Determinism in a reversible polarizer experiment

In summary: What's indeterministic is the outcome of measurements, given this complete state preparation. That's simply because the meaning of the quantum state in the standard minimal interpretation is probabilistic, i.e., all you can say about the outcomes of measurements of any observable are the probabilities for measuring one of the possible values of this observable. The outcome of the measurement of some observable is certain only if the system at the time of the measurement is in an eigenstate of the self-adjoint operator representing the measured observable. Then you get always the corresponding eigenvalue as a measurement outcome.What's indeterministic is the outcome of measurements, given this complete state preparation. What's indeterministic is the outcome of
  • #1
kurt101
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TL;DR Summary
How is a polarizer experiment where changes to the polarization state are restored by following the reverse process understood by different quantum mechanical interpretations?
If light at a known polarization goes through a beam splitting polarizer that changes the light's polarization and then goes through the reverse orientation of that polarizer it will exit with the same polarization that it entered with. See the following picture:

analyzer-loop-2.png


If the polarization state can be changed and then later changed back to its original state by following the reverse process, doesn't this imply these changes are deterministic?

How is this experiment understood by someone who supports the interpretation that our universe is fundamentally probabilistic?
 
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  • #2
kurt101 said:
If the polarization state can be changed and then later changed back to its original state by following the reverse process, doesn't this imply these changes are deterministic?
These particular ones, yes. But that's a much, much weaker statement than saying that all quantum processes are deterministic.

kurt101 said:
How is this experiment understood by someone who supports the interpretation that our universe is fundamentally probabilistic?
No interpretation that I'm aware of says that nothing can be deterministic. Every interpretation must be consistent with the basic math of QM, and the deterministic nature of the particular process you describe is a consequence of the basic math of QM. But again, as above, that is very different from claiming that all quantum processes must be deterministic.
 
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  • #3
kurt101 said:
Summary:: How is a polarizer experiment where changes to the polarization state are restored by following the reverse process understood by different quantum mechanical interpretations?

If light at a known polarization goes through a beam splitting polarizer that changes the light's polarization and then goes through the reverse orientation of that polarizer it will exit with the same polarization that it entered with. See the following picture:

View attachment 293363

If the polarization state can be changed and then later changed back to its original state by following the reverse process, doesn't this imply these changes are deterministic?

How is this experiment understood by someone who supports the interpretation that our universe is fundamentally probabilistic?
Its technically in a superposition of output beams, so you're not really changing a 'definite' state (ie definitely horizontally polarized photon). The fundamental Schrodinger equation is deterministic.
 
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  • #4
That's a very important point! The dynamics of quantum theory is deterministic, i.e., if you have determined the state of the quantum system completely at time ##t=0##, i.e., if you have prepared a quantum system at time ##t=0## in a pure state, then the unitary time evolution of this state (using the Schrödinger picture of time evolution) delivers the precise state at any later time ##t>0##. That means quantum theory is causal. Since the unitary time evolution is reversible, it's also possible (in principle) to get back the the original state.

What's indeterministic is the outcome of measurements, given this complete state preparation. That's simply because the meaning of the quantum state in the standard minimal interpretation is probabilistic, i.e., all you can say about the outcomes of measurements of any observable are the probabilities for measuring one of the possible values of this observable. The outcome of the measurement of some observable is certain only if the system at the time of the measurement is in an eigenstate of the self-adjoint operator representing the measured observable. Then you get always the corresponding eigenvalue as a measurement outcome.
 
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  • #5
vanhees71 said:
What's indeterministic is the outcome of measurements, given this complete state preparation.
My understanding is a measurement is a local interaction and once that interaction occurs, in principle we know where the particle is, and so the previous quantum system that described the particle is gone and a new one is created. Furthermore I am able to treat the two quantum systems as independent because their probabilities can be added in a classical way.

Is it correct to say that the quantum system is deterministic prior to the measurement and is deterministic right after the measurement and only at some instant in time and space where the measurement takes place is it indeterministic?
 
  • #6
No, what's "deterministic" is the state evolution, but that's rather called "causality", because "deterministic" means that all observables always take sharp values as in classsical physics, but that's not the case according to quantum theory, becausing having complete knowledge of the state, i.e., having prepared the system in a pure state, only implies probabilities for the outcome of measurements. With the work of Bell we have pretty strong evidence that this is indeed the case, i.e., that observables don't take determined values, if the system is not prepared in a state for which an observable (or at most a complete set of compatible observables) take a determined value.

I'm a proponent of the minimal statistical interpretation, and for me the collapse hypothesis is not part of this minimal interpretation of QT. To the contrary usually a measurement does not prepare the system in a state where the measured observables take the measured value. E.g., if you register a photon with a detector or a photoplate it's simply absorbed and destroyed.

Last but not least, it's important to keep in mind that photons are the least particle like object you can imagine. It doesn't even have a position observable to begin with. In other words you cannot localize a photon at all. All there is are probability distributions for detecting a photon at the place the detector is located, given the state the photon is prepared in.
 
  • #7
vanhees71 said:
"deterministic" means that all observables always take sharp values as in classsical physics, but that's not the case according to quantum theory
My understanding is that a particle with mass has a probability amplitude that is made up of a summation of simple waves and these waves are essential in defining the state of the particle. If particles with mass are fundamentally made up of waves then the probabilistic nature is ultimately dependent on something that is deterministic and not just causal. Put in another way, it very much seems like the ingredients are all deterministic and it is only the result of making a calculation using deterministic ingredients that gives a probability.

Also when you square the sum of the probability amplitudes you are incorporating entanglement. If I understand correctly entanglement requires local preparation which is deterministic. So it seems as if the recipe for a statistical outcome is made up of ingredients that are deterministic.
vanhees71 said:
what's "deterministic" is the state evolution
Would it be correct to say the quantum state in a quantum system is "deterministic" prior to the measurement and is "deterministic" right after the measurement and only at some instant in time and space where the measurement takes place is the quantum state indeterministic?
 
  • #8
kurt101 said:
Would it be correct to say the quantum state in a quantum system is "deterministic" prior to the measurement and is "deterministic" right after the measurement and only at some instant in time and space where the measurement takes place is the quantum state indeterministic?
In QM (Schrodinger equation), there is no concept of measurement. A superposition continues to evolve as a superposition.
 
  • #9
kurt101 said:
My understanding is that a particle with mass has a probability amplitude that is made up of a summation of simple waves
Where are you getting this understanding from? It would be helpful if you gave a reference.

Expressions for probability amplitudes are often basis dependent, and one should not be trying to construct one's viewpoint of the physical meaning of an amplitude from basis dependent expressions. I suspect that is what you are doing here.
 
  • #10
StevieTNZ said:
In QM (Schrodinger equation), there is no concept of measurement. A superposition continues to evolve as a superposition.
Can you explain the "In QM there is no concept of measurement" part? Are you saying that during a measurement the superposition continues to evolve? If that is what you are saying that seems very reasonable to me as I would not expect something special to happen with regards to superposition when a measurement happens.
 
  • #11
No one knows when "measurement" occurs, nor is it defined in qm.
 
  • #12
PeterDonis said:
Where are you getting this understanding from? It would be helpful if you gave a reference.

Expressions for probability amplitudes are often basis dependent, and one should not be trying to construct one's viewpoint of the physical meaning of an amplitude from basis dependent expressions. I suspect that is what you are doing here.
What do you mean by "basis dependent"? The probabilities for the outcome of measurements calculated in quantum theory are of course not basis dependent in any sense, as it must be for observable quantities.
 
  • #13
vanhees71 said:
What do you mean by "basis dependent"?
When @kurt101 talks about a probability amplitude being "a summation of simple waves", I think he is talking about the expression for the amplitude (or the quantum state) in a particular basis. But it's possible I'm misinterpreting; that's why I asked him for a reference.
 
  • #14
vanhees71 said:
The probabilities for the outcome of measurements calculated in quantum theory are of course not basis dependent
Yes, agreed. But neither are those observable quantities "a summation of simple waves". They're just observables.
 
  • #15
PeterDonis said:
Where are you getting this understanding from? It would be helpful if you gave a reference.

Expressions for probability amplitudes are often basis dependent, and one should not be trying to construct one's viewpoint of the physical meaning of an amplitude from basis dependent expressions. I suspect that is what you are doing here.
I am reading this from Quantum Mechanics (second edition) Concepts and Applications by Nouredine Zettili. Chapter 1.8.1. "Localized wave packets can be constructed by superposing, in the same region of space, waves of slightly different wavelengths, but with phases and amplitudes chosen to make the superposition constructive in the desired region and destructive outside it."
Later on it says "We can construct the packet psi(x, t) by superposing plane waves (propagating along the x-axis) of different frequencies (or wavelengths):" and the equation is displayed which I will post, but it will take me a bit to do as I am not great with Latex math equations.
 
  • #16
kurt101 said:
Localized wave packets can be constructed by superposing, in the same region of space, waves of slightly different wavelengths, but with phases and amplitudes chosen to make the superposition constructive in the desired region and destructive outside it."
Ok, so I was right, this claim is basis dependent; the "waves" (i.e., pure waves of a single wavelength) are the basis. (A "basis" is just a set of states that are linearly independent and span the Hilbert space, i.e., any state can be written as a linear combination of them. The pure waves of a single wavelength are just one of an infinite number of possible sets of basis states.)

But physically, we have no reason to believe that those particular states have any special status; the only reason we do the kind of construction being described is that it's easier for us mathematically. In other words, the fact that we do this particular mathematical trick in our models does not mean that the particles we actually observe in the real world really are "a summation of simple waves".
 
  • #17
kurt101 said:
"We can construct the packet psi(x, t) by superposing plane waves (propagating along the x-axis) of different frequencies (or wavelengths):" and the equation is displayed which I will post
You don't need to post it, it's something that anyone at all familiar with QM will have seen. There is nothing wrong with it mathematically; it just doesn't have the physical meaning you are trying to give it.
 
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  • #18
PeterDonis said:
But physically, we have no reason to believe that those particular states have any special status; the only reason we do the kind of construction being described is that it's easier for us mathematically. In other words, the fact that we do this particular mathematical trick in our models does not mean that the particles we actually observe in the real world really are "a summation of simple waves".
Is it wrong to think of particles as having the general shape and characteristics of the graph created from the summation of the simple waves like in Figure 7.3.1 of this link wave packet 7.3.1 ? Does this fall under interpretation?

I realize that thinking of a particle this way would imply the particle is infinitely spread out, which leads me to my next question. How do we know that a particle is not physically spread out as a direct interpretation of the function might suggest versus something more abstract?

If the math does not change between the two interpretations how could a direct interpretation be wrong?
 
  • #19
kurt101 said:
Is it wrong to think of particles as having the general shape and characteristics of the graph created from the summation of the simple waves like in Figure 7.3.1 of this link wave packet 7.3.1 ?
If a "wave packet" state accurately predicts the probabilities of various measurement results, then that "wave packet" state can be reasonably thought of as describing the particles being measured (or more precisely the preparation process that produced those particles).

But that is not what you are doing. You are not thinking of the wave packet state as describing the particles. You are thinking of the "simple waves" as describing the particles--different particles being different sums of them. There is no physical justification for that, since there are an infinite number of different ways to express the "wave packet" states that actually predict measurement results as sums of states from some set of basis states. The only physically meaningful thing is the probabilities of measurement results, and the state that predicts those is the "wave packet" state. So the "wave packet" state is the only thing in the math that has any claim to physical meaning.

kurt101 said:
Does this fall under interpretation?
I'm not aware of any QM interpretation that assigns any special physical meaning to the "simple wave" basis states. So unless you can find a reference to some QM interpretation that does that, it's your personal theory or speculation, which is off topic here.

kurt101 said:
How do we know that a particle is not physically spread out
Because we don't measure particles to be physically spread out.

kurt101 said:
If the math does not change between the two interpretations how could a direct interpretation be wrong?
Your claim that the "simple wave" basis states have physical meaning is not a "direct interpretation". It is a very indirect interpretation, taking one particular way of expressing the "wave packet" states, the ones that actually predict the probabilities of measurement results, as sums of states from a set of basis states, and claiming that that particular way of doing the summation has some special physical meaning, even though there is nothing in our actual observations to support such a claim.
 
  • #20
PeterDonis said:
But that is not what you are doing. You are not thinking of the wave packet state as describing the particles. You are thinking of the "simple waves" as describing the particles--different particles being different sums of them.
I am confused because I thought I had banished any mention that the wave packet was actually a summation of simple states after you said the first time not to think that way. I only referred to wave packet in the figure 7.3.1 and not the simple graphs.
 
  • #21
kurt101 said:
I thought I had banished any mention that the wave packet was actually a summation of simple states
You thought wrong:

kurt101 said:
the graph created from the summation of the simple waves
 
  • #22
kurt101 said:
My understanding is a measurement is a local interaction and once that interaction occurs, in principle we know where the particle is, and so the previous quantum system that described the particle is gone and a new one is created.
Yes, if you choose to use (for example) a collapse interpretation. Other interpretations, most notable MWI, will describe measurements differently.
 
  • #23
PeterDonis said:
You thought wrong:
You are taking me out of context. This is what I wrote: "Is it wrong to think of particles as having the general shape and characteristics of the graph created from the summation of the simple waves like in Figure 7.3.1 of this link". I am asking about the shape and characteristic of the graph created from the summation of simple waves. I say nothing about the simple waves being real. As far as I know the graph of the wave packet is still composed of simple waves like my book says. When you told me the simple waves were not to be thought of as real, my next question was is the complex wave (i.e. the wave packet) that is the summation of simple waves real?
 
  • #24
kurt101 said:
You are taking me out of context.
No, I'm not. You made a straightforward claim:

kurt101 said:
I thought I had banished any mention that the wave packet was actually a summation of simple states
"Banished any mention" does not mean "only mentioned it in a particular context". It means not mentioning it at all. If you are going to abandon the mental model of a wave packet as "a summation of simple states", then abandon it.

kurt101 said:
I am asking about the shape and characteristic of the graph created from the summation of simple waves.
Notice what I crossed out in the quote above. The quote still asks the same question and is a basis for the same substantive discussion. If you think "the graph" is not specific enough without some kind of qualifier, you could say "the wave packet graph".

kurt101 said:
As far as I know the graph of the wave packet is still composed of simple waves like my book says.
You're missing the point. Mathematically, there are an infinite number of different ways of making a wave packet by a summation of some set of states. Not only that, you could just as well write a single "simple wave" as a summation of different wave packets. None of this tells you anything whatever about the actual physics. It's just math. If you want to talk about math, you can start a separate thread in the math subforum. Here in this subforum, we are talking about physics.

kurt101 said:
When you told me the simple waves were not to be thought of as real, my next question was is the complex wave (i.e. the wave packet) that is the summation of simple waves real?
Again, you're missing the point. As I said above: None of this math about "summation" tells you anything whatever about the actual physics. You don't figure out the actual physics by looking at what can be mathematically expressed as a summation of what; that's an infinite rabbit hole that will lead you nowhere. You figure out the actual physics by starting with what we actually observe.
 
  • #25
PeterDonis said:
"Banished any mention" does not mean "only mentioned it in a particular context". It means not mentioning it at all. If you are going to abandon the mental model of a wave packet as "a summation of simple states", then abandon it.
Here is what I wrote "I am confused because I thought I had banished any mention that the wave packet was actually a summation of simple states after you said the first time not to think that way." In other words, I abandoned the idea that it has reality associated with it, not that it is the mathematical way to calculate the graph of the wave packet.

PeterDonis said:
Notice what I crossed out in the quote above. The quote still asks the same question and is a basis for the same substantive discussion. If you think "the graph" is not specific enough without some kind of qualifier, you could say "the wave packet graph".
Ok.

PeterDonis said:
You're missing the point. Mathematically, there are an infinite number of different waysof making a wave packet by a summation of some set of states. Not only that, you could just as well write a single "simple wave" as a summation of different wave packets. None of this tells you anything whatever about the actual physics. It's just math. If you want to talk about math, you can start a separate thread in the math subforum. Here in this subforum, we are talking about physics.

Again, you're missing the point. As I said above: None of this math about "summation" tells you anything whatever about the actual physics. You don't figure out the actual physics by looking at what can be mathematically expressed as a summation of what; that's an infinite rabbit hole that will lead you nowhere. You figure out the actual physics by starting with what we actually observe.
Here is what this link says about a wave packet https://phys.libretexts.org/Bookshelves/University_Physics/Book:_University_Physics_(OpenStax)/Book:_University_Physics_III_-_Optics_and_Modern_Physics_(OpenStax)/07:_Quantum_Mechanics/7.03:_The_Heisenberg_Uncertainty_Principle

"Similar statements can be made of localized particles. In quantum theory, a localized particle is modeled by a linear superposition of free-particle (or plane-wave) states called a wave packet. An example of a wave packet is shown in Figure 7.3.17.3.1. A wave packet contains many wavelengths and therefore by de Broglie’s relations many momenta—possible in quantum mechanics! This particle also has many values of position, although the particle is confined mostly to the interval ΔxΔx. The particle can be better localized (ΔxΔx can be decreased) if more plane-wave states of different wavelengths or momenta are added together in the right way (ΔpΔp is increased). According to Heisenberg, these uncertainties obey the following relation."

Is this a bad description? Am I not supposed to take the characteristics of the wave packet graph as an accurate model of a particle? The two characteristics I notice in the graph is that it has infinite length and diminishing sinusoidal amplitude. Are those not accurate characteristics?
 
  • #26
kurt101 said:
Here is what this link says about a wave packet
Nothing in there contradicts anything I said. That article, like innumerable articles, books, and papers about QM, uses plane waves as a set of basis states because it is mathematically convenient. That is all there is to it.

kurt101 said:
Is this a bad description?
Is what a bad description? The description of the properties of the particular kind of curve that is called a "wave packet"? No. But that's a mathematical description, not a description of physics.

kurt101 said:
Am I not supposed to take the characteristics of the wave packet graph as an accurate model of a particle?
It depends on what kind of model you think it is. If you think it's a model that gives you predictions of the probability of finding a particle in various places, yes, it's fine (but with some caveats--see below). But if you think it's something else, that might not be fine.

kurt101 said:
The two characteristics I notice in the graph is that it has infinite length and diminishing sinusoidal amplitude. Are those not accurate characteristics?
Taken literally, no, the first one is not. If I have an electron in a lab on Earth, and I model it with one of these wave packets, taken literally, the mathematical model is telling me there is a nonzero probability of finding that electron in a galaxy a billion light years away. Physically, nobody ever makes such a claim based on that mathematical model, at least not in legitimate sources (textbooks and peer reviewed papers).

The second one is OK if we confine our attention to the region of space where the particle has a physically realistic chance of being (for example, somewhere inside the cathode ray tube or accelerator we're using to produce electrons).
 
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FAQ: Determinism in a reversible polarizer experiment

What is determinism in a reversible polarizer experiment?

Determinism in a reversible polarizer experiment refers to the idea that the outcome of the experiment is predetermined and can be predicted with complete accuracy. This is based on the belief that all events and actions in the universe are determined by previous causes and the laws of physics.

How does a reversible polarizer experiment work?

In a reversible polarizer experiment, a polarizer is used to filter light and only allow light waves with a specific orientation to pass through. The light then passes through a second polarizer, which can be rotated to change the orientation of the light waves. By measuring the intensity of the light that passes through the second polarizer at different orientations, the experiment can determine the original orientation of the light waves.

What is the significance of determinism in a reversible polarizer experiment?

The idea of determinism in a reversible polarizer experiment is significant because it challenges the concept of free will and suggests that all actions and events are predetermined. It also has implications for our understanding of the nature of reality and the laws of physics.

Can determinism be proven in a reversible polarizer experiment?

No, determinism cannot be proven in a reversible polarizer experiment. While the experiment may suggest that the outcome is predetermined, it cannot account for all variables and there may still be some level of uncertainty in the results.

What are the ethical implications of determinism in a reversible polarizer experiment?

Determinism in a reversible polarizer experiment raises ethical questions about the role of personal responsibility and accountability. If all actions and events are predetermined, then individuals may not be fully responsible for their actions. This can have implications for the justice system and our understanding of moral responsibility.

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