0
$\begingroup$

I'm currently learning Quantum Mechanics from online video lectures and resources. In most of the web articles and videos, the wave functions are shown as circular waves $e^{i\omega t}$ instead of planar waves $\sin{\omega t}$.

[Note: I'm considering a fixed position and hence the equation $e^{i(k\cdot r + \omega t)}$ reduces to $e^{i\omega t}$]

Some examples from the web:

This video shows the wave amplitude to be rotating around the position (i.e. a circular wave in accordance with $e^{i\omega t}$): Quantum Wave Function Visualization

The Wikipedia Article on Schrödinger equation describes the plane wave using $e^{i(k\cdot r + \omega t)}$ instead of $\sin{\omega t}$ even though they call it a planar wave: Schrödinger equation

In this Video the derivation of Probability density is based on a circular wave: Quantum Mechanics 1 Lecture 3

$\endgroup$
  • $\begingroup$ A (complex) plane wave is defined $e^{\mathbf {k}\cdot \mathbf{r}+\omega t}=\cos(\mathbf {k}\cdot \mathbf{r}+\omega t)+i\sin(\mathbf {k}\cdot \mathbf{r}+\omega t)$. Notice the bold font, which means that the position and momentum are vectors. Wikipedia is correct. A circular wave is something else. $\endgroup$ – sintetico Jun 1 at 12:04
  • $\begingroup$ sorry there is a typo in my comment. I mean $e^{i(\mathbf k \cdot \mathbf r + \omega t)}$ $\endgroup$ – sintetico Jun 1 at 12:14
  • $\begingroup$ I'm considering the amplitude at a given position. So you can ignore the $k \cdot r$ . If I understood correctly, $\cos \omega t + i \sin \omega t$ represents a rotating amplitude vector on YZ page with Real component pointing in Y axis and Imaginary component pointing in z axis. (according to the video youtube.com/watch?v=KKr91v7yLcM ). $\endgroup$ – NiKS001 Jun 1 at 12:15
  • $\begingroup$ why y and z axes? what is special about y and z? The wave function in this case is not a vector, it is a scalar. $\endgroup$ – sintetico Jun 1 at 12:17
  • $\begingroup$ where X Axis is the direction of wave propagation. $\endgroup$ – NiKS001 Jun 1 at 12:43
1
$\begingroup$

There's a misunderstanding what the word "plane" represents in the term "plane wave". A plane wave is a wave in which the surface of constant phase (wavefront) is a plane:

enter image description here

(image source)

What is shown as a circular thing that rotates for $e^{i\omega t}$ is the phasor that represents the value of the wavefunction at a given (single!) point of space. Phasors are used not only for quantum mechanical wavefunctions: this concept originated in the theory of electric circuits, and is also useful for treatment of other types of waves—even real-valued—e.g. electromagnetic.

What makes quantum mechanical wavefunction special is that it's not usually observable, only its absolute value is. But the effect of interference of quantum particles, like in the double-slit experiment, makes it necessary to introduce an additional parameter to capture this kind of effects. This parameter is the phase, and it's the thing that makes the phasor rotate in the animations you see in the resources on quantum mechanics.

Note that phasor is a vector not in the ordinary physical space: it's a vector in the complex plane, and it doesn't point to any direction in the real physical space, rather being a mathematical abstraction.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Can't the Two-slit experiment be explained with an amplitude vector perpendicular to the direction of motion and having a fixed direction? How are complex numbers necessary? $\endgroup$ – NiKS001 Jun 1 at 12:39
  • $\begingroup$ Side note: From what I understood, Phasors in circuits are only used to simplify the calculations of Voltage and current having a phase difference. And, only the real part of Phasor is considered at any given time t. Please, correct me if I'm wrong. $\endgroup$ – NiKS001 Jun 1 at 12:42
  • $\begingroup$ @NiKS001 you could of course introduce such a model. But you'd need some special way of computing the probability density from your wavefunction (which is not $\Re\psi$, but rather $|\psi|^2$). And then a question arises: in which direction is this wave polarized? You'll never answer this for the phase of quantum wavefunction, because it's a scalar, and however you rotate your apparatus, this (supposed) polarization doesn't change. (I'm ignoring spin here, which is a completely separate thing.) $\endgroup$ – Ruslan Jun 1 at 13:09
  • $\begingroup$ @NiKS001 "only used to simplify the calculations of Voltage and current having a phase difference" — not only. They are useful whenever your differential equation admits wave-like solutions. Maxwell's equations, for example, are also often written in complex form, although in plane-wave solutions electric and magnetic fields are in phase with each other. $\endgroup$ – Ruslan Jun 1 at 13:12
  • $\begingroup$ @NiKS001 There's a fundamental difference in how the solutions to differential equations are interpreted in electromagnetism vs. quantum mechanics. In electromagnetism, we assume that the real part of the solution of the differential equation corresponds to the physical voltage or current, since in this context, complex exponentials are tools to simplify the solution of differential equations. However, in quantum mechanics, we assume Born's rule, which tells us that the squared magnitude, not the real part, of the wavefunction corresponds to the physical probability density. $\endgroup$ – probably_someone Jun 1 at 13:52
0
$\begingroup$

Basically, the reason we use complex waves in quantum mechanics is because the mathematics is simplest when we do it that way. It is technically possible to formulate quantum mechanics using only real functions, but it is more complicated, and it gives exactly the same results (see the first page of Adler's book Quarternionic Quantum Field Theory for reference: https://projecteuclid.org/download/pdf_1/euclid.cmp/1104115172).

There are many reasons why it is simpler to use complex numbers. For example, differential equations are generally easier to solve using complex waves instead of real waves. Another reason is that the Fourier transform is cleaner with complex waves. The Fourier transform is very important because it relates the "position" of a particle to its "momentum" (although, in general, a single particle will be spread out over many "positions" and "momenta"). Another reason is that the mathematical operations that represent measurements of observables like energy, position, and momentum can be written in a simple form using complex numbers.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thanks for the pointer. Looks like I need to delve a little on Quantum Mechanics using real functions. $\endgroup$ – NiKS001 Jun 1 at 14:14
0
$\begingroup$

Remember, that when using rotating phasors to represent out of phase voltages in an AC circuit, only one component of the phasor has physical significance. It does not matter which you choose as long as you keep in mind that that is what you are doing. If working with phasors in the complex number plane, I would be more comfortable using the real components to represent my physical system.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.