[MontavonM]

The linear momentum operator is (^ on top of) P, which is -ih(d/dx), where h is h bad, and the d's are partials... Now you operate on your function, easy enough. But this function is a bit complex, f(x) = e^ickx, where c is a constant, and im assuming k is the kinetic energy operator. So you operate and get ie^ickx(c) from the chain rule, e.g is c = 5 then it would be 5ie^i5kc... Is this correct, or is K also pulled out as a constant which would make it 5kie^i5kc? Also, is K(operator) is -(h^2/2m)(del^2), what would become of the function? Thanks in advance

[Aeon]

What is beautiful about is that . This should simplify greatly your problem.

Also, chances are K is not an operator. I'm pretty sure you are assuming wrong. The kinetic energy operator is usually denoted .

Try the derivative again.

[Aeon]

Can you describe the setting of your problem? I'm pretty sure I can help you, but you need to give me some more information.

[juanrga]

The linear momentum operator is (^ on top of) P, which is -ih(d/dx), where h is h bad, and the d's are partials... Now you operate on your function, easy enough. But this function is a bit complex, f(x) = e^ickx, where c is a constant, and im assuming k is the kinetic energy operator. So you operate and get ie^ickx(c) from the chain rule, e.g is c = 5 then it would be 5ie^i5kc... Is this correct, or is K also pulled out as a constant which would make it 5kie^i5kc? Also, is K(operator) is -(h^2/2m)(del^2), what would become of the function? Thanks in advance

$$\hat{p}/$$ is the momentum of the particle, not only its linear component $$\hat{\pi}/$$. For charged particles without magnetic interactions and/or for massive particles without 'gravitomagnetic' interactions .

You do not explain what is your function but it looks like a free wavefunction. For free wavefunctions, k denotes the wave number.

The derivative rule for the exponential of a function is

$$ \frac{\partial e^{g(x)}}{\partial x} = e^{g(x)} \frac{\partial g(x)}{\partial x} /$$

Try again.