To find the linear momentum of a function

MontavonM
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The linear momentum operator is (^ on top of) P, which is -ih(d/dx), where h is h bar, and the d's are partials... Now you operate on your function, easy enough. But this function is complex, f(x) = e^i5kx, and I'm assuming k is the kinetic energy operator. So the simple derivative of this function is 5kie^i5kx, where is K(operator) is -(h^2/2m)(del^2). This is where I don't know where to go, considering you have to operate K within the function and the e^(...) part. Especially since the operator includes the del^2... Any help? Thanks in advance!
 
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MontavonM said:
The linear momentum operator is (^ on top of) P, which is -ih(d/dx), where h is h bar, and the d's are partials... Now you operate on your function, easy enough. But this function is complex, f(x) = e^i5kx, and I'm assuming k is the kinetic energy operator. So the simple derivative of this function is 5kie^i5kx, where is K(operator) is -(h^2/2m)(del^2). This is where I don't know where to go, considering you have to operate K within the function and the e^(...) part. Especially since the operator includes the del^2... Any help? Thanks in advance!

Here k is the wave number (k=2*pi/lambda), not the kinetic energy operator. The argument of the exponential has to be non-dimensional. The deBroglie relation is p = hbar*k. Hope this helps.
 
Yep, your right... I figured it out. Always nice when you're making it way more complicated than it is
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. Towards the end of the first lecture for the Qiskit Global Summer School 2025, Foundations of Quantum Mechanics, Olivia Lanes (Global Lead, Content and Education IBM) stated... Source: https://www.physicsforums.com/insights/quantum-entanglement-is-a-kinematic-fact-not-a-dynamical-effect/ by @RUTA
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