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It is known that coordinates $C_k\in\mathbb{C}$ of the QM-state vectors $|\psi\rangle$ has an interpretation as probability weights $p_k$ in the whole state through the formula like $|C_k|^2=p_k$. We thus have the two (independent) math operation on the coefficients: making a modulus $C_k\to |C_k|$ and their squaring $|C_k|\to|C_k|^2$. Does anybody know a derivation of that transition, i.e., $C_k\to |C_k|^2=p_k$, from plausible/forceful physical arguments? The key words in the question are 'derivation' and 'forceful'. In other words, can we throw out the word 'interpretation' in the context and derive the rule $|C_k|^2$ is a probability weight? And reversely, the probability weight is exactly $|C_k|^2$ and only this formula?

I've knew right now about the 2005 paper by Aaronson but observed there the following. He begins there with a norm of the space, not with a scalar product of two independent vectors of $H$. The problem here, in my view, is that I prefere the object $(\psi,\phi)$ to be more fundamental (in the QM context) than $(\psi,\psi)$. This is because the derivation of the former requires some extra/exotic arguments on properties of the norm (polarization procedure). Consideration in the reverse direction does not. So, one could rephrase the question about 'interpretation/derivation' of $|(\psi,\phi)|^2$, not about $|(\psi,\psi)|^2$.

valerio
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maximav
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  • The word "interpretation" shoudn't be there at all. The principle is known as the "Born rule". And it is one operation - the squared absolute value is more natural "single operation" than the absolute value itself (which is the square root of the squared absolute value, and therefore needs two steps). One needs the Born rule because the operations must be expressed by nice groups and they must preserve the total probability,and the bilinear invariants of the unitary groups are the only mathematically rich solution. – Luboš Motl Jun 21 '16 at 09:00
  • It's neither derivation nor interpretation but observation. One can ask whether nature has "natural" alternatives to this and I would answer in the positive. The stochastic interpretation as independent probabilities only makes sense if we assume that QM systems are free of memory, otherwise there would be correlations between multiple repetitions of the same experiment (i.e. between members of the ensemble). Such memory is not known to exist on any scales we have observed, so far, but that is not an assurance that it doesn't exist on scales that we have not observed, yet. – CuriousOne Jun 21 '16 at 09:00
  • There are other constraints from observation. Is the sum of all probabilities really one? If we keep measuring a proton spin, for instance, it won't just produce ups and downs, but eventually, so we believe, the proton may disappear because of proton decay (but it should not disappear tracelessly). There is, however, no "fundamental" law that e.g. photons can't "disappear". I am not certain we could construct a high precision experiment to test that hypothesis and disappearance of e.g. .1% of low energy photons during the past 13 billion years may not make a dent into standard cosmology. – CuriousOne Jun 21 '16 at 09:28
  • to CuriousOne. Your comment on unitarity tells me that justification of probability appearance comes (should involve) arguments on $t$-evolution in QM. But it seems to me that we may construct QM (Hilbert space + linearity+operators+probability) with no appeal to quantum dynamics; just quantum statics, i.e. measurements. The time $t$ is, as for me, is just an external parameter to QM-dynamics.... Or? – maximav Jun 21 '16 at 09:31
  • We can construct whatever we want, that's not the problem. The problem is that we have no assurances that nature gives as much as a penny for our thoughts and constructions. I haven't looked into the experimental limits for unitarity, but I am pretty sure that they wouldn't be great if we zoomed in on free low energy field states. Let's also not forget that part of the problem with general relativity is that it is not even close to unitary in the strong field case - classical black holes take but don't give back. – CuriousOne Jun 21 '16 at 09:36
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    See the answers in Born rule and unitary evolution. – Stéphane Rollandin Jun 21 '16 at 11:02

1 Answers1

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There is no derivation of the probabilistic nature of the wavefunction: it is an interpretation (postulate), the only one that makes the theory consistent.

From Sakurai, Modern Quantum Mechanics:

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Schrödinger published his famous wave equation in February 1926 in the famous paper Quantisierung als Eigenwertproblem (Quantization as an Eigenvalue Problem), but at the beginning no one know how to interpret the function $\psi(\vec r,t)$ which appeared in it.

Quoting from Wikipedia:

At first, Schrödinger and others thought that wave functions represent particles that are spread out with most of the particle being where the wave function is large. This was shown to be incompatible with how elastic scattering of a wave packet representing a particle off a target appears; it spreads out in all directions. While a scattered particle may scatter in any direction, it does not break up and take off in all directions.

The same year, in June, Born published the paper Zur Quantenmechanik der Stroßvorgänge (On the quantum mechanics of collisions), in which he solved the Schrödinger equation for a scattering problem and concluded that the probabilistic interpretation of the solution was the only possible one.

The point is that if you want to build a consistent theory starting from the Schrödinger equation, then the object that appears in it must be interpreted in a way such that

$$\mid \psi(\vec r,t) \mid ^2 dV = \text{Probability that the particle is found in the volume}\ dV \ \text{after a measurement}$$

valerio
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