I Coefficients in the Schrodinger equation and the momentum operator

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The discussion centers on the coefficients in the Schrödinger equation and the momentum operator, specifically questioning why they are set to specific values like ##\hbar## and ##-i\hbar##. Participants emphasize that altering these coefficients would lead to results inconsistent with experimental observations, highlighting the importance of the commutation relations in quantum mechanics. The conversation also touches on the relationship between particle spin and these coefficients, although it is noted that spin is a more advanced topic that should not complicate initial learning. Ultimately, the thread is closed due to forum rules against discussing personal theories. Understanding the established principles of quantum mechanics is deemed essential before proposing modifications.
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How are the coefficients in the Schrodinger equation and the momentum operator determined?
The Schrodinger 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]
 
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How did you get the idea of fraction as alternative ? Commutation relation of x and p would be helpful.
 
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You might start with the early chapters of Ballentine.
 
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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 Schrodinger did from Borowitz (1967).

IMG_0013.jpegIMG_0014.jpegIMG_0015.jpegIMG_0016.jpeg
 
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anuttarasammyak said:
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 Schrodinger 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 Schrodinger 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?
 
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Shen712 said:
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!"
 
Shen712 said:
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.
 
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The forum rules do not allow discussion of unpublished personal theories, so this thread has been closed.
 
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