Understanding the Momentum Operator in Quantum Mechanics

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SUMMARY

The momentum operator in quantum mechanics is defined as P = -iħ/(2π) * d/dx, where ħ represents the reduced Planck constant. This operator is derived from the relationship between momentum and wave number, specifically through De Broglie's relation p = ħk/(2π). The operator K is introduced as K = -id/dx, which is Hermitian, ensuring real eigenvalues. The discussion emphasizes the connection between the wave function and the momentum operator, particularly in the context of eigenstates represented by plane waves.

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  • Understanding of quantum mechanics fundamentals
  • Familiarity with wave functions and eigenstates
  • Knowledge of De Broglie's relation
  • Basic concepts of Hermitian operators
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Dathascome
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Ahoy hoy...I'm having some trouble understanding exactly where the momentum operator comes from. The momentum operator is P=-ih/(2*pi)*d/dx
I know that according to the DeBroglie relation p=kh/(2*pi)
and in the first chapter of my book we introduce the operator K=-id/dx
which is hermitian (which is necessary for getting real eigenvalues). So they say in the book that the P opertor is just P=hK/(2*pi)...but I don't see why. Is there some sort of relation between the wave # k and the operator K?
When I asked my professor all he said was something about the units of K being right, but I don't even see that.
Any help would be greatly appreciated.
 
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I don't know the argument used in the book, but if you consider a plane wave (state with a well defined momentum):

\psi(x)=Ae^{ikx}, then this is an eigenstate of the momentum operator with eigenvalue p=\hbar k.
Does that clear it up a bit?
 
What book do you use...?I bet it's not a good one.A proof that in the coordinate representation the momentum operator (for a general axis 'i') \hat{P}_{i} has the form

\hat{P}_{i}=-i\hbar\frac{\partial}{\partial x_{i}} *

would use 3 things:
1.De Broglie's relation \vec{p}=\hbar\vec{k}
2.Parseval identity;
3.Fourier transformation of the wavefunction.

Daniel.
------------------------------------------------------------
* valid in the Schrödinger picture in Dirac's formulation.
 
You can derive the momentum operator by calling the change in time of the position expectation value the "velocity":

<p> = m \frac{d}{dt} \int \Psi^*(x,t) \ x \ \Psi(x,t) \ dx

This can be put into the form of an operator between the wavefunction and its conjugate. You'll need to first move the time derivative inside the integral, use the schroedinger equation to get d\Psi(x,t)/dt, and integrate by parts a couple times, getting rid of the boundary term since the wavefunction must go to 0 at infinity. You'll end up with:

<p> = \int \Psi^*(x,t) (-i \hbar )\frac{d\Psi(x,t)}{dx} \ \ dx
 
Thanks everyone, I think that cleared things up for me. It's an undergrad class and the book I'm using is "principles of quantum mechanics" by R.Shankar. It's probably not the book, it's most likely me :cry:
Does anyone have any suggestions of a good book to compliment the one I have already?
 
I thought Shankar's book is pretty good.It's advanced,compared,let's say to an introductory text like Griffiths or Blokhintsev,but pretty good on the ensemble...

Cohen-Tannoudji is very calculative,though the structure of the material is not appealing with those "Compléments".If u like more Messiah's book,be my guest.

Daniel.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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