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As far as I've read online, there isn't a good explanation for the Born Rule. Is this the case? Why does taking the square of the wave function give you the Probability? Naturally it removes negatives and imaginary numbers, but why is it the square, not the fourth or some higher power?

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    $\begingroup$ possible duplicate of Born rule and unitary evolution $\endgroup$ – Ali Aug 6 '13 at 6:00
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    $\begingroup$ This is discussed in Exercises 4 and 5 here as well as in this paper. Also see this answer and the approving comment to it. And finally, see this answer to a related question. $\endgroup$ – Keep these mind Aug 6 '13 at 7:32
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    $\begingroup$ It's the same as the relationship between amplitude and energy in the double-slit experiment. $\endgroup$ – Mike Dunlavey Aug 6 '13 at 21:18
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    $\begingroup$ You don't have to think of squaring it as being a special operation. For any observable in QM, the mean value is given by <Ψ|A|Ψ> , where A is the linear operator corresponding to the observable. In the case of position, the operator is R. All that R does is return the same function, so you end up with <Ψ|Ψ>. $\endgroup$ – Nick Aug 6 '13 at 22:29
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    $\begingroup$ Deutsch's "derivation" is nonsensical... There is a definition of rational behavior according to which the rational choice is that which maximizes your "expected utility". The utility of an outcome is the payoff you obtain from that outcome, and then the expected utility of a choice, is obtained by considering all the possible outcomes that might arise from a choice, and then giving them each a weight which is (utility of the outcome) x (probability of the outcome)... $\endgroup$ – Mitchell Porter Aug 7 '13 at 8:08
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Suppose you want to describe the quantum mechanical behaviour of a system, building from scratch the wave equation it should satisfy. Consider first the diffraction pattern obtained with a double slit by a monochromatic light beam and compare it to the one by a monoenergetic beam of electrons.

In optics, the total amplitude $\Phi$ for two coherent incident light beams on a plane is the sum of the individual amplitudes, $\Phi=\Phi_1+\Phi_2=A_1e^{i\theta_1}+A_1e^{i\theta_2}$ and the intensity $I$ of the beam will be proportional to $|\Phi|^2$, $$I\sim|\Phi|^2=A_1^2+A_2^2+2A_1A_2\cos(\theta_1-\theta_2)$$ which is real and positive, and in fact by the definition of $I$ it is proportional to the number of photons in each point of the screen.

The pattern with the beam of electrons is entirely similar, so you can define a complex amplitude $\psi$ with the properties

  1. It may satisfy a wave equation.
  2. The density of electrons $\rho(x)$ is proportional to $|\psi|^2=\psi^*\psi$ in each point.

This way you guarantee that the density of particles will be positive and that it may manifest interference by means of superposition of amplitudes. Now, let's denote $A$ the factor of proportionality in property 2, then the total number of particles $N$ is given by $$N=\int\rho\,dx=A\int|\psi|^2dx\;\Longrightarrow\;\int|\psi|^2dx=\frac{N}{A}$$ Now, the number $N$ in general is big, unknown and knowing it is irrelevant, also as may be seen, the wave equation is homogeneous, so that $\psi$ is determined up to an arbitrary constant. This way, it is accustomed to take $A=N$, i.e. to take $|\psi|$ as a normalized function, $$\int\psi^*\psi\,dx=1$$ So you have that $\rho=N|\psi|^2$, then you can define $$\tilde{\rho}\equiv\frac{\rho}{N}$$ as a relative density of particles, that tells you what fraction of the total of particles is contained in the element $dx$, from here then $$\int\tilde{\rho}\,dx=1$$ So here it is: suppose you do the experiment with just one electron. Then $\tilde{\rho}\,dx$ may be interpreted as the probability that the electron is contained in the element $dx$ and $\int\tilde{\rho}\,dx=1$ tells you that the particle is somewhere in space with all confidence.

This is why in general, $\tilde{\rho}=\psi^*\psi$ may be interpreted as a probability density for localization of particles that hence implies conservation of probability.

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  • $\begingroup$ Did Born reason this way? $\endgroup$ – Guerlando OCs Jul 7 '18 at 7:22
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For an electromagnetic wave, the Energy is proportional to the Electric/Magnetic Field (i.e. the wave) squared. This is a classical result which can be derived from the Maxwell equations.

When photons were discovered, the intensity of photons, or number of photons arriving at a certain place (for example, on a screen behind a double slit) was seen to be proportional to this squared field. However, it was now given a probabilistic interpretation: The intensity of the light is the probability of a photon impinging on that location.

The extension to "electron waves" is of course a wild guess that then proved to be true.

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  • $\begingroup$ The extension to "electron waves" is of course a wild guess that then proved to be true. I would argue that the extension to electrons follows from the fact that electrons couple to electromagnetic waves: physics.stackexchange.com/a/73388/4552 $\endgroup$ – Ben Crowell Aug 8 '13 at 20:18
  • $\begingroup$ Since the Born rule has to hold for the electromagnetic "wavefunction," and electromagnetic waves can interact with matter, it clearly has to hold for material particles as well, or else we wouldn't have a consistent notion of the probability that a photon "is" in a certain place and the probability that the photon would be detected in that place by a material detector. Could you explain this in more detail? $\endgroup$ – yippy_yay Aug 8 '13 at 20:40
  • $\begingroup$ Would it make more sense to discuss this in comments on my answer rather than yours? $\endgroup$ – Ben Crowell Aug 8 '13 at 20:50
  • $\begingroup$ Yes. I reposted the comment on your answer. $\endgroup$ – yippy_yay Aug 8 '13 at 20:56
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The "square" follows from the Schroedinger equation in a scattering problem setup. For a given flux of incident particles $J$ it gives the number of scattered particles per second per unit of solid angle $\frac{d^2N}{dtd\Omega}\propto J$, and thus for one particle it is a probability per second per unit of solid angle.

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    $\begingroup$ This is circular reasoning. A justification for the Born rule should explain why the square appears in the solution to the scattering problem. $\endgroup$ – Ben Crowell Aug 6 '13 at 21:51
  • $\begingroup$ @BenCrowell This isn't really circular. I think this is how Born himself justified it. You have to remember that Schrodinger's Equation came before the Born rule; people knew the eigenvalues of the equation gave the correct Hydrogen energy levels, but had no idea what the wavefunctions themselved signified. Born solved Schrodinger's equation for scattering problems using perturbation theory, and noticed that $|\psi|^2$ corresponded to the classical scattering amplitude. So this is really the most correct answer, at least historically. $\endgroup$ – Jahan Claes Oct 5 '16 at 2:14
  • $\begingroup$ As an interesting sidenote, the classical scattering amplitude agrees EXACTLY with first-order perturbation theory in quantum mechanics. This is a fortunate coincidence, since that's what allowed Born to deduce the Born rule. $\endgroup$ – Jahan Claes Oct 5 '16 at 2:15
  • $\begingroup$ Also, for advancing a probability interpretation, the experimental fact of charge quantization (we observe one electron at once) is very important. In other words, at low intensity fluxes the intensity is quantized. Thus for the quantized entity we use the probability to be observed. $\endgroup$ – Vladimir Kalitvianski Oct 24 '16 at 17:08