How Do You Calculate the Expectation Value of Momentum in Quantum Mechanics?

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Homework Help Overview

The discussion revolves around calculating the expectation value of momentum in quantum mechanics, specifically focusing on the mathematical expression and the application of derivatives within that context.

Discussion Character

  • Mathematical reasoning, Conceptual clarification

Approaches and Questions Raised

  • The original poster questions the application of the derivative in the expectation value formula, specifically whether it should be applied to the wave function ψ. Some participants confirm this approach and clarify the order of operations regarding the complex conjugate.

Discussion Status

Participants are actively discussing the correct application of derivatives in the expectation value formula. There appears to be a shared understanding of the mathematical operations involved, although explicit consensus on the broader implications of these operations is not stated.

Contextual Notes

There is a focus on the mathematical details of the expectation value calculation, with references to the order of operations and comparisons to other expectation values, such as that of position.

atomicpedals
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I've managed to get myself confused on a seemingly simple point of mathematics. When I calculate the expectation value of momentum in quantum mechanics

[tex]<p>=\int{\psi* \frac{\hbar}{i} \frac{d}{dx} \psi dx}[/tex]
To what should I be applying the derivative? [itex]\psi[/itex]?
 
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Of course you would apply it to ψ. What else?
 
Order of operation would rule out complex conjugate of [itex]\psi[/itex]. So, instead of being like <x> which can simplify to

[tex]<x>=\int{|\psi|^{2}xdx}[/tex]
it would instead be
[tex]<p_x>=\int{\psi * \frac{\hbar}{i} \frac{d \psi}{dx} dx}[/tex]
 
atomicpedals said:
Order of operation would rule out complex conjugate of [itex]\psi[/itex]. So, instead of being like <x> which can simplify to

[tex]<x>=\int{|\psi|^{2}xdx}[/tex]
it would instead be
[tex]<p_x>=\int{\psi * \frac{\hbar}{i} \frac{d \psi}{dx} dx}[/tex]

Yes. That's it.
 
Right on, thanks.
 

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