
#1
Nov1511, 10:01 PM

P: 18

Hi!
I would like to ask everyone's opinion about this wavefunction in the momentum representation: ψ(p) = N[θ(p)exp(ap/hbar) + θ(p)exp(ap/hbar)], where N is a normalization constant, a > 0, and θ(p) is a function defined as θ(p) = 0 for p > 0 and also θ(p) = 0 for p < 0. I think the θ function has been written incorrectly, right? It is just zero all over the momentum space. What I did is I assume it to be a step function, replacing θ(p) = 0 for p > 0 with θ(p) = 1 for p > 0. Now, when calculating for the probability density of finding the particle at x, I used Fourier transform to do it. But to my surprise, the exponential terms were cancelled and I am left with only dx in the integration. What do you think did I miss? Thanks everyone and I am hoping for your suggestions! 



#2
Nov1511, 10:16 PM

HW Helper
P: 1,391

[tex]\Psi(p) \propto \exp(ap/\hbar)[/tex] 



#3
Nov1511, 10:31 PM

P: 18





#4
Nov1611, 03:33 AM

Sci Advisor
Thanks
P: 2,139

Typo error or correct wavefunction?
To go from the momentum representation to the position representation you have to take the Fourier transform of the wave function, not the probability:
[tex]\psi(t,x)=\langle x\psi \rangle=\int_{\mathbb{R}} \mathrm{d} p \langle xp \rangle \langle p  \psi \rangle.[/tex] Now you have (setting [itex]\hbar=1[/itex]) [tex]\langle x  p \rangle=\frac{1}{\sqrt{2 \pi}} \exp(\mathrm{i} p x).[/tex] That means [tex]\psi(t,x)=\int_{\mathbb{R}} \mathrm{d} p \frac{1}{\sqrt{2 \pi}} \exp(\mathrm{i} p x) \tilde{\psi}(t,p).[/tex] In your case it's a quite simple integral. You just have to split the integration in the ranges [itex]p<0[/itex] and [itex]p>0[/itex] and just calculate the integral. 



#5
Nov1711, 06:39 AM

P: 18

Thanks for the detailed note. I did it but it turns out that the total integral vanish! What does it implies when the position representation is zero? I am expecting to get a Gaussian like solution. Or do you think I need to use Dirac delta function here instead of the exp(ipx/hbar) term? 



#6
Nov1811, 10:38 AM

Sci Advisor
P: 1,185

That integral does not vanish.



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