believe to be true for classical mechanics as well here) is that the probability of particle a going into detector 1 (and, hence, particle b going into detector 2) is, ultimately, determined by some complex-valued wave function *f* whose argument is the angle θ. Accordingly, we believe that the probability of particle a going into detector 2 (and, hence, particle b going into detector 1) is determined by the very same wave function, but then we have to use π–θ (instead θ) of as the argument. To be precise: we have to take the absolute square of these wave functions, and then we have to *add* them to find the *combined** *probability for finding *some* particle (i.e. particle a *or* b) in *some* detector (i.e. detector 1 *or* 2). So we write:

P = P_{a→}_{1 and b→}_{2} + P_{a→}_{2 and b→1} = P(θ) + P(π–θ) = |*f*(θ)|^{2 }+ |*f*(π–θ)|^{2}

The following remarks should be made here:

- While an elastic collision is a classical situation, the use of a wave function to model this situation integrates uncertainty, and that’s why we call it a scattering process, rather than a collision. Indeed, while it looks like we’re analyzing two billiard balls colliding here, we’re not: we can’t predict anything. We’re calculating
*probabilities*only. - In my previous post, I pointed out that we should actually not use the very same wave function for situation (b). We should incorporate an arbitrary phase factor
*e*^{iδ}and, hence, assume that the wave function describing situation (b) is*not**f*(π–θ), but*e*^{iδ}*f*(π–θ).

That’s a fine point, but a crucial one, I think. Indeed, while Feynman argues that situation (b) is basically “equivalent to situation (a) with counter 1 moved over to the angle π–θ” and, hence, that there’s no need for a phase factor when analyzing the above-modeled situation for non-identical particles, I argue that exchanging the role of the detectors is equivalent to exchanging the role of the particles. Hence, we should introduce the very same phase factor also when analyzing the scattering process for non-identical particles. While introducing this phase factor in the analysis of the above-modeled scattering process *for non-identical particles* does *not *change the probability (|*e*^{δ}*f*(π–θ)|^{2 }equals |*e*^{δ}|^{2}|*f*(π–θ)|^{2 }= |1|^{2}|*f*(π–θ)|^{2 }= |*f*(π–θ)|^{2}), it does change the *interpretation* of the event, because it introduces another degree of freedom, and here I use the term the way it’s used in statistics indeed: δ is an additional variable in the final calculation of a statistic (i.e. the probability P(π–θ) in this case).

That being said, choosing a specific functional form for *f*(θ) obviously also determines the functional form of *e*^{iδ}*f*(π–θ) and, because of the symmetry of the situation, we know that δ has to be equal to 0 or 180 degrees (in radians: δ = 0 or* *±π). Indeed, if we move over the counters once more, we’re back where we were, with counter 1 at θ and counter 2 at π–θ, and so we should then find our original *f*(θ) function once again. Hence, *e*^{iδ}*e*^{iδ }= [*e*^{iδ}]^{2 }must equal 1, which implies that *e*^{iδ} = ± 1 and, hence, δ = 0 or* *±π.

Now that we’re here, I should ask a second critical question: why should we apply the *same *phase factor *e*^{iδ}? Why can’t we have some phase factor *e*^{iα }for the first exchange of roles (with exchange of roles, we mean the switching of the position of the detectors) and some other (different) phase factor *e*^{iβ }for the second exchange of roles? The answer to that question has to be philosophical, I am afraid: Nature doesn’t care about our analysis and, hence, Nature cannot keep track of whether or not this is our first or second switch. Therefore, we invoke the symmetry of the situation once more to argue that the phase factor should be the same. [You'll say I've gone mad in my attempt of fully specify all of the assumptions underlying Feynman's analysis and... Well... You're probably right. :-)]

Now, what can we say about *f*(θ)? There surely must be some restrictions on its functional form? Of course. Let’s look at them.

**Well-behaved wave functions**

We might think that *f*(θ) should be a symmetric function, in the sense that we might think that *f*(θ) should equal *f*(–θ). Why? Well… No. First, I should note *f*(θ) is a complex-valued function, so what does symmetry mean in this case? If we’d take *f*(θ) = *e*^{iθ} = cosθ + *i*sinθ as a functional form, then its real part (Re[*f*(θ)] = cosθ) is symmetric (about the y-axis, that is), but its imaginary part (Im[*f*(θ)] = sinθ) isn’t. Second, I should note that a ‘symmetric function’ usually means something different in math: it’s a concept related to functions of two or more variables, and it means *f*(x_{1}, x_{2},… x_{n}) = *f*(x_{2}, x_{1},… x_{n}) = … = *f*(x_{n},…, x_{2 }, x_{1}). What we really mean to say here is that *f*(θ) should be a so-called *even function*. But so the example of *f*(θ) = *e*^{iθ}, which is a perfectly legitimate wave function (as we’ll show in a minute), should make it clear there’s no reason to impose a condition of evenness. Likewise, the concept of oddness (*f*(θ) = –*f*(θ)) doesn’t apply either.

The function should be periodic though: adding or subtracting 2nπ (n = 1, 2,…) should not make any difference, because we’re talking the same angle: θ ± 2nπ = θ. That remark leads us to *the *condition we have to impose, and that’s that the wave function f(θ) should be a solution to the wave equation which, as you know, is (from a mathematical point of view at least) a second-order linear differential equation describing… Well… Waves. What waves? All waves: sound waves, water waves, light waves and, in this case, *de Broglie *waves. These wave equations all have the same shape:

** Huh? **Relax. The first expression is the most general form. In this equation,

*u*is, quite simply, the displacement (or amplitude) of the wave, and c is the speed of the wave (i.e. the speed of sound, light, etcetera). As for the ∇

^{2}symbol, you’ve seen that before: it’s the

*Laplace*operator, which tells you to take the second (partial) derivative of u with respect to each

*spatial*variable, i.e. x, y and z in three dimensions. The second and third functional form of the wave equation show how that goes for one respectively three dimensions. Indeed, in the second formula,

*u*is substituted for χ (i.e. the Greek letter

*chi*), and

*c*

^{2}has been brought over to the other side, but it’s essentially the same equation. In the third formula,

*u*is substituted for

*P*, which is the air pressure, and the formula basically represents a sound wave in three dimensions, with the speed of the wave (

_{e}*c*

_{s}_{, }i.e. the speed of

*s*ound) being determined by the physical characteristics of the medium (air density, pressure, temperature, etcetera).

The fourth wave equation is the Schrödinger equation, and it’s the one that’s relevant for our discussion here, with *ϕ* represents the (complex) amplitude for finding a particle at position (x, y, z) at the time t. In this equation, *c *represents the speed of light and, hence, the left-hand side of this equation gives us x, y, z and t “in the nice combination which relativity usually involves”, as Feynman puts it.

The thing to note is that, because we’re not going into the detail of the underlying spatial and time variables as we merge everything into one variable only (θ), this Schrödinger equation becomes particularly simple for the situation that we’re analyzing here. Its shape will be something like:

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 λ^{2}) will do. Now, there is plenty of nice stuff around (see, for example, the second chapter of a Princeton University course on complex numbers) showing that the general solution for this equation will be

*f*(θ) = A*e*^{iωθ} + B*e*^{–iωθ}

It’s important to note that, in this general solution, A and B represent complex numbers too. However, one can show this general solution is *real-valued *if the initial conditions (i.e. the value of the function and its first derivative at θ = 0) are expressed as real numbers (see the above-mentioned reference). So, yes, in theory, we could have a real-valued wave function but, generally speaking, the real-valued solution is only one of an infinite number of complex-valued functions. Indeed, you can check for yourself that both *e*^{iωθ }and *e*^{–iωθ }are solutions (d^{2}*e*^{iωθ}/dθ^{2} = *i*ω*i*ω*e*^{iωθ} = (*i*ω)^{2}*e*^{iωθ} = –ω^{2}*e*^{iωθ} and d^{2}*e*^{–iωθ}/dθ^{2} = (–*i*ω)(–*i*ω)*e*^{–iωθ} = (*i*ω)^{2}*e*^{–iωθ} = –ω^{2}*e*^{–iωθ}) and, hence, any linear combination (cf. the principle of superposition) is a solution as well!

**The examples revisited**

This allows us to re-examine the examples we gave (we’ll again not take into account the normalization rule, but that doesn’t matter). You’ll remember that, for reasons of simplicity, I wanted to find some real-valued wave function first. Well… The general solution above yields such real-valued wave function if we equate A with B with one: A = B = 1. We then get:

*f*(θ) = *e*^{iωθ} + *e*^{–iωθ }= cosθ + *i*sinθ + cos(–θ) + *i*sin(–θ)

= cosθ + *i*sinθ + cosθ – *i*sinθ = 2cosθ

Now, *this *example is a lot more interesting than the one I gave in my previous post, because it does *not *yield a constant for the total probability P = P_{a→}_{1 and b→}_{2} + P_{a→}_{2 and b→1} = P(θ) + P(π–θ) = |*f*(θ)|^{2 }+ |*f*(π–θ)|^{2}. Indeed, if *f*(θ) = 2cosθ, then *f*(π–θ) = 2cos(π–θ) = –2cosθ. That yields the following graph of (1) *f*(θ) (blue), (2) *f*(π–θ) (red), and (3) |*f*(θ)|^{2 }+ |*f*(π–θ)|^{2 }(green). Here we have two *minima and maxima* for the total probability, instead of just one constant over the function’s domain [–π, π]. It basically says that the particles are most likely to either go in a straight line (θ ≈ 0) *or *bounce back ((θ ≈ ±π), rather than *not *favoring any angle.

Now, how does this compare to Bose-Einstein and Fermi-Dirac statistics, according to which the wave functions might interfere, either positively or negatively? Well… Let’s just go through the math. The result is shown below: the blue graph is the one we obtained above, from adding probabilities rather than amplitudes, while the red one gives us the Bose-Einstein interference pattern, and the green one the Fermi-Dirac interference. All graphs have the same maxima and minima, and the areas under the graphs do *not *add up to the same. Hence, this result is *not *plausible.

I must be doing *something* wrong here! But what can it be?

[...] Well… We forgot about the initial conditions, of course. The *f*(θ) = A*e*^{iωθ} + B*e*^{–iωθ }gives us a general solution but the actual solution will depend on the initial conditions, and it’s *only* when the initial conditions are expressed in real numbers that our solution will be real-valued. However, because we’re talking *de Broglie *waves here, we must assume we’re talking waves that have both a real as well as an imaginary part. Hence, equating *f*(θ) with *f*(θ) = *e*^{iθ }= cosθ + *i*sinθ is, most probably, a better bet. That yields the following graphs for Bose-Einstein and Fermi-Dirac probabilities:

Let me remind you how I got these. The imaginary part of f(θ) and f(π–θ) is the same, because sin(π–θ) = sin(θ). The real part of f(θ) and f(π–θ) is also the same except for a phase difference equal to π: cos(π–θ) = cos[–(θ–π)] = cos(θ–π). The absolute square of both f(θ) and f(π–θ) yields the same constant, and so their sum P = |f(θ)|^{2 }+|f(π–θ)|^{2 }= 2|f(θ)|^{2 }= 2|f(π–θ)|^{2 }= 2P(θ) = 2P(π–θ) was also some constant (not shown here). Now, for Bose-Einstein (left-hand side), the sum of the real parts of f(θ) and f(π–θ) yields zero (blue line), while the sum of their imaginary parts (i.e. the red graph) yields a sine-like function but it has double the amplitude of sin(θ). That’s logical: sin(θ) + sin(π–θ) = 2sin(θ). The green curve is the more interesting one, because that’s the total probability we’re looking for. It has *two *maxima: at +π/2 and at –π/2. That’s good.

Likewise, the Fermi-Dirac probability density function looks good as well (right-hand side). The imaginary parts of f(θ) and f(π–θ) ‘add’ to zero: sin(θ) – sin(π–θ) = 0 (I put ‘add’ between brackets because, with Fermi-Dirac, we’re subtracting of course), while the real parts ‘add’ up to a double cosine function: cos(θ) – cos(π–θ) = cos(θ) – [–cos(θ)] = 2cos(θ). We now get a minimum at +π/2 and at –π/2, which is also good, because it’s in line with the general result we’d expect. The (final) graph below summarizes our findings. It gives the three ‘types’ of probabilities, i.e. the probability of finding *some *particle in *some *detector as a function of the angle –π < θ < +π using:

- Maxwell-Boltzmann statistics: that’s the green constant (non-identical particles, and probability does not vary with the angle θ).
- Bose-Einstein: that’s the blue graph below. It has
*two*maxima, at +π/2 and at –π/2, and two minima, at 0 and at ±π (+π and –π are the same angle obviously), with the maxima equal to*twice*the value we get under Maxwell-Boltzmann statistics. - Finally, the red graph gives the Fermi-Dirac probabilities. Also two maxima and minima, but at different places: the maxima are at θ = 0 and θ = ±π, while the minima are at at +π/2 and at –π/2.

Of course, the above examples assume that ω = 1, which is, obviously, not the type of frequency you’d expect to see in an example. I showed you how it looks for, for example, ω = 25. In that case, *f*(θ) = cos(25·θ) + *i*sin(25·θ), and we get the following interference patterns:

The explosion of color hurts the eye, doesn’t it? :-) But, apart from that, do you now see why physicists say that, at high frequencies, the interference pattern gets smeared out? Indeed, if we move the detector just a little bit (i.e. we change the angle θ just a little bit) in the example below, we hit a maximum instead of a minimum, and vice versa. In short, the *granularity *may be such that we can only *measure* that green line, in which case we’d think we’re dealing with Maxwell-Boltzmann statistics, while the underlying reality may be different.

So… There we are. I’d say this analysis is somewhat more correct, from a theoretical point of view, than the intuitive approach which characterized my previous post. I hope you enjoyed it.

**Post scriptum**: While I noted that the (complex-valued) f(θ) function was not a symmetric function (or, to use somewhat more precise language, not an *even* function) because *f*(θ) ≠ f(–θ), the (real-valued) probability function itself actually is, as shown in the table below. The table below also shows that the Bose-Einstein and Fermi-Dirac probability functions are actually the same *except for a phase difference of π/2. *This phase difference matches the difference between the sine and cosine function, as cos(π/2 – θ) = cos(θ – π/2) = sinθ. Hence, the sine function also *follows *the cosine function with a phase delay equal to π/2. Likewise, the Fermi-Dirac probability function (P) follows the Bose-Einstein function also with a phase delay equal to π/2. * *