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I know the Schrödinger equation is bascially the "quantized" Hamiltonian formalism from classical mechanics, and the Dirac equation is the special-relativistic version. But these equations do not describe deterministic "eye-visisble hard steel-ball-like" objects anymore, like the classical hamiltonian, but just probabilities about "when to probably find them where" instead (until you measure, when one of the possible probability amplitudes is "selected", so the collapse of the wave function happens).

Both equations contain the mass. What does the mass has to do with probabilities?

An often stated argument why big objects do not show quantum properties is that they consist of $10^{23}$ particles and thus lose their quantum behaviour because of the interaction between these many particles. OK, accepted, but no mass-based argument here, either (only a number argument). Now let's consider a single (hypothetical) elementary particle with a mass of one ton. Would the Schrödinger or Dirac equation be considered valid for this? In the Schrödinger equation one term would "go to zero" (considered neglectable, mass in denominator) and in the Dirac equation the mass would make the left side contain a huge, dominating $-mc$ term.

What would this mean for the probablities? So, how would this single hypthetical particle behave in a double slit experiment, for example?

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    $\begingroup$ Mass $m$ is a parameter in the Schroedinger equation. You can insert any positive value for $m$. For example, hydrogen atom size comes out proportional to $\frac{h^2}{me^2}$, where $m$ is mass of the electron. So the larger the mass $m$, the smaller the atom would be. For muon $m$ is 207 times smaller than for electron, so a muonic hydrogen atom should be 207x smaller than the ordinary hydrogen atom. $\endgroup$ Mar 14 at 23:17
  • $\begingroup$ @JánLalinský Identifying that parameter with the mass matches the velocity of an unbound classical particle with the quantum wave group velocity. This is necessary for the quantum theory to yield results consistent with classical experiments. $\endgroup$
    – John Doty
    Mar 15 at 0:52
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    $\begingroup$ Voting to reopen. A perfectly clear question, as comments already show. $\endgroup$
    – gandalf61
    Mar 15 at 7:27
  • $\begingroup$ I agree. I don't see why this was closed. $\endgroup$ Mar 15 at 8:59
  • $\begingroup$ It's really the momentum in the equation rather than the mass. Classically the KE is given by $E = p^2/2m$, and in QM the momentum operator is $\hat{p} = -i\hbar d/dx$. Changing the classical $p$ to the operator $\hat{p}$ gives the KE term in the Schrodinger equation. $\endgroup$ Mar 15 at 9:15

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I think you are confusing the concepts of evolution and collapse of the wavefunction:

  • The collapse of the wavefunction is that non-deterministic effect that everyone finds so un-intuitive and non-classical, as it is probabilistic. As you might know, the probability of the wavefunction collapsing to a given state/region of space/momentum interval/whatever else depends on the square of its amplitude. This is not described by Schrödinger's equation.

  • The temporal evolution of the wavefunction (i.e. how the wavefunction changes with time) is described by Schrödinger's equation (or other relativistic equations) and is perfectly deterministic. This equation has nothing to do with collapse. The evolution of your physical state may depend on many parameters, among which you might find its mass, which is just a measure of its inertia, as is the case of Newton's second law (but keeping the distances). As a rule of thumb, a bigger mass implies a "more classical" behaviour, which is translated to a more localized wavefunction. This is seen as the kinetic term of the Schrödinger equation is reduced as the mass increased, as it is $\propto \frac{1}{m} \nabla^2$.

With respect to the second part of your question, for a very massive particle, it would have a very small de Broglie wavelength, so its wavefunction would be extremely localized. In the limit $\hbar/m \to 0$ we would recover a completely localized wavefunction, so we could talk about its position/trajectory in classical terms. It would also be more convenient to describe it's trajectory with Newton's second law (or equivalents).

I hope this helps you clear some concepts.

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