On Saturday, September 13, 2025 at 6:24:16 AM UTC-6 John Clark wrote:
On Sat, Sep 13, 2025 at 7:19 AM Alan Grayson <[email protected]> wrote: *>> Schrodinger's Equation describes how fermions behave quantum mechanically. * *> I've solved it many times and never saw any reference to fermions.* *I find that extremely difficult to believe. * *In college level and graduate courses, there are standard problems requiring the solution of S's equation, such as the potential well, the free particle, etc. The equation is fairly easy to solve, which is one reason it's so well known and popular. But S wasn't happen that he has been known for that, since like E, he didn't like QM. One thing of interest is that one of the so-called weirdnesses of QM isn't really weird at all, such as the claim that a system in a superposition of states, is in all states of that superposition simultaneously. It's just a basic result of linear algebra. For example, a vector in a plane can be written as a linear sum of two (non unique) basis vectors, so that vector can be legitimately interpreted as being partly in each of the basis vectors. No big mystery here. So when Von Neumann used Hilbert spaces to interpret QM, which is linear algebra, QM got stuck with an undeserved "weirdness". AG* *Schrodinger's equation describes how non-relativistic fermions behave. Dirac's Equation describes how fermions behave if special relativity is taken into account, and Dirac can be simplified down to Schrodinger's equation in cases where Special Relativity is not important. And all the solutions to the Dirac equation, and thus Schrodinger's Equation too, must obey Fermi-Dirac statistics and the Pauli Exclusion Principle.* *>> A macroscopic object, such as a polarizer, is made out of fermions. * *> No bosons in a polarizer? AG* *For massless spin-1 bosons, such as photons, you can use Maxwell's Equations, and they enable you to derive Malus's Law which says that the transmitted intensity of a beam of light that can make it through a polarizer is I = I₀ cos²(θ) where θ is the angle between the incident polarization and the polarizer axis. And if a beam of light is made out of photons then the probability of a single photon making it through a polarizer must be cos²(θ).* *I presume you mean that after a photon passes through a polarizer set at an arbitrary angle, which has a 50% probability of doing so, and then is passed through a second polarizer, the cosine probability applies where the second polarizer is offset from the first at some arbitrary angle. If the angle is 90 degrees, it won't pass, whereas if the angle is zero, it will pass 100%, and for intermediate offsets, the cosine probability applies. (I migh have gotten something wrong here. Please correct if necessary.) AG * *To summarize, for a beam of light with N photons:* - *Quantum prediction: On average, N × cos²(θ) photons pass through* - *Classical prediction: cos²(θ) fraction of the intensity passes through* - *These are the same thing.* * John K Clark See what's on my new list at Extropolis <https://groups.google.com/g/extropolis>* tst -- You received this message because you are subscribed to the Google Groups "Everything List" group. To unsubscribe from this group and stop receiving emails from it, send an email to [email protected]. To view this discussion visit https://groups.google.com/d/msgid/everything-list/ee532563-8eb0-4fa4-9e6b-8f6dde1706e3n%40googlegroups.com.
