What is the justification for the substitution p_x -> (h/i)d/dx in physics?

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Discussion Overview

The discussion revolves around the justification for the substitution of momentum operator \( p_x \) with the expression \( (h/i)d/dx \) in the context of quantum mechanics. Participants explore the physical and mathematical reasoning behind this common practice, touching on its implications for wave functions and momentum conservation.

Discussion Character

  • Exploratory, Technical explanation, Conceptual clarification

Main Points Raised

  • One participant questions the physical justification for the substitution \( p_x \to (h/i)d/dx \) and seeks a deeper understanding of its significance.
  • Another participant explains that for a plane wave described by \( \psi(x) = e^{i k x} \), the operator \( \frac{1}{i} d/dx \) effectively extracts the wave vector \( k \), linking it to de Broglie's relation to derive \( \hbar \).
  • A third participant provides a mathematical perspective, noting that the quantum Lagrangian's invariance under spatial translations leads to momentum conservation, with Dirac demonstrating that \( d/dx \) serves as the generator of these translations.

Areas of Agreement / Disagreement

Participants present various viewpoints on the justification for the substitution, with some focusing on physical interpretations and others emphasizing mathematical foundations. No consensus is reached on a singular justification.

Contextual Notes

The discussion highlights different interpretations and applications of the substitution, indicating potential limitations in understanding its broader implications or assumptions underlying the mathematical framework.

gn0m0n
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Hi, I am wondering what justifies the substitution p_x -> (h/i)d/dx ? I know it is very common but I have not seen a reason for it anywhere. Why does it make physical sense?
 
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For a plane wave, that is [itex]\psi(x) = e^{i k x}[/itex], [itex]\frac{1}{i} d/dx[/itex] extracts the wave vector or wavenumber [itex]k[/itex]of the wave. From there we use de Broglie's relation to get the [tex]\hbar[/tex].
 
There's rather a good mathematical reason. The quantum Lagrangian is invariant under spatial translations, which guarantees momentum conservation. As Dirac shows, d/dx is the generator of translations ( page 100 'Princples..'). Dirac calls translation 'displacement' and demonstrates that the action of the operator is to return the momentum.

Clever fellow, that Dirac.
 
Thanks once again, all.
 

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