18
$\begingroup$

In thinking how to ask this question (somewhat) succinctly, I keep coming back to a Microwave Oven.

A Microwave Oven has a grid of holes over the window specifically designed to be smaller in diameter than the wavelength of the microwaves it produces, yet larger than the wavelengths of the visible light spectrum - this is so you can watch your food being heated without getting an eye full of microwaves.

The "realness" of electromagnetic waves seems indisputable - both from the microwave example above, and also because if I want to broadcast a radio wave with a certain wavelength, then I need to make sure I have an antenna of corresponding length to produce the wave I'm looking for. Furthermore, we discuss and treat these waves as real, measurable "objects" that exist and can be manipulated.

Now, if I want to describe the behavior of my Microwave Oven in the framework of QM (let's pretend my oven is going to only produce 1 photon of energy corresponding to a wavelength of the microwave spectrum for simplicity) then I'll describe the behavior of that photon as a wavefunction that evolves over time and gives a probability distribution within my microwave that similarly does not allow the photon to pass through the safety grid and exit the oven cavity giving me a retinal burn.

The difference is, the wavefunction is never treated as something "real" in this description. When the safety grid is described as working to protect you because it has holes smaller than the wavelength of the classical waves it's blocking, this is a useful description that seems to describe "real" objects/be-ables. While it is possible to describe why an individual photon has low probability of passing through the same grid but extended physical properties such as wavelength (in space) are treated as non-real because we're dealing with a point particle and with behavior described by something we also treat as non-real (the wavefunction); it seems unclear to me why we insist this wave function which predicts behavior of physical measurements so well is somehow "non-real."

Put another way, if we have no problem treating EM waves as "Real," then why do we insist on treating the wavefunction that describes the same behavior as "unreal?"

I understand there is recent research (Eric Cavalcanti and his group for one) trying to argue this point, but as every respectable physics professor, I've ever encountered, has treated the wavefunction as an indisputably non-real mathematical tool, I needed to ask this community for an answer.

$\endgroup$
  • $\begingroup$ see eg toy models of QM $\endgroup$ – vzn Feb 4 '15 at 16:05
  • 1
    $\begingroup$ Having read the answers I would summarize with " the the QM wavefunction does not correspond to a field, in the way the classical electromagnetic wavefunction does (E,B)". field in the definition for physics en.wikipedia.org/wiki/Field_%28physics%29 "A field is a physical quantity that has a value for each point in space and time.". physical here means real number. Psi's are complex. $\endgroup$ – anna v Feb 5 '15 at 5:44
  • $\begingroup$ First you need to define "real". $\endgroup$ – Hot Licks Feb 5 '15 at 17:43
  • $\begingroup$ @annav I understand the real number argument, however a classical EM field can only be measured by an instrument, which reacts in a QM way with the field, and can be described too in QFT as interaction mediated by virtual photons - the interaction of field charge and measuring device can be described by a wave function, making the "object" of measurement in this case (what we thought was a classical field E) seem just as "unreal" as the wave function we treat as a mathematical fiction. My point is either they're both unreal, or they're both real, but I see no reason to declare one special $\endgroup$ – JPattarini Feb 5 '15 at 18:08
  • 2
    $\begingroup$ You are mixing up two frameworks in this argument. The electric fields are classical and macroscopic. Classical electrodynamics works to high accuracy . You could go to the details of how classical fields emerge from photon wavefunctions but it is a collective emergence, analogous to the way temperature emerges from average kinetic energies: real numbers are involved in averages. You would have to get the "psi*psi" from the individual photon entries to define what we measure as an electric field macroscopically, imo. $\endgroup$ – anna v Feb 5 '15 at 19:45
17
$\begingroup$

When dealing with a single quantum mechanical particle, both the wavefunction and the electric field appear to belong to the familiar class of "fields", both $\mathbf{E}(x)$ or $\psi(x)$. This analogy completely breaks down when you consider multiple particles, in which case the wavefunction depends on all of the particle coordinates, i.e. $\psi(x_1,x_2,x_3,\ldots,x_N)$. This is totally different from the behaviour of the "physical fields" such as the electromagnetic fields, which can be described by a function of a single coordinate, no matter how many particles one has in the system. These physical fields are included in our theories precisely because they enable us to describe physics in a local way. On the other hand, taking the object $\psi(x_1,x_2,x_3,\ldots,x_N)$ to be a physical field, which now propagates in a $3N$-dimensional configuration space, leads to a grossly non-local description that is philosophically abhorrent to many physicists.

$\endgroup$
  • $\begingroup$ At least one answer which conveys an easy to understand important point. $\endgroup$ – Thomas Klimpel Feb 4 '15 at 19:40
  • $\begingroup$ @Mark I think I've misunderstood you so please correct my misunderstanding: If I have a setup that generates photons in a state Y, say a standard double slit experiment, then the wave function for the system describing the expected coordinates doesn't describe the results of a SINGLE run of the experiment (though I can predict that the photon will hit my plate at one of the maxima of the squared probability function) but is instead a summation of an infinite number of runs - i.e. I only see the predicted interference pattern after many runs. This seems to run counter to what you're saying. $\endgroup$ – JPattarini Feb 4 '15 at 21:07
  • $\begingroup$ @JamesPattarini I think we are talking about different things. First, about your setup: the wavefunction does describe the state of each photon, but it gives only a probabilistic description. In order to reconstruct the wavefunction you therefore need to do many experiments on identically prepared photons, which are each individually described by the same single-particle wavefunction. $\endgroup$ – Mark Mitchison Feb 5 '15 at 0:05
  • 2
    $\begingroup$ @vzn I believe what you have just described is a local hidden variable theory, which has already been entirely ruled out by experiments. Bohmian mechanics is not at all close to doing what you ask, since there the wavefunction is explicitly a "physical field" living in $3N$-dimensional configuration space. Both Bohm and Bell went to lengths to point this out. Bell wrote that the advantage of Bohm's theory was that it made the non-locality of $\psi$ manifestly obvious. I believe that the extreme non-locality of Bohm's theory was one of Bell's inspirations for his subsequent no-go theorem. $\endgroup$ – Mark Mitchison Feb 5 '15 at 9:06
  • 1
    $\begingroup$ (The extreme non-locality of Bohm's theory was certainly the inspiration for my answer!) $\endgroup$ – Mark Mitchison Feb 5 '15 at 9:10
14
$\begingroup$

You can, in principle, measure the electric and magnetic field strength at every point in space and time. Thus, the EM field is real in the the sense that its value can be determined uniquely by measurements, and thus, also excitations of it - the EM waves - are real.

You cannot, in principle, measure the wavefunction at any point. The true quantum state $\lvert \psi \rangle$, and hence the complete wavefunction $\psi(x)$, is inaccessible to experiment even in principle, since there is at least a global phase we cannot determine. Also, even the probability $\lvert \psi(x) \rvert^2$ cannot really be measured for a single state - you need to prepare and ensemble of equal states, and then you can go about and approximately reconstruct the probability density out of the myriads of measurements, but there is no device that measures the values $\psi(x)$ or even $\lvert \psi(x)^2 \rvert$ exactly. (Exact here not in the sense of "without experimental errors" but in the sense of there the isn't even an idealized, error-free device that could exactly determine $\psi(x)$ of a single state (or even by finitely many measurements on finitely many state of the same kind), while, for the EM field, it's rather easy in principle - just take test charges and measure the forces exacted upon them. Thus, the wavefunction is not real in the sense that it is not an emprically accessible property of a single state.

$\endgroup$
  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – David Z Feb 9 '15 at 7:57
3
$\begingroup$

There is no correct criticism of treating the wave function as real if by that you mean treating the equations of motion of quantum systems as if they describe how those systems actually work. There are many criticisms of treating the wave function as real that are no good: some examples follow.

If quantum mechanics is a correct description of how macroscopic systems work, then it implies the existence of multiple versions of every object you see around you. For example, every time you measure the radiation from a decaying atom in an unsharp state, there would be one version of the detector that goes off and another that doesn't. The criticism runs as follow: we don't see such alternate versions of those objects, so they don't exist, so quantum mechanics is broken and we will just treat it as a But this argument is garbage because it doesn't work out what we would expect to see if those other versions of the detector did exist. In reality, we would expect to see just one version because the other versions cannot interact with one another as a result of decoherence, see

http://arxiv.org/abs/quant-ph/0306072.

The critic then switches to a different argument. "Well, we can't see those other versions of the detector, so it is extravagant to postulate their existence". This argument is silly for two reasons. First, if we applied this standard consistently then we would have to say that the core of the sun doesn't exist since nobody has seen it. And nobody has ever seen a dinosaur, only a dinosaur, only dinosaur skeletons, so then dinosaurs don't exist, right? But there is another much worse problem with this argument. The mere fact that you can't directly detect the existence of the other version of the detector doesn't imply that it plays no role in explanations of experimental results. For example, in general the results of experiments on entangled systems can only be explained by realising that decoherent systems can carry quantum information that can't be detected by measurements on that system alone:

http://arxiv.org/abs/quant-ph/9906007.

"But if we apply the equations of quantum mechanics to all systems then quantum mechanics is deterministic and doesn't predict probabilities," might be the next criticism. The only trouble is that the Born rule has been explained by taking quantum mechanics as true, see:

http://arxiv.org/abs/quant-ph/9906015

http://arxiv.org/abs/0906.2718

http://arxiv.org/abs/quant-ph/0405161.

You might be interested in "The Fabric of Reality" and "The Beginning of Infinity" by David Deutsch.

$\endgroup$
  • 1
    $\begingroup$ I incline to believe that the w.f. is a real thing, just because the QM works extremely well. As to single photons, it's not on the no. of photons in a wave-packet that the reality of the w.f. depends. The problem of the wave-function is why we speak of a wave, but get a single point on the screen, how do the entanglements work the non-locality. I.e. the doubts stem from there where the QM cannot offer explanations. $\endgroup$ – Sofia Feb 4 '15 at 17:29
  • $\begingroup$ I answered those objections in the first and second paragraphs respectively. $\endgroup$ – alanf Feb 4 '15 at 17:49
  • $\begingroup$ The point is that the wave function isn't an observable of any quantum theory -- it is only observable up to a non-observable phase. $\endgroup$ – Jerry Schirmer Feb 5 '15 at 3:11
  • $\begingroup$ @Jerry: I don't think that is the point. If it were, Bell type experiments would be explained as described above. $\endgroup$ – alanf Feb 5 '15 at 9:57
3
$\begingroup$

To me, the whole subject feels somewhat like a comedy of errors, and your question reveals some misconceptions. Hopefully, I'll be able to clear up some of them without just spreading my own misconceptions ;)

Let's start off with de Broglie, who wanted to represent matter as physical waves. But when quantum mechanics emerged, we did not end up with matter waves on top of spacetime, but wavefunctions living in phase space. They represent the state of the system and are not physical fields.

De Broglie did not like that, and what he came up with when he tackled this issue for a second time was the idea of the double solution: He wanted to represent particles as singular solutions of physical fields (or possibly solitons - I'm not sure if that was his idea or just my interpretation, and I'd have to make a trip to the library to find out). Wavefunctions would be secondary solutions that describe the motion of these singularities.

What happened, though, is quantum field theory, which, personally, I failed to 'get' at first. Take the Klein-Gordon equation. Its solutions are not wavefunctions at all - instead, they represent classical configurations of the theory. We can make a proper quantum theory out of it if we take these classical solutions and use them to construct a Hilbert space. After choice of basis of that space, wavefunctions emerge again when we expand arbitrary superpositions.

So the solution of the Klein-Gordon equation at a time $t=t_0$ is not a single-particle wave function - instead, it's the system's configuration, ie the equivalent of a position eigenstate $x=x_0$ in quantum mechanics.

The traditional approach via second quantization, Fock space and field operators paints a somewhat different picture, but we again have a clear distinction between physical fields represented by field operators on one side and quantum states on the other side.

On top of all this, we still have to tackle open questions about the 'reality' of the 'non-physical' wave function: Are quantum states objective or subjective, ie do they represent the state of the system or just our incomplete knowledge of its state? Does quantum theory describe how nature really works, or is it more like thermodynamics - an effective theory that emerges due to large number effects (there are lots of powers of 10 until we reach the Planck scale, and who's to say that's the end of it)?

If anyone has good answers to these questions, feel free to tell me about them ;)

$\endgroup$
3
$\begingroup$

If the wave-function is a real thing or not, that doesn't depend on how many particles you put in a single wave-packet. If the wave-packet is of very low intensity, you should produce many copies of it. If the wave-function is a real thing, so it is for a thousand particles in a wave-packet, or for a single one particle.

If the wave-function is a real-thing or not, there is a big dispute, and if you'll ask scientists you may have answers in favor and answers in disfavor. Though, when we do measurements on quantum systems, there is something running through our apparatuses. The question is only whether the wave-function that we obtain from our equations (Schrodinger, Dirac, etc.) are a faithful description of what runs in our apparatuses, or not so faithful. And to this issue it is hard to answer, because when we measure these quantum particles, they are so frail that we disturb them.

But it's not true that "extended physical properties such as wavelength (in space) are treated as non-real because we're dealing with a point particle".

I'll give you an example: imagine that we produce a beam of photons, what we call "a wave-packet". This wave-packet you can detect, especially if you produce it repeatedly and measure each time, you can end with details of its length or time-duration, most probable wave-length, etc.

But now, assume that you can control the intensity of the source and reduce the intensity more and more, until you can say that in each wave-packet exiting your source there is, on average, one photon. Well, the properties of the wave-packet will remain the same. Your photon will have the same most-probable wave-length, length, etc.

If my words "most-probable wave-length" raise a question mark, then yes, a photon is not necessarily of a precise wave-length, its properties depend on the preparation procedure, of how we prepare the wave-packets.

$\endgroup$
3
$\begingroup$

The view that wave function is only a mathematical object rather than a real object is, unfortunately, only a (probably) majority view but it is not agreed upon universally. This is in part because not all people agree that mathematical object is not a real object. There are people who prefer to think mathematical objects can be real objects.

This is partly because already defining criteria for what "real object" is, is a difficult task which probably nobody has found solution convincing to everybody.

One possible view on this mess:

Electromagnetic field $F$ surely is a mathematical concept - a function of position in physical space. There is an equation for it.

  • Some people call it real, because one use of this function is to calculate estimation of a force on a charged body. This force can be measured and compared to estimation and one can calculate which EM field $F'$ should have been used to produce the measured forces. In this sense, by measurement of force we can calculate "the real" EM field. But there is no necessity to say this, since we do not have any universally accepted criterion for being "real object". There are mathematical theories (Tetrode, Fokker, Frenkel, Feynman-Wheeler, ...) where EM field is used as a visualizable mathematical tool and notational device and no idea of its reality is needed.

Wave function $\psi$ surely is a mathematical concept - a function of position in configuration space. There is an equation for it.

  • One use of this function (in one possible view of it) is to calculate probability (there is no "estimation" of "real" probability) that given all the data that lead to $\psi$ the actual physical state is in some definite region of configuration space (Born's rule saying $|\psi|^2$ is proportional to probability density in cfg. space). The actual state mentioned (positions, momenta of the particles) cannot be found from it. The probability calculated cannot be compared to any measurement of probability, since probability cannot be measured, only calculated from data; and "the real" wave function cannot be inferred from the measurements either.

This is only one possible view, because people do not even agree on what the concept of probability means. To some it is a degree of belief that something is a fact and always depends on data at hand. To other people, in some cases probability is an absolute, the most fundamental description of the world and every use of idea that the world is actually in some definite state devoid of probability concept is only a simplification of The Description of The True Probabilistic Nature.

To sum up, neither EM field nor wave function needs to be called real. Study physics, study probability, study how it is used, how it is misused, what makes sense and what leads to fruitless arguments. Make up your own mind.

One very good resource: http://bayes.wustl.edu/etj/node1.html

(please add any other you know)

$\endgroup$
2
$\begingroup$

This is not a "real" answer to your question but rather points you should consider while making a mental picture of the wavefunction.

"as every respectable physics professor I've ever encountered has treated the wavefunction as an indisputably non-real mathematical tool"

Quantum Mechanics, as taught in undergrad and early graduate courses, deals with artifacts that are as real as dry water, inextensible strings, single photons, etc. They are mathematical limits that we use to simplify the calculations. Can you imagine how many would-be physicists would have left the field if we required them to include the weight of the strings and find the correct catenary shape for simple free body diagrams in Physics 101? Thus, we lie.

A classical electromagnetic wave seems real only because of the large number of photons involved. You can find in some graduate textbooks the "derivations" proving how the quantum number and phase are canonical pairs. That is another "nuanced truth" albeit qualitatively correct. The classical electromagnetic wave is as real as the wavefunction in any macroscopic quantum phenomena (superfluidity, superconductivity).

After you are done with the elementary QM, move on to Quantum Field Theory. There, you will find that the electric field is not modeled with a Hermitian operator. That means that your preciously real electric field is not (gasp) a measurable quantity. Yikes, a premise in your question, the "realness" of the electromagnetic field is not quite correct.


This is a comment to

When dealing with a single quantum mechanical particle, both the wavefunction and the electric field appear to belong to the familiar class of "fields" ... This analogy completely breaks down when you consider multiple particles,

The analogy does not break. The analogy goes farther than one would expect.

The square of the amplitude of the electromagnetic field gives you the energy density, which in turn gives you the number of particles present at that point.

The square of the wavefunction gives you the probability density, which is basically proportional to the number of particles present at that point (when the particles do not interact with each other).

The stronger the electromagnetic field, the larger the number of photons, the closer the classical and quantum descriptions are.

If the electromagnetic field is so that you have about 1 photon, you cannot use the classical description. The electromagnetic wave is then no more real than the wavefunction.

You can decompose the electromagnetic wave into orthogonal components (say right and left handed polarization) and collapse one of them by having the light pass through a polarizer, similar to wavefunction collapse in QM.

The equations of motion of the electromagnetic wave preserve the number of photons (energy) just like the equation of motion of the wavefunction preserve the number of particles.


This is a comment to

One cannot put an instrument at (x,y,z) and measure the wavefunction as it passes, the way one can measure the electric field.

First, you do not measure the electric field. You measure the energy and momentum exchanged between the electric field and your probe. Second, you can measure the phase of the wavefunction. From Wikipedia

The supercurrent $I_s$ through a conventional Josephson junction (JJ) is given by $I_s = I_c \sin(φ)$, where φ is the phase difference of the superconducting wave functions of the two electrodes, i.e. the Josephson phase.


Keeping up with comments to other answers.

You can, in principle, measure the electric and magnetic field strength at every point in space and time.

How can you do that without disturbing the field that you are trying to measure? Your measurement apparatus seems to consist of infinitely many probes of infinite accuracy and precision. I would expect you use infinite energy to run it and it will become really hot really fast; thus, emanating em field themselves.

$\endgroup$
  • $\begingroup$ In quantum optics the electric field is modelled as a Hermitian operator. The positive and negative frequency components are not Hermitian, however. Where have you seen that the electric field operator is not Hermitian in QFT? $\endgroup$ – Mark Mitchison Feb 5 '15 at 14:41
  • $\begingroup$ @MarkMitchison: I read it 20 years ago. I do not remember the textbook. I think it used a weird gauge. I will certainly make a trip to the library. $\endgroup$ – Hector Feb 5 '15 at 21:40
  • $\begingroup$ I'm pretty sure it is always possible to define the electric field as Hermitian. Certainly that is the usual convention in my quantum optics texts. However, just as in classical electrodynamics, maybe it is also possible to define the electric field to have complex eigenvalues. It is often said that one can only really measure $|E|^2$ (although I've never really understood why in a classical context; certainly this is what photodetectors measure), in which case it wouldn't matter whether $E$ itself is Hermitian. $\endgroup$ – Mark Mitchison Feb 5 '15 at 21:46
  • 1
    $\begingroup$ Welcome to Physics.SE by the way :) I realise I have given you rather more argumentation and rather less welcome so far! $\endgroup$ – Mark Mitchison Feb 5 '15 at 21:48
  • $\begingroup$ @MarkMitchison: I am sorry I had to erase my other answer and with it your valid comments on the JJ. $\endgroup$ – Hector Feb 5 '15 at 23:09
0
$\begingroup$

There are many issues.

Firstly, in nonrelativistic quantum mechanics, there is a wavefunction that is a function not of spacetime, but of time plus configuration space. That is the true sense in which it is not real, because it's not a thing that has a value at every point in space.

Secondly, there is an actual quantum field, that only some very particular limits and special situations looks and acts enough like a classical wave that it could be called an EM wave. Specifically it is a limit with many photons, all in phase with each other. Individually, different photons have individual complex phases, whereas a classical EM wave only has a phase in the sense that sometimes it is at its peak, sometimes at its trough, and sometimes at its average value. When many photons are in phase with each other, the common relative complex phase can act very similarly to the phase of an EM wave. But they are still different.

Thirdly, there are quantum fields (in Quantum Field Theory) that are real in most reasonable senses of the word, but unlike a classical wave that could be a scalar field or a vector field, a quantum field is operator valued.

$\endgroup$
0
$\begingroup$

I will expand on my comment, answering the title:

“Reality” of EM waves vs. wavefunction of individual photons - why not treat the wave function as equally “Real”?

What does Real mean in physics

It is instructive to look at the definition of fields for physics:

"A field is a physical quantity that has a value for each point in space and time."

It is evident here that "physical" means real number:

What is a physical quantity? It is a measurement that is represented by real numbers, measured with rulers etc . Humans started counting on their fingers, invented rulers and geometry. The contrast between "real" and "complex" numbers came with algebra and the solutions of higher order equations. Those solutions needed two numbers to be codified, and thus were contrasted to the real numbers accessible to measurements directly, with rulers and their higher order technological products.

Psi's are complex. So even though Psi is the solution of a wave equation, it is intrinsically described by two numbers and cannot be directly measurable. In contrast the waves described by electromagnetic equations are directly measurable by the effect of the real fields on other real fields.

Thus the the QM wavefunction does not correspond to a field, a mathematical field for physics.

Why was Psi invented? Because its " complex square" represents a real number that does describe measurements if interpreted as a probability, very well.

Psi itself as a complex number is unmeasurable, thus not real in any physics sense.

$\endgroup$
  • $\begingroup$ The real and imaginary parts of the wavefunction are real numbers, do these then correspond to physical fields? You can describe the classical electric field as a complex number, does that mean that the electric field is not a physical field? $\endgroup$ – Mark Mitchison Feb 5 '15 at 9:16
  • $\begingroup$ @MarkMitchison The electric field is a real number, otherwise it would not be measurable. Measurable quantities are in real numbers. The real and imaginary parts of the wavefunction are not separately measurable . they come in a couple. $\endgroup$ – anna v Feb 5 '15 at 10:34
  • $\begingroup$ The magnitude and phase of the wavefunction are real numbers which separately measurable, assuming that one has a sufficiently large number of identically prepared systems (see ACuriousMind's answer), and that the multi-particle observables required to reconstruct the many-body wavefunction are accessible in the experiment (see my answer). You seem to be interpreting the "real" in the question title to mean the same thing as "real number". I am sorry but I just don't see why the mathematics of real vs. complex numbers is relevant. I think that the answer to the question has to do with physics. $\endgroup$ – Mark Mitchison Feb 5 '15 at 11:12
  • $\begingroup$ @MarkMitchison it is the way physical quantities are modeled with mathematical quantities. Note that the phase of the wf is not measureable, which is the same as saying that only the "complex square" is measurable. One cannot put an instrument at (x,y,z) and measure the wavefunction as it passes, the way one can measure the electric field. $\endgroup$ – anna v Feb 5 '15 at 11:38
  • 2
    $\begingroup$ Relative phases are of course measurable. One simply has to perform more than one set of measurements to reconstruct the state. This is so-called quantum state tomography. Only the global phase degree of freedom is unmeasurable. Your final point in your comment seems to be that one cannot make a measurement without destroying the state, i.e. the wavefunction is only a probabilistic description. This is exactly the content of ACuriousMind's (correct) answer. But this has nothing to do with whether the wavefunction is complex or real. $\endgroup$ – Mark Mitchison Feb 5 '15 at 14:12

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.