# Why is a wave packet normalizable?

I'm doing some reading in Zettili's quantum mechanics book and came across this passage:

Now the thing I'm left dumbfounded on is why it can be considered normalizable? I tried finding some explanation in the book but that didn't yield anything other than some vague verbal argument for it. Now I tried just plugging the wave packet into the normalization condition and got some nasty looking integral expression that I'm not sure what to do with.

Maybe this is something trivial but as I'm trying to learn this on my own I'd really like to leave nothing unclear in my understanding so if someone could tell me why defining the wave function like this makes it normalizable that would be appreciated.

• Show your work. What is the normalization you find? Can you normalize a Gaussian? – Cosmas Zachos Nov 26 '20 at 22:11
• Well the thing is that the book really doesnt cover how you normalize such a function at all from what i can find. The examples that ive found it covers only shows it the other way around, like how you find some normalization constant and thats just for simple functions. Now when i try to just show that this function equals 1 i just get like an expression that contains multiple exponential integrals that i dont know what to do with. – Ornate Nov 26 '20 at 22:39

Actually, you're right questioning this. In general, a "wave packet" with unspecified $$\psi(x,0)$$ doesn't have to be normalizable. E.g. what if you have $$\psi(x,0)=(x^2+1)^{-1/4}$$? This function decays at infinity, is somewhat localized near $$x=0$$, but is not normalizable.
So, take the third point in the book to be part of a definition of a wave packet, a constraint on $$\psi$$.
The reason is that the square of amplitude of the wave function should give you the probability of the existence of that particle (scaled to a factor $$c$$ ). The sum of all these probabilities should go to $$c$$. So if there is no constant factor ($$c=$$ normalisation constant) with which I can divide the sum (precisely the integral) to get the actual probability $$\in [0,1]$$, then the wave function cannot be used to model any physically meaningful wave.