Register to reply

Typo error or correct wavefunction?

by Thunder_Jet
Tags: correct, error, typo, wavefunction
Share this thread:
Thunder_Jet
#1
Nov15-11, 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!
Phys.Org News Partner Physics news on Phys.org
A new, tunable device for spintronics
Watching the structure of glass under pressure
New imaging technique shows how cocaine shuts down blood flow in mouse brains
Mute
#2
Nov15-11, 10:16 PM
HW Helper
P: 1,391
Quote Quote by Thunder_Jet View Post
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]
Thunder_Jet
#3
Nov15-11, 10:31 PM
P: 18
Quote Quote by Mute View Post
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?

vanhees71
#4
Nov16-11, 03:33 AM
Sci Advisor
Thanks
P: 2,471
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 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.
Thunder_Jet
#5
Nov17-11, 06:39 AM
P: 18
Quote Quote by vanhees71 View Post
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?
Avodyne
#6
Nov18-11, 10:38 AM
Sci Advisor
P: 1,205
That integral does not vanish.


Register to reply

Related Discussions
How to correct for random measurement error? Set Theory, Logic, Probability, Statistics 3
Infinite potential well wavefunction contribution to 'classical' initial wavefunction Advanced Physics Homework 1
Quaternions angle error - correct? General Math 7
Electromagnetic Wave Prob. (Is this a typo/error?) Introductory Physics Homework 3
Tension, weight and forces Introductory Physics Homework 2