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

[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.

 

[PeterDonis]

Calling the mixture a super-position is just a matter of choice.

No, it isn't. Saying that "the value of spin-x is definite" is a much stronger statement than saying "there is a 50-50 probability of measuring spin-z up or down". There are an infinite number of possible states that can give the latter probabilities. But there are only two states (the spin-x eigenstates) in which the value of spin-x is definite.

Saying that a state is a "mixture" of 50-50 spin-z up and spin-z down only tells you that you are in one of the infinite number of possible states that can give those probabilities. Saying that you are in a "superposition" of spin-z up and spin-z down implies that you know which of those infinite number of possible states is the one that was actually prepared.

In other words, a "superposition" is a pure state, whereas a "mixture" is not. At least, that's the standard terminology. If you are going to use non-standard terminology, that's your choice, but it doesn't change the physical distinction between pure states and mixed states.

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

That's a matter of interpretation. Some interpretations limit the meaning of the state to this, and some don't. But the distinction between pure states and mixed states is not interpretation dependent.

 

[Mentz114]

No, it isn't. Saying that "the value of spin-x is definite" is a much stronger statement than saying "there is a 50-50 probability of measuring spin-z up or down". There are an infinite number of possible states that can give the latter probabilities. But there are only two states (the spin-x eigenstates) in which the value of spin-x is definite.

Saying that a state is a "mixture" of 50-50 spin-z up and spin-z down only tells you that you are in one of the infinite number of possible states that can give those probabilities. Saying that you are in a "superposition" of spin-z up and spin-z down implies that you know which of those infinite number of possible states is the one that was actually prepared.

In other words, a "superposition" is a pure state, whereas a "mixture" is not. At least, that's the standard terminology. If you are going to use non-standard terminology, that's your choice, but it doesn't change the physical distinction between pure states and mixed states.

Thanks for the explanations. If I understand correctly - suppose we prepare 2 beams A (spin-z up) and B(thermal) and subject them to measurements in the three orientations. For beam A we get 100% z-up, x-thermal, y-thermal, For B all three will be thermal. The first x and y measurements are on a superposition and the second two on a mixture, but they give the same results. Thus I find it hard to distinguish physically between the cases.

However, this has never been a serious issue, but the next one is.

That's a matter of interpretation. Some interpretations limit the meaning of the state to this, and some don't. But the distinction between pure states and mixed states is not interpretation dependent.

Do you know in what way (other than probabilistic) can the state $\alpha |u\rangle + \beta | d \rangle$ be interpreted, according to those who interpret it differently ?
(that is a terrible sentence - I hope you understand)

 

[PeterDonis]

Do you know in what way (other than probabilistic) can the state $\alpha |u\rangle + \beta | d \rangle$ be interpreted

As an eigenstate of a different operator. In other words, as the real, physical state of the system, which leads to probabilistic results with the operator you originally chose, but which leads to perfectly definite, non-probabilistic results with another operator.

To put this another way: there is a theorem which says that given any state in a Hilbert space, there will be some Hermitian operator that has that state as an eigenstate. To some people, that theorem is sufficient to show that the state the actual, physical state of the system, not just something probabilistic. You can agree or disagree with that interpretation, but it is a permissible interpretation at our current state of knowledge.

 

[PeterDonis]

The first x and y measurements are on a superposition and the second two on a mixture, but they give the same results. Thus I find it hard to distinguish physically between the cases.

In other words, you ignore the crucial measurement--spin-z--that actually does easily distinguish physically between the cases. Of course that will make it hard to distinguish physically between the cases.

The fact that the spin-z eigenstate is not actually a superposition in the spin-z basis is a red herring. The crucial point, as I have said a couple of times now, is that the spin-z eigenstate is a pure state, whereas your "thermal" state is a mixed state.

 

[Mentz114]

As an eigenstate of a different operator. In other words, as the real, physical state of the system, which leads to probabilistic results with the operator you originally chose, but which leads to perfectly definite, non-probabilistic results with another operator.

To put this another way: there is a theorem which says that given any state in a Hilbert space, there will be some Hermitian operator that has that state as an eigenstate. To some people, that theorem is sufficient to show that the state the actual, physical state of the system, not just something probabilistic. You can agree or disagree with that interpretation, but it is a permissible interpretation at our current state of knowledge.

Thank you.

 

[Mentz114]

In other words, you ignore the crucial measurement--spin-z--that actually does easily distinguish physically between the cases. Of course that will make it hard to distinguish physically between the cases.

More important is that the state of z-spin will not affect orthogonal directions. It gives no information about anything but z. So the state is perfect ie.it does not change under a measurement - in a chosen direction only. But this is a mere quibble.

 

[stevendaryl]

More important is that the state of z-spin will not affect orthogonal directions.

What do you mean by "affect"? If you start with a particle that is spin-up in the x-direction, and then measure the spin in the z-direction, it will no longer be spin-up in the x-direction. So it's unclear what it means to say that it's z-spin will not affect its x-spin.

 

[PeterDonis]

state of z-spin will not affect orthogonal directions. It gives no information about anything but z.

This statement is much too strong. An eigenstate of spin-z gives no information (in the sense of giving 50-50 probabilities) about spin-x or spin-y or linear combinations of those two states. But it does give information (i.e., does not give 50-50 probabilities) about any other observable, i.e., any observable that includes any amplitude at all for spin-z.

 

[Mentz114]

What do you mean by "affect"? If you start with a particle that is spin-up in the x-direction, and then measure the spin in the z-direction, it will no longer be spin-up in the x-direction. So it's unclear what it means to say that it's z-spin will not affect its x-spin.

Sorry that was loosely worded, I mean 'affect predictions about'.

 

[Mentz114]

This statement is much too strong. An eigenstate of spin-z gives no information (in the sense of giving 50-50 probabilities) about spin-x or spin-y or linear combinations of those two states. But it does give information (i.e., does not give 50-50 probabilities) about any other observable, i.e., any observable that includes any amplitude at all for spin-z.

Sure.

The first question I asked about superposition of the measurement apparatus has been answered and I now understand more about how one may choose to interpret a state.

 

[gmalcolm77]

I've recommended this book so many times that I'm beginning to think I should get a commission on the sales... But if you can get hold of a copy of David Lindley's "Where does the weirdness go?", give it a try. It's a pretty good layman-friendly treatment of how our world emerges from the quantum world.

Ditto on that.