Typo error or correct wavefunction?

  • #1

Main Question or Discussion Point

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!
 

Answers and Replies

  • #2
Mute
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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. [itex]\theta(-p)[/itex] 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

[tex]\Psi(p) \propto \exp(-a|p|/\hbar)[/tex]
 
  • #3
Looks fine to me. [itex]\theta(-p)[/itex] 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

[tex]\Psi(p) \propto \exp(-a|p|/\hbar)[/tex]
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?
 
  • #4
vanhees71
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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 x|p \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
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 x|p \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.

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
Avodyne
Science Advisor
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That integral does not vanish.
 

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