# Why isn't Mathematica outputting the inverse fourier transform of this function?

I'm brand new to Mathmematica, so please be nice. I don't understand where I'm going wrong. I just need to use the method of inverse fourier transform on this wavefunction to get the momentum-space wavefunction.

The original wavefunction:

The output, which doesn't seem right

The code:

InverseFourierTransform[(m*w/(Pi*\[HBar]))^(1/4) E^(-m*w*x^2/(2*\[HBar])) E^(-I*w*t/2), x, t]

• 'Pi' versus 'pi' and'E' versus 'e' and 'I' versus 'i'. All built-in names start with a capital. – Bob Hanlon Oct 5 '16 at 2:15
• Ok I fixed that but it still gives me the same output as the image above. – whatwhatwhat Oct 5 '16 at 2:17
• Then enter code in code blocks vice images so people can can work with it and correct/edit your question. – Bob Hanlon Oct 5 '16 at 2:19
• @BobHanlon Done. Good idea, I don't know why I didn't do that initially. I normally do that on SO. – whatwhatwhat Oct 5 '16 at 2:21
• Why are the transform and original variables both present in your wavefunction? – J. M.'s ennui Oct 5 '16 at 2:32

J.M.' s comment related to both w and t appearing in the expression to be transformed. Assuming that the t in E^(-I*w*t/2) was actually meant to be an x then

InverseFourierTransform[(m*w/(Pi*ℏ))^(1/4) E^(-m*w*
x^2/(2*ℏ)) E^(-I*w*x/2), w, t]

(*  (((m/ℏ)^(1/4)*
(-((x*(m*x + I*ℏ))/ℏ))^
(1/4) - (-(m/ℏ))^(1/4)*
((x*(m*x + I*ℏ))/ℏ)^(1/4))*
ℏ*Gamma[1/4]*
(Abs[t]*Cos[(5/4)*ArcTan[
(2*t*ℏ)/(m*x^2 +
I*x*ℏ)]] -
I*t*Sin[(5/4)*ArcTan[
(2*ℏ*Abs[t])/(m*x^2 +
I*x*ℏ)]]))/
(2*2^(1/4)*Pi^(3/4)*x*
(m*x + I*ℏ)*
((x^2*((-I)*m*x + ℏ)^2)/ℏ^2)^
(1/4)*(1 + (4*t^2*ℏ^2)/
(x^2*(m*x + I*ℏ)^2))^(5/8)*
Abs[t])  *)


Assumptions on any of the parameters might simplify this.