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.

  • $\begingroup$ 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


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.


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


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