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Consider this Phys.SE question which spells out the energy of a vibrating string as a variant of Hooke's law, and thereby explains why it's proportional to the square of the amplitude (here: the transversal speed).

To what extent can one make the analogy to Born's rule where, this time, it's the probability that's the square of the amplitude?

I.e., could one consider that

  • quantum mechanical probability $\equiv$ classical energy, and
  • amplitude of wave function $\equiv$ transversal velocity?
Qmechanic
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Tfovid
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  • Could you precise a bit. Since the wave function must be normalized to one, you can't have directly "amplitude $\propto$ velocity" since increasing velocity would give a non-normalized wavefunction. How do you take this into account? – Ewen Bellec Oct 05 '18 at 08:29
  • @E.Bellec I completely agree that's a problem, and I can't account for it. My motivation for seeing an analogy between the two formalisms comes from the fact that I always struggled to find an intuitive explanation for Born's rule. I agree that the analogy I propose here is a complete shot in the dark, and that's why I put it out there for people to criticize it. It's just that the identical "power-two" formulation of the classical and quantum waves seems neat and compelling. – Tfovid Oct 05 '18 at 09:17
  • ... as for the normalization, I don't see it much as a problem. It's the dynamics of the waves I'm interested in, not so much whether we can force-fit normalization due to auxiliary physical constraints. – Tfovid Oct 05 '18 at 09:19
  • Ok, I think I see what you mean. I had the same (and still have) trouble with QM. This is the same question about why we need a wave function collapse. Think of the electron double-slits experiment (or any scattering experiment), you can describe your electron as a wave while it propagate. BUT when the wave hits the screen, you only see a dot, not a full wave. This is where Born rule comes from, when the wave hits the screen, the wave function collapses and you get only 1 dot with probability for his position given by Born rule. – Ewen Bellec Oct 05 '18 at 09:29
  • There's no intuitive reason for this wavefunction collapse. It just a rule which fits experiments. This is why you can find so many different QM interpretations https://en.wikipedia.org/wiki/Interpretations_of_quantum_mechanics – Ewen Bellec Oct 05 '18 at 09:31
  • Sure, but why power-of-two? I'm familiar with most interpretations out there, but my problem is that power-two, so I'm on the hunt for any hints from classical mechanics---hence the vibrating string... – Tfovid Oct 05 '18 at 09:33

1 Answers1

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Ok, if your problem is "why the probability density is given by $|\psi|^2$?", one can show that the integral of this probability density is conserved for all time.

Starting from the Schrodinger equation (I take $\hbar=1$)

$\frac{\partial \psi}{\partial t} = -iH\psi$

$\frac{\partial \psi^*}{\partial t} = +iH\psi^*$

$\begin{align} \frac{\partial |\psi|^2}{\partial t} &= \psi\frac{\partial \psi^*}{\partial t}+\psi^*\frac{\partial \psi}{\partial t} \\ &=i\psi H \psi^* -i \psi^* H \psi \end{align}$

Taking here a 1D case and and Hamiltonian given by $H = -\frac{\partial^2}{\partial x^2}+V(x)$ and integrating the equation above over x, we get

$\begin{align} \frac{\partial}{\partial t}\int\limits_{-\infty}^{+\infty}|\psi(x)|^2 \, dx &= i[\int dx \, \psi(x)(-\psi''^*(x)+V(x)\psi^*(x))-\int dx \, \psi^*(x)(-\psi''(x)+V(x)\psi(x))] \\ &= i[-\int dx \, \psi''(x)\psi^*(x)+\int dx \, \psi^*(x)\psi''(x)] \\ &= 0\end{align}$

Where I used 2 integration by parts in the first integral and used the fact that $\psi(\pm \infty)=0$

Now this does not prove that it is the only possibility for a probability density. If you can show that $\frac{\partial}{\partial t}\int\limits_{-\infty}^{+\infty}|\psi(x)| \, dx=0$ for example then my demonstration is useless.

Ewen Bellec
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  • Another way would be to start from the Path integral formulation of QM, then you can define a wavefunction which satisfy the schrodinger equation and show that the probability density is given by $|\psi|^2$. But then you get the same kind of problem on how to justify the path integral formulation. – Ewen Bellec Oct 05 '18 at 12:00