1
$\begingroup$

My book has started using the wave packet definition as follows (time independent form):

$$\Psi(x) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} A(k) \ e^{ikx}dx$$

I do not understand where the $1/\sqrt{2\pi}$ comes from in this definition. First, I thought it has something to do with normalization however, I can't seem to prove this to myself.

$$\Psi'(x) = N \int_{-\infty}^{\infty} A(k) \ e^{ikx}dx$$ $$\Psi'(x)^{\ast} = N \int_{-\infty}^{\infty} A^{\ast}(k) \ e^{-ikx}dx$$

$$\Psi'(x) \Psi'(x)^{\ast}= N^2 \int_{-\infty}^{\infty} A(k) \ e^{ikx}dx \int_{-\infty}^{\infty} A^{\ast}(k) \ e^{-ikx}dx = N^2 \int_{-\infty}^{\infty} A(k) A^{\ast}(k)dx = N^2 A(k) A^{\ast}(k)$$

The last step I justify by the conditions that the wave functions must approach zero as you go from $\pm \infty$.

$$P = 1 = N^2 \int_{-\infty}^{\infty} A(k) A^{\ast}(k)dk$$

I am not sure where to go from here. Does this term actually come from the normalization? If so, how can I show this.

$\endgroup$
3
  • $\begingroup$ Have you studied Fourier transforms yet? You should look into that. $\endgroup$
    – joseph h
    Commented Dec 29, 2021 at 2:27
  • $\begingroup$ Ok I can see the parallel, so is the multiplication of $1/\sqrt{2pi}$ just so we can write the Fourier transform for $A(k)$? $\endgroup$ Commented Dec 29, 2021 at 2:37
  • $\begingroup$ yep, that's right. see rob's answer below. $\endgroup$
    – joseph h
    Commented Dec 29, 2021 at 2:41

2 Answers 2

1
$\begingroup$

The Fourier-transform operators

$$ \hat F = \int \mathrm dx \frac{e^{ikx}}{\sqrt{2\pi}} \qquad \hat F{}^{-1} = \int \mathrm dk \frac{e^{-ikx}}{\sqrt{2\pi}} $$

are prettier if you attach a $\sqrt{2\pi}$ to them both, instead of the asymmetric

$$ \hat F_\text{ugly} = \int \mathrm dx \ {e^{ikx}} \qquad \hat F{}_\text{ugly}^{-1} = \int \mathrm dk \frac{e^{-ikx}}{{2\pi}} $$

You should think of $\Psi(x) = \hat F {}^{-1} A(k)$ as your wavefunction, but $A(k) = \hat F\Psi(x)$ as also your wavefunction, in momentum space. You’re currently stuck because the normalization of $\Psi$ is related to the normalization of $A$.

Perhaps abetting your confusion: in the current version of your question (v3), you erroneously have $\Psi(x) = \hat F A(k)$. That’s impossible, because the integral on the right side eats up the dummy variable $x$, so the left side should be a function of $k$.

Likewise you have to be careful about the dummy variables when you are computing your overlap integrals. You expect that

\begin{align} 1 = \left< \Psi | \Psi \right> &= \int \mathrm dx\ \Psi^*(x) \Psi(x) \\ &= \int \mathrm dx \left( \int\mathrm dk A^*(k) \frac{e^{+ikx}}{\sqrt{2\pi}} \right) \left( \int\mathrm dk’ A(k’) \frac{e^{-ik’x}}{\sqrt{2\pi}} \right) \end{align}

You’re going to exploit $ \int\mathrm dx\ e^{i(k-k’)x} \sim \delta(k-k’) $ and get rid of two of the integrals, leaving you with

$$ 1= \left<\Psi|\Psi\right>_x = \left<A|A\right>_k $$

which is what I meant about the normalizations being related. But you’re going to look up and/or prove that there are the correct amount of $\sqrt{2\pi}$ involved, rather than trusting my one-squiggle relationship.

If you have been putting off reading about Fourier transforms, today is a good day.

$\endgroup$
0
$\begingroup$

For normalization, you want to look at $\int\psi^*\psi\,\text dx=1$. Try this and the result will be much better.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.