# Typo error or correct wavefunction?

• Thunder_Jet
In summary: It's just that the exp(ipx/hbar) term cancels out with the inverse Fourier transform. So in the end you just have dx.
Thunder_Jet
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 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!

Thunder_Jet said:
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 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

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

Mute said:
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?

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.

vanhees71 said:
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?

That integral does not vanish.

## 1. What is a typo error in a wavefunction?

A typo error in a wavefunction refers to a mistake or error that has been made in the mathematical representation of a wavefunction. This can occur due to human error or a mistake in the calculation process.

## 2. How can a typo error affect a wavefunction?

A typo error in a wavefunction can lead to incorrect calculations and results. This can affect the accuracy and reliability of any predictions or conclusions drawn from the wavefunction.

## 3. How can a typo error in a wavefunction be identified and corrected?

A typo error can be identified by carefully reviewing the calculations and comparing them to the expected results. Once identified, the error can be corrected by re-doing the calculations or using mathematical tools to correct the mistake.

## 4. Can a typo error in a wavefunction be detrimental to a scientific study?

Yes, a typo error in a wavefunction can significantly impact the validity and reliability of a scientific study. It can lead to incorrect conclusions and hinder the progress of research in a particular field.

## 5. How can scientists prevent typo errors in wavefunctions?

To prevent typo errors, scientists should double-check their calculations and ensure that they are following the correct mathematical procedures. Collaborating with other scientists and peer-reviewing can also help catch any potential errors before publishing their findings.

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