Find the expectation value of the linear momentum

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SUMMARY

The discussion focuses on calculating the expectation value of linear momentum for the wave function Psi(x,t) = A exp^(-x/a)^2 * exp^(-iwt) * sin(kx). The relevant momentum operator is defined as p^ = -ih(bar) d/dx. Participants emphasize the importance of analyzing the integrands to determine if they are odd or even functions, which simplifies the integration process. Specifically, they note that odd functions integrate to zero over symmetric intervals, while even functions can be simplified by doubling the integral from 0 to the upper boundary.

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Homework Statement


For a given wave function Psi(x,t)=Aexp^-(x/a)^2*exp^-iwt*sin(kx) find the expectation value of the linear momentum.



Homework Equations


<p>=integral(-inf,inf) psi* p^ psi dx
p^=-ih(bar) d/dx
sin x = (exp ix - exp -ix)/2i
cos x = (exp ix + exp -ix)/2

The Attempt at a Solution


I understand the technique of sandwiching the operator between the wave function and it's complex conjugate. Then the integral is a mess of sines cosines and exponentials. I was wondering if anyone had any advice to simplify the expression or to solve this type of infinite integral. Your suggestions would be very helpful. Thanks in advance.

 
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Look at the individual integrands and determine if they are odd or even functions. If the integrand is odd and if the integration interval is symmetric with respect to some origin (i.e. negative infinity to positive infinity or [-a,a]), you can exploit the fact that integrand integrates to zero. For even functions with symmetric intervals, you can multiply the integral by 2 and run the integral from 0 to your upper boundary value.

[tex]\int^{a}_{-a}f_{odd}\left(x\right) dx = 0[/tex]

[tex]\int^{a}_{-a}f_{even}\left(x\right) dx = 2 \int^{a}_{0}f_{even}\left(x\right) dx[/tex]

The formula for even functions is useful for exponential terms, like e^(x), when the integration interval runs from negative to positive infinity.
 

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