Typo error or correct wavefunction?

[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 canceled 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!

 

[Mute]

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 canceled 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)

 

[Thunder_Jet]

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?

 

[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&lt;0 and p&gt;0 and just calculate the integral.

 

[Thunder_Jet]

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&lt;0 and p&gt;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?

 

[Avodyne]

That integral does not vanish.