# Working with Fourier Transforms

I am in a quantum mechanics class and we are working with integrating mathematica and our problem sets. I have tried to do do a Fourier transform with the k function given but can get it to my match my working on paper.

I apologize in advance in that I don't know how to add my code as an attachment, I am still new to Mathematica and coding, so here is a snapshot of my work:

Edit, added absolute value of a but it still looks funky numerator wise, I have a 2a in my numerator, I thought maybe multiplying by inverse sqrt of two pi would help but it still looks off.

• You forgot the absolute value in the exponent. A*Exp[-a*Abs[k]-I*b*k]. – Roman Sep 20 at 19:29
• Please show us the code text rather than the screenshot so we can easily test your code. – xzczd Sep 21 at 2:31

There is an obvious mistake in your code. You write f[x_] instead of f[k_], because your variable is k, not x, according to how you write things.

On the second place, you should modify your code and write f[k_]:= A Exp[- a Abs[k] - I b k];. If you do so, the Fourier transform becomes

g[x_] := Assuming[a > 0 && A \[Element] Reals && b \[Element] Reals,
FourierTransform[f[k], k, -x, FourierParameters -> {0, 1}]];

In[48]:= g[x]

Out[48]= (a A Sqrt[2/\[Pi]])/(a^2 + (b + x)^2).


The Inverse fourier transform is exactly the initial function

In[47]:= Assuming[a > 0 && A > 0 && b \[Element] Reals,
FourierTransform[g[x], x, k]]

Out[47]= A E^(-(a + I b) k) (E^(2 a k) HeavisideTheta[-k] +
HeavisideTheta[k]).


Of course the shape is different, but this because instead of the absolute value, the output changes accordingly to the sign of k.

• Thankyou, this will help a lot – Biometric09 Sep 22 at 20:42