Coefficients in the Schrodinger equation and the momentum operator

In summary, the coefficients in the Schrödinger equation represent the probability amplitudes of a quantum state, while the momentum operator is a key component in quantum mechanics, defined as the generator of translations in space. The momentum operator acts on wave functions to yield information about the momentum of a particle, and its relationship with the Schrödinger equation helps describe the evolution of quantum systems. Together, they illustrate the interplay between position, momentum, and the probabilistic nature of quantum mechanics.
  • #1
Shen712
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TL;DR Summary
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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  • #2
How did you get the idea of fraction as alternative ? Commutation relation of x and p would be helpful.
 
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  • #3
You might start with the early chapters of Ballentine.
 
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  • #4
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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  • #5
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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  • #6
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?
[tex]\frac{\partial }{\partial x}xf - x\frac{\partial }{\partial x}f = f[/tex]
for any f so
[tex]\frac{\partial }{\partial x}x - x\frac{\partial }{\partial x} = 1[/tex]
as an operator so
[tex]-i\hbar \frac{\partial}{\partial x}x - x(- i\hbar) \frac{\partial }{\partial x} = -i\hbar[/tex]
[tex]px - xp = -i\hbar[/tex]
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!"
 
  • #7
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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  • #8
The forum rules do not allow discussion of unpublished personal theories, so this thread has been closed.
 
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FAQ: Coefficients in the Schrodinger equation and the momentum operator

What is the role of coefficients in the Schrödinger equation?

Coefficients in the Schrödinger equation often represent physical constants or parameters specific to the system being studied. For example, the coefficient \( \hbar \) (reduced Planck's constant) appears in the time-dependent Schrödinger equation and is fundamental in relating the wavefunction to physical observables such as energy and momentum.

How do coefficients affect the solutions to the Schrödinger equation?

Coefficients can affect the solutions to the Schrödinger equation by scaling the wavefunction and influencing the energy levels and eigenstates of the system. For instance, changing the potential energy coefficient in the equation can alter the shape of the potential well, thereby modifying the allowed energy levels and the corresponding wavefunctions.

What is the momentum operator in quantum mechanics?

The momentum operator in quantum mechanics is an operator that corresponds to the momentum observable of a particle. In the position representation, it is given by \( \hat{p} = -i\hbar \frac{\partial}{\partial x} \) for a single particle in one dimension. This operator acts on the wavefunction to yield the momentum of the particle.

How are the coefficients in the Schrödinger equation related to the momentum operator?

The coefficients in the Schrödinger equation, such as \( \hbar \), directly relate to the definition of the momentum operator. For example, the presence of \( \hbar \) in the momentum operator \( \hat{p} = -i\hbar \frac{\partial}{\partial x} \) ensures that the operator has the correct units of momentum and correctly scales the derivative of the wavefunction to give the physical momentum.

Can the coefficients in the Schrödinger equation change with different systems?

Yes, the coefficients in the Schrödinger equation can change depending on the system being analyzed. For example, in a system with different masses or potential energies, the coefficients that appear in the kinetic and potential energy terms of the Schrödinger equation will differ. These changes reflect the unique physical properties and constraints of each system.

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