quantum mechanics - Why do wave functions need to be normalized? Why aren't the normalized to begin with?

Before I started studying quantum mechanics, I thought I knew what normalization was. Just pulling off Google, here's a definition that matches what I've understood normalization to mean:

Normalization -- to multiply (a series, function, or item of data) by a factor that makes the norm or some associated quantity such as an integral equal to a desired value (usually 1).

Most often I have seen normalization that normalizes to 1 or 100% or something like that. For instance, isn't putting things in percentages a kind of normalization? If I take a quiz and get 24 / 25 points, then I "normalize" this by saying I got 96%. That's what I understood normalization to be.

Why I am Confused Now

Ever since I started studying quantum mechanics, I have felt confused by the term normalization. Let me quote this portion from Griffiths to illustrate an example of how he uses the term:

We return now to the statistical interpretation of the wave function, which says taht $|\Psi(x,t)|^2$ is the probability density for finding the particle at point $x$, at time $t$. It follows that the integral of $|\Psi|^2$ must be 1 (the particle's got to be somewhere. \begin{equation} \int^{+\infty}_{-\infty} |\Psi(x,t)|^2 dx = 1 \end{equation} Without this, the statistical interpretation would be nonsense.

However, this requirement should disturb you: After all, the wave function is supposed to be determined by the Schrödinger equation --- we can't go imposing an extraneous condition on $\Psi$ without checking that the two are consistent. Well, a glance at [the time-dependent Schrödinger equation] reveals that if $\Psi(x,t)$ is a solution, so too is $A\Psi(x,t)$, where $A$ is any (complex) constant. What we must do, then, is pick this undetermined multiplicative factor so as to ensure $\int^{+\infty}_{-\infty} |\Psi(x,t)|^2 dx = 1 $ is satisfied. This process is called normalizing the wave function.

I get the idea we need the probability distribution $\rho$ to be 1 over the whole position space. That makes sense and is obvious. So the integral makes sense. But I don't understand a couple things:

1. What was the wave function like prior to normalization? Why did it need to be normalized in the first place? To use my quiz analogy, why wasn't the test out of 100 points to begin which in which case no normalization would be needed. 96% would be 96 points.


2. Why if $\Psi(x,t)$ is a solution, so too is $A\Psi(x,t)$?

Perhaps an answer could comment on how my initial definition of normalization relates to normalizing the wave function. Also, if you like to write, adding a comment or two about Dirac normalization would be awesome.

Answer

Let us take a canonical coin toss to examine probability normalization. The set of states here is $\{|H\rangle,|T\rangle\}$. We want them to occur in equal amounts on average, so we suggest a simple sum with unit coefficients: $$\phi=|H\rangle+|T\rangle$$ When looking at probabilities, we fundamentally care about ratios. Since the ratio of the coefficients is one, we get a 1:1 distribution. We simply define the unnormalized probability as $$P(\xi)=|\langle\xi|\phi\rangle|^2$$ Plugging the above state in, we see we get a probability of 1 for both states. The probability (as we normally think of it), is the unnormalized probability divided by the total probability: $$P(\xi)=\frac{|\langle\xi|\phi\rangle|^2}{\langle\phi|\phi\rangle}$$ If we make the conscious choice of $\langle\phi|\phi\rangle$ every time, we don't have to worry about this normalized definition.

For your 2., note that the SE is linear. Thus $A\Psi$ is also a solution.