Inverse Fourier Transform of X(w): Real Values for x(t)

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SUMMARY

The inverse Fourier transform of the function X(w) = 1/(j*(w*hbar-Ek)+(hbar/T2)) - 1/(j*(w*hbar+Ek)+(hbar/T2)) yields x(t) = 2*j/hbar*heaviside(t)*sin(t/hbar*Ek)*exp(-t/T2) in MATLAB, resulting in imaginary values. This outcome is consistent with the antisymmetry of X, indicating that the time evolution is 90 degrees out of phase with a real solution. An alternative solution found in literature presents x(t) = -2/hbar*heaviside(t)*sin(t/hbar*Ek)*exp(-2*pi/T2*t), which is real-valued and includes a negative sign and an additional 2*pi in the exponential term.

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X(w) = 1/(j*(w*hbar-Ek)+(hbar/T2)) - 1/(j*(w*hbar+Ek)+(hbar/T2))

The inverse Fourier transform of the above equation using MATLAB will obtain the following:

x(t) = 2*j/hbar*heaviside(t)*sin(t/hbar*Ek)*exp(-t/T2)

We can see that the values of x(t) are all imaginary values, however this shouldn't be the case, should have real values for x(t) instead.

Does anyone knows what should be the correct inverse Fourier transform?

Thanks!
 
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The result given by Matlab will be correct, assuming you have adopted the same sign conventions when setting up the analytic FT. In particular the antisymmetry of X results in imaginary x.

Also, this looks like a QM problem, and as far as I know there's no reason you can't have an imaginary solution: it just means the time evolution is 90 degrees out of phase with that of a real solution.
 
I've found an answer from a book which states that

x(t) = -2/hbar*heaviside(t)*sin(t/hbar*Ek)*exp(-2*pi/T2*t)

I'm wondering how to get this answer, because it's a bit different from what I got from MATLAB (the difference is that the solution obtained from the book is in real values, and now there's a negative sign in front, and lastly the exponential term has an additional 2*pi)

Hope that someone would be able to assist me in my queries! Thanks in advance!
 

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