High School The Wave Function of Our World: How Does It Emerge?

[stevendaryl]

Thanks for this. I'll think about it but I doubt it will change my views. In fact the last sentence just reinforces the fact that the 'state' you talk about is only a statistical statement - however it came about.

I don't think that's a consistent interpretation of quantum probabilities. To say that "spin-up in the x-direction" means "50/50 probability of being spin-up or spin-down in the z-direction" doesn't make a lot of sense to me.

 

[Mentz114]

For microscopic objects, we can definitely demonstrate the difference. As I said already, an interference pattern is due to a superposition of two different sources of light (or particles).

Classical light is a wave and can form super-positions so it's not very relevant and is an exception where super-position actually happens.

For macroscopic objects, there is no way in practice to demonstrate interference between macroscopically distinguishable states, so the distinction between superpositions and mixtures can't be demonstrated.

OK.

However, it seems strange to me, or even inconsistent, to assume that macroscopic objects can only be in proper mixtures, while microscopic objects can be in superpositions. It doesn't make sense to me to have different rules for macroscopic systems.

If you insist on using the same rules for macro/micro contexts then you should use only statistical mechanics when doing a classical problem.
Also, you should never assume the position has a value etc.

I don't think that's a consistent interpretation of quantum probabilities. To say that "spin-up in the x-direction" means "50/50 probability of being spin-up or spin-down in the z-direction" doesn't make a lot of sense to me.

Nor me. But you said it. Maybe a typo ?

 

[Nugatory]

Nor me. But you said it. Maybe a typo ?

No matter who said it, it sounds to me more like one of the innumerable pitfalls of using natural language than anything else. '"Spin-up in the x-direction" means "50/50 probability of being spin-up or spin-down in the z-direction"' doesn't make a lot of sense to me either because of the problem with the word "being"... But "A particle prepared in the state 'spin-up in the x-direction' will, if measured in the z-direction, yield spin up or spin down with 50/50 probability" works a lot better.

 

[stevendaryl]

Classical light is a wave and can form super-positions so it's not very relevant and is an exception where super-position actually happens.

But you would agree that a classical notion of superposition is not statistical. Classically, light doesn't have a certain probability of going through one slit or another. Superposition is NOT about statistics, classically or quantum-mechanically.

 

[stevendaryl]

If you insist on using the same rules for macro/micro contexts then you should use only statistical mechanics when doing a classical problem. Also, you should never assume the position has a value etc.

I really don't understand what you're talking about. At the microscopic level, superpositions are NOT about statistics. So the mystery is how do they become about statistics at the macroscopic level?

 

[stevendaryl]

I said:

I don't think that's a consistent interpretation of quantum probabilities. To say that "spin-up in the x-direction" means "50/50 probability of being spin-up or spin-down in the z-direction" doesn't make a lot of sense to me.

@Mentz114 responded:

Nor me. But you said it. Maybe a typo ?

Then I have no idea what you mean by saying that superpositions are statistical. You previously said (about what it might mean for a macroscopic detector be in a superposition of an "Up" state and a "Down" state):

It means that each run in the ensemble produces a Up or Down, and the relative frequencies will converge to some probabiities.

It seems to me that you are saying two things that seem mutually inconsistent:

1. For a macroscopic device, being in a superposition of an "Up" state and a "Down" state means that for every run, it's either in the "Up" state or the "Down" state, with probabilities given by the relevant coefficients.
2. For a single particle, being in a superposition of "spin-up in the z-direction" and "spin-down in the z-direction" does NOT mean that for every run, it's either in the "spin-up in the z-direction" state or "spin-down in the z-direction" state, with certain probability give by the relevant coefficients.

You are explain macroscopic superpositions as being statistical, but microscopic superpositions are NOT statistical.

 

[stevendaryl]

No matter who said it, it sounds to me more like one of the innumerable pitfalls of using natural language than anything else. '"Spin-up in the x-direction" means "50/50 probability of being spin-up or spin-down in the z-direction"' doesn't make a lot of sense to me either because of the problem with the word "being"... But "A particle prepared in the state 'spin-up in the x-direction' will, if measured in the z-direction, yield spin up or spin down with 50/50 probability" works a lot better.

I said it because that was how I interpreted @Mentz114 when he said that the meaning of a superposition was statistical. There is nothing statistical about spin-up in the x-direction being a superposition of spin-up in the z-direction and spin-down in the z-direction.

 

[PeterDonis]

The fact that super-position can be created by a change of basis shows that super-position has no physical significance.

This is true, but it does not imply this:

FAPP a super-position is a statistical mixture.

You can do experiments that distinguish pure states from mixed states. A superposition is a pure state (more precisely, a pure state can be said to be a superposition with an appropriate choice of basis). It is not a mixed state, which is what the term "statistical mixture" denotes.

 

[stevendaryl]

You can do experiments that distinguish pure states from mixed states. A superposition is a pure state (more precisely, a pure state can be said to be a superposition with an appropriate choice of basis). It is not a mixed state, which is what the term "statistical mixture" denotes.

Yes, but the miracle of decoherence, if you want to consider it that, is that a superposition involving an astronomical number of degrees of freedom is FAPP indistinguishable from a statistical mixture. So once a system has interacted sufficiently with the environment, it can be treated as if it were in a statistical mixture.

 

[Mentz114]

It seems to me that you are saying two things that seem mutually inconsistent:

1. For a macroscopic device, being in a superposition of an "Up" state and a "Down" state means that for every run, it's either in the "Up" state or the "Down" state, with probabilities given by the relevant coefficients.
2. For a single particle, being in a superposition of "spin-up in the z-direction" and "spin-down in the z-direction" does NOT mean that for every run, it's either in the "spin-up in the z-direction" state or "spin-down in the z-direction" state, with certain probability give by the relevant coefficients.

You explain macroscopic superpositions as being statistical, but microscopic superpositions are NOT statistical.

I don't understand.

To me - a single particle being in a superposition of "spin-up in the z-direction" and "spin-down in the z-direction" means EXACTLY and ONLY that it will be detected in either of those states with probability 1/2.

I guess I'm missing the point.

Post #26 by VanHees71 in the other topic https://www.physicsforums.com/threads/are-superposition-states-observable.909502 cleared some things for me - but not in a way you would necessarilly agree with.

 

[stevendaryl]

To me - a single particle being in a superposition of "spin-up in the z-direction" and "spin-down in the z-direction" means EXACTLY and ONLY that it will be detected in either of those states with probability 1/2.

Well the state |u_x\rangle (spin-up in the x-direction) is equal to \frac{1}{\sqrt{2}} (|u_z\rangle + |d_z\rangle). So that would seem to mean that

• Being spin up in the x-direction means EXACTLY and ONLY that it will be detected in either the state "spin-up in the z-direction" or "spin-down in the z-direction".

But it seemed that you were denying that that was the meaning of "spin-up in the x-direction".

 

[mike1000]

I don't understand.

To me - a single particle being in a superposition of "spin-up in the z-direction" and "spin-down in the z-direction" means EXACTLY and ONLY that it will be detected in either of those states with probability 1/2.

Well the state |u_x\rangle (spin-up in the x-direction) is equal to \frac{1}{\sqrt{2}} (|u_z\rangle + |d_z\rangle). So that would seem to mean that

• Being spin up in the x-direction means EXACTLY and ONLY that it will be detected in either the state "spin-up in the z-direction" or "spin-down in the z-direction".

But it seemed that you were denying that that was the meaning of "spin-up in the x-direction".

I hope you two clear this up because I would very much like to know what the correct answer is. Does \frac{1}{\sqrt{2}} (|u_z\rangle + |d_z\rangle) mean 50% probability of observing spin up in the +/- z direction or does it mean you will measure spin up in the x direction?

 

[stevendaryl]

To me - a single particle being in a superposition of "spin-up in the z-direction" and "spin-down in the z-direction" means EXACTLY and ONLY that it will be detected in either of those states with probability 1/2.

That can't possibly be right. It's true that the state |u_z\rangle = \frac{1}{\sqrt{2}} (|u_z\rangle + |d_z\rangle) implies that a measurement of spin in the z-direction will give spin-up with probability 1/2 and spin-down with probability 1/2. But it also means that a measurement of spin in the x-direction will give spin-up with probability 1. In contrast, the state |d_x\rangle = \frac{1}{\sqrt{2}} (|u\rangle - |d\rangle) has the same probabilities--1/2 and 1/2--of being measured spin-up or spin-down in the z-direction, but it isn't the same state. And it has probability 0 of being measured to be spin-up in the x-direction.

 

[stevendaryl]

I hope you two clear this up because I would very much like to know what the correct answer is. Does \frac{1}{\sqrt{2}} (|u_z\rangle + |d_z\rangle) mean 50% probability of observing spin up in the +/- z direction or does it mean you will measure spin up in the x direction?

It implies both:

1. if you measure the spin in the z-direction, you'll get either spin-up or spin-down with 50/50 probability
2. if you measure the spin in the x-direction, you'll get spin-up with 100% probability

Clearly #1 can't be the entire meaning of the superposition, because #1 does not imply #2.

 

[Mentz114]

... It's true that the state |u_z\rangle = \frac{1}{\sqrt{2}} (|u_z\rangle + |d_z\rangle) implies that a measurement of spin in the z-direction will give spin-up with probability 1/2 and spin-down with probability 1/2.

''

OK, that settles it. I don't see the relevance of the other (counterfactual) remarks.

 

[stevendaryl]

OK, that settles it. I don't see the relevance of the other (counterfactual) remarks.

Who says it's counterfactual? The question is: What is the meaning of the state \frac{1}{\sqrt{2}} (|u_z\rangle + |d_z\rangle)? You seemed to be saying that its meaning is EXACTLY and ONLY that it has probability 1/2 of being measured to be spin-up in the z-direction and probability 1/2 of being measured to be spin-down in the z-direction. That's not true.

Perhaps you're saying that IF I plan to measure the spin in the z-direction, then those probabilities are all I care about. But I can choose what direction to use to measure the spin after the particle is prepared in the above state. So it doesn't make any sense to say that the state's meaning changes depending on what I do afterward.

 

[stevendaryl]

No offense, but prolonged discussions have just confirmed that you really don't have a coherent idea of the meaning of quantum amplitudes. That's not your fault, because anybody else has a completely satisfying understanding, either. I guess it's just a matter of taste whether you consider the incoherence to be an intriguing mystery or much ado about nothing.

 

[vanhees71]

To understand this issue is really important. It's at the essence of quantum theory in fact to understand the meaning of how the state of a system in QT is described. Of course, stevendaryl is right, and it's not contradicting anything I wrote.

Let's keep this must simple case of a single spin ($s=1/2$). Now, indeed if the spin is prepared in the state, represented by
$$|\psi \rangle=\frac{1}{\sqrt{2}} (|u_z \rangle+|d_z \rangle),$$
the probality to measure $\sigma_z=\pm 1/2$ is both 1/2.

That doesn't rule out to measure the spin in any other direction. Let's take, e.g., the $x$ direction. Then you know (from diagonalizing the Pauli spin matrix $\hat{\sigma}_x$, which is given in the representation where $\hat{\sigma}_z=\mathrm{diag}(1,-1)$, i.e., in the eigenbasis of $\hat{s}_z$) that
$$|u_x \rangle = \frac{1}{\sqrt{2}} (|u_z \rangle+|d_z \rangle), \quad |u_x \rangle = \frac{1}{\sqrt{2}} (|u_z \rangle-|d_z \rangle).$$
So the probability to measure $\sigma_x=+1$ is 1 and for $\sigma_x=-1$ it's 0. Indeed the state is just represented by the eigenvector of $\hat{s}_x$ with the eigenvalue $+\hbar/2$. So $\sigma_x$ is determined to be $\hbar/2$ in this case.

 

[Mentz114]

''
''
Perhaps you're saying that IF I plan to measure the spin in the z-direction, then those probabilities are all I care about.

Yes ! That is what I mean.

But I can choose what direction to use to measure the spin after the particle is prepared in the above state.

When did I deny that ? You can choose to do anything you like.

So it doesn't make any sense to say that the state's meaning changes depending on what I do afterward.

What ? When did I say that. I don't even know what it means

 

[Mentz114]

No offense, but prolonged discussions have just confirmed that you really don't have a coherent idea of the meaning of quantum amplitudes. That's not your fault, because anybody else has a completely satisfying understanding, either. I guess it's just a matter of taste whether you consider the incoherence to be an intriguing mystery or much ado about nothing.

All I wanted to assert is that something cannot have a spin property that is both +1 and -1. Nor can an electron be in two places at once. Super-position is a mathematical artefact that results from going from exact Euler-Lagrange/Hamiltonian dynamics to statistical mechanics. Super-positions of amplitude do not have any observable physical correlate. There is no 'super-position' operator.

If that is misunderstanding then I'll live with it.

 

[ftr]

It seems that there are many threads and discussions elsewhere. I think the OR but we don't know which until we measure plus AND plus undefined all are pretty much the same thing.

 

[rubi]

It might be interesting to pinpoint precisely, where the mathematical formalism of quantum mechanics differs from the formalism of classical probability theory in order to understand what parts have a clear interpretation and where the interpretational difficulty arises. Let's first recall what the basic settings of these two formalisms are:

• Classical probability theory:
  • The state space is a measurable space $(\Lambda,\Sigma)$ consisting of a set $\Lambda$, a sigma-algebra $\Sigma \subseteq \mathcal P(\Lambda)$ of subsets of $\Lambda$, equipped with a probability measure $P:\Sigma\rightarrow\mathbb R$ with $P(\Lambda)=1$.
  • Observables are given by random variables $O:\Lambda\rightarrow\mathbb R$.
  • The probability to find the value of $O$ to lie in the Borel set $B$ is given by $P_O(B) = \int_{O^{-1}(B)} \mathrm d P$.
  • The expectation value of $O$ is given by $\left<O\right> = \int_\Lambda O\, \mathrm d P$.
• Quantum theory
  • The state space is a complex Hilbert space $\mathcal H$ and it is equipped with a quantum state $\rho$, which is given by a self-adjoint trace-class operator with $\mathrm{Tr}(\rho)=1$.
  • Observables are given by densely defined self-adjoint operators $O:\mathcal D\rightarrow \mathcal H$ ($\mathcal D\subseteq\mathcal H$, $O^\dagger=O$). They have an associated projection-valued measure $\pi_O$.
  • The probability to find the value of $O$ to lie in the Borel set $B$ is given by $P_O(B) = \mathrm{Tr}(\rho \pi_O(B))$.
  • The expectation value of $O$ is given by $\left<O\right> = \mathrm{Tr}(\rho O)$.

At first, these two formalisms look completely different, but we can find a common basis. We will construct a probability space for each (classical or quantum) observable in such a way that the full list of these probability spaces contains the same information as the big (classical or quantum) state space we started with.

• Classical probability theory: Given a random variable $O$, we construct the probability space $(\Lambda_O,\Sigma_O,P_O)$ by setting $\Lambda_O = O(\Lambda)$, $\Sigma_O=\mathcal B(\Lambda_O)$ (the Borel sigma algebra) and $P_O : \Sigma_O\rightarrow\mathbb R$, $P_O(B) = P(O^{-1}(B))$.
• Quantum theory: Given a self-adjoint operator $O$, we construct the probability space $(\Lambda_O,\Sigma_O,P_O)$ by setting $\Lambda_O = \mathrm{spec}(O)$, $\Sigma_O = \mathcal B(\Lambda_O)$ and $P_O : \Sigma_O\rightarrow\mathbb R$, $P_O(B) = \mathrm{Tr}(\rho \pi_O(B))$.
• The observables $O$ are represented by the identity function $\mathrm{id}_{\Lambda_O}$ on $\Lambda_O$.

In both cases, we find:

• The probability to find the value of $O$ to lie in the Borel set $B$ is given by $P_O(B)$.
• The expectation value of $O$ is given by $\left<O\right> = \int_{\Lambda_O} \mathrm{id}_{\Lambda_O}\,\mathrm d P_O$.

So actually, both formalisms can be reduced to a common framework, namely the list $(\Lambda_O,\Sigma_O,P_O)_O$ of probability spaces associated to each observable $O$. In fact, this list contains all information that can be computed in each formalism, so it is exactly as good to have this list of probability spaces as having the traditional information as written in the beginning. In the quantum case for instance, this list of probability spaces already contains all that can be known about superpositions.

Now, since the probability spaces $(\Lambda_O,\Sigma_O,P_O)$ are just classical probability spaces, we can also interpret them exactly this way. The computed probabilities are just relative frequencies. So what is the difference between the quantum formalism and classical probability theory? The difference is precisely that the probability spaces $(\Lambda_O,\Sigma_O,P_O)$ that arise from the classical formalism can be combined back into a huge probability space $(\Lambda,\Sigma,P)$, while this is not possible for the probability spaces that arise from the quantum formalism. This is exactly the only difference between these formalisms and all interpretational questions of quantum mechanics (that go beyond those of classical probability theory) are just instances of the single question: "How should we interpret the fact that the probability spaces $(\Lambda_O,\Sigma_O,P_O)$ can't be combined into one big probability space?" This is the only question that we don't currently have a definite answer to.

 

[jaydnul]

I'm going to interject with a question haha. So decoherence is just an explanation of how the superpositions of large groups of particles are removed, and turned into classical probabilities?

If so, I am still confused. Are those classical probabilities still representative of the possible outcomes of a measurement? Or do those particles have a definite position, and the probabilities just reflect our lack of knowledge, rather than a physical reality?

 

[rubi]

I'm going to interject with a question haha. So decoherence is just an explanation of how the superpositions of large groups of particles are removed, and turned into classical probabilities?

If so, I am still confused. Are those classical probabilities still representative of the possible outcomes of a measurement? Or do those particles have a definite position, and the probabilities just reflect our lack of knowledge, rather than a physical reality?

In decoherence you are looking at a system consisting of the environment $E$, some measurement apparata $A$ and some quantum system $S$ itself. If you only consider the observables $O^A$ corresponding to the pointers measurement apparata, then decoherence means that the probability spaces $(\Lambda_{O^A},\Sigma_{O^A},P_{O^A})$ that I constructed in post #82 can be combined into a big classical probability space, contrary to what is usually the case in quantum theory. This means that if you ignore the environment and the quantum system itself, you obtain a description of the measurement apparatus in terms of a bona-fide classical probability theory, i.e. the pointers seem to acquire definite positions that are distributed according to an ordinary classical probability distribution. This is how the classical world arises from quantum mechanics. However, you must be aware that you have ignored lots of observables concerning $E$ and $S$. Those still can't be combined with the probability space you obtained from the $A$ observables. The reason for why we don't notice this problem in our daily life is that we usually only observe things that are described by observables of type $A$, such as the position of a ball. If we manage to isolate the system from decoherence well enough, also observables of type $A$ may have the peculiar feature that they can't be combined into a big probability space. This is the case for instance for the pointer positions in a Bell test experiment.

 

[jaydnul]

The measurement apparatus is explained by classical probability theory, but what do the probabilities represent? That the apparatus could be in a particular state?

 

[PeroK]

The measurement apparatus is explained by classical probability theory, but what do the probabilities represent? That the apparatus could be in a particular state?

85 posts later, we may have established or confirmed the impossibility of explaining the macroscopic world using QM without first understanding the microscopic world using QM!

 

[rubi]

The measurement apparatus is explained by classical probability theory, but what do the probabilities represent? That the apparatus could be in a particular state?

The probabilities represent what probabilities always represent: The relative frequency to find the pointer position in a certain range.

 

[PeterDonis]

Super-positions of amplitude do not have any observable physical correlate.

Sure they do. @stevendaryl gave a good example: the states $| u_x \rangle$ and $| d_x \rangle$ are both superpositions of spin-z up and spin-z down, with the same probabilities of detecting each (1/2). But they are not the same state--they are easily distinguished by making a spin-x measurement. So you can certainly physically distinguish different superpositions even though they have the same probabilities for a particular observable.

 

[Mentz114]

Sure they do. @stevendaryl gave a good example: the states $| u_x \rangle$ and $| d_x \rangle$ are both superpositions of spin-z up and spin-z down, with the same probabilities of detecting each (1/2). But they are not the same state--they are easily distinguished by making a spin-x measurement. So you can certainly physically distinguish different superpositions even though they have the same probabilities for a particular observable.

You don't have to invoke super-position to make that statement. If the value of the x-component is definate then by the uncertainty principle nothing is known about the other spins, which means they will be measured as 50/50 mixtures of up and down. Calling the mixture a super-position is just a matter of choice.

Anyway, it isn't relevant to my assertion that the state $\alpha |u\rangle + \beta |d\rangle$ can only be interpreted as a prediction of the reletave frequencies of certain observations.

 

[ftr]

Anyway, it isn't relevant to my assertion that the state α|u⟩+β|d⟩α|u⟩+β|d⟩\alpha |u\rangle + \beta |d\rangle can only be interpreted as a prediction of the reletave frequencies of certain observations.

sure. But the OR does not clarifying anything fundamentally. For example, what do we get conceptually if we say the electron in the Hydrogen atom is here, or here or here... nothing.