To find the linear momentum of a function

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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