Simple Quantum Mechanics question about the Free particle, (part2)
The first part about velocities says that we're looking at a function
$$\psi(x,t) = \psi(u)$$
for $u=x-vt$. For example, $\psi = \cos(x-vt)$.
Now pick some fixed value for $\psi$, say $0.4$. Find a place where $\psi = 0.4$ such as $x=1.16$. If you let some small amount of time $\textrm{d}t$ go by, then look at $x=1.16$ again, $\psi$ has changed a little, but if you look at the point $x = 1.16 + v\textrm{d}t$, you will find $\psi = 0.4$ there, so we could say that the place where $\psi$ is $0.4$ is moving at speed $v$.
This is not the same as saying that the particle is moving at that speed. It is saying that $v$ is the phase velocity of the wave.
The second part about normalizability says that a wavefunction must be an element in a vector space called a Hilbert space (by physicists; I think mathematicians call it $L_2$). The Hilbert space consists of wavefunctions that are square-normalizable; you can square them, integrate from negative infinity to infinity, and get a finite value. Things that die off exponentially on their tails do this, for example.
The sine function doesn't die off exponentially, or at all. If you square it and integrate from negative infinity to infinity, you get something infinite. Thus, the sine wave doesn't represent a reasonable probability density function for the location of a particle. The particle would be "infinitesimally likely to be observed everywhere in an infinite region", which physically does not make sense. Instead, for a real particle, we must have a normalizable wavefunction. Since a free particle with definite energy would have a pure sinusoidal wavefunction, a free particle with definite energy is physically not possible.
I will answer this part of the question since it is important to distinguish what we are doing as physicists.
"A free particle cannot exist in a stationary state; or, to put it another way, there is no such thing as a free particle with a definite energy."
As Mark says in his comment above:
it's pretty standard to find a solution to an equation and then ask whether that solution is physically meaningful.
Which is fine. When the answer is "no", as in this case, one goes ahead to find physically meaningful solutions that will describe a free particle.
There is ample experimental evidence that particles exist, from atomic physics to nuclear physics to particle physics. Solving Schroedinger's equation for a potential well not only gives well defined state functions, the solutions describe the behaviour of the particle quite well to excellently. One then solves the "no potential" equation and finds the plane waves .
First thought: no potential means free particle.
second thought: are these plane wave functions well behaved so that they can be used to describe quantum mechanically a free particle?
Third thought should be: No, they cannot be normalized to a probability function.
Fourth thought: Could one use these solutions as a complete set to find a function that can be normalized and its probability describes a physical free particle ?
Bingo: wave packets.
The statement "there is no such thing as a free particle with a definite energy" is wrong for a physical particle. It should be : plane waves cannot describe free particles.
Free particles ( experiment) trump plane waves ( theory).
You were right to be confused.
I would normally just add this as a comment, but I still can not comment.
I would say the author is just saying that because the arguments are effectively the same, and the modulus of each number is arbitrary, one can use the distributive law to simplify the expression. If one tries to normalize using standard procedure, one gets: $$A^2$$ If one does not specify finite boundary conditions, one must find the integral of this constant value from -infinity to infinity which gives an infinite value, and thus is not normalizable.