25
$\begingroup$

I've read every thread on StackExchange (and Quora and reddit...) that I can find about a physical intuition for the phase in the quantum wave function, and I still Just. Don't. Get. It. (Yes, I've seen this thread--didn't help!)

As a jumping off point, I've been watching this terrific visualization of the quantum wave function. According to this video, for a particle in an infinite square well, the "phase" will rotate in the complex plane. Okay...what does that mean physically? Let's focus on just the ground state wave function. If it's not "rotating" in real space (right?), then what exactly is changing to make the phase "rotate"? If I could "see" the wave function with my eyes, what would I see?

I understand the mathematical argument that the phase doesn't matter: the complex exponential cancels out when you calculate the probability distribution, etc.

Maybe my confusion stems from a misunderstanding of what phase even is in quantum mechanics. When I visualize phase, I think of a sine wave and how much it has been shifted to the left or right (relative to some origin). But when I watch that visualization of the ground state wave function, nothing is sliding left or right, the wave isn't going anywhere. So what information does the phase encode here? I'm clearly missing something...

As background, I'm fairly new to quantum physics. I've always wanted to understand it beyond the usual "pop sci" descriptions, so I've been following the MIT OpenCourseware lectures on quantum physics. The physical nature of this phase is just really tripping me up and I've yet to find any explanation anywhere that goes beyond "well it works out in the math." Or is that all it is: just a convenient mathematical bookkeeping trick that physicists keep around because it happens to match observations?

$\endgroup$
  • $\begingroup$ physics.stackexchange.com/q/177588/23615 does this help a little? $\endgroup$ – Triatticus Jul 21 at 22:22
  • $\begingroup$ Have you read Feynman's little book QED? Or have you watched representations of phase as color? $\endgroup$ – Pieter Jul 21 at 22:51
  • 2
    $\begingroup$ Commenters, please keep in mind that comments should be used for suggesting improvements and requesting clarification on the question, not for answering. $\endgroup$ – David Z Jul 22 at 9:12
  • $\begingroup$ Hi @Pieter. It's been a while since I've read QED, but I do recall discussion of 'little clocks'. I'm just having trouble envisioning what those little clocks actually are. I have seen phase as color visualizations. They're very cool to watch, but I don't get what's physically changing in the wavefunction, if that makes sense. $\endgroup$ – QuantumFumbler Jul 23 at 15:24
  • $\begingroup$ @QuantumFumbler you may find the 'Aharonov–Bohm effect' and the 'principle of local gauge invariance' intersting reads. Though they might be a bit advanced for you at this stage, I presume, they illustrate how even though the phase of a wave function has no physical meaning, the relative phase does. $\endgroup$ – lineage Jul 25 at 1:11

13 Answers 13

12
$\begingroup$

While the "devil is in the details", as they say, the concept itself is actually much simpler than you think. There are two related ideas to consider - one is the phase ($\theta$ - just a real number), and the other one is a phase factor ($e^{i\theta}$ - a complex number). Now, I know you asked for "physical — not mathematical — intuition", but what I think you meant is that you're after something that is more tangible than the abstract equations, and not something necessarily physical. So here goes.

The phase $\theta$, in a very general sense, is really just an angle (or, at least that's one way to think about it, and one that happens to work well with the visualization you've posted). This is not quite the same as the overall phase of a wavefunction, but we'll get there.

enter image description here

In the image above, you see a point revolving around the unit circle. Here, the angle $\theta$, a.k.a. the phase, increases in what's, by convention, the positive direction. The projection of the point on each axis traces the cosine/sine function. This works the other way around as well, if you combine the cosine and the sine functions to describe the position of a point in 2D space in this particular way, you get a point going around a circle - and that's basically the Euler's formula ($e^{i\theta} = \cos \theta + i \sin \theta$).

Aside:

A complex number is, superficially, not unlike a 2D vector. Roughly speaking, what sets complex numbers apart is their properties (how they "behave", i.e. the kinds of operations possible with them, what they do, etc.) BTW, "real" and "imaginary" are just labels, like x & y, and they don't have any special meaning regarding reality of things.

If you multiply a complex number with a scalar, you can, well, scale it to any size. In other words, by setting the phase (the angle, a.k.a the argument) to some fixed value, and by scaling, you can get any complex number whatsoever (exponential version of the polar form: $z = Ae^{i\theta}$, with $A$ being the magnitude (the scaling factor)).

The more interesting thing is what happens in complex multiplication; more specifically, when you multiply by another complex number that is of unit length (i.e by one that lies on the unit circle), having the angle (argument) $\theta$. Such a multiplication results in a rotation of the first complex number by $\theta$.

In the visualization you've posted, there's essentially a "string" of complex numbers placed along a line, describing a wavefuntion in 1D space. That is, the "base space" is 1D, but to each point there's a complex number attached.

enter image description here

(Note that there's infinitely many of these arrows, but since that's hard to depict, a selection of representative arrows is shown.)

At this point, there are two kinds of phases we can talk about - the phase of each individual complex number, and the overall phase of the wavefunction. As for the complex numbers themselves, in this case they are all in phase (have the same angle). The phase of the wavefunction is just the overall "rotation" of the whole wavefunction about the central axis; here it is in a different phase:

enter image description here

"Let's focus on just the ground state wave function. If it's not "rotating" in real space (right?), then what exactly is changing to make the phase "rotate"? If I could "see" the wave function with my eyes, what would I see?"

It's not rotating in real space. Instead, you can think of it like this: each point in space has a complex value attached to it. You've seen this kind of thing before. E.g., with temperature, each point in space has a single real number attached to it, describing the temperature at that point; and these values change over time. With a gravitational field, each point in space has a vector attached to it. This is the same basic concept, except it's complex numbers, and the way they evolve in time over all that space is "wave-like" in some (formal and informal) sense. To "see" the wavefunction in 3D space "with your own eyes", you would have to have the sensory ability to independently sense/judge/estimate the size of the two components of the complex number at every point throughout 3D space. Imagine that at every point there's a little piece of paper with the complex plane depicted on it, and a little arrow drawn. Or, perhaps, a tiny digital screen displaying a 2D grid with a complex number drawn on it, that can be updated in real-time. The visualization you linked to limits itself to 1D physical space, and essentially uses the other two dimensions to represent the complex plane at each point. It's rotating by having all these arrows (complex numbers) rotate in sync - imagine the little screens updating in sync. For a more complicated situation, there would be some more complicated relationship between the arrows; e.g., the screens could update in some wave-like pattern.

Here's another screenshot from the video you've posted. The blue wavefunction is the superposition of the two others; that just means that the red and green arrows add up (pretty much like vectors) at each point, to form the blue arrows. I'm guessing that you already understand this, but just for clarity, the quantum state is just the blue wavefunctions (there aren't three sets of arrows rotating around, the other two are just shown as the "building blocks" of the blue one).

enter image description here

When the narrator says "when the phasors are in phase", he just means that the arrows of the two independent components have, in some region, roughly the same angle and point in the same direction, so they add up to a big arrow pointing in that same direction.

But, the wavefunction itself doesn't give you the probability, its square does. And that's what we're physically interested in. It is represented in the video by the solid green surface (it's the probability associated with the blue (superposed) wavefunction):

enter image description here

The probability itself at any point is just a real number; this green 3D surface is just a visualization aid. The probability is really the distance of the surface from the central axis (i.e., the radius of the cross section at a given point - that's why it's axially symmetrical).

Now, as you know, the thing about the overall phase (of the wavefunction) is that it doesn't affect the probability. If you freeze time, and rotate the whole thing, the probability distribution (the shape of this solid green surface) doesn't change at all (i.e., the relative relationships stay fixed, you're just spinning the axis, as if everything was glued to a stick that you're spinning between your fingers). That's why absolute phase has no meaning. In a sense, it's just an artifact of the particular mathematical formalism used.

Aside:
Sometimes, a mathematical description of something can be useful and can have desirable properties, but can still give you more than you need. E.g., you can use the mathematical formalism of vectors to describe directions in space - they are arrows, after all. However, many vectors describe the same direction, e.g. $(1, 0, 0)$ and $(5, 0, 0)$, and any $s(1, 0, 0)$, with $s$ being the scaling factor. Then you could say that the $s$ doesn't make a difference for the notion of a direction. But if you need to do things like vector addition (to combine directions for some reason), you have to be careful about $s$ or you may get wrong results - it's what software engineers would call a "leaky abstraction".

The back and forth "sloshing" of the probability distribution happens because the relative relationship of the constituent (red and green) wavefunctions changes in time (because they are rotating at different rates), so the overall shape of the superposed (blue) wavefunction changes, and in turn, so does its square. In other words, it's because the overall phases of the red and green wavefunctions change independently, at different rates, so the shape of the probability distribution dependes on the difference in the overall phases of the two.

"Maybe my confusion stems from a misunderstanding of what phase even is in quantum mechanics. When I visualize phase, I think of a sine wave and how much it has been shifted to the left or right (relative to some origin)."

I'd say that's the core of the issue; the phase is not how much it is shifted left or right, although it can often look like that. The phase is the overall rotation in the sense discussed above. If you disregard time evolution (stop time), you can describe the (overall) phase by picking a value of the wavefunction at some point to serve as a reference. If you then change the phase, you can get the phase difference by comparing the angular displacement of the phasor at the same point. This works well when you have a "wiggly" wavefunction. That way, you can talk about the phase with respect to some reference orientation.

Here's a trickier situation; this is the quantum version of the plane wave, and you can't really tell the difference between an overall phase change (rotation of the overall shape) and wave propagation just by looking at it:

enter image description here

The reason is that its mathematical formula is this (the minus sign is a matter of convention and is not important):

$$\Psi(r, t) = Ae^{i(f(\vec{r}) - g(t))}$$

with, $f(\vec{r})$ giving you the "local" phase of the phasor at point $\vec r$ (its orientation at $t=0$), and $g(t)$ providing a time-based offset from that (both are real-valued functions). The $-g(t)$ essentially works by offsetting the phases of each individual phasor from an "initial" value given by $f(\vec{r})$ for $\vec{r}$ (a particular point in space).

I know this can be confusing, but, again, one way to think about it is that the overall phase change is what happens if you freeze time and rotate the overall wavefunction around its axis, and then "resume playback". It's a mathematical thing - a quirk of the mathematical formalism, rather then something of physical significance. In isolation, a wavefunction with a different phase is technically (mathematically) not the same function, but it is the same physical state that gives you the same probability distribution; there's a redundancy to the mathematical description.

Aside: The more standard version of the above formula is
$$\Psi(r, t) = Ae^{i(\vec{k}\vec{r} - \omega t)}$$

Finally, let me wrap up by coming back to the idea of a phase factor. A change to the overall phase in the above case can be described like this - you just rotate everything by some angle $\alpha$:

$$Ae^{i(f(\vec{r}) - g(t) + \alpha)}$$

Now, due to properties of exponentiation, this is the same as

$$e^{i\alpha}Ae^{i(f(\vec{r}) - g(t))}$$

In other words, rotating everything by $\alpha$ is the same as multiplying by a unit-length complex number $e^{i\alpha}$ (called the phase factor). It's just another way to write things down, one that exploits the properties of complex multiplication.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ This is a really wonderful and patient explanation, thank you. Lots of food for thought here. I never realized that a rotation of the phase ends up looking just like propagation of the wave, for example. I have a feeling I'll be reading this a few more times through... $\endgroup$ – QuantumFumbler Jul 24 at 19:49
24
$\begingroup$

The reason that the overall phase of a wavefunction is hard to gain to intuition for is that it's really not real, in the sense that it's just an artifact of a particular choice of quantum formalism and doesn't appear - even purely mathematically - in other formalisms.

In particular, in the density-matrix and projective-space formalisms of quantum mechanics (which are for some purposes more useful than the state-vector formalism), you don't have the freedom to multiply the overall state by an unobservable phase factor. In these formalisms, the states $|\psi\rangle$ and $e^{i \theta} |\psi\rangle$ are represented by literally the exact same mathematical object (a rank-one projection operator or a projective ray, respectively). And energy eigenstates don't pick up an unobservable phase under time evolution, but instead remain completely unchanged.

The reason that you're having a hard time picturing the overall phase of a quantum state is that there is truly nothing to picture.

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ So phase is just a tool that works for a particular way of describing QM mathematically? I'll have to take a look at those formalisms you mention. Perhaps seeing QM described in different ways will help me wrap my brain around this... $\endgroup$ – QuantumFumbler Jul 23 at 15:40
  • 1
    $\begingroup$ Physics uses mathematics to describe the physical world. In Newtonian physics, the mathematics makes an intuitive connection to the physical world, so it seems obvious what position, velocity, and acceleration mean. However, mathematics is a tool. Going forward in history, physics accumulates more abstract mathematics for description. In modern physics, 1900 until today, the mathematics is more obviously a tool. Quantum mechanics employs functions in complex space to encode information, from which we calculate physical quantities. You can also use matrices, etc. to encode the information. $\endgroup$ – wdb Jul 23 at 17:38
  • $\begingroup$ @QuantumFumbler Correct. $\endgroup$ – tparker Jul 24 at 3:05
  • $\begingroup$ The whole of QM is just a bunch of math that empirically seems to describe the world. There are many quotes that describe this. It is better to just smile and nod...No one has intuition re QM. $\endgroup$ – Aron Jul 24 at 9:33
  • 3
    $\begingroup$ This strikes me as a very odd point of view. Certainly we understand that subsystems can have relative phases. And unless our system is the entire universe, we certainly understand that our system could be considered a subsystem of a larger system. Therefore, we must understand that it could have a phase in its entirety relative to some other subsystem of a greater system. That we can sleight-of-hand the global phase away by imagining our system to be its entire universe merely obscures these facts rather than enlightening some hidden truth. $\endgroup$ – Xerxes Jul 24 at 16:45
22
$\begingroup$

Visualizing the phase

There are different tricks to visualize the phase. The one you hint at in your answer, amounts to taking the real (or imaginary) part of the wavefunction. For instance, ${\rm Re}[e^{i (k x + \phi_0)}]=\cos(kx + \phi_0)$, which is a sinusoid. The phase determines the location and distance between the crests and troughs. Another trick is to use color -- if you google "complex plane color" you will find many pictures and explanations of this way of looking at this. Yet another visualization is to imagine a little clock face at every point in space, and the position of the minute hand on the clock tracks the phase of the wavefunction. Keep in mind that all of these are simply different representations and are not "what the phase really is"; it's useful to know understand several different ways to visualize the phase, and use the one that is the most convenient or insightful in a particular instance.

Note added, thanks to jgerber's comment: There are some very nice visulations here: vqm.uni-graz.at

Consequence of the phase: interference

The fact that complex probability amplitudes, rather than real-valued probabilities, add in quantum mechanics leads to interference effects. Here we can think about the classic double slit experiment.

Classically, the probability that the particle goes through one slit simply adds to the probability that the particle goes through the other slit. Since probabilities are positive, there are no points with zero probability on the screen which catches the particles.

On the other hand, in quantum mechanics, there are points on the screen where the probability amplitude to go through slit 1 is $a$, and the probability amplitude to go through slit 2 differs by a phase, $e^{i \pi}a=-a$, so the sum of the probability amplitudes is zero and there is zero probability to find the particle at this location. At other locations on the screen, the relative phase will be +1, and the probability amplitudes will add constructively, leading to a "large" probability to find the particle at these locations. In general, the relative phase of the probability amplitude in passing through the two slits is what determines the shape of the interference pattern.

Variations of the phase encode physical information

In general, the phase of the wavefunction $\Psi$ appears in the definition of the "probability current", $\vec{j}\propto i (\Psi^\star \nabla \Psi - \Psi \nabla \Psi^\star)$. A state which is purely real has a vanishing probability current, which is easy to see from the definition. Thus having a non-zero phase allows the state to have a non-zero current, and thus allows the state to change.

We can give this a more direct meaning in some special cases, where the phase of the wavefunction (or more exactly the derivative of the phase with respect to some parameter) encodes an observable quantity. (This is related to the idea of action angle variables in classical mechanics).

This is exactly true when eigenvalue problems for the wave function $\Psi$ of the following form appear \begin{equation} i \frac{\partial \Psi}{\partial z} = \lambda \Psi \end{equation} where $z$ is some parameter and $\lambda$ is an associated eigenvalue. Note that if we write $\Psi=A e^{i \phi}$, where $A$ is a constant, then the above equation can be written \begin{equation} -\frac{\partial \phi}{\partial z} = \lambda \end{equation} If $i \partial/\partial z$ is an operator, and $\lambda$ is a possible measurable outcome of $\lambda$, then the above equation says that the variation of the phase with respect to $z$ gives us the observable value $\lambda$.

There are many examples of type of equation:

  • If $z$ is position then $\lambda$ is the momentum; for plane wave states, the derivative of the phase with respect to position is the momentum.
  • If $z$ is momentum then $\lambda$ is the position.
  • If $z$ is time then $\lambda$ is the energy.
  • If $z$ is the azimuthal angle (angle in the $x-y$ plane) then $\lambda$ is the component of the angular momentum parallel to the $z$ axis.

Furthermore, this kind of equation shows up in an approximate form in the WKB approximation. Then, there is a similar type of relation between, say, the derivative of the phase with respect to position, and the momentum, will approximately hold.

Having said that, don't go overboard; the interpretation is a little bit fuzzier for states which are superpositions of eigenstates, and not all observables have the form $i \partial/\partial z$. However, this seed of intuition is useful to keep in mind when considering more complicated observables, which have eigenfunctions more complicated than a plane wave.

TL;DR

  • There are many tricks for visualizing complex numbers -- learn several.
  • A non-zero phase means the wavefunction is complex, which:
    • allows interference phenomena to occur,
    • means the probability current is non-zero and therefore means the wavefunction can change in time.
  • The "more wiggly" the state is with respect to position, the more momentum it has. The "wiggliness" is encoded in the phase. Similar "wiggliness" relationships exist between energy and time, and angle and angular momentum.
| cite | improve this answer | |
$\endgroup$
  • 4
    $\begingroup$ Readers may enjoy the visualizations found here: vqm.uni-graz.at and in the books referenced therein. These utilize the colormap method of visualizing complex numbers which is by far my favorite. As pointed out in the OP and the nice answer by @Andrew, visualization doesn't mean you have physical intuition but it can help a lot. For single particle real space wavefunctions the phase tells you something about the flow of the wavefunction through space and for multiple particles the phase is related to interference phenomena. $\endgroup$ – jgerber Jul 22 at 0:08
  • $\begingroup$ Absolutely, having a good visualization is very important. The main message I wanted to convey that any visualization is a representation and there are multiple possible representations, so it's better to learn many of them than to treat one as "the truth." Since sometimes comments get deleted, I added a link to the site you recommended to my answer (attributing you for pointing it out). $\endgroup$ – Andrew Jul 22 at 0:14
  • $\begingroup$ Thanks for the link to those visualizations, @jgerber. Unfortunately the movies don't work in any of my browsers on any of my devices! But I found this applet that I suspect has similar visualizations: phet.colorado.edu/sims/cheerpj/bound-states/latest/…. I'm focused on trying to understand just the ground state in a square well. The phase keeps changing color over time while the magnitude at each location remains constant. What I'm stuck on is: what is physically changing in the wave? It's not 'moving', it's not rotating in real space... $\endgroup$ – QuantumFumbler Jul 23 at 15:31
  • 1
    $\begingroup$ @QuantumFumbler "The phase... is just a convenient way of encoding information" -- You could make the same kind of point about any mathematical formulation of physics. All of our theories are (typically approximate) representations of some underlying reality. Nevertheless, within the mathematical framework of quantum mechanics, the phase encodes an enormous amount of information (interference, observables...), and is really what makes quantum mechanics work. $\endgroup$ – Andrew Jul 23 at 18:28
  • 1
    $\begingroup$ I think maybe a pitfall here is to try and connect the phase of the wavefunction directly to something you are more familiar with from your intuition dealing with macroscopic objects. Ultimately ,I think this is a mistake. Our brains have not evolved to have the right intuitions, which is not surprising given that we don't deal with quantum phenomena every day. I think basically the best we can do is accept that there is this strange quantity called the phase of the wavefunction, which encodes many things we want to know about the state of a system. $\endgroup$ – Andrew Jul 23 at 18:31
7
$\begingroup$

I love the pictures from the book of Feynman "QED: The Strange Theory of Light and Matter". There, the phase of a wavefunction traveling though the space is depicted as a clock with a moving hand: when the hand makes a full round it corresponds to $2 \pi$ of the complex phase.

This can be easily used to explain the effect of interference: different paths between A and B lead to different travel times and thus to different final positions of the clock hand. Now, the hands obtained from all paths are added like vectors (clock hands look like arrows!). Similar angles add up constructively, opposite angles add up destructively.

The book contains many more examples based on this brilliant visualization idea.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ It's been a while since I've read QED, but I do remember that description. It's wonderful, simple, and classic Feynman! I'm having trouble picturing what the clock physically, actually is, though from reading other responses, I'm suspecting that it's just a mathematical convenience and not something "real" $\endgroup$ – QuantumFumbler Jul 23 at 15:42
  • $\begingroup$ Od course, the clocks are not real. They just visualize the changing phase along a path. Finally, instead of adding up complex numbers one deals here with vectors (represented by clock hands). $\endgroup$ – Nikodem Jul 23 at 15:52
  • $\begingroup$ @QuantumFumbler But nothing is “real”, as I understand you’re using the word. You don’t have direct access to observe what the universe is really doing. All you can do is interpret the interactions of one quantum system (your brain) with other quantum systems (everything else). Your brain has evolved to understand some of those interactions intuitively, but that doesn’t make them more real. $\endgroup$ – Mike Scott Jul 24 at 8:41
  • $\begingroup$ @QuantumFumblern : Of course the phase is real. Distribute the wavefunction along a circular track. If the change in clocks around one orbit isn't a multiple of $2\pi$, the corresponding momentum isn't observed. (I've heard this reduced to the imprecise "either the phases interfere to give net probability zero or they are in-phase and give positive probability".) Phase forces a consistency condition in systems with "loops" (more precisely, nontrivial homotopy). Loops need not be in space, e.g., time crystal. $\endgroup$ – Eric Towers Jul 24 at 21:35
5
$\begingroup$

Since you asked for non-mathematical intuition for the phase, I'd say that a classical wave is indeed a good analogy except for the fact that the global phase---as opposed to the relative phase---between two modes is completely indeterminate. Phenomena such as as interference between two modes are fully explained by visualizing them as classical waves that add up or cancel out, with the only caveat that the starting point of the "wiggling" for any of the individual modes isn't possible to determine.

Consider for example a Mach-Zehnder interferometer where only a single photon is fed in at one of the modes. Before it enters the first beam splitter, its (global) phase is unknown. However, as soon as it delocalizes after the first beam splitter, the relative phase between the two arms of the interferometer is well defined and fully accounts for which output mode the photon is likely to re-emerge from.

As a simple exercise, you could try to calculate how interaction-free measurements come about. You'll see that the classical wave picture is perfectly adequate to explain what happens between the two beam splitters, including the precise visualization of the relative phase, but it does not say anything about the global one outside the interferometer.

EDIT:

In a nutshell, if the two arms of the interferometer make up the two modes, you can start with a photon in a single mode only, i.e., the input state could be $e^{i\theta}\mid10\rangle$ where $\theta$ is the (unknowable) global phase. After the first 50/50 beam splitter, we end up with a "delocalized" photon $\frac{1}{\sqrt{2}}\left(\mid10\rangle+e^{i\phi}\mid01\rangle \right)$ where $\phi$ is the relative phase between the two arms. After you pass the second 50/50 beam splitter, you get a probability of $\frac{1}{2} \left( 1+\cos\phi \right)$, hence the "discretized" interference pattern between the two modes modulated by $\phi$. As you can see, nowhere in this discussion has the global phase entered---and that's the main difference with classical waves. (The addition of an obstacle in one of the arms, as in the interaction-free experiment, is just a bonus scenario to better understand the concept of phase.)

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Could you sketch out in more detail how to do (and how to read your interpretation from) that proposed exercise? $\endgroup$ – benjimin Jul 23 at 8:24
  • $\begingroup$ @benjimin I've added an edit above. (For the full derivation, you can find equations on how beam splitters work in most introductory quantum optics textbooks.) The takeaway is that only the relative phase is physically observable and can give rise to interference. The global phase, though obvious from classical mechanics, is irrelevant---and I'd dare to say, unknowable. $\endgroup$ – Tfovid Jul 23 at 11:49
  • $\begingroup$ I think you've hit on what I suspect is my main stumbling block: I keep picturing a classical wave, like a wave on a rope, and trying to map it to QM. But it sounds like QM waves, while having somethings in common with classical waves, are just a different beast entirely. I'll have to take some time and dig into the M-Z interferometer a bit more, thanks for that reference. $\endgroup$ – QuantumFumbler Jul 23 at 15:47
  • 1
    $\begingroup$ @QuantumFumbler The first part of this video touches on the notion of quantum phase and its relation to the complexity of the wave function: [youtube.com/watch?v=V5kgruUjVBs] . The visualization of minute 4:21 was somewhat of an aha-moment for me as well. $\endgroup$ – Tfovid Aug 9 at 10:45
2
$\begingroup$

The short answer is no. No one can give a physical intuition for the meaning of phase, because it has no physical meaning. To clarify this, it may help to recognise that the wave function of a plane wave state in quantum mechanics is not simply a sin wave. It is a helix in the complex plane.

enter image description here

When the helix rotates (as it does under time evolution), the apparent waves on the real and imaginary parts appear to move like waves. But the rotation does not take place in physical space. It takes place only mathematically in a complex configuration space, and the absolute angular position of the helix has no physical meaning.

Wave functions are better called probability amplitudes. They are not real, but are part of the calculation of probabilities for measurement results. The reason we need them is buried deep in the mathematical foundations of quantum mechanics and is not covered in undergraduate (or even most graduate level) courses of quantum theory, which are concerned with applications not conceptual foundations. They are solutions of the Schrodinger equation, and it can be shown that the general form of the the Schrodinger equation is required in a probability theory for indeterminate processes, as distinct from classical probability theory in which outcomes are determined by unknown variables.

(figure from The Mathematics of Gravity and Quanta)

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Do you know where that figure is from? Would love to dig into that text a bit. $\endgroup$ – QuantumFumbler Jul 31 at 19:45
  • $\begingroup$ The figure was from my third book. I have added a link. This is a rigorous mathematical treatment. My second book (see profile) has the same figure, and gives a conceptual treatment, without mathematics. $\endgroup$ – Charles Francis Jul 31 at 20:00
1
$\begingroup$

Besides what Andrew already said, I want to add that you cannot expect to have human intuition about every physically meaningful concept or quantity or property. For example, you cannot have human intuition about the energy states of an electron in a hydrogen atom; it just falls out of the mathematics. That said, you can think of phase of a wavefunction as analogous to the phase of the electric field vector in the classical description of a light beam. It is there, and may rotate over time (as in circularly polarized light), and can be observed using detectors that are specifically designed to react differently to different polarization (such as a polarizing filter followed by a light detector).

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

I understand the mathematical argument that the phase doesn't matter: the complex exponential cancels out when you calculate the probability distribution, etc.

This is not true always, as you've seen in this thread, when the particle is in superposition of two or more wavefunctions.

Maybe my confusion stems from a misunderstanding of what phase even is in quantum mechanics. When I visualize phase, I think of a sine wave and how much it has been shifted to the left or right (relative to some origin).So what information does the phase encode here? I'm clearly missing something...

This shift can be also determined for a $e^{-i(kx -\omega t)}$ which is independent of the imaginary part and it can be used to calculate the group velocity and phase velocity of the wavefunction. And as many other have stated the physical meaning of the phase is not so important, even some wavefunctions are imaginary, the most important part is the physical meaning they represent.

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

That's because the absolute phase is not physically meaningful.

It only acquires any meaning when compared to a reference, another path that also has a phase, for instance another path from a beamsplitter, or reflected from a boundary. Then the two paths can interfere, constructively or destructively, as dictated by their phase difference.

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

tl;dr You can think of how sounds can combine to form louder sounds or cancel out. Whether sounds add or cancel depends on their phase.


Phase is a more general mathematical concept. It comes up in all sorts of stuff beyond quantum-mechanics, e.g. sounds, electric-current, and radio-waves. Probably easier to understand the concept itself outside of quantum-mechanics.


Example: Phase determines if sound-waves add or cancel.

Say you set up a microphone that listens for sounds and then replays them at the same volume on a speaker. Does this result in louder sounds or noise-cancellation?

Depends on the phase!:

  • If the sound-waves are lined up to peak together, the sounds'll add.

  • If the sound-waves have their peaks in opposition, then they'll tend to cancel out. Which is how active noise-cancellation works.


Some observations...

Contrary to the above descriptions, sound-waves don't actually have phases in an absolute sense. We can just talk about sound-waves as though they had phases as one mode of thinking about them.

If we do choose to think of two sets of sound-waves as having phases, then the big thing is that we correctly consider their phases relative to each other. But if we choose to say that one's in the default-phase while the other's in an anti-phase or something, well, that'd be pretty arbitrary.

Sound-waves were probably the simplest example. Once you feel comfortable with phases with respect to them, the next analogy to consider might be alternating electric current (AC electricity), e.g. in three-phase electric systems.

As for math, probably best to start by learning about the Fourier transform. The gist is to redescribe things in repeating signals of different frequencies (often described as the frequency domain). In plain English, this basically just means that instead of directly describing the value of a function, we instead describe the function as a bunch of waves that add up to that value. For example, instead of describing a sound in terms of its pressure at any given moment at each point in space, we might describe it as a sum of various sound-waves with their own frequencies and amplitudes.

As for the connection to quantum-mechanics, it'd all go back to the double-slit experiment:

The experiment belongs to a general class of "double path" experiments, in which a wave is split into two separate waves that later combine into a single wave. Changes in the path lengths of both waves result in a phase shift, creating an interference pattern.

"Double-slit experiment", Wikipedia

One may've expected light-sources to add up, sorta like how one might assume that adding more sound always creates a louder noise (rather than noise-cancellation), but historically they were surprised by the interference pattern that heavily implicated wave mechanics, with a connection to the concept of phase as in the quote above.


In short..

When combined, waves can add or cancel. They add when they largely share the same phase, while they can cancel if they're anti-phased.

So "phase" describes how a wave'll interact with others when combined.

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

The OP clarified:

It's been a while since I've read QED, but I do recall discussion of 'little clocks'. I'm just having trouble envisioning what those little clocks actually are. I have seen phase as color visualizations. They're very cool to watch, but I don't get what's physically changing in the wavefunction, if that makes sense.

I liked Feynman's way of explaining that. I don't have the book here but as I remember he explains that the dials are just a way to describe they behavior of particles in the quantum world. There is no way to say what those little clocks "actually are". Feynman deliberately chose a very unphysical representation. In the video that the OP linked to these arrows are turning. It is difficult to make such movies, and I find it hard to watch them.

In the same way one cannot say what is "physically changing" in the color visualizations. It is much easier to make such movies and they can also help create intuitions about the behavior of wave functions.

Both these visualisations represent the phase factor $e^{iEt/\hbar}$ but that is also just a mathematical description of what is observed: diffraction and interference. The empirical observation is that particles behave like waves at the quantum scale, can interfere destructively. We are not used to that in our everyday experience. That is why quantum mechanics is weird.

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

Consider Zeno's paradoxes in light of the Einstein's thought experiment of the two lightnings hitting a train. The "paradoxes" show that we can't really talk about objects outside the context of observation and the problems that arise if we postulate existence of object outside of an observer's frame of reference and thus not subjected to the mechanics of perception.

To understand perceived objects we must understand how perception works because anything we think about as a real object is included in the scope of perception, is perceived and is absolutely subjected to it's mechanics.

Zeno's paradoxes are given in terms of what is seen or otherwise perceived, such as a race seen by the eye. When we observe a race where some fast moving object passes by [overtakes] a slower moving object at some point we can't tell who is ahead but after that point the faster moving object is perceived to be ahead, measurement is a matter of precision but is likewise an act of observation.

As I understand it, the wave function is a predictive model describing a to-be-perceived collapse of the function rather than an object's non perceived state.

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

The short answer is that nobody really knows. To understand what the pahse physically means would required a detailed physical understanding of the wave function, that is, of quantum mechanics. Alas, it is not known why point particles such as electrons are described, in a probabilistic manner, by - complex - wave functions. We only understand the phase in terms of the wave function. If that is enough for you then the answers above should be sufficient.

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

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