July 1, 2015

Demonstration of necessity of the Born Rule

The Born Rule in quantum mechanics states that probability of detection of a particle etc. is proportional to the square of the absolute value of the net wavefunction at that place and time. Despite inviting comparison to energy density being proportional to field amplitude squared, the BR is often presented as mysterious--as if it were a free parameter of nature rather than something that makes logical sense. I came up with a simple way to show that the known form of the BR is necessary, if we neglect complicated and unusual alternatives. We also reasonably assume simple additive superposition of amplitudes, basic linearity (e.g. of filter response), and that exponents must be positive or zero (to avoid the zero-amplitude crisis.)

My proof derives from the need to conserve the total number of particles transiting a Mach-Zehnder interferometer with asymmetrical beamsplitters. The total is normalized as unity. An ABS splits an incoming beam into unequal outputs. Hence a ≠ b, where a is transmitted amplitude and b is reflected amplitude. These may have different phases and thus complex values, but the proof can proceed because of the equal phases that combine for the fully constructive output from the relevant channel. This demonstration may not show universal applicability of squared moduli, but it does rule out alternatives.

We know from the BR that the corresponding intensities equal a² and b², and hence in the idealized case of no absorption used for modeling: a² +  b² = 1 . But did that have to be true, instead of say,  cubed amplitudes; such that a³ +  b³ = 1? If we simplify by considering one-term exponent laws, then consistency says "yes." (Further exploration is welcome, but the case implies any alternative would be contrived.) So, consider an MZI with asymmetrical BS at each end. The first, ABS₁, has transmitting amplitude a, and reflecting as b. Considering simple exponents (which don't have to be integers), we need aⁿ + bⁿ = 1. So far, we have no way to narrow that. These beams are recombined in ABS₂. This latter follows typical practice of outputting maximum constructive interference (no phase difference) in the lower, "A" Channel. However, it reverses transmission/reflection amplitudes relative to ABS₁. So: the originally transmitted beam is reflected at ABS₂ into Ch. A for a final output amplitude there of a². The originally reflected beam is transmitted at ABS₂ into Ch. A for a final output amplitude there of b². Superposition gives the total as a² +  b².

That already looks promising but we aren't done yet. First, we have to ensure that the output at Channel B must be zero. We can: since the Ch. A output is already the maximum output, pairing it with other than zero amplitude would be a contradiction. Suppose zero output was paired with less than the maximum possible amplitude. If so, then pairing the maximum with any value zero or over, would produce a larger total than before. But the totals must always be the same, so zero and maximum are paired. (It may seem obvious, but it's good to show the formal necessity.)

Now, we can proceed to satisfy the following equation:

(a² +  b²)ⁿ  =  1

a² +  b²  =  1⁻ⁿ  =  1

That is basically it. If the rule had been say, the amplitude itself or the cube; it could not be so that
 a³ +  b³  =  1 and a² +  b²  =  1, as well. Note: this whole argument only makes sense if we assume or accept, that there really is a number of particles output according to some rule, and not just two "branches" of arbitrary relative amplitudes. The whole idea of probability falls apart in the latter case, despite awkward attempts by MWI supporters to contrive an equivalence.

2 Comments:

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