Continuing from my first question titled, Simple Quantum Mechanics question about the Free particle, (part1)
Griffiths goes on and says,
"A fixed point on the waveform corresponds to a fixed value of the argument and hence to x and t such that,"
x $\pm$ vt = constant, or x = $\mp$ vt + constant
What does this mean? I am so confused.
He goes on and says that psi(x,t) might as well be the following:
$$\psi(x,t) = A e^{i(kx-(hk^2/2m)t}$$
because the original equation,
$$\psi(x,t) = A e^{ik(x-(\hbar k/2m)t)} + B e^{-ik(x + (\hbar k/2m)t)}$$
only differs by the sign in front of k. He then says let k run negative to cover the case of waves traveling to the left, $k = \pm \sqrt{(2mE)}/\hbar$.
Then after trying to normalize psi(x,t), you find out you can't do it! He then says the following,
"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."
How did he come to this logical deduction. I don't follow. Can someone please explain Griffith's statement to me?
Best Answer
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.