Coefficients in the Schrodinger equation and the momentum operator

[Shen712]

TL;DR

How are the coefficients in the Schrodinger equation and the momentum operator determined?

The Schrödinger equation is

$$
i\hbar \frac{\partial\Psi}{\partial t} = -\frac{\hbar^{2}}{2m} \frac{\partial^{2}\Psi}{\partial x^{2}} + V \Psi
$$

Why is the coeffient on the left-hand side $\hbar$, not $\frac{\hbar}{2}$ or $i\frac{\hbar}{3}$ or something like these
Besides, in quantum mechanics, the momentum operator is defined to be

$$
p \rightarrow -i\hbar \frac{\partial}{\partial x}
$$

Again, why is the coefficient $-i\hbar$, not $-i\frac{\hbar}{2}$ or $-i\frac{\hbar}{3}$ or something like these?

[Mentor's note: post edited to fix some Latex formatting]

 

[anuttarasammyak]

How did you get the idea of fraction as alternative ? Commutation relation of x and p would be helpful.

 

[Nugatory]

You might start with the early chapters of Ballentine.

 

[Frabjous]

If you stick in different constants, you would get different answers which would not agree with experiment.

The SE was inspired by classical equations. Here is a brief description of what Schrödinger did from Borowitz (1967).

 

[Shen712]

How did you get the idea of fraction as alternative ? Commutation relation of x and p would be helpful.

I got the idea of fraction because I believe electrons have substructure, and each component of the electron must have a spin smaller than $\frac{\hbar}{2}$, say, $\frac{\hbar}{4}$ or $\frac{\hbar}{6}$. But this would violate our convention that fermions have spin $\frac{\hbar}{2}$. As I try to trace out the origin of the fermion spin $\frac{\hbar}{2}$, I found that it has to do with the constant coefficients in the Schrödinger equation (and/or the Dirac equation) and the momentum operator (the commutation relation of x and p can be traced to the definition of the p operator). So I am thinking that the Schrödinger equation (and/or the Dirac equation) and the momentum operator might be modified in order to describe the spin of the components of the electron. Am I on the right track?

 

[anuttarasammyak]

Am I on the right track?

In OP you did not state what the particle is but you are wondering about electrons in #5. Why don't you forget about spins now? Spin appears later in advanced treatment and it does not harm the first lessons.

Have you followed my suggesion?
$$\frac{\partial }{\partial x}xf - x\frac{\partial }{\partial x}f = f$$
for any f so
$$\frac{\partial }{\partial x}x - x\frac{\partial }{\partial x} = 1$$
as an operator so
$$-i\hbar \frac{\partial}{\partial x}x - x(- i\hbar) \frac{\partial }{\partial x} = -i\hbar$$
$$px - xp = -i\hbar$$
which meets commutation relation which is one of QM principles. Ref. (20.74) in https://www.feynmanlectures.caltech.edu/III_20.html "If Planck’s constant were zero, the classical and quantum results would be the same, and there would be no quantum mechanics to learn!"

 

[weirdoguy]

Am I on the right track?

No, and if you don't know where does the spin come from (eg. you don't know what is the double cover of $SO(3)$) then surely you are not prepared to do any reasearch on a structure of electron. Besides, PF rules forbid talking about personal theories.

 

[Nugatory]

The forum rules do not allow discussion of unpublished personal theories, so this thread has been closed.