So, I'm trying to write down the passages required to create a wave function (from first to second quantization, to just wave functions).
From the Wikipedia page, basic behaviour, non-dispersive wave section: https://en.wikipedia.org/wiki/Wave_packet
I understand this:
-
you want a wave packet, so you start from writing the function fourier transformed (e.g: you convolve an unknown function with a moving sinusoidale plane wave, which is the eigenfunction of the conjugate basis (I need to understand this better though)):
\begin{equation}
f(x,t) = \int^{\infty}_{-\infty} A(k) e^{i(kx-\omega(k)t} dk
\end{equation} -
you want to know $A(k)$, so you fourier transform again
-
you make an ansatz in the initial wave at $t=0$: it has to be a wave packet, so you multiply a gaussian with a sinusoidal plane wave. You choose the sinusoidal plane wave as it satisfies the wave equation. You choose the gaussian cause it has a finite support and you like it(?)
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you find $A(k)$
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you use a dispersion relation
-
you now have your wave function
What I don't understand is where to put the solution of the wave equation it needs to satisfy.
Is it in the initial ansatz?
In the non dispersive section the solution is a simple sinusoidal plane wave.
In the dispersive section it uses the Schrödringer equation, however as initial ansatz it uses a sinusoidal plane wave again.
If it satisfies both equations, then why one is dispersive and the other not?
Just the normalization changes. Is it the dispersion relation?
Then why using different wave equations, if the result (the wave function) is the same?
Where does the dispersion relation of the schrödringer eq. stems from?
We know:
\begin{equation}
E =\hbar \omega \quad \quad
p = \hbar k.
\end{equation}
I would naively have used the relativistic result:
\begin{equation}
E = \sqrt{m^{2} + p^{2}}
\end{equation}
To connect $E(p)$. I would now use the relations I wrote before to finally connect $\omega(k)$.
In the ultra relativistic limit we indeed recover $\omega = kc$.
However in my lessons we used $E=p^{2}/2m$ which leads to $\omega = \frac{\hbar k^{2}}{2m}$.
And my hypothesis would lead to $\omega = \sqrt{k^{2} + (m/\hbar) ^{2}}$.
And Wikipedia used $\omega = k^{2}/2$.
Best Answer
It feels to me like OP is mixing a few concepts. I'll try to disentangle the "mess". I will write in more detail the explanation of the non-dispersive case, and hope that the other can be understood from the concepts introduced there.
The wave equation is $$\partial_t^2 u(x, t) = c^2\partial_x^2 u(x, t).$$ A common procedure to solve this equation is known as separation of variables, that is, using the Ansatz $u(x, t) = X(x)T(t)$. Plugging it into the wave equation we get $$T''(t)/T(t) = c^2 X''(x)/X(x),$$ which has to be true for all $x$ and $t$. Hence, both sides must be constant, lets call this constant $-(kc)^2$: $$ \begin{align} T''(t) &= -(kc)^2 T(t) \\ X''(x) &= -k^2 X(x).\end{align} $$ The solution to these equations is then easily expressible as $$ \begin{align} T(t) &= \exp(\pm ikc t) \\ X_\pm(x) &= \exp(ik x)\end{align} \implies u_k(x, t) = \exp(i(kx \pm kc t)),$$ depending on the separation constant $k$. Note that there is not just "one" solution, but rather we have found a family of solutions that can be described through the continuous parameter $k$, which can take any positive or negative value. Additionally, for each $k$ we have 2 solutions, one with $\omega=kc$ and another with $\omega=-kc$. The first describes a wave whose profile propagates to the left, while the second propagates to the right. Incidentally, notice that the previous relation implies $\omega(k)=\pm kc$, where $\omega$ is the oscillation frequency in time of the solution. This kind of relation is usually known as the "dispersion relation" of the wave equation. It is also easy to check that the original wave equation is linear, such that any linear combination of particular solutions is a solution. Here is where wave packets become relevant, as the most general wave that can be formed through a superposition of the fundamental $u_k$ is $$ u(x, t) = \int_{-\infty}^\infty \mathrm{d}k \big(A(k) \exp (i(kx -kct)) + B(k) \exp (i(kx -kct))\big).$$ Indeed, $A(k)$ and $B(k)$ indicate how much of the fundamental functions corresponding to $k$ is present in the final wave.
Let us recap briefly: from the wave equation, we have obtained a family of independent functions that allowed us to write a generic wave, for any sensible amplitude function. However, we were also able to extract from the wave equation another important piece of information, the dispersion relation. It tells us how fast a fundamental solution $u_k$ oscillates in time, as a function of the continuous label that characterizes the solutions, $k$. In this case, the relation was very simple and gives rise to the so-called non-dispersive regime. This is because in the argument of the exponential we have $(\pm x - ct)$, and thus the phase between different points in the wave is preserved along the spatial profile.
This has "nothing" to do with particles and antiparticles, because everything has been classical so far. If one is interested in the antiparticle interpretations of the negative frequency solutions and so on, there are other posts on the site, such as Corresponding particle-antiparticle solutions for Klein-Gordon equation that address this issue.
We can now look at the dispersive regime induced by the Schrödinger equation: $$ i\partial_t \psi(x, t)= \frac{-\hbar}{2 m}\partial^2_x \psi(x, t).$$ Here, the same separation of variables procedure yields $\omega = \frac{\hbar k^2}{2m}$, and a similar expression for the wave packets, with the only difference that now $\omega$ is proportional to $k^2$ rather than to $k$. This implies that the phase between different the different fundamental functions building the wave evolves differently depending on $k$, which is called dispersive propagation, in contrast to the previous. You correctly observed that some constants differed from your lessons. The reason for this, as addressed in the comments to the question, is that in WP, they use units in which $m=1$ and $\hbar=1$. Equivalently, you can imagine that you measure every mass in terms of the mass of the electron and every angular momentum in terms of $\hbar$, instead of kg and kg·m${}^2$·s${}^{-1}$, respectively.
Notice that the Schrödinger equation is only one of many possible dispersive equations. In particular, I have discussed the Schrödinger equation in the non-relativistic regime. For massless, relativistic particles, another wave equation has to be used that gives rise to the $E=|p|c \implies \omega=|k|c$ we derived in the non-dispersive regime. In other cases, $E=\sqrt{(pc)^2 + (mc^2)^2} \implies \omega=\sqrt{(kc)^2 + (mc^2)^2/\hbar^2}$ would be the final dispersion relation.