# Most basic question about psi ever

1. Aug 21, 2014

### fluidistic

Hi guys,
I thought that |psi|^2 was a probability density function and that when integrated over a region this would give me the probability to find the particle in that region. (assuming that psi was normalized)
But some ph.d. in physics told me that no, that's wrong and when I use the word particle I'm already using an interpretation of QM.
I notice that Landau never mention the word "particle" when he defines psi.

Now that I think about it, if I have a system of many particles and a single psi representating the probability density funtion of all those particles, I do not really know what the integral of |psi|^2 would represent in that case. The prob. to measure a single one of these many particles inside the integrated region?

Am I completely off?

2. Aug 21, 2014

### atyy

The Born rule says that the probability to find the system with state |ψ> in a state |ø> is |<ø|ψ>|2.

Let's do a system with one particle first. The state |x> means the particle has a definite position x. So the probability to find the particle at position x is the probability to find the particle in state |x>. By the Born rule, that probability is |<x|ψ>|2 = |ψ(x)|2.

For a system with two non-identical particles, the state |x,y> means particle 1 has position x and particle 2 has position y. Then the probability to find patricle 1 at position x and particle 2 at position y is |<x,y|ψ>|2 = |ψ(x,y)|2

http://galileo.phys.virginia.edu/classes/252/symmetry/Symmetry.html
http://www.eng.fsu.edu/~dommelen/quantum/style_a/complex.html

Last edited: Aug 21, 2014
3. Aug 22, 2014

### microsansfil

Here is an example about "A wave function for a single electron". Only shows the places where the electron's probability density is above a certain value (here 0.02 nm−3) : this is calculated from the probability amplitude.

They are also very careful in the dialectic.

Patrick

4. Aug 22, 2014

### Staff: Mentor

Not so much wrong as incomplete. The way you were thinking about it is fine for a system consisting of a single particle constrained to move only on the x-axis, but it doesn't work for anything more complicated - as you've discovered when you found yourself asking:

Consider the two-particle case: we have two particles, both constrained to move only on the x-axis. Now we write the wave function as $\psi(p,q)$ where $p$ and $q$ are the positions of the two particles. The probability of finding the first particle in the region bounded by $A$ and $A+\Delta{A}$ and the second particle in the region bounded by $B$ and $B+\Delta{B}$ is given by
$$\int_A^{A+\Delta{A}}\int_B^{B+\Delta{B}}\psi^*(p,q)\psi(p,q)\;\; dpdq$$
There are two things to notice here.
First, this is not an integration across physical space. Instead, it's an integration across a region of an abstract two-dimensional space in which the two axes correspond to the two positions of the two particles on the physical x-axis.
Second, this solution is formally identical to the solution for a single particle that is free to move in about in two dimensions. In fact, it works for any system with two degrees of freedom; $p$ and $q$ don't even have to represent physical positions. They can be whatever dynamical variables are needed for the problem at hand.

5. Aug 22, 2014

### microsansfil

Perhaps when you read this kind of articule "There are no particles, there are only fields" on arxiv. Calculate the position is not it already an interpretation ?

This is how I understand the sentence : "But some ph.d. in physics told me that no, that's wrong and when I use the word particle I'm already using an interpretation of QM".

However, I may be wrong.

Patrick

6. Aug 22, 2014

### bhobba

Its not an interpretation - its an assumption - namely position is an observable.

The link you gave is correct - there are no particles - only fields.

From those fields position may or may not be an observable:
http://www.mat.univie.ac.at/~neum/physfaq/topics/position.html

Thanks
Bill

7. Aug 22, 2014

### Staff: Mentor

It might be time to start a new thread here - we're moving way beyond "the most basic question about psi ever"

8. Aug 22, 2014

### microsansfil

Is it the same in QFT ? Is there a position operator in second quantization ?

Patrick

9. Aug 22, 2014

### bhobba

Well, second quantisation is a rather bad name.

There is a basic difficulty with normal QM and relativity.

Position is an operator, and time a parameter, yet relativity says they should be treated on the same footing.

QFT treats both as a parameter ie a field.

So, no, position is not inherently an observable in QFT, and that is by design.

Thanks
Bill

10. Aug 22, 2014

### stevendaryl

Staff Emeritus
That was something that occurred to me just recently, and was briefly discussed in some thread. It seems that the physics of 3-dimensional space with N particles is almost the same as the physics of 3N-dimensional space with 1 particle. The difference is locality of interactions. For some interactions that seem local in 3N-dimensional configuration space, the corresponding interaction in 3-dimensional space would be nonlocal.

Last edited: Aug 22, 2014
11. Aug 22, 2014

### microsansfil

In this article the author write : "Localization via position operators is in conflict with causality".

He give also comments on the question whether these mathematical differences have significant consequences for the physical interpretation of basic concepts of QM.

Patrick

12. Aug 22, 2014

### atyy

13. Aug 22, 2014

### bhobba

I had a quick look - it looked fine to me.

What was your issue.

Thanks
Bill

14. Aug 22, 2014

### atyy

In non-relativistic quantum mechanics, there are certainly systems with a fixed number of particles. These are not classical particles. They are quantum particles, which do not simultaneously have definite position and momentum. Thus they do not have classical trajectories.

The matter of interpretation is whether they have non-classical trajectories. In Bohmian Mechanics, they do.

In exact relativistic quantum field theory, it can be argued that fields are more fundamental. Here the quantum particle is an excitation of the field. However, no one can guarantee that relativity is exact, so even to explain relativistic phenomena, quantum mechanics with a fixed number of entities is viable, as in lattice gauge theory.

15. Aug 22, 2014

### atyy

The abstract claims "As this paper shows, experiment and theory imply unbounded fields, not bounded particles, are fundamental." That is wrong. Experiment only goes up to a certain energy, and has a certain precision. Claiming to know the final theory is as much pseudoscience as all the things he criticizes.

For example, can he distinguish exact relativistic theories defined in infinite volume from a lattice gauge theory with fine but finite lattice spacing in large but finite volume? The Hilbert space of the latter is finite dimensional, and so more like quantum mechanics than quantum field theory.

Last edited: Aug 22, 2014
16. Aug 22, 2014

### bhobba

Yea - fair point.

But I am sure the author would agree and point out you are reading more into it than intended.

QFT is a deeper more exact theory than standard QM - that's all that's meant - not that its the final theory.

Thanks
Bill

17. Aug 22, 2014

### atyy

If he is talking about QFT, then it is roughly a theory with an infinite number of particles, or at least a theory in which particles can be created and destroyed. So it is a theory with more than one particle. But he writes "Thus the Schroedinger field is a space-filling physical field whose value at any spatial point is the probability amplitude for an interaction to occur at that point." and "It follows that the Schroedinger matter field, the analogue of the classical EM field, is a physical, space-filling field." That is of course fine for one non-relativistic particle, and a very useful mental image. But given his concern about quantum field theory, this picture of the single particle wave function is one that does not generalize to more than one particle. The wave function of two particles is not in general "space filling" in the sense that he means.

18. Aug 22, 2014

### Demystifier

Two (hopefully non-controversial) points:

First, in non-relativistic QFT (used e.g. in solid state physics), wave function is still the probability amplitude for the position of the particle.

Second, in fermionic QFT, the field is not even an observable.

In addition, 3 different concepts are often mixed up: relativistic QM, QFT (which may be non-relativistic), relativistic QFT.

19. Aug 22, 2014

### bhobba

Indeed, as the Fock space formalism shows. And even deeper, when deriving QFT, one often uses a model of a large number of interacting particles that you take the limit in the continuum.

That does not contradict the infinite particle picture.

A quantum field can be viewed as a field of operators. The space it operates on is spanned by the creation and annihilation operators. Its two sides of exactly the same coin.

I don't think its fine for a single particle. A matter field, just like any quantum field, is a field of quantum operators acting on the creation and annihilation operators. You cant speak of single particles in that formalism.

The paper is not meant to be a rigorous development of QFT. It is meant to be a review article of the idea that quantum fields are a deeper theory than normal QM, and as such explains many of the issues people get worried about with QM.

He freely admits it doesn't solve the measurement problem, but many things like the double slit experiment become clearer.

Also it soon will be published, if it already hasn't been, in the AJP. Articles with glaringly obvious issues are normally not published there, they are vetted thoroughly to ensure that doesn't happen. Of course no process is perfect and some incorrect things can creep through - but I don't think that's what's going on here. Since its explaining highly complex ideas that should really be expressed mathematically in English subtleties are bound to arise. IMHO that all that's going on - overall it is correct.

It's at a bit more advanced level to its beginner companion book - Fields Of Color:
https://www.amazon.com/Fields-Color-theory-escaped-Einstein/dp/0473179768

But if you want a more advanced look again check out:
https://www.amazon.com/An-Introduction-Realistic-Quantum-Physics/dp/9812381767

The author may have a copy on his site as well - but don't hold me to it.

That QFT explains many of the issues (not all - but many) with standard QM is for me not controversial at all.

Thanks
Bill

Last edited by a moderator: May 6, 2017
20. Aug 22, 2014

### atyy

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook