Do Entangled Quantum Particles Remain Aligned After Initial Preparation?

[rasp]

TL;DR

Please clarify the procedure after 2 (or more) particles are prepared in a (let’s say contrary) entangled state and then measured independently.

I am asking a very basic question. Asking for clarification on the procedure for preparing quantum particles in an entangled state. My question asks if once the particles are prepared along a certain axis, is it then true that the intervention is then removed so that each particle can become random (I.e. explore the full opportunity presented to it.) The opposing assumption, which I think results in a trivial answer, is that the particle prepared into a particular state (for example spin up) continues in that state. In that case, it would not seem to be paradoxical that we should find it spin as prepared and it’s partner spin contrary.

 

[PeterDonis]

if once the particles are prepared along a certain axis, is it then true that the intervention is then removed so that each particle can become random (I.e. explore the full opportunity presented to it.) The opposing assumption, which I think results in a trivial answer, is that the particle prepared into a particular state (for example spin up) continues in that state.

Neither of these are correct, and neither of them are relevant to your question about how particles get prepared in entangled states.

The correct answer for a single particle, which is what you are talking about in the quote above (you say "particles", but everything you say is about what happens to a single particle), is that, between measurements, the time evolution of its state is unitary. Unitary evolution is not random; it is deterministic. So while it does not mean the particle just stays in the same state (for some states that will be true, for others not), it does mean that if we know the result of a particular preparation process that was done on a particle, we can predict what its quantum state will be right up until the next measurement that is made on it. All of the randomness in QM comes from measurements; none of it comes from time evolution in between measurements.

Answering your question about how particles get prepared in entangled states requires you to consider preparation processes that produce multiple particles at once; such processes will typically produce the particles in an entangled joint state that is an eigenstate of some observable corresponding to a conserved quantity (usually spin). For example, pairs of entangled photons are often produced by a process called parametric down conversion; the spins of the two photons in the pair are entangled because their joint state is an eigenstate of total spin.

 

[PeterDonis]

it would not seem to be paradoxical that we should find it spin as prepared and it’s partner spin contrary.

The difficult part is not that the spins are correlated; that can happen in classical physics.

The difficult part is that the results of spin measurements always end up correlated, even though the individual measurement results are random. In other words, the two spins end up correlated even though neither particle has a definite spin at all when considered on its own. (The alternative hypothesis, that the two particles do have definite spins on their own, is ruled out by the fact that the correlations violate the Bell inequalities.)

 

[vanhees71]

One "preparation procedure" for particles is to use the decay of an unstable scalar particle, e.g., $\pi^+ \rightarrow \mu^{+}+\nu_{\mu}$. In the rest frame of the pion due to the conservation of angular momentum the spins of the muon and the neutrino together are in the singlet state $(|1/2,-1/2 \rangle-|-1/2,1/2 \rangle)/\sqrt{2}$.

 

[WernerQH]

In that case, it would not seem to be paradoxical that we should find it spin as prepared and it’s partner spin contrary.

"Property" is kind of a taboo word, because it can be very misleading. In quantum mechanics one must not think of the property of a "particle" without the environment, the measurement to which it is subjected. (If you decide to measure the x- or y-component of spin, or position or momentum, for example.) The outcome of a measurement depends on the detector settings; one could argue that what's measured are "properties" of the detector as well as those of the particles. Unfortunately this is rarely said clearly.

 

[WernerQH]

All of the randomness in QM comes from measurements

That's a rather extreme statement. Surely you are not claiming that the randomness of radioactive decay is due to "measurements"?

 

[vanhees71]

I think what @PeterDonis wanted to express with this thought-provoking formulation is the fact that quantum theory is causal but not deterministic.

Indeed according to quantum theory if you know the state of a closed system at some time $t_0$ as well as the Hamilton operator you know the state of the system at an time $t>t_0$ due to unitary time evolution. That's causality.

What's random then indeed are the outcome of measurements of observables of the system since the state only describes precisely these probabilities according to Born's rule. In this sense randomness comes into the game when measuring an observable. The outcome of the measurement is usually random (except in the case if the state is such that with 100% probability the measurement results in one of the possible outcomes; for a pure state this means that the corresponding state vector is an eigenvector of the self-adjoint operator representing the measured observable).

 

[WernerQH]

Indeed according to quantum theory if you know the state of a closed
system at some time $t_0$ as well as the Hamilton operator you know the state of the system at an time $t>t_0$ due to unitary time evolution.

That's a big "if". For an experimentalist, a radioactive atom that has not decayed (after, say, 1 hour) is absolutely indistinguishable from an atom that is newly "prepared". For him it is in the same "state". But for the theorist, if he believes in the reality of the wave function, the state has changed after 1 hour. Continuous and deterministic evolution according to Schrödinger's equation is clearly at odds with the abruptness and randomness of what is happening in the real world. And talk about "measurements" doesn't really clarify things.

The OP's question was clearly about something physical, "properties" of particles, resulting from misguided thinking. It's important to emphasize that atomic particles (whether entangled or not) do not have properties in the same sense as ordinary classical objects.

 

[vanhees71]

There is no contradiction between your right statement and my statement. If you have an radioactive atom, of course the state of this atom has changed an hour later, because the atom is not in an Eigenstate of the Hamiltonian (that's why it's instable and thus "radioactive"). The probability to find it in the same state it was prepared in is approximately given by the usual exponential radioactive decay law $P(t)=\exp(-\lambda t)$, where $\lambda=1/\tau$ is the decay constant and $\tau$ the average lifetime of the atom. Quantum mechanically this follows from the Wigner-Weisskopf approximation.

I don't precisely understand your 2nd paragraph. What do you mean by "properties"? I think it's such vague statements that confuse people most about quantum theory. I'd rather say that quantum systems' observables do not necessarily take determined values but it depends on the state the system is prepared in which observables take determined values and which don't.

 

[PeterDonis]

Surely you are not claiming that the randomness of radioactive decay is due to "measurements"?

Yes. "Measurements" in QM does not just mean measurements made by humans in a lab. A better term would probably be "irreversible decoherent interaction". The radioactive decay of an atom involves an irreversible decoherent interaction between the atom and its environment. In an experiment in a lab, such as the OP is thinking of, things are set up very carefully so that such irreversible decoherent interactions do not take place with the environment (which is happening all the time with ordinary objects outside of labs), but only with the specific measuring devices that the experimenter has set up.

 

[vanhees71]

The radioactive decay of an atom is of course due to some interaction. Take $\beta$ decay. There you have an atomic nucleus bound by the strong interaction. If you neglect the weak interaction the nucleus were stable. You define the ground state of this nucleus neglecting the weak interaction as "the nucleus", i.e., neglecting the weak interaction you have an energy eigenstate which is just stable. But now there is the weak interaction, and the nucleus decays by converting one of the neutrons and emitting an electron and an antielectron neutrino. That's how "unstable states" or "resonances" are defined. The nucleus doesn't decay due to a measurement though. It would decay with some probability without any interaction with anything outside the nucleus. There's also no decoherence as long as you consider the total system of all involved particles and fields. After all it's described by a unitary S-matrix.

 

[PeterDonis]

The nucleus doesn't decay due to a measurement though.

Not due to an intervention by a human experimenter, you mean. But the decay is, as I said, an irreversible decoherent interaction with the environment. Even if there's no human there to register the decay with, say, a Geiger counter, that's still true.

There's also no decoherence as long as you consider the total system of all involved particles and fields.

"The total system of all involved particles and fields" includes a potentially unbounded number of degrees of freedom in the environment. That is what makes the process irreversible. And that is what "decoherence" means. So yes, there is decoherence in a radioactive decay.

After all it's described by a unitary S-matrix.

Decoherence, in principle, is a unitary process; the irreversibility is due to the degrees of freedom in the environment not being trackable.

 

[vanhees71]

Yes, but the "degrees of freedom in the environment not being trackable" make the description an open quantum system. Decoherence usually means that a coherent superposition of a pure state goes into a incoherent sum, e.g., something like a pure state $\hat{\rho}=|psi \rangle \langle \psi|$ with
$$|\psi \rangle = \sum \lambda_i |\psi_i \rangle$$
to
$$\rho'=\sum_i |\lambda_i|^2 |\psi_i \rangle \langle \psi_i|$$
and this can not described by a unitary transformation, i.e., it is achieved by some effective description of a subsystem of some larger system.

If you wish you can consider the field-degrees of freedom as the additional "environment", but that's rather uncommon.

 

[rasp]

Neither of these are correct, and neither of them are relevant to your question about how particles get prepared in entangled states.

The correct answer for a single particle, which is what you are talking about in the quote above (you say "particles", but everything you say is about what happens to a single particle), is that, between measurements, the time evolution of its state is unitary. Unitary evolution is not random; it is deterministic. So while it does not mean the particle just stays in the same state (for some states that will be true, for others not), it does mean that if we know the result of a particular preparation process that was done on a particle, we can predict what its quantum state will be right up until the next measurement that is made on it. All of the randomness in QM comes from measurements; none of it comes from time evolution in between measurements.

Answering your question about how particles get prepared in entangled states requires you to consider preparation processes that produce multiple particles at once; such processes will typically produce the particles in an entangled joint state that is an eigenstate of some observable corresponding to a conserved quantity (usually spin). For example, pairs of entangled photons are often produced by a process called parametric down conversion; the spins of the two photons in the pair are entangled because their joint state is an eigenstate of total spin.

 

[rasp]

Thank you. Obviously I Need to learn more. Although I’m fascinated, I’m finding I’m in sufficiently prepared even to grasp the basic concepts.

 

[WernerQH]

I’m insufficiently prepared even to grasp the basic concepts.

Surely it's very confusing. Even among physicists there are strong disagreements about the terminology. One could think of "z-component of angular momentum" as a "property" of an atomic system. But as @vanhees71 pointed out, it is better to speak of an "observable", even though to many people this will sound synonymous. But a statement like "the electron has been prepared in a spin up state" is always a statistical statement characterizing possible measurement results. Even if the measurements yield 100% in special cases, it is misleading to think that an electron has an axis of rotation pointing in a definite direction.

 

[rasp]

Lol about the confusing part. I’m glad I share at least that with physicists. I especially like the quote from John Wheeler.
https://Rhys.wordpress.com/2006/06/09/quantum-mechanical-quotes/

 

[PeroK]

Even if the measurements yield 100% in special cases, it is misleading to think that an electron has an axis of rotation pointing in a definite direction.

Especially since the magnitude of the measured z-spin squared is $\frac {\hbar^2}{4}$ and the magnitude of the total spin squared is $\frac {3\hbar^2}{4}$, regardless of the spin state.

 

[PeterDonis]

the "degrees of freedom in the environment not being trackable" make the description an open quantum system.

Yes, you can treat it that way, by simply ignoring the environment degrees of freedom altogether (or tracing over them, or at least some of them, if necessary).

Decoherence usually means that a coherent superposition of a pure state goes into a incoherent sum

As an approximation for practical purposes, yes.

 

[vanhees71]

Especially since the magnitude of the measured z-spin squared is $\frac {\hbar^2}{4}$ and the magnitude of the total spin squared is $\frac {3\hbar^2}{4}$, regardless of the spin state.

This simply reflects the fact that only one spin component and $\vec{s}^2$ are compatible observables, i.e., you can prepare one spin component of a particle (usually $s_z$) and $\vec{s}^2$ to have simultaneously determined values. $\vec{s}^2$ is a property of the particle used (e.g., for an electron the eigenvalue is $s(s+1)\hbar^2=3 \hbar^2 4$, because $s=1/2$ for an electron. So that's fixed anyway. So if you prepare the electron in, say, the $m_1=+1/2$ such that $s_z=\hbar/2$, the other spin components $s_x$ and $s_y$ cannot be simultaneously determined.

 

[vanhees71]

Yes, you can treat it that way, by simply ignoring the environment degrees of freedom altogether (or tracing over them, or at least some of them, if necessary).
As an approximation for practical purposes, yes.

Well it's a damn good approximation for the said "tracing out of the environment".

 

[PeroK]

This simply reflects the fact that only one spin component and $\vec{s}^2$ are compatible observables, i.e., you can prepare one spin component of a particle (usually $s_z$) and $\vec{s}^2$ to have simultaneously determined values. $\vec{s}^2$ is a property of the particle used (e.g., for an electron the eigenvalue is $s(s+1)\hbar^2=3 \hbar^2 4$, because $s=1/2$ for an electron. So that's fixed anyway. So if you prepare the electron in, say, the $m_1=+1/2$ such that $s_z=\hbar/2$, the other spin components $s_x$ and $s_y$ cannot be simultaneously determined.

Yes, but my point was that this is not analagous to a classical particle with an axis of rotation in the z-direction. The spin in the z-direction is still only about a third of the total spin. All we know is the direction of spin in the z-direction, not that the z-direction represents an axis of rotation. Even in the eigenstates of $s_z$, there is still no sense of a well-defined axis of rotation.

 

[EPR]

. A better term would probably be "irreversible decoherent interaction". The radioactive decay of an atom involves an irreversible decoherent interaction between the atom and its environment. In an experiment in a lab, such as the OP is thinking of, things are set up very carefully so that such irreversible decoherent interactions do not take place with the environment (which is happening all the time with ordinary objects outside of labs), but only with the specific measuring devices that the experimenter has set up.

Doesn't this mean that decoherence has solved the measurement problem?

 

[vanhees71]

Yes, but my point was that this is not analagous to a classical particle with an axis of rotation in the z-direction. The spin in the z-direction is still only about a third of the total spin. All we know is the direction of spin in the z-direction, not that the z-direction represents an axis of rotation. Even in the eigenstates of $s_z$, there is still no sense of a well-defined axis of rotation.

Of course, because a classical particle with an axis of rotation in the $z$ direction means that all the spin components have determined values, namely $\vec{s}=(0,0,s_z)$. That's impossible for a particle with spin $\neq 0$. The one exception where all angular-momentum components take determined values is if $J=0$.

 

[vanhees71]

Doesn't this mean that decoherence has solved the measurement problem?

For me yes, but there's no consensus about this. I'm anyway not thinking that there is any measurement problem in QT, because QT makes predictions about the outcome of measurements given the state of the measured system which are in complete agreement with all observations. For me there's no problem but just amazing success of QT. The only physically unsolved problem is a consistent description of the gravitational interaction within QT.

 

[andrew s 1905]

I have been reading this https://arxiv.org/abs/quant-ph/0312059 on the decoherence and measurement problem. It may be out of date but it makes the issues and progress to 2005 fairly clear.
Regards Andrew

 

[vanhees71]

Yes, as Schlosshauer (one of the leading experts in this field), writes already in the abstract: "yet their implications for the foundational problems of quantum mechanics, most notably the quantum measurement problem, have remained a matter of great controversy."

He has also written a book on the subject:

Schlosshauer, Decoherence and the quantum-to-classical transition, Springer (2007)

 

[PeterDonis]

Doesn't this mean that decoherence has solved the measurement problem?

No. Decoherence can explain why there is never any interference observed between different irreversible measurement results. But it cannot explain why we only observe one result when there are multiple possible ones.

 

[vanhees71]

I'd say that's because we construct our measurement apparati such that we get one definite result. Otherwise we'd not take it as a well-functioning device.

 

[stevendaryl]

I'd say that's because we construct our measurement apparati such that we get one definite result. Otherwise we'd not take it as a well-functioning device.

I don't think that's a good way to describe it. What I would say is that we design a measurement process so that distinct values of the observable that we are trying to measure become correlated with macroscopically distinguishable properties of the measurement device. We don't do anything additional to handle the case that the microscopic system we are measuring is in a superposition of eigenvalues of the observable being measured.