I How to "derive" momentum operator in position basis using STE?

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The discussion centers on deriving the momentum operator in the position basis using a more general form of the Schrödinger time evolution (STE) equation, as opposed to the specific form presented in standard texts. The user encounters technical difficulties with LaTeX, opting to share handwritten equations instead. Suggestions for resolving the LaTeX issues include logging out, clearing cookies, and trying a different browser. Additionally, clarification is sought regarding the meaning of STE, and the importance of including the kinetic energy term in the Hamiltonian for the derivation is highlighted. The conversation emphasizes the need for a clear understanding of the underlying principles in quantum mechanics.
LightPhoton
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I ask about how one can use generalized STE to motivate momentum operator in position basis using the approach of Griffiths and Schroeter
I am not able to use Latex for some reason. It is very glitchy and if I do one backspace then it fills my whole screen with multiple copies of the same equation. Thus I am pasting a screenshot of handwritten equations instead. Apologies for any inconvenience.

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In Introduction to Quantum Mechanics by Griffiths and Schroeter, the IMO motivates the form of momentum operator in position basis in a very nice manner. However, the problem is that they use a very specific form of STE (1)

Instead, I want to work in a much more general setting by writing STE as (2)

Now, the authors motivate it by taking time derivative of the expectation value of the position, which leads me to (3).

However, I am not sure how to proceed from here.
 
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LightPhoton said:
I am not able to use Latex for some reason.
You might try logging out, clearing cookies, and then logging in again.

You might also try a different browser.
 
For the derivation by integration you would need the fact that H includes "kinetic energy" part of ##-\hbar^2/2m \ \nabla_x^2##.
 
Sorry, what does it mean STE ?
 
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For the quantum state ##|l,m\rangle= |2,0\rangle## the z-component of angular momentum is zero and ##|L^2|=6 \hbar^2##. According to uncertainty it is impossible to determine the values of ##L_x, L_y, L_z## simultaneously. However, we know that ##L_x## and ## L_y##, like ##L_z##, get the values ##(-2,-1,0,1,2) \hbar##. In other words, for the state ##|2,0\rangle## we have ##\vec{L}=(L_x, L_y,0)## with ##L_x## and ## L_y## one of the values ##(-2,-1,0,1,2) \hbar##. But none of these...

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