Is the quantum wave function a real object or a mathematical tool?

[DanteKennedy]

I've read that there are a few interpretations about this (the ontic and epistemic view), but I'm curious about you guys. What's your opinion?

 

[martinbn]

When you say a real obejct, what do yo mean by that?

 

[Paul Colby]

Hum, is the number 3 a real object? I would say no but that’s just me.

 

[FactChecker]

Subatomic particles present wave-like behavior that is real and measurable but is not classical particle behavior. It's not just a mathematical trick.

 

[timmdeeg]

Whether or not the wave function real is in the sense of ontic is interpretation dependent.

 

[pines-demon]

I've read that there are a few interpretations about this (the ontic and epistemic view), but I'm curious about you guys. What's your opinion?

What is yours?

 

[gentzen]

Is quantum wave function a real object?​

I've read that there are a few interpretations about this

Independent of whether the wave function is real, it is certainly not an object. Just like an electric field is not an object either, despite being quite real.

 

[timmdeeg]

Independent of whether the wave function is real, it is certainly not an object. Just like an electric field is not an object either, despite being quite real.

What exactly is the difference between the wave function being an object and being ontic in case it is ontic according to a certain interpretation?

 

[PeterDonis]

What's your opinion?

Physics is not about opinions. Different QM interpretations say different, and mutually inconsistent, things about what the wave function is or is not, but they all make the same predictions for all experimental results, so there's no way of testing their claims. That's all we can say at this point.

 

[Demystifier]

Schrodinger equation is similar to the Hamilton-Jacobi equation, so I find intuitive the interpretation in which the wave function is analogous to the Hamilton-Jacobi function.

 

[javisot]

I've read that there are a few interpretations about this (the ontic and epistemic view), but I'm curious about you guys. What's your opinion?

My opinion is: "shut up and calculate"

 

[weirdoguy]

My opinion: I don't care.

See, that's why @PeterDonis wrote:

Physics is not about opinions.

 

[selfsimilar]

its real in the sense that you can derive the density matrix from it which gives the probabilistic interpretation or derive the quiding wave in bohmian interpretation.

 

[PeterDonis]

its real in the sense that you can derive the density matrix from it which gives the probabilistic interpretation or derive the quiding wave in bohmian interpretation.

How does that make it real?

 

[selfsimilar]

How does that make it real?

all variables used in equations are used to indicate some aspect of the behavior of the system.

 

[PeterDonis]

all variables used in equations are used to indicate some aspect of the behavior of the system.

But that doesn't mean variables used in equations are real. Equations and their variables are mathematical tools we use to make predictions. Which, according to the thread title and the OP's question, means they're not real.

 

[bhobba]

It can't be real, only a very good approximation to something that many think of as real.

That is because the current best theory we have is not ordinary, non-relativistic Quantum Mechanics (QM) as found in textbooks such as Ballentine, but rather Quantum Field Theory (QFT).

Nearly everyone, including me, assumed that QM is the non-relativistic limit of QFT. It came as a shock out of the blue when I read a paper showing that this is not the case. It is at the graduate level, but here is the link to the paper:

https://arxiv.org/abs/1712.06605

The conclusion is:
In the non-relativistic limit, our system is described by two fields, a and b, which actually represent the particle and anti-particle of the original system. Both of them satisfy the non-relativistic Schroedinger equation in operator form. So, anti-particles do not go away when you take the non-relativistic limit if you do it correctly.

Two wave functions are required in the classical domain because the possibility of an antiparticle can never be made to go away. One of the great mysteries of modern science is why antiparticles are so rare, especially considering that they must be included even in ordinary QM. Ignoring it in the vast majority of cases leads to no issues. If the fields of QFT are considered real, then yes, those two fields are real, but both are required. The single one in QM is wrong and hence can not be real. However, as an approximation used in QM, it is of great utility. But when considering foundational issues, its reality is 'problematic'

If you want to learn more about QFT as an interpretation, see the following book (expensive, but it has the advantage of being (mostly) correct):
https://www.amazon.com.au/Fields-Their-Quanta-Quantum-Foundations-ebook/dp/B0DLNLLG7Y

When I say "mostly", even it assumes QM is the non-relativistic limit of QFT. However, fortunately, it makes little difference to the interpretation. But it has many advantages, e.g., no wave-particle duality, and there are problems applying relativity directly to a QM 'particle' (see page 96 for a detailed explanation).

Thanks
Bill

 

[bhobba]

all variables used in equations are used to indicate some aspect of the behavior of the system.

Are the probabilities of the Black–Scholes model real?

Thanks
Bill

 

[gentzen]

its real in the sense that you can derive the density matrix from it which gives the probabilistic interpretation or derive the quiding wave in bohmian interpretation.

"Density Matrix Realism" by Eddy Keming Chen (https://arxiv.org/abs/2405.01025)

 

[selfsimilar]

But that doesn't mean variables used in equations are real. Equations and their variables are mathematical tools we use to make predictions. Which, according to the thread title and the OP's question, means they're not real.

That is why the answer is qualified by some sense which I mean that something "measurable" is the outcome. As far as I know the equation is generally considered correct that is why we say "shut up and calculate" the dispute is what that answer means, hence interpretations which all claim to make the same predictions. On the other hand, the question as to what do we mean by "real" is a more deeper philosophical enigma as all our equations are models that describe the systems but does not tell us what is out there in clear ontology, as is well known.

Hum, is the number 3 a real object? I would say no but that’s just me.

I say it is REAL. That is my only consolation in this existence

 

[Paul Colby]

I say it is REAL. That is my only consolation in this existence

Yeah, in multiple editing passes I added “object” to side step your double meaning style reply. In retrospect “is the number $3+i1$ a real object” would have been a more complete safe guard.

 

[gentzen]

Yeah, in multiple editing passes I added “object” to side step your double meaning style reply. In retrospect “is the number $3+i1$ a real object” would have been a more complete safe guard.

No, my opinion is that the number $3+i1$ is not an element of reality.
And if it would occur as an entry in a density matrix? Still no, because it would only be an artifact of a specific choice of basis.

 

[Paul Colby]

No, my opinion is that the number 3+i1 is not an element of reality.

An element of reality isn’t an object either, right? Physics is a human activity. Humans use tools to do physics. These tools include devices we fabricate as well as intellectual tools like mathematics. Fragments of intellectual tools aren’t objects in the usual sense of object. A brick on my patio is a real object, the number characterizing it’s measured length isn’t.

 

[FactChecker]

Are we talking about the thing that is represented? Or about the means of representing it?