Appearance of $e^{i(kx-\omega t)}$ in the derivation of Schroedinger Equation

Question

One of the derivations of Schroedinger's Equation that I came across assumed the wave function for the particle to be $$\Psi = e^{i(kx-\omega t)} ,$$ and I hardly understand why.
This equation implies that $$\Psi = cos{(kx-\omega t)}+ isin(kx-\omega t). $$ I came across this answer and I believe what it is meant to convey is that the complex expression of the wave function works like a coordinate system that largely simplifies calculations.
I'm trying to understand the physical meaning of this. Does this mean that the wavefunction has a travelling cosine wave along, say, an $x$-axis and a travelling sine wave along the $y$-axis, and the actual wavefunction is a superposition of these?
Sorry if all this sounds absurd. I am an absolute beginner to this course. I'm looking for an intuitive answer more than a mathematical one. I did my research, but it requires some heavy mathematics and I haven't started off with any rigorous math courses at college(I am familiar with basic calculus and linear algebra but not Fourier Transforms or Hilbert Spaces, which often popped up when I looked for explanations.)

Answer 1

The real and imaginary axes have nothing to do with the x, y axes. A wavefunction is a function which asigns a complex number to every point in its domain. Consider a one-dimensional system. The wavefunction asigns to every point $x$ a complex number. You can visualize this a large row of clocks placed on the x-axis. The angle that the clock makes is the phase of the complex number and the length of the hand is the magnitude of the complex number. In your case the magnitude is everywhere 1 so the clocks only rotate.

I made a visualization of the 1D Schrödinger equation a long time ago. In the following image the left-to-right axis represents the x-axis. The other two axes show the real and imaginary component of the complex number that is situated at $x$. The distance to the x-axis shows the magnitude of the complex number. Each individual "clock" rotates with a speed that is determined by the Schrödinger equation. In the first animation you can see a wavepacket, a wavefunction that looks like a normal distribution. In the second animation you can see a plane wave which corresponds to your example. In both examples the potential is zero so they are "free particles".

Answer 2

Quantum mechanical waves are complex in nature, unlike electromagnetic or water waves which are modelled by real waves. And this is why we focus on complex waves.

Now, the expression you have pointed out is an example of a complex plane wave. Super-posing some combination of them will give you any wave. Hence we can focus on this as the simplest possible wave.

Its rather like analysing a molecule by focusing on its constituent atoms.

Answer 3

From the de Broglie's matter wave hypothesis $\lambda = h/p$ we note that position and momentum are conjugate variables in a particle-like wave form. If we have a physical matter wave (i.e. a wave packet made of de Broglie plane waves) in a position space, the wave must be two dimensional (i.e. complex) to have the corresponding "inverse wave" in the momentum space, since the wave-like particle has its distribution in the both position and momentum spaces. Thus, the waves in position and momentum space form a Fourier transformation pair. This wave phenomenon is related to the fundamental uncertainty principle that is just a property of the Fourier transformation.

See: https://en.wikipedia.org/wiki/Sine_and_cosine_transforms https://en.wikipedia.org/wiki/Fourier_transform#Uncertainty_principle

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The free de Broglie plane wave $\psi (x,t)=e^{i(px-Et)/\hbar}$ is a prototype of a quantum mechanical wave. It can be assumed from optics, since photons behave like free electrons in the double-slit experiment and photons also carry momentum and energy (here we have applied the de Broglie's matter wave and Planck's energy–frequency relations $p=\hbar k$ and $E=\hbar \omega$). From optics we also know that the probability of finding a photon on the screen is proportional to the wave intensity $|\psi (x)|^2$. The previous intensity function illustrates the superposition principle and wave interference. In order to have a physical wave that has a finite size, so that probability density function $|\psi (x)|^2$ has its integral over the entire space equal to 1, we must add all the plane waves of different momentums $p_j$, energies $E_k$ and amplitudes $A(p_j,E_k)$. That is, we have the superposition of all the possible experimental outcomes:

$$\psi (x,t)=\sum_{j}\sum_{k}A(p_j,E_k)e^{i(p_jx-E_kt)/\hbar}.$$

Here we assume that our wave-like quantum particle is subjected to an external potential $V(x)$ that vanishes at infinity. Now we can submerge the sum over energy (both discrete and continuous ranges) into the amplitude function. We also note that the sum over momentum becomes an integral, and thus we can conclude that

$$\psi (x,t)=\int_{-\infty}^{\infty}A(p,t)e^{ipx/\hbar}dp$$

with the probabilistic condition $\int_{-\infty}^{\infty}|\psi (x,t)|^2dx = 1,$ for all $t \geq 0$. If we set $\phi(p,t)=A(p,t)\sqrt{2\pi\hbar}$, then we have the wave packet in its familiar form

$$\psi (x,t)=\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi\hbar}}\phi(p,t)e^{ipx/\hbar}dp,$$

and also its inverse Fourier transformation that represents the momentum wave function

$$\phi (p,t)=\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi\hbar}}\psi(x,t)e^{-ipx/\hbar}dp$$

that leads to the Heisenberg's uncertainty relation $\Delta(\psi)\Delta(\phi) \geq \hbar/2.$

Next we assume that a physical wave function $\psi$ satisfies the same total energy equation as the free de Broglie plane wave, i.e.

$$i\hbar \frac{\partial}{\partial t}e^{i(px-Et)/\hbar} = Ee^{i(px-Et)/\hbar}$$

(The general case can be justified as a weighted sum over many different values of $E$ if we assume that $\psi$ has a basis representation in the Hilbert space $L^2$.)

and

$$-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}e^{i(px-Et)/\hbar} = \frac{p^2}{2m}e^{i(px-Et)/\hbar} = E_{kin}e^{i(px-Et)/\hbar}.$$

(The general case can be justified with the momentum probability density function $|\phi (p)|^2$ and Plancherel's theorem.)

To consider a particle in a potential $V(x)$, we simply note that the averaged total energy equation

$$\langle E\rangle=\langle E_{kin}\rangle+\langle E_{pot}\rangle$$

can be written as

$$\int_{-\infty}^{\infty}E|\psi|^2 dx = \int_{-\infty}^{\infty}E_{kin}|\psi|^2 dx + \int_{-\infty}^{\infty}E_{pot}|\psi|^2 dx;$$

$$\int_{-\infty}^{\infty}(E\psi)\overline{\psi} dx = \int_{-\infty}^{\infty}(E_{kin}\psi)\overline{\psi} dx + \int_{-\infty}^{\infty}(E_{pot}\psi)\overline{\psi} dx;$$

$$\int_{-\infty}^{\infty}(i\hbar \frac{\partial}{\partial t}\psi)\overline{\psi} dx = \int_{-\infty}^{\infty}(-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\psi)\overline{\psi} dx + \int_{-\infty}^{\infty}(V\psi)\overline{\psi} dx;$$

$$\int_{-\infty}^{\infty}[i\hbar \frac{\partial}{\partial t}\psi+\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\psi - V\psi]\overline{\psi}dx = 0.$$

The calculus of variations allows us to infer that

$$i\hbar \frac{\partial}{\partial t}\psi(x,t) = -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\psi(x,t) + V(x)\psi(x,t).$$