Meaning of localization of wave function? What is a wave packet and how is it formed?


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*I found out on Wikipedia that wave packets are formed by localisation and localisation occurs when different wavelengths are superposed. But it's unclear and it's not stated about how exactly it's formed in experiment.

*I also found out about normalization and it's stated that all wave function must be normalized or else its probability density function does not exist. If it's so, does it mean that all wave function of particle are already somehow sort of normalized and hence localized in the first place? 

*If it's so, what makes it localised and possess an uncertainty in its de Broglie wavelengths? 
 A: "Wave packets are formed by localisation" is a weird statement. Let me try to make sense of it.
If you have a system (such as a particle in non-relativistic quantum mechanics), quantum mechanics dictates that it be described by a wave function.
What's a wave function? 


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*Well, first of all, a wave function is an abstract mathematical object, a function that to every point $x$ in space associates some finite complex number $\psi(x)$. 

*In addition, $|\psi(x)|^2$ must be probability distribution, where the $|\cdot |$ denotes the norm of complex numbers. Basically, this probability distribution tells you the probability to find the particle at position $x$ in space. 
The thing is that if you want that your wave function is a probability distribution, you need
$$ \int |\psi(x)|^2\, dx = 1.$$
This property is called normalisation The integral here is a Lebesgue integral, but that doesn't really matter. What you need to know is that this integral is basically a measure for the area under the curve. If your particle was at one point $x_0$ only, then your function $\psi(x)$ would be nonzero only at $x_0$. However, an integral over a function that is nonzero but finite only at one point is always zero, since there is no area under the curve. In other words, your particle will never only be at one point only. 
Now comes the interesting part: Mathematics tells you that you can build any such function with the properties described above by adding up plane waves (sinusses and cosinusses in multiple dimensions if you want). A plane wave only has one frequency, but the sum of such plane waves has multiple frequencies of course. You can even build things like Gaussian functions only from plane waves. Here is an animation that might be useful to see what happens.
This "building up the wave packet with plane waves" isn't done in experiments - everything in your experiment is already a wave function - it's just a mathematical trick called the Fourier decomposition/transformation. There is something remarkable about the Fourier decomposition: Since the number of possible frequencies $k$ is a space with the same dimension as the space you start out with, the decomposition into plane waves defines a function $\psi(k)$ just like your original function $\psi(x)$ - and this function has the same properties as a wave function. In particular, it defines a probability function over frequencies and it also implies that no wave-function can have a single frequency (this is also impossible, because a wave with only one frequency would be a sinus or cosinus and if you add the area under the curve of the square of that function, it'll never be finite). 
What can we know about where the particle can be? 


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*Have a look at a sinus wave and you imagine that there was a particle with this wave-function: Clearly, you will never have certainty that the particle is at one point, since it could be anywhere where the function is nonzero, which is basically everywhere. 

*What can happen, however, is that the bulk of the wave function is at some point. In other words, there is a region $C$ in space so that the probability of finding the particle in that region is almost one (in other words $\int_C |\psi(x)|^2dx \approx 1$). You can call such a particle localized since you know it's in this region (but beware: this is not the same as Anderson localisation). Does a particle have to be localised just because it needs to be normalised? That depends on what you want to call localised. The bulk of the probability cannot be spread over an infinite region, but if your space is finite (like our universe seems to be), the particle could well have equal probability to be anywhere. Surely, if the particle could be anywhere in the universe, you wouldn't call it localised, so being "localised" very much depends on your context. 

*Looking at Wikipedia gives you plenty of wave functions that seem perfectly localised in some area: https://en.wikipedia.org/wiki/Wave_packet.

*Finally, let's get back to the Fourier transform: Your wave packet also has a Fourier transform, so you can also think of it as a function of frequencies. As I said, no wave function can have just one frequency, so there must a variance to the frequency (and thus the wave-length of the wave) of a particle. This is basically the variance in the de Broglie wavelength, which would basically be the expectation value of the frequencies that frequency function.
What about wave packets?
I'm not going to go into wave packets, because then I'd need to introduce time into the picture. So far, I haven't done that. In principle, a wave packet is just a wave function that is localised in some region and over time, that region travels as a "packet" and doesn't spread out much. But in truth, that's a different question.
NB: I'm glossing over a lot of details, but my basic points are that 
a) A particle (for instance) is described by a wave function and this is such that you cannot say with complete certainty that the particle is at one specific point or have one specific frequency. 
b) Seeing the wave function as a superposition of wavelengths should be thought of as a mathematical property. A quantum mechanical system is not "built up" in experiment by adding superpositions of plane waves.
A: To adress the question as clarified in the comments, I will explain what the standard formalism of quantum Mechanics gives rise to for the observables of position and momentum.
I assume that the standard mathematical tools (Hilbert space theory) used to talk about quantum mechanics are unfamiliar to you, so I will try to only use high school mathematics. I will also not attempt to justify the formalism of quantum mechanics, just explain its consequences.
We are concerned with "quantum particles" (e.g. electrons) moving around in 3-dimensional space in a non-relativistic setting (the particles must be "moving" much slower than the speed of light). We also ignore some slight subtelties such as spin, which can be added later keeping the following discussion unmodified. In that setting:
Every particle has a complex valued function associated to it called its wave function $\psi(\vec{x},t)$. This wave function contains all information we can possibly know about the state of the particle. Two wave functions, one of which is a complex multiple of the other, represent the exact same physical state and every wave function has to be normalizable. This is to say that the integral (called the norm of the wave function) $$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}|\psi(x,y,z,t)|^2\ dx\ dy\ dz $$
is finite at all times. Given the wave function at some point in time, this integral will not change with time as long as no measurement is performed. Since two wavefunctions represent the same state if one is a complex multiple of the other, we can divide any given wave function by the above integral (which is a number) to obtain a wave function of norm 1. Such wave functions are called normalized. Note that there is still the freedom to multiply such a wavefunction by a complex number with absolute value one, so there still are many wavefunctions that represent the same state.
If we know the wave function at some initial time $t_0$, the wave function at all later (and earlier) times can be calculated by solving the Schroedinger equation (which may be very hard but still possible in principle) $$i\hbar\partial_t\psi=-\frac{\hbar^2}{2m}(\partial^2_x\psi+\partial^2_y\psi+\partial^2_z\psi)+V(\vec{x})\psi$$
where $V(x)$ is the potential seen by the particle (For example the Coulomb potential seen by an electron) and $m$ is its mass.
If we measure the position of a particle, this means answering the question "Is the particle in the region $A\subseteq \mathbb{R}^3$ at the time $t_0$?" where $A$ is the size of our measurement apparatus. Note that the formalism of quantum mechanics assumes that you know which processes are measurements and which are not. The fact that it's not obvious in general (particularily outside of experiments) what processes should be considered measurements is called the "measurement problem". It does not make sense to ask "where exactly is the particle" and expect to get fixed coordinates because that would mean measuring the position of the particle with infinite precision, which we cannot do. On the other hand, in our setting the region $A$ can be arbitrarily small in principle.
Knowing the wave function of a particle, we cannot predict the outcome of such a measurement (which is a hard fact to accept, even knowing the state of a particle COMPLETELY does not allow you to know where it is!). We can only predict the probability that the particle will be found inside the region. That probability is the integral of $|\psi|^2$ over the region $A$ divided by the norm of the wavefunction. Suppose for simplicity that $A$ is a box, $A=\{(x,y,z): a<x<b, c<y<d, f<z<g\}$ for real numbers $a,\dots, g$. The probability to find the particle inside region $A$ if measured at time $t_0$ is now
$$P=\frac{\int_a^b\int_c^d\int_f^g |\psi(x,y,z,t_0)|^2 dx\ dy\ dz }{\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}|\psi(x,y,z,t)|^2\ dx\ dy\ dz}$$
If the wave function is normalized, the denominator is just equal to one making the calculation simpler, which is the main reason one wants to use normalized wavefunctions.
It is a postulate of quantum mechanics that after a measurement is performed, the state of the particle changes and the new state depends on the outcome of the measurement. More precisely, if we have a wave function $\psi(\vec{x},t)$ and measure if the particle is inside $A$ at time $t_0$, if we get the result "yes" the new state of the particle at time $t_0$ becomes $$\psi_{\text{new}}(\vec{x},t_0)=\begin{cases}
     \psi(\vec{x},t_0) & \text{if } \vec{x}\in A \\
    0 & \text{otherwise}  
   \end{cases}$$
On the other hand, if we get the result "no", $$\psi_{\text{new}}(\vec{x},t_0)=\begin{cases}
     0 & \text{if } \vec{x}\in A \\
    \psi(\vec{x},t_0) & \text{otherwise}  
   \end{cases}$$
that is to say, the function is cut off at the boundary of the region $A$ and the part where it is not is set to zero while the other part is kept as is. This of course changes normalization, but that's not a problem. We can just compute the norm of the new wave function and divide it by that norm to make sure it now has norm one, if we prefer working with normalized wavefunctions. Knowing the wave function immediately after our measurment, we can again solve the Schroedinger equation to obtain the wave function for all later times (actually only until another measurment is performed on that particle).
Note that, if the region $A$ really has only one single point (which you might want to do if you thought that infinitely precise measurements and infinitely small measurement devices exist) the probability for a particle to be at any single individual point is zero, since $\int_a^af(x)dx=0$ for any function $f$.
The above is standard modern quantum mechanics. Let's try to answer some of your questions, now that the terminology is somewhat clearer. The probability density function of the particle is by definition "the thing you integrate over a region of space to get the probability to find the particle in that region of space". From above we see that this probability density $p$ is given by
 $$p(\vec{x},t):=\frac{|\psi(\vec{x},t)|^2}{\text{norm of }\psi}$$
and clearly doesn't exist, unless the norm is finite. On the other hand, the probability to find the particle somewhere in space is one (we don't need to consider the destruction and creation of particles in non-relativistic contexts) so the integral of $p$ over all of space is equal to one, which is clear from the formula.
The solutions of the Schroedinger equation look like waves, which is why we talk about waves all the time in quantum mechanics. Suppose there are no forces acting on the particle, that is to say $V(\vec{x})=0$. Given some initial wave function $\psi(\vec{x},t_0)$ it is always possible to find a function $\tilde{\psi}(\vec{k})$ such that
$$\psi(\vec{x},t_0)=\frac{1}{(\sqrt{2\pi})^{3}}\int_{-\infty}^{\infty} \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} e^{i\vec{k}\cdot \vec{x}}\tilde{\psi}(\vec{k})\ dk_1\ dk_2\ dk_3$$
where $\vec{k}=(k_1,k_2,k_3)$ is called a wave vector and the scalar product is $\vec{k}\cdot\vec{ x}=k_1 x+k_2 y+ k_3 z$. The function $\tilde{\psi}$ is called the Fourier transform of the function $\psi(\vec{x},t_0)$ and is given by the formula. 
$$\tilde{\psi}(\vec{k})=\frac{1}{(\sqrt{2\pi})^{3}}\int_{-\infty}^{\infty} \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} e^{-i\vec{k}\cdot \vec{x}}\tilde{\psi}(\vec{k})\ dx\ dy\ dz$$
This is perhaps a bit overwhelming but it's just math and has a priori nothing to do with quantum mechanics. The factors of $\pi$ are convenctional. Now why we need this:

Claim: If the potential $V$ is zero, given at time $t_0=0$ the initial wave function $\psi(\vec{x},0)$ the solution $\psi(\vec{x},t)$ of the Schroedinger equation for all times  is given by $$\psi(\vec{x},t)=\frac{1}{(\sqrt{2\pi})^{3}}\int_{-\infty}^{\infty} \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} e^{i\vec{k}\cdot \vec{x}} e^{-i\omega(\vec{k}) t}\tilde{\psi}(\vec{k})\ dk_1\ dk_2\ dk_3$$
  where $\omega(\vec{k})=\frac{\hbar}{2m}\vec{k}\cdot\vec{k}$ and $\tilde{\psi}(\vec{k})$ is the Frouier transform of $\psi(\vec{x},0)$ defined as above.

This means that this solution (the wave function of a "free" (=not acted uppon by forces) particle is an infinite superposition (=sum) of functions $e^{i(kx-\omega(\vec{k}) t)}$ (these are called plane waves) weighted (=multiplied) by $\tilde{\psi}(\vec{k})$. You see there is a huge amount of terminology here, mostly due to historic reasons. This means that $\psi(\vec{x},t)$ is in fact a wave packet (refer to the wiki page on wave packets). As you can see, the idea of Fourier transforms is that every function is in fact a superposition of plane waves. A such a superposition is usually called a wave packet, if the function $\psi(\vec{x},t)$ is small in magnitude outside some region of space where it "is localized".

Proof of claim.
  Compute both sides of the equation individually.
  $$i\hbar\partial_t\psi(\vec{x},t)=\frac{i\hbar}{(\sqrt{2\pi})^{3}}\int_{-\infty}^{\infty} \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} e^{i\vec{k}\cdot \vec{x}} (-i\omega(\vec{k}))e^{-i\omega t}\tilde{\psi}(\vec{k})\ dk_1\ dk_2\ dk_3= $$ $$=\frac{\hbar ^2}{(\sqrt{2\pi})^{3}}\int_{-\infty}^{\infty} \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} e^{i\vec{k}\cdot \vec{x}} ( \frac{\vec{k}\cdot\vec{k}}{2m})e^{-i\omega(\vec{k}) t}\tilde{\psi}(\vec{k})\ dk_1\ dk_2\ dk_3$$
  Right side:
  $$-\frac{\hbar^2}{2m}(\partial^2_x\psi(\vec{x},t)+\partial^2_y\psi(\vec{x},t)+\partial^2_z\psi(\vec{x},t))=$$ $$=-\frac{\hbar^2}{2m}\frac{1}{(\sqrt{2\pi})^{3}}\int_{-\infty}^{\infty} \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} (-\vec{k}\cdot \vec{k})e^{i\vec{k}\cdot \vec{x}} e^{-i\omega t}\tilde{\psi}(\vec{k})\ dk_1\ dk_2\ dk_3$$

