Talk:Wave function

so the wave function...

Is it the probability of a particle hitting a space. Like in the double slit experiment is it the probability of the electron hitting any of the points on the electron buildup? — Preceding unsigned comment added by 86.185.99.78 (talk) 18:18, 11 December 2012 (UTC)

The second paragraph says:

"Although ψ is a complex number, |ψ|² is real, and corresponds to the probability density of finding a particle in a given place at a given time, if the particle's position is measured."

which is how the wavefunction is interpreted. Strictly the probabilities are not "at" any points, but within an interval of points (see the integrals). For the double-slit experiment in QM, you could see Schrödinger equation (Particles as waves). Does that help? Maschen (talk) 18:27, 11 December 2012 (UTC)

Yes thank you! :) — Preceding unsigned comment added by 86.185.99.78 (talk) 18:41, 11 December 2012 (UTC)

article contains no mention of wavefunction's importance to charge conservation

This article makes no mention of the deep connection between the gauge invariance of the wavefunction and conservation of electric charge. It would seem to me to be important to note that connection in this article. On a lesser note, don't particles without electric charge have wavefunctions? If so, then how does that work to conserve charge?

BrianStanton61 (talk) 22:42, 17 December 2012 (UTC)

I haven’t come across this particular point in the literature to understand in detail. Are you referring to this? Do you have any secondary sources (textbooks, papers, research monographs etc) that could cite this point your making in the article? Thanks and welcome, M∧Ŝc²ħεИ_τlk 22:58, 17 December 2012 (UTC)

The source I was referring specifically to was the Charge conservation article. It states: This (i.e. changes in the phase of a wavefunction being unobservable) is the ultimate theoretical origin of charge conservation.

67.81.217.212 (talk) 23:18, 17 December 2012 (UTC)

Wikipedia articles don't count as sources or references; it's still necessary to provide sources as said to back up the statement. Of course you can link to the article. M∧Ŝc²ħεИ_τlk 23:52, 17 December 2012 (UTC)

It is surely worth mentioning somewhere in this article that the wave function of a charged particle is not gauge-invariant. It is important in practice, and it's also important conceptually to understand that a wavefunction is at least partly an arbitrary convention (even beyond the famous global phase factor).

But I don't think the connection to charge conservation needs to be mentioned in this article. The "main plot" is the (Noether's theorem) relationship between gauge invariance and charge conservation. Yes, changing the wavefunction happens to be one aspect of changing the electromagnetic gauge. But I don't think that adds up to a particularly direct connection between wavefunctions in general and charge conservation.

(I am not particularly knowledgeable about this and could be wrong.) --Steve (talk) 03:42, 18 December 2012 (UTC)

In light of this thread, would it be worth mentioning the Aharonov-Bohm effect? M∧Ŝc²ħεИ_τlk 11:16, 31 March 2013 (UTC)

Recent edits (late March → early April 2013)

Much of the vector spaces section has been almost completely trimmed/rewritten. The silly mix of inline LaTeX and HTML maths will be fixed later, no time right now. M∧Ŝc²ħεИ_τlk 12:28, 5 April 2013 (UTC)

Wave function as probability of feeling a sensation

On April 18, I changed the words "…the probability density of finding a particle…" to "…the probability density of sensing a particle…." This was reverted by Maschen with his personal comment, "'sensing'? 'finding' was clearer."

Another member of the priesthood, 52 years ago, had made the following statement about wave functions: "…the wave function permits one to foretell with what probabilities the object will make one or another impression on us if we let it interact with us either directly or indirectly." Also, in discussing the communicability of a wave function, he wrote, "If someone else somehow determines the wave function of a system, he can tell me about it and, according to the theory, the probabilities for the possible different impressions (or 'sensations') will be equally large…." (Eugene P. Wigner, "Remarks on the Mind-Body Question," The Scientist Speculates, I. J. Good, ed., pp. 284-302, Heinemann, London)

Professor Wigner clearly asserted that a wave function tells "one" the probability that "one" will see a flash, or a dark spot on a photographic plate, or some other sensation. This is of considerable importance because it relates a wave function's probability to "one's" subjective and inter-subjective private sensations or impressions, not to external, public, objective particles that can be found. The "one" who is mentioned by Professor Wigner is the observing subject in whose nervous system the sensations are felt.Lestrade (talk) 12:50, 19 April 2013 (UTC)Lestrade

"Sensing" has a connotation of human sensory perceptions like hear/see/taste/touch/feel... "Finding the particle" seems far clearer, more scientific and less philosophical, and direct, to me.

But if you feel strongly about this - then add it - I will not revert next time. If someone comes to this page in the future to post a section heading "what does "sensing" have to do with it?", be sure to explain to them.

P.S. In case you say it - I would never disparage/disregard Wigner, a crucial contributor to quantum theory. M∧Ŝc²ħεИ_τlk 17:26, 19 April 2013 (UTC)

Wigner seems to have been dragging science into the realm of the subjective. This is totally unacceptable to any scientist. Science’s supreme goal is to be totally objective. It wants to describe all observed objects without regard to the condition of the observing subject. But this may not be fruitful. It may be impossible to have an observed object without an observing subject. Einstein took the relativity of the object to the subject into account. Quantum mechanics seriously considers the role of the observer. Wigner’s definition of the wave function with regard to the probability of subjective sensation may not seem "far clearer, more scientific and less philosophical, and direct," but it might be a necessary part of the definition of "wave function."Lestrade (talk) 20:56, 19 April 2013 (UTC)Lestrade

Real valued wave function proposed by de Broglie

It should be mentioned in the article that a real valued wave function has been formulated by Louis de Broglie.--188.26.22.131 (talk) 12:44, 2 August 2013 (UTC)

You need a reliable secondary source for this before mentioning it in the article. M∧Ŝc²ħεИ_τlk 07:03, 27 October 2013 (UTC)

Limited by the speed of light?

I'm not a quantum mechanic (can't handle those teeny, tiny wrenches) so forgive me if I'm not putting this question correctly, but I wonder if someone could say (and maybe put in the article) something about how causality and the speed of light relate to the time-dependent wavefunction. I assume the wavefunction is causal? If the boundary conditions or the potential are altered in one place, do the resulting changes in the wavefunction propagate to other places at the speed of light? And what about the delayed quantum eraser experiment? Thanks. --Chetvorno^TALK 06:38, 27 October 2013 (UTC)

An interesting question.

Quantum entanglement itself does not violate special relativity because information is not transmitted "in" the wavefunction. For two observers to agree on the outcome of an experiment, one observer has to send a signal to the other, which is limited by c. For time-dependent potentials, presumably there would be time-dependent phase factors in the wavefunction related to the potential, like the Aharonov–Bohm effect.

In any case it would be a good addition to the article, thanks for raising this, will look into it. M∧Ŝc²ħεИ_τlk 07:03, 27 October 2013 (UTC)

In non-relativistic QM causality isn't a problem, while the speed of light certainly is a problem. In relativistic QFT, this can dealt with using the cluster decomposition principle, which is weaker (I believe) than a strict causality requirement, but has the advantage that one need not care about whether fields are measurable in any sense. This principle (roughly) implies that quantum fields commute at space-like distances, in turn yielding relativistic field equations, both for states and field operators. YohanN7 (talk) 17:12, 3 April 2014 (UTC)

units of psi

From the lede:

The SI units for ψ depend on the system. For one particle in three dimensions, its units are m^–3/2. These unusual units are required so that an integral of |ψ|² over a region of three-dimensional space is a unitless probability (i.e., the probability that the particle is in that region).

Question: Why isn't the ψ unit trivially L^–3/2? To integrate the square of this with L³ (volume) over all space, to get unitless P = 1?

What you just said is exactly the case for one particle in 3d, which the example already is. What would the units be for two particles (with positions r₁ and r₂) in n (= 1, 2, 3...) dimensions? We need:

$\int d^{n}\mathbf {r} _{1}d^{n}\mathbf {r} _{2}\left|\psi (\mathbf {r} _{1},\mathbf {r} _{2})\right|^{2}=1$

so for this case ψ has dimensions of L^–2n/2. In general for N particles (with positions r₁, r₂, ..., r_N) in n dimensions we have:

$\int d^{n}\mathbf {r} _{1}d^{n}\mathbf {r} _{2}\ldots d^{n}\mathbf {r} _{N}\left|\psi (\mathbf {r} _{1},\mathbf {r} _{2},\ldots \mathbf {r} _{N})\right|^{2}=1$

so ψ has dimensions of L^–Nn/2. In momentum space the dimensions would be p^–Nn/2 or (MLT^–1)^–Nn/2 = (M^–1L^–1T)^Nn/2. The results would be the same for particles with or without spin because the summations are over the dimensionless spin quantum number.

The units in the lead have been queried before. I think it's time we add a specialized section on this somewhere. The alternative would be to add the units after each normalization condition, although IMO it would be better to describe the units collectively. M∧Ŝc²ħεИ_τlk 10:19, 24 November 2013 (UTC)

You are using m for "metre" which even though linked is bad because so easily misunderstood to mean mass. This is a dimensional argument. For that we use "L" or length which cannot be misunderstood. Insisting on meters is like insisting that the dimensions of m in E =mc^2 is "kg". That's just bad writing. Fix it please. SBHarris 20:17, 24 November 2013 (UTC)

(edit conflict)Yes, m is for the SI unit metre (as the sentence "For one particle in three dimensions, its SI units are m^–3/2." in the lead says), while L is for the dimension of length. Standard symbols for two different things. The reason for not discussing dimensions (in the context of units) and using associated symbols in the lead is to prevent any confusion/conflation with dimensions in the context of space. I'll try and clarify again. M∧Ŝc²ħεИ_τlk 20:27, 24 November 2013 (UTC)

I restored your [length]^–3/2 in the lead and units section. Better? It should get the point across to the reader. Apologies for being slow on that particular uptake. M∧Ŝc²ħεИ_τlk 20:32, 24 November 2013 (UTC)

Thank you! SBHarris 01:11, 25 November 2013 (UTC)

Warning: idiot’s dashes in the thread (11 items). Also, several idiot’s dashes in the article as of now: unfortunately, I did not notice those then edited. Incnis Mrsi (talk) 21:31, 11 January 2014 (UTC)

If you're referring to my edits which used the minus sign the "special characters" -> "symbols" character map atop the edit window, I'm not insulted and will not argue. M∧Ŝc²ħεИ_τlk 22:09, 11 January 2014 (UTC)

Firefox has a nice feature: it can highlight all occurrences of a search term on the page. Can your browser do it? Enter the “–” symbol into the search term bar and look where does it appear in the article and here, on the talk page. Incnis Mrsi (talk) 08:36, 12 January 2014 (UTC)

Maschen deems I’m a great expert in wiki typography and waited for me to have abominations in the article slaughtered. Thanks, I fixed 8 instances of an idiot’s dash. Do not paste the code from this section to articles. Incnis Mrsi (talk) 16:48, 12 January 2014 (UTC)

Wave function #Requirements

First of all, “continuously differentiable in the sense of distributions” is an oxymoron. The continuous differentiability (C¹) means that the derivative is a continuous function. Not more and not less. If one uses generalized function spaces for such definitions, such as D′, then a requirement to be differentiable certain number of times is meaningless. Any D′ “function”, by construction of the differentiation operator, is differentiable any finite number of times.

Second, if one source on the wave function formalism requires that it must be continuous, then it does not mean yet that the continuity as an absolute requirement represents a mainstream view. Incnis Mrsi (talk) 12:32, 10 January 2014 (UTC)

The line:

"It must everywhere be a continuous function, and continuously differentiable (in the sense of distributions, for potentials that are not functions but are distributions, such as the dirac delta function). "

seems to suggest delta potentials, but I agree it is badly worded to be useless (even if it was technically correct). (I didn't write it by the way, if that helps). The original line read:

"It must everywhere be a continuous function, and continuously differentiable (at least up to all possible first order derivatives)."

I may be able to add more sources to the section later, the Eisberg and Resnick book is a respected one, but agreeably not enough. M∧Ŝc²ħεИ_τlk 22:42, 10 January 2014 (UTC)

So, did you agree with letting W^2,1W^1,2 in? Incnis Mrsi (talk) 21:31, 11 January 2014 (UTC)

Yes, thanks for that. M∧Ŝc²ħεИ_τlk 22:09, 11 January 2014 (UTC)

Compress

Currently the article makes laborious reading in introducing wavefunctions as functions, then vectors, then several sections on wavefunctions as functions, then another several sections on wavefunctions as vectors, concluded by ontology. That's fine, but the article could be streamlined by explaining the two side by side, making more room for sections on examples and other interesting facts. I'll try, anyone is free to revert. M∧Ŝc²ħεИ_τlk 17:07, 11 January 2014 (UTC)

What facts about vector space do you deem very relevant? BTW, look at my recent change in quantum state: the article contains a stuff that this article misses, and it is relevant here more than there. Incnis Mrsi (talk) 21:31, 11 January 2014 (UTC)

For this article: linear combinations, the inner product (should be obvious why), normalization (again obvious), and completeness condition (useful for representing states and operators in an orthogonal basis), for the discrete and continuous bases, which should be orthonormal.

The edits to streamline were not finished, something else came up. I'll carry on now. M∧Ŝc²ħεИ_τlk 22:09, 11 January 2014 (UTC)

It hasn't worked... The article is top-heavy with abstractions... But rather than reverting, could we manually move the sections back (move the Dirac notation after the function space examples, then the Dirac notations for the individual examples like "one spin-0 particle in 1d", "one spin-0 particle in 3d", "one spin particle in 3d" etc...)? M∧Ŝc²ħεИ_τlk 01:11, 12 January 2014 (UTC)

I'll move things back. M∧Ŝc²ħεИ_τlk 12:41, 12 January 2014 (UTC)

Uncountably infinite dimensional Hilbert spaces

This section has problems; at the very least its title.

"Uncountable many components" is not the same thing as "Uncountably infinite dimensional Hilbert spaces". The Hilbert spaces involved here are countably infinite dimensional. YohanN7 (talk) 08:18, 12 January 2014 (UTC)

I agree: somebody learned poorly that the cardinality of the set of xes in ψ(x) counts dimensions only in the discrete case (i.e. “function” spaces built atop of an atomic measure). Paradoxically, the probability amplitude article now explains it better than this one. Incnis Mrsi (talk) 08:36, 12 January 2014 (UTC)

Fortunately, Maschen reduced it by now, AFAIK. Incnis Mrsi (talk) 16:48, 12 January 2014 (UTC)

Pauli

Pauli and the Pauli equation belong in the history section. YohanN7 (talk) 08:21, 12 January 2014 (UTC)

What makes Pauli equation less relevant than Schrödinger equation? Incnis Mrsi (talk) 08:36, 12 January 2014 (UTC)

Not sure I understand what you mean. Let me rephrase; Pauli and the Pauli equation are missing from the history section.YohanN7 (talk) 11:28, 12 January 2014 (UTC)

Obviously, and anyone could add him. I'll do it now. M∧Ŝc²ħεИ_τlk 12:41, 12 January 2014 (UTC)

I briefly mentioned it. Incnis Mrsi (talk) 16:48, 12 January 2014 (UTC)

The "off-topic" section...

...is inside here...

Two-state Simple examples can be found from a two-state quantum system, two energy eigenstates: $|\psi \rangle =\psi _{1}|E_{1}\rangle +\psi _{2}|E_{2}\rangle .$ where the basis vectors are the energy eigenstates $|E_{1}\rangle$ for the first energy level, $|E_{2}\rangle$ for the second energy level. (Common alternative notations are simply |1⟩ for and |2⟩, respectively). Another example is two spin states of a spin-½ particle, neglecting spatial degrees of freedom: $|\psi \rangle =c_{1}|\uparrow _{z}\rangle +c_{2}|\downarrow _{z}\rangle ={\begin{bmatrix}c_{1}\\c_{2}\end{bmatrix}}={\begin{bmatrix}\langle \uparrow _{z}|\psi \rangle \\\langle \downarrow _{z}|\psi \rangle \end{bmatrix}},$ where the basis vectors are the spin states $|\uparrow _{z}\rangle$ for "spin up" or sz = +1/2, $|\downarrow _{z}\rangle$ for "spin down" or sz = −1/2, (Common alternative notations include |+⟩ for and |−⟩, respectively). In these examples, the particle is not in any one definite or preferred state, but in a superposition of them: in both states at the same time.

Just moved it to here after this edit so it is not sat in the article in the way of anything. M∧Ŝc²ħεИ_τlk 17:46, 12 January 2014 (UTC)

Non-interacting particles

The multi-particle free wave functions are usually considered to be tensor products, each particle living in a private Hilbert space, and the total wave function residing in the tensor product of these. The way of writing it as an ordinary product is just a matter or notation. Interpreting it as an ordinary product is wrong, because the "natural guess" as to what a multi-particle free wave function ought to be is a sum (by linearity of the governing equation) if it is supposed to be residing in a "single" Hilbert space. (Naturally, (at least some, haven't investigated in detail) operators are (sums of) tensor products too with the identity operator in appropriate places.) YohanN7 (talk) 21:04, 8 March 2014 (UTC)

Yes I'm aware of that, but the section is not about abstract vectors and tensor products only wavefunctions (The symmetrized/antisymmetrized tensor products are linked to through Identical particles#Wavefunction representation). M∧Ŝc²ħεИ_τlk 22:18, 8 March 2014 (UTC)

I am, of course, aware of that you are aware of that, but it still deserves mention in the article because it's borderline whether the average reader (taking QM101) is aware. The literature usually (very vaguely) introduces the tensor product notation (without much (if any) elaboration on tensor products) and then immediately scraps it. One sentence would suffice. I'll try to think of something. Cheers! YohanN7 (talk) 22:27, 8 March 2014 (UTC)

The edits are an improvement, but it is still problematical. The wave functions are state vectors, and that passage (state vectors and Dirac notation) is irrelevant. I'll sleep on the matter and see if I can reformulate it tomorrow. Nightie! YohanN7 (talk) 01:17, 9 March 2014 (UTC)

---

The section is blatantly wrong not for any of YohanN7’s reasons, but because it conflates concepts of interaction and quantum entanglement. We know that an interaction usually leads to entanglement, but it is egregiously wrong to assert that an entangled system of two particles always owes its entanglement to their interaction. Should I clarify the distinction with examples or you already understood me? Incnis Mrsi (talk) 13:55, 9 March 2014 (UTC)

No, please do not do that because no such assertion is made. Also, the quantum entanglement article is confused as to what the definition of entanglement is. On the on hand (consider a pair of identical particles), it's entangled if the state of one particle can't be described without mention of the other; Quantum entanglement is a physical phenomenon that occurs when pairs or groups of particles are generated or interact in ways such that the quantum state of each particle cannot be described independently – instead, a quantum state may be given for the system as a whole., (first sentence in lead). Using this definition we do not have entanglement here. On the other hand, the state of the two-particle system cannot be factorized (either for bosons or fermions) and using the alternative definition (given in quantum entanglement#concept); If the quantum state of a pair of particles is in a definite superposition, and that superposition cannot be factored out into the product of two states (one for each particle), then that pair is entangled, we do have entanglement in the present case.

Thus, let us keep entanglement at arms-length distance here. YohanN7 (talk) 23:25, 9 March 2014 (UTC)

The phrase in question is

For N non-interacting distinguishable particles, which do not interact mutually and move independently, the wavefunction can be separated into a product of separate wavefunctions for each particle…

I do not know what the 3 books referenced below this text say about interactions actually. But any person with a basic notion of quantum information knows that a state of the form |ψ₁⟩⊗|ψ₂⟩ is called non-entangled. Not any state of a composite system of two non-interacting constituents is non-entangled. An entangled but (currently) non-interacting composite system can be produced by a decay, for example. In short, the quoted text is false. Also, I am not interested in what wiki cranks and waste-makers wrote about the quantum entanglement; I just expected that user:YohanN7 might be familiar with the concept. Incnis Mrsi (talk) 18:07, 10 March 2014 (UTC)

Seems like over-explanation. I reduced it. M∧Ŝc²ħεИ_τlk 18:20, 10 March 2014 (UTC)

Oops, I might have been jumping the gun above. I don't know much about this. Is Ψ₁(x₁, ...)Ψ₂(x₂, ...) + Ψ₁(x₂, ...)Ψ₂(x₁, ...) entangled? Does not Ψ₁(x₁, ...) describe one of the involved particles? If so, according to one of the definitions above, there is no entanglement, but according to the other definition (the one also given by Incnis) there is entanglement. I'm too confused at the moment to contribute with anything useful. YohanN7 (talk) 20:54, 10 March 2014 (UTC)

An interesting question that I initially ignored. Classification of Ψ₁ ⊗ Ψ₂ + Ψ₂ ⊗ Ψ₁ depends on whether the particles are identical. If they are not, then the state is entangled. If they are identical bosons, then the state is not. If they are identical fermions, then such state is not possible. Note that the definition of indiscernibility is tricky: you must ensure that all freedoms are accounted for before symmetrizing or antisymmetrizing the tensor square space. Incnis Mrsi (talk) 10:43, 15 March 2014 (UTC)

Thank you for an enlightening answer. YohanN7 (talk) 20:51, 15 March 2014 (UTC)

---

The second paragraph (idential bosons) is, at best, confusing. YohanN7 (talk) 23:25, 9 March 2014 (UTC)

What is confusing?

In response to both of you - the section was originally to sharpen up the previous statement of separability for non-interacting particles, for the cases of any number of distinguishable particles, bosons, and fermions. To have general formulae kept to a minimum of mathematics, and which almost always appear in QM books (hence the citations...). No entanglement. No tensor products or linear algebra, which are linked to Identical particles#Wavefunction representation and Bra–ket notation#Composite bras and kets, and could be described later in the sections which at least attempt to deal with linear algebra in abstract generality.

Would you give an example where particles are entangled and not interacting? Even so this is not the article on quantum entanglement. M∧Ŝc²ħεИ_τlk 17:47, 10 March 2014 (UTC)

It says that the Ψ_i, 1 ≤ i ≤ N all have distinct quantum numbers. Moreover, N_i, 1 ≤ i ≤ N is the number of particles with the quantum number of Ψ_i. This means that either some of the Ψ_i have the same quantum numbers or that there are fewer than N Ψ_i in case N_i > 1 for some i. (Also that reciprocal of the square root had me confused for a while, more immediate would be to say square root of the reciprocal.) YohanN7 (talk) 20:39, 10 March 2014 (UTC)

It's taking up more space than contributing useful content. So it has been temporarily deleted.

Here it is for now:

Non-interacting particles The idea of "non-interacting" is hypothetical, since a system of real particles always interact. For N non-interacting distinguishable particles, the wavefunction can be separated into a product of separate wavefunctions for each particle: $\Psi (\mathbf {r} _{1},\mathbf {r} _{2},\cdots ,\mathbf {r} _{N},t)=\prod _{i=1}^{N}\psi _{i}(\mathbf {r} _{i},t)=\psi _{1}(\mathbf {r} _{1},t)\psi _{2}(\mathbf {r} _{2},t)\cdots \psi _{N}(\mathbf {r} _{N},t)\,.$ This separation of variables is a simple method for solving partial differential equations like the Schrödinger equation. If the particles interact, then the wavefunction cannot be separated into the separate wavefunctions of the particles. For N non-interacting identical bosons, the wavefunction is a sum over permutations (indicated by σ) of one-particle wavefunctions ψ1, ψ2, ..., ψN (each of these has a different set of quantum numbers): $\Psi (\mathbf {r} _{1},\mathbf {r} _{2},\cdots ,\mathbf {r} _{N})={\sqrt {\frac {N_{1}!N_{2}!\cdots N_{N}!}{N!}}}\sum _{\sigma (i_{1},i_{2},\ldots ,i_{N})}\psi _{i_{1}}(\mathbf {r} _{1})\psi _{i_{2}}(\mathbf {r} _{2})\cdots \psi _{i_{N}}(\mathbf {r} _{N})\,,$ where N1 is the number of particles all with the same quantum numbers of ψ1 (the number is called the multiplicity), etc. and the sum is taken over all the permutations of i1, i2, ..., iN which lead to different terms. The number of terms is N! / N1!N2!...NN!, so the reciprocal of the square root of this is the normalization constant. The equivalent for N non-interacting identical fermions is (again each of ψ1, ψ2, ..., ψN has a different set of quantum numbers): $\Psi (\mathbf {r} _{1},\mathbf {r} _{2},\cdots \mathbf {r} _{N})={\frac {1}{\sqrt {N!}}}\sum _{\sigma (i_{1},i_{2},\ldots ,i_{N})}\mathrm {sgn} \sigma \left(i_{1},i_{2},\ldots ,i_{N}\right)\psi _{i_{1}}(\mathbf {r} _{1})\psi _{i_{2}}(\mathbf {r} _{2})\cdots \psi _{i_{N}}(\mathbf {r} _{N})\,,$ where "sgn" is the sign of the permutation: −1 for an odd number of permutations and +1 for an even number of permutations. This expression is the Slater determinant. Note for fermions the multiplicities of all the one-particle wavefunctions can only be 1, by the antisymmetric property. The number of terms in the sum is N!, so the reciprocal of the square root of this is the normalization constant. In these equations, the wavefunctions are complex scalars multiplied according to ordinary complex number arithmetic. Using the formalism of quantum state vectors in Bra–ket notation, the formulae are similar but the ordering of single particle states |ψ1⟩, |ψ2⟩, etc. in each term is important as they are tensor products - a non-commutative operation unlike ordinary complex number multiplication. The single particle states each live in a Hilbert space, and tensor products of the single particle states give new states which live in the corresponding tensor products of the Hilbert spaces (see Identical particles#Wavefunction representation, and below). The tensor product nature of multi-particle states is usually not explicit in the notation the literature (or here).

M∧Ŝc²ħεИ_τlk 21:05, 10 March 2014 (UTC)

P(Ψ₂→Ψ₁)

This piece of Copenhagen currently resides in the “position space” section, that is misleading. The formula for the transition probability is correct, but it describes the situation where Ψ₁ is an eigenvector of the measurement. This depends on observable and doesn’t depend on the presentation (the position space or whichever). Interpretation of |Ψ(x)| as the probability density corresponds to a measurement of the position operator when in position space (it would be another coordinate operator in other presentations). These are two different scenarios, that usually are incompatble: in the non-discrete case the position operator has a continuous spectrum only. It is sad that my work on probability amplitude did not result in significant improvements of QM literacy here. Incnis Mrsi (talk) 10:43, 15 March 2014 (UTC)

Better?

The new section Inner product should not be a subsection of “position space” section, but I don't have time a t m. YohanN7 (talk) 12:34, 15 March 2014 (UTC)

The additional material on the inner product was placed in that section for concreteness. The other inner products for more complicated systems are given later. Please keep the inner product section there at least for now.

What's missing from this article is a section on how the wavefunction is used to calculate the possible results of observables (eigenvalues of operators), as well as expectation values. Perhaps a general statement of the formula

$P(\Psi _{2}\rightarrow \Psi _{1})=|\langle \Psi _{1},\Psi _{2}\rangle |^{2}$

could go in a section like this somewhere? M∧Ŝc²ħεИ_τlk 13:23, 15 March 2014 (UTC)

I thought to contain all the inner products and terminology together in its own section. M∧Ŝc²ħεИ_τlk 13:42, 15 March 2014 (UTC)

Yup, structurally, that looks good. With the inner product, we have an Inner product space. Then appeal to completeness of the set of eigenfunctions of Hermitean operators and (what more?) conclude that we have a Hilbert space of solutions to the Schrödinger equation, i.e. an inner product space with a complete metric. Just thinking out loud. Not much is needed here, a sentence or two. YohanN7 (talk) 14:48, 15 March 2014 (UTC)

Prior to yesterday, the reason for including superposition followed by the inner product in the first section (1d one particle states position representation) was to provide the very basic features of an inner product space in a concrete setting. This flow is a bit disjointed now, which is why I'm not sure if it was an improvement. M∧Ŝc²ħεИ_τlk 08:58, 16 March 2014 (UTC)

Ironically, the [complete] inner product space argument discredits all C¹ and Sobolev space stuff presented above in the article. By the way I do not see where this completeness has actually any use. Although one can’t rely, without completeness, on the spectral theorem for an arbitrary self-ajoint operator, it will likely hold for practically important operators. In any case, the article must clarify why the metric completeness in one part of the article does not contradict to differentiability restrictions in another. Possibly, we should hint that the concepts of a wave function and of an abstract state vector have certain subtle differences. Namely, one gets a wave function from the Schrödinger equation, but operates with state vectors doing various measurements. Incnis Mrsi (talk) 09:20, 16 March 2014 (UTC)

Complex valued function?

"Typically, its values are complex numbers and, for a single particle, it is a function of space and time."

Is this true of any but a scalar particle? I know I'll probably spark a debate because this is not my area, but surely this is oversimplification to the point of confusion? For example, I'd expect any fermion to have a wavefunction that is spinor-valued, in which case complex numbers might come into it, depending on the representation chosen, but also might not. Shouldn't this description be changed? —Quondum 21:48, 15 March 2014 (UTC)

Yes, in general a spinor,or other things I don't know about, but we need to start from somewhere, and the first introduction to wavefunctions is always through complex numbers as far as I know. Presumably your'e thinking of multivector-valued wavefunctions in geometric algebra which have the "spinorial" behavior (according to Hestenes' use of the term "spinor")? For now, probably best left as complex numbers with a pointer saying other quantities are occur, maybe have a section on this too. M∧Ŝc²ħεИ_τlk 08:58, 16 March 2014 (UTC)

This, being an encyclopaedia and not a textbook, has rather strict guidelines about how it can present things. As such, the lead completely violates the guidelines. Even if one chose to take a pedagogical approach in the body of the text, I would have considered it necessary to make it clear that the approach is not the general one, and also to omit the invalid simplification from the lead. For example, one could say "Typically, for a single particle, it is a function of space and time."

I was thinking of spinors of any representation, be they Dirac's complexified vectors or either of the two most suitable GA representations, neither of which is an algebra over the complex numbers. From all perspectives (pedagogic and encyclopaedic), I think that it is important to maintain a clear distinction between what is being represented and the representation. I see the confusion in thinking that extends quite high (to some postgrads) when they use only one representation because they fail to disentangle these, for example with tensors and the Ricci calculus, where people are taught in effect that the collection of components is a tensor, and that the definition of a tensor is a set of components that transforms appropriately. Especially pedagogically, I feel that this is highly counterproductive because people are being taught the "arithmetic" of QM while glossing over its "mathematics", and encyclopaedically it is just plain wrong. —Quondum 16:05, 16 March 2014 (UTC)

Representation of what? Schrödingerian mechanics isn’t Lorentz/Poincaré invariant. It is Galilean-invariant. Usually, things that are Lorentz/Poincaré invariant are not called wave functions. These are fermionic fields and so. Incnis Mrsi (talk) 17:22, 16 March 2014 (UTC)

I can see this bogging down in undefined terminology. Would you care to put that statement ("things that are Lorentz/Poincaré invariant are not called wave functions") into the article? If true, it would certainly be helpful. Also, is the concept restricted to scalar particles? It would also help if this were made clear in the article. Coming from an outside perspective, I cannot even determine what the term wavefunction covers, and what it doesn't cover. With regard to representation, a particle is not a complex function of space and time, but it could be represented by a complex function. Or is the wavefunction by definition the complex value representation of a particle? The article seems to leave all these questions unanswered. —Quondum 06:03, 17 March 2014 (UTC)

Physics bogged down in terminology for a long time, but this is your “represented” verb that is undefined here. When I read about representations, I thought we talk about representation theory of symmetry groups. No, I won’t put the Lorentz/Poincaré invariance speculation into the article. The relativistic quantum mechanics uses hyperbolic (Lorentz-covariant) Hamiltonians with wave functions. Also, Wikipedia states:

Nevertheless, RQM is only an approximation to a fully self-consistent relativistic theory of known particle interactions because it does not describe cases where the number of particles changes; for example in matter creation and annihilation.

If you replace QM with wave function, then it will be consistent with my understanding. When you know how many (and which) particles do you have, then you have a wave function. When you don’t know it, then you have a field. Also, you can read here on a thought experiment that shows that “how many and which particles” concept is blurry, and it is related to your spinors. You can speak about spinor-valued wave functions only if you have some kind of Spin(3) symmetry. If you haven’t, then values are merely complex (finite-dimensional) vectors that represent some opaque states of the system. Yes, a particle, or a system of particles, is not obliged to have only one internal state. Incnis Mrsi (talk) 07:40, 17 March 2014 (UTC)

My use of the term representation is clearly confusing the matter, so let's drop the term. Whether we are dealing with (non-R)QM or RQM does not seem to be significant to the concept of a wavefunction. I agree that the concept of a wavefunction does not really apply to QFT. From what you say, you appear to consider solutions the single-particle Dirac equation as wavefunctions in the sense of the article, even if regarded as complex vector fields. This alone is enough to suggest that the description "its values are complex numbers" in the lead should be made more general. I'm suggesting that in the lead it should rather say something like "It is generally vector-valued and, for a single particle, it is a function of space and time." Or even simply "For a single particle, it is a function of space and time." The discussion of whether the vectors are considered to be over complex numbers or to be elements of a Clifford algebra need not be addressed in the lead. Scalar particles are simply a special case of vectors. —Quondum 18:40, 17 March 2014 (UTC)

Quondum, you refer to WP policies. One of the most important is to write from what most sources say: of course not like a textbook but as a plain English summary. Well, does anyone know how many introductory QM sources which do not use write wavefunctions as complex numbers or complex-valued vectors? Complex numbers are exclusively used in almost every single introductory QM book I've ever seen (anyone is welcome to contradict with examples). The only exception I can think of is when QM is written in the language of geometric algebra, which uses the field of real numbers. The restricted scope "introductory QM" is not meant to be pedagogic - this is one important level of sources we should use for this article.

As for:

"With regard to representation, a particle is not a complex function of space and time, but it could be represented by a complex function. Or is the wavefunction by definition the complex value representation of a particle? The article seems to leave all these questions unanswered"

the article has an ontology section linking to the main ontology article - saying the meaning of the wavefunction is unclear, and always has been from day 1. The sections on wavefunctions for one particle in 1d, and more particles in higher dimensions, all mention the Copenhagen interpretation, the most introductory interpretation of the wavefunction.

It's a bit unfair to say the lead totally violates the guidelines. It can be improved, but it gets the main points across. I'm not denying the article still needs work. But this discussion confuses me. M∧Ŝc²ħεИ_τlk 16:40, 22 March 2014 (UTC)

A sensible definition of what a wave function is might be that it solves a wave equation. This holds true in all cases I can think of. What a wave equation is is well defined as far as I know.

It is not true that wave functions aren't present in QFT. They are necessary to construct the Hilbert space. (Any complete set will do.) They just don't occupy the center of the stage; the dynamics (time evolution of the system) is moved to the operators on the Hilbert space. In addition, new operators are introduced to handle creation and destruction of particles. YohanN7 (talk) 23:40, 22 March 2014 (UTC)

QFT is beyond scope. I said nothing about the interpretation. I have evidently failed to communicate my point about the space from which the wave function may take its values (codomain), without which my comment about guidelines is also meaningless. I see little point in pursuing this. —Quondum 00:27, 23 March 2014 (UTC)

Quondum and YohanN7 - I was still thinking we could describe the role of the wavefunction in non-RQM, RQM, and QFT, all in a section for comparison. In all this time it just hasn't been done. Currently the article biases the Schrödinger picture. As I mentioned above, we could have a brief section on how operators act on the wavefunction in the Schrödinger picture (and related, briefly mention how statistical quantities like averages can be found), then go onto the Heisenberg picture, and possibly the interaction picture, again all in a section for comparison. No, we will not turn the article into a second Dynamical pictures (quantum mechanics), just an overview.

Quondum, no, you've raised a good point. There is a bias with complex numbers, sure, and having a variety of mathematical languages would remove the bias. Since you raised the point, others will probably ask the same question. I'm simply not sure how to do this yet.

Thanks. (P.S. YohanN7 sorry about not checking in Griffiths yet, I couldn't find it last week). M∧Ŝc²ħεИ_τlk 07:54, 23 March 2014 (UTC)

Wave functions take their values in Cⁿ for n ≥ 1, and this covers the rare cases (e.g Majorana fermions, and certain potentials in one dimension (forgot details, see Landau and Lifshits)) when they can be taken as real-valued. Why talk about guidelines? I can't honestly see the point in having a huge debate about this. The GA representations are clearly fringe (unless something revolutionary has happened the last few years), so they can be left out. YohanN7 (talk) 09:40, 23 March 2014 (UTC)

This isn't just about guidelines, it's about content.

If it helps, I saved up enough to order Wienberg's vol 1 Quantum theory of fields, Landau and Lifshitz's vol 2 Classical theory of fields, and Doran and Lasenby's Geometric algebra for physicists (which includes QM formulated using geometric algebra). All fairly expensive, but there is no way round that, and they're top quality books so it's just about tolerable. Next week it will be the easter break too. Within the next week the books hopefully arrive and I get more time, it will be easier to make edits. M∧Ŝc²ħεИ_τlk 10:03, 23 March 2014 (UTC)

Excellent choices. The L&L book also covers general relativity (guess you knew this, much much better than MTW). YohanN7 (talk) 10:20, 23 March 2014 (UTC)

You might want to have a look at Field Quantization by Walter Greiner et al. Weinbergs attitude is that he doesn't quantize any pre-existing theory. But is quite useful to see that view too. It contains hundreds of detailed calculations, and it has a highly unusual description of classical field theory (Poisson-bracket formulation). Only $20 at amazon. YohanN7 (talk) 12:46, 23 March 2014 (UTC)

Greiner's books are OK (the classical mechanics, QM, RQM, QED, QCD, and Field quantization (if I recall correctly) are all at the library). But for now Weinberg's first volume is enough. Thanks for the suggestion. Yes, while MTW is a well-known and widely used classic, some bits are inadequate (e.g. the angular momentum sections are a bit vague, the relativistic heat conduction is not thorough, and there doesn't seem to be any description of the relativistic D, H, M, P fields in matter, all these are remedied either by Pauli or Tolman). LL's classical theory of fields seems to be loaded: SR, EM, GR using Lagrangian mechanics and field theory. M∧Ŝc²ħεИ_τlk 14:11, 23 March 2014 (UTC)

Mentioning that Cⁿ may be the space is reasonable, mainly just not implying a limitation to C. No need (and it may be unwise especially in the lede) to mention GA or any specific space (vectors, matrices etc.), but preferably don't imply their exclusion by the wording. And forget my mention of guidelines: I was hoping to aid understanding of rather than to push my point. And my point does not imply any effort in adding detail about spaces, only wording that does not create misconceptions. We know that isomorphisms exist; we don't need them to be enumerated. —Quondum 16:33, 23 March 2014 (UTC)

I'll have a very quick go at tweaking the lead along these lines. M∧Ŝc²ħεИ_τlk 16:56, 23 March 2014 (UTC)

Not so quick, please. First, wave mechanics is a dab page. Second, although it is true that a particle with spin has a spinor-valued wave function, I’d not assert that a spinless particle is always scalar. Which spin has a (non-rotating) ammonia molecule? With ¹⁴
₇N
¹
₁H
₃ it certainly isn’t zero (even number of electrons and odd number of nucleons), but I do not see a reason for it not to be spinless with an appropriate isotope composition. So, is it a scalar particle, is it really? Incnis Mrsi (talk) 18:24, 23 March 2014 (UTC)

Nice example. As far as I can make out it is a superposition of two states ("left" and "right" handed).

So what is the best wording for the lead then? M∧Ŝc²ħεИ_τlk 22:29, 23 March 2014 (UTC)

I hope the lead is better now. Anyone should boldly edit themselves. M∧Ŝc²ħεИ_τlk 23:15, 23 March 2014 (UTC)

Thank you: your edit is a massive improvement, and makes the lede both understandable and essentially complete in its description. —Quondum 18:10, 23 March 2014 (UTC)

Relation between wave functions

I tried to explain a little. There are probably bugs in the new equations, possibly a missing minus sign (the plane wave), and possible a missing factor of one over two pi (delta function normalization). These things depend on conventions, including for the Fourier transform. Can anyone check what Griffith (the ref in the section) says? YohanN7 (talk) 22:47, 15 March 2014 (UTC)

Nice work. Except for one quibble: we now have the inner product of momentum states before the concepts of the inner product is introduced even in the position representation, so these should be moved down into the inner product section. I don't have Griffith's QM book now but may be able to check sometime in the week. M∧Ŝc²ħεИ_τlk 08:58, 16 March 2014 (UTC)

I have noticed the mentioned quibble. Will make an attempt later today to fix it. YohanN7 (talk) 10:33, 16 March 2014 (UTC)

Oops, I missed the time dependence. But, should we instead state that it is the solution of the time-independent Schrödinger equation? Time dependence is mostly disturbing in this section. YohanN7 (talk) 03:22, 24 March 2014 (UTC)

I moved the momentum inner product content down, and indicated the time-independent Schrödinger equation. M∧Ŝc²ħεИ_τlk 06:58, 24 March 2014 (UTC)

Hydrogen wave animation

May I destroy [1]? This flickering picture that does not, actually, “animate” anything useful, only distracts attention and wastes the processor time. Incnis Mrsi (talk) 13:36, 29 March 2014 (UTC)

It does not add anything; it is merely an attempt at a different presentation of what is already there. From the description "hydrogen wave animation" I had expected the evolution of a wavefunction in time, but it is nothing of the sort. I agree with Incnis that it should be removed. —Quondum 16:10, 29 March 2014 (UTC)

Inner product (again)

The definition of the "inner product" suffers from an annoying and confusing error omission that is present in most basic QM texts(, and some math texts too). The "inner product" is not an inner product, it is a semi-inner product. The resulting "norm" is a pseudo-norm semi-norm, and its "metric" is a pseudo-metric (not semi-metric).

There are standard ways (all equivalent I believe) of dealing with this. One way is to declare points at a zero distance apart to be equivalent. Then by passing to the quotient w r t this equivalence relation, one obtains a true inner product space. It's elements are equivalence classes of functions from the original "inner product space".

Before I (possibly) do anything, I'd like your opinion on whether this conceptually important, but practically unimportant, issue is worth mention in the article. IMO, QM books being generally crappy mathematically should not be a reason for this article to make the same mistakes.

There are probably three categories of students (of QM books or this article). The first category never notices the problem. The third category does notice the problem and also its solution. The second category will be left puzzled. I belonged once to the second category. The instructor did some hand-waving when I asked him (he really didn't know I think). YohanN7 (talk) 12:20, 3 April 2014 (UTC)

I'm not too sure how you're applying the metric. We're dealing with a Hilbert space, so the vectors are not 4-vectors to which a metric tensor would apply for this inner product. On a scalar (multi-particle) field, this seems to be unambiguously an inner product (it is always positive for a nonzero field used for both arguments, and the equivalence class issue should not occur), and it is not clear to me that for a spinor field it will be any different. But I'm not familiar with the area at all, just looking at it from the outside, so really I'm just asking. —Quondum 20:00, 3 April 2014 (UTC)

We are not dealing (yet) with a Hilbert space (or metrics for that matter). Hilbert spaces have inner products. What is defined here on a function space is a semi-inner product, since

$\left\langle \Psi ,\Psi \right\rangle =0$

does not imply

$\Psi =0.$

In other words, you have missed the axiom that distinguishes an inner product from a semi-inner product. It has nothing to do with spin or 4-vectors. (This puts you in category one of above :D)YohanN7 (talk) 20:19, 3 April 2014 (UTC)

What reference are you using for the definition of a semi-inner product outside of WP? Do you know any refs for semi-inner products in the context of QM? (Admittedly I'm in your categories 1 and 2 above). M∧Ŝc²ħεИ_τlk 20:24, 3 April 2014 (UTC)

See A course in Functional analysis by John B Conway or General topology by Willard. The latter explains how to systematically obtain a metric space from a pseudo-metric space. There is probably a note too in our article on L^p-spaces. (L² isn't a Hilbert space before the mentioned identification and taking quotients.) YohanN7 (talk) 20:34, 3 April 2014 (UTC)

And oh, the integration must be taken as Lebesgue-integration. Riemann-integration will fail in some technical aspects to yield what we need, but this is probably definitely outside of the scope. YohanN7 (talk) 20:34, 3 April 2014 (UTC)

OK, thanks, I'll look into these. This article really has always been in turmoil... M∧Ŝc²ħεИ_τlk 20:52, 3 April 2014 (UTC)

The omission is no disaster, it is systematically made in the literature. The difference between the math literature and the QM literature is that the math literature takes it as an obvious standard thing to do (the metric identification procedure, that is, and it usually gets a parenthetical remark, math students are supposed to have seen it done). The QM literature should not be able to do the same thing, but it does, leaving some of us, including me, confused for a while. YohanN7 (talk) 21:17, 3 April 2014 (UTC)

Remark on Lebesgue vs Riemann integration: My ref is Byron and Fuller, Mathematics of Quantum and Classical Physics, a book I don't have at where I am now. I think Lebesgue integration is required for completeness of the inner product space (so that it is a Hilbert space), but the details have faded from my memory. At least, it is crucial for some proofs. Might perhaps be found in Conway, but that book by default uses Lebesgue-integration, so that it passes without mention. YohanN7 (talk) 21:36, 3 April 2014 (UTC)

For the purposes of this article, the following should be enough:

• A remark, and link to Lp space#Lp spaces
• A clarification that this is of no practical concern, i.e. which function one takes to represent an equivalence class doesn't matter.

Only the latter point has a problem. Such a statement needs to be sourced. YohanN7 (talk) 21:51, 3 April 2014 (UTC)

Then again, is it really possible for two continuous functions to differ from each other on a set of Lebesgue measure zero? A continuous function is completely determined by its values on a dense set (e.g the rationals). This requires some thought. I'd say no, so if attention is restricted to continuous square-integrable functions at the outset, then we do have a Hilbert space. Edit: Probably no, see below. YohanN7 (talk) 23:45, 3 April 2014 (UTC)

No, this procedure will probably not work either. I think I can imagine a Cauchy-sequence of continuous square-integrable functions with the limit being a discontinuous function. (Modify an example given here: Inner product space#Examples.) We get an inner product space, but not a Hilbert space. Adding further smoothness requirements probably wont help. YohanN7 (talk) 02:21, 4 April 2014 (UTC)

I think the conclusion may be this: With this choice of inner product, the space of functions should be the space of square-integrable functions period. No continuity assumptions allowed. Only then will we get a Hilbert space (upon taking an appropriate quotient). There are other possible Hilbert spaces to make use of, e.g. the Sobolev spaces. These are mentioned in the article. But, for these, the inner product given here is not the right one. YohanN7 (talk) 13:54, 4 April 2014 (UTC)

Careful authors would say that elements of L^2 are equivalence classes of functions, although this is seldom emphasized when it's expedient not to do so. I recently read a paper where the author placed "equivalence classes of" in parentheses at one point. That seems to be about as far as I'd be willing to push things. Of course, for continuous functions such as those that appear in quantum mechanics, it makes no difference whether you identify functions if they are equal outside a set of measure zero. I think a rather enlightening perspective is that in quantum mechanics it isn't really the Hilbert space itself that one sees, but rather its dual. But the dual of the continuous dual space of the space of all square integrable Lebesgue measurable functions (which is a nonhausdorff topological vector space with semi norm give by the usual hilbertian inner product) *is* a Hilbert space. No magic identifications are needed. This also explains why distributions have such a big role to play in the theory. Sławomir Biały (talk) 14:44, 4 April 2014 (UTC)