# Questions about properties of wave-functions

1. Jun 26, 2012

### San K

Had some questions about wave-functions. Some of the questions might be invalid due to limited knowledge of the OP.

1. Are wave-functions real? i.e. do they exist in reality?

Wave-functions are mathematical/probabilistic constructs.

However if they can modify/change the behavior/path of a photon then something real must exist, that is taking into account/calculating both/all the slits/paths.

2. Are wave-functions ("travelling") instantaneous?

7. When a photon is travelling from the sun towards the earth, is there a wave-function existing at all times between the photon and its final destination? and does that final destination keep changing to the first object of opaque obstruction?

a wave-function exists at all times when a particle is not being detected/measured.
regarding the length/reach of the wave-function I don't know, someone from/in the forum might have information.

3. Are wave-functions the same concept/thing that are used in Quantum Entanglement as well?

4. Do wave-functions have energy?

Pro argument - well if they can change the path of a photon/particle.....

Con argument - if they have energy then it would be split at the slits, however when the photon is finally detected at the detector there is no loss in energy detected.

5. Is the collapse of the wave-function instantaneous?

a collapse happens when we try to detect the photon. at that point the photon is now "entangled" with the detector (?)

6. Are wave-functions (existing) within space-time?

"The new fad/fashion for reducing weight is Raspberry ketones..;)"

Last edited: Jun 26, 2012
2. Jun 26, 2012

### Staff: Mentor

We calculate the expectation value of the energy of a system by using the energy operator, in the same way that we can calculate the expectation value of momentum by using the momentum operator, or the expectation value of the position by using the position operator:

$$\langle E \rangle = \int {\Psi(x,t)^* \hat E \Psi(x,t) dx} = \int {\Psi(x,t)^* \left( i \hbar \frac {\partial}{\partial t} \right) \Psi(x,t) dx}$$

$$\langle p \rangle = \int {\Psi(x,t)^* \hat p \Psi(x,t) dx} = \int {\Psi(x,t)^* \left( -i \hbar \frac {\partial}{\partial x} \right)\Psi(x,t) dx}$$

$$\langle x \rangle = \int {\Psi(x,t)^* \hat x \Psi(x,t) dx} = \int {\Psi(x,t)^* \left( x \right) \Psi(x,t) dx}$$

Whether this means the wave function "has" energy, or momentum, or position, depends on what you mean by the word "has".

What does this mean?

3. Jun 26, 2012

### Fredrik

Staff Emeritus
1. It's a mathematical concept that's a part of a theory that makes very accurate predictions. The question of whether it's also "real" doesn't seem to be a scientific question. Can you think of an experiment that can answer it?

2. No, they change with time as described by the Schrödinger equation.

7. I don't fully understand the question, so I'll just say that wavefunctions for massless particles is a tricky concept, and that typically, when a particle is emitted from a source, its wavefunction spreads out in all directions. The particle can't be described as localized until it interacts with something else (e.g. molecules in the air).

3. To discuss entanglement, you need a state vector for a two-particle system. You can think of it as a wavefunction that depends on twice as many variables.

4. A wavefunction represents the state of a particle. We will only say that the state has a well-defined energy if knowledge of the state is enough to predict the result of an energy measurement with certainty. This is the case e.g. after an energy measurement that doesn't destroy the particle. (A second measurement will have the same result). If the state doesn't have a well-defined energy, then it's a superposition (linear combination) of many states that do have well-defined energies.

5. The question only makes sense if we view collapse as a physical process. I don't.

6. They are functions from ℝ4 (="spacetime") into ℂ that satisfy the Schrödinger equation. Another way of looking at it is to say that they are functions from ℝ3 (="space") into ℂ, and given such a function, i.e. a point in the space of such functions, the Schrödinger equation determines a unique curve through that point.

4. Jun 26, 2012

### San K

Great answers, Fredrik and jtbell, thanks!. You have clarified some difficult/deep questions.

first the following need to be understood:

what is - R4, R3, Hilbert space, relationship between time and space, 4D Vs 3D

sounds reasonable.... perhaps it is a swap...

Last edited: Jun 26, 2012
5. Jun 26, 2012

### audioloop

.-psi epistemic ontic:
reflect something of the reality.
.-psi epistemic epistemic:
nothing to do with reality.
.-psi ontic:
they are the real.

10^10 - 10^15 secs
around attoseconds, zeptoseconds, yoctoseconds.

as for macroscopy objects, time depend on the number of atoms.

if time is relational, it does not.

Last edited: Jun 26, 2012
6. Jun 26, 2012

### Jazzdude

And which mechanism do you suggest creates that time dependence? I saw in a different thread that you rightly stated that decoherence does not lead to collapse, so this can't be the decoherence time you're referring to. So what time scale do you mean?

7. Jun 26, 2012

### audioloop

a nonlinear schrodinger equation.

8. Jun 26, 2012

### Jazzdude

There are several nonlinear modifications of the Schroedinger equation. Which one do you have in mind? GRW? In any case, I find added nonlinearities to be highly speculative and probably not suitable for a general authoritative answer to a beginner's question.

9. Jun 26, 2012

### nonequilibrium

I disagree, they are only defined on $\mathbb R^3$ (spatially speaking) for one particle systems. A general N particle wavefunction is defined on $\mathbb R^{3N}$. And since the latter obviously cannot be interpreted as being defined on "space", it indicates that even for the one-particle the $\mathbb R^3$ should not be identified with space in any way.

Shortly put: no, the wavefunctions are not defined on spacetime

EDIT:

I believe this question can even be answered without giving the wavefunction a physical meaning, again in disagreement with Fredrik (although note that due to my comment above I don't really see how you would even give the wavefunction a physical meaning, since it can't be defined on spacetime), all you need is that the wavefunction is an objective property.

Although Copenhagen treats it instananeously, there are other interpretations which describe that as a limit of the real process which takes a finite length of time. One interpretation that I know of that says collapse is not instantaneous, is the de Broglie-Bohm (which uses the concept of decoherence).

Last edited: Jun 26, 2012
10. Jun 26, 2012

### Fredrik

Staff Emeritus
I was talking about one-particle states. It didn't seem necessary to consider anything more complicated here. The space of complex-valued functions defined on $\mathbb R^{3N}$ is isomorphic to the the tensor product of N 1-particle Hilbert spaces. So a "wavefunction" for an N-particle system can also be thought of as an N-tuple of 1-particle wavefunctions, each of which is defined on "space".

My answer about the speed of collapse was a bit lazy. So maybe I should elaborate on it, or even change it. But I'm not going to spend any significant amount of time on that right now.

I think Dr Chinese has answered questions about the speed of collapse in several other threads, so the OP might want to search for those posts.

11. Jun 27, 2012

### nonequilibrium

I do think that it's important to look at the more general case, since it indicates to me that even the 1-particle case wavefunction can not be interpreted as being defined on space. Sure, you might take a wavefunction for a system of N particles and integrate out N-1 variables, giving a one-particle wavefunction defined on $\mathbb R^3$ and do this N times, giving N one-particle wavefunctions, but this leaves out most information (entanglement!) and thus is in no way equivalent to the original wavefunction, which simply cannot be embedded in ordinary space in any sensible way.

12. Jun 27, 2012

### Fredrik

Staff Emeritus
I don't follow your argument for why we shouldn't never think of wavefunctions as functions defined on ℝ3. Why shouldn't we do that when we're dealing with the non-relativistic theory of a single spin-0 particle under the influence of a potential?

13. Jun 27, 2012

### nonequilibrium

No sure, that is defined on $\mathbb R^3$, but I'm saying we shouldn't identify this with ordinary space but acknowledge that it is just a coincidence of the one-particle case. And my motivation for that assertion is the general N-particle case.

14. Jun 27, 2012

### audioloop

hi jazz, i hate likewise, ad hoc propositions.
i mean inherently nonlinear like, Dobner-Goldin Class (not GRW/CSL) or from an invariance principle.
with testable predictions.

by the way, jazz,
what do you think about a derivation of the Born probability rule, from a nonlinear standpoint ?

15. Jun 27, 2012

### Jazzdude

That's interesting. I haven't looked into nonlinearities due to non commutative geometry very deeply, but my impression was that there is no complete theory that would predict collapse including the Born rule. I could very well be wrong. Do you have an literature pointers?

I'm quite certain that some kind of nonlinearity is needed to fully explain the observation of single unique measurement outcomes and as a consequence the Born statistic. The question is where such a nonlinearity might come from. As you said, something that doesn't have to be artificially added but is inherent to the theory is surely preferable.

But what if this nonlinearity is already present in the linear structure of quantum theory as we know it? If we eradicate the measurement postulate, the nonlinearity seems to arise as a subjective artifact of trying to reconstruct the state of the universe by interacting with it.

I know this sounds quite unlikely, but please check http://arxiv.org/abs/1205.0293 for the exact argument and a thorough discussion. The paper derives the measurement postulate including the Born rule and makes predictions for observable effects. I'd be happy to have someone to discuss it with.

16. Jun 27, 2012

### HallsofIvy

I think you meant 10^{-10} - 10^{-15} sec. What you wrote would run to several million years.

17. Jun 27, 2012

### nonequilibrium

The pilot-wave interpretation can also explain/deduce it. I would presume there are other (nonlinear) interpretations which can also do that.

18. Jun 27, 2012

### Jazzdude

The introduction of particles that are piloted by the field is a nonlinearity of the theory. The fields stay linear, but the properties of the particles are not subject to superposition. So it's the lack of a linear structure in the point particle representation that breaks the linearity of the theory.

As for other interpretations that are able to actually derive the Born rule, I'd be interested in hearing which you have in mind. The only derivations I know of are in the context of MWI (with artificial assumptions that are practically equivalent to stating the Born rule directly) or Zurek's envariance, which I consider circular.

19. Jun 27, 2012

### audioloop

yes sir, i forgot the minus symboll.
thanks.

20. Jun 28, 2012

### audioloop

well, related arguments.

Schroedinger equation from an exact uncertainty principle
J. Phys. A 35 (2002) 3289
http://arxiv.org/abs/quant-ph/0102069

Exact uncertainty approach in quantum mechanics and quantum gravity
Gen. Relativ. Gravit. 37 (2005) 1505
http://arxiv.org/pdf/gr-qc/0408098v1.pdf

and
http://arxiv.org/pdf/gr-qc/0610142v2.pdf

21. Jun 30, 2012

### audioloop

unifying quantum and thermodinamics

Nonlinear extensions of Schrödinger-von Neumann quantum dynamics: a list of conditions for compatibility with thermodynamics.
http://arxiv.org/pdf/quant-ph/0402180v1.pdf

Nonlinear Quantum Evolution Equations to Model Irreversible Adiabatic Relaxation with Maximal Entropy Production and Other Nonunitary Processes.
http://arxiv.org/pdf/0907.1977v1.pdf

....In contrast with the epistemic framework, where the linearity of the dynamical law is a requirement, the assumed ontic status of the density operator emancipates its dynamical law from the restrictive requirement of linearity....

22. Jul 3, 2012

### Simon Bridge

San K is still wrestling with these concepts if recent threads are anything to go by...
Depends what you mean by real - it is useful to think of them as mathematical tools for helping make predictions.
They cannot do this. You do not get a photon interacting with a wavefunction - the photon and wavefunction are different descriptions of the same thing.[also see answer to #4]
Some wave functions describe travelling waves, so they can be thought of as travelling. This basically means that the time evolution of a wave-function means that the peaks and troughs are in different places at different times.
Wavefunctions can be used to describe quantum entanglement.