## Typo error or correct wavefunction?

Hi!

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!

Recognitions:
Homework Help
 Quote by Thunder_Jet 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!
Looks fine to me. $\theta(-p)$ is 1 when p is negative because of the minus sign, so the first term is non-zero when p < 0 and the second term is non-zero when p > 0. The whole thing could be written

$$\Psi(p) \propto \exp(-a|p|/\hbar)$$

 Quote by Mute Looks fine to me. $\theta(-p)$ is 1 when p is negative because of the minus sign, so the first term is non-zero when p < 0 and the second term is non-zero when p > 0. The whole thing could be written $$\Psi(p) \propto \exp(-a|p|/\hbar)$$
Thanks for your suggestion. My problem now is on converting this momentum representation into its x representation. The probability density in x can be written as ∫<ψ(p)|x><x|ψ(p)> dx. Since I have here a complex conjugate of the Fourier transform term exp(ipx/hbar), those Fourier terms will just cancel (i.e., exp(-ipx/hbar)exp(ipx/hbar) is just 1). And there will be no integration anymore except ∫dx. What do you think of this?

Recognitions:

## 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:

$$\psi(t,x)=\langle x|\psi \rangle=\int_{\mathbb{R}} \mathrm{d} p \langle x|p \rangle \langle p | \psi \rangle.$$

Now you have (setting $\hbar=1$)

$$\langle x | p \rangle=\frac{1}{\sqrt{2 \pi}} \exp(\mathrm{i} p x).$$

That means

$$\psi(t,x)=\int_{\mathbb{R}} \mathrm{d} p \frac{1}{\sqrt{2 \pi}} \exp(\mathrm{i} p x) \tilde{\psi}(t,p).$$

In your case it's a quite simple integral. You just have to split the integration in the ranges $p<0$ and $p>0$ and just calculate the integral.

 Quote by vanhees71 To go from the momentum representation to the position representation you have to take the Fourier transform of the wave function, not the probability: $$\psi(t,x)=\langle x|\psi \rangle=\int_{\mathbb{R}} \mathrm{d} p \langle x|p \rangle \langle p | \psi \rangle.$$ Now you have (setting $\hbar=1$) $$\langle x | p \rangle=\frac{1}{\sqrt{2 \pi}} \exp(\mathrm{i} p x).$$ That means $$\psi(t,x)=\int_{\mathbb{R}} \mathrm{d} p \frac{1}{\sqrt{2 \pi}} \exp(\mathrm{i} p x) \tilde{\psi}(t,p).$$ In your case it's a quite simple integral. You just have to split the integration in the ranges $p<0$ and $p>0$ and just calculate the integral.

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?
 Recognitions: Science Advisor That integral does not vanish.