I What happens to the other particle in an entanglement state?

[accdd]

if I have two particles in an entangled state, I make them travel in different directions, and I measure the state of only one of them then I know the outcome of the measurement of the other.
But when I take a measurement on the first particle, what happens to the second? Does it undergo a measurement? Does the measurement of one particle force the measurement of the other?

 

[DrChinese]

Does it undergo a measurement? Does the measurement of one particle force the measurement of the other?

Surprisingly, quantum mechanics does not supply a definitive answer to this question. What QM does is make statistical predictions that are verified by experiment.

The QM statistical predictions tell us the average relationship between measurement outcomes on pairs of entangled particles. No one, however, can be sure of the mechanics of what is happening prior to the measurement. When considering "normal" cause and effect relationships, you would have to believe something *like* the following occurs.

a. At T=1: Alice measures particle A on some basis, and records an outcome.
b. When that happens, particle B instantaneously takes on a value on that same measurement basis (that of Alice) which is consistent with the type of the entanglement.
c. At T=2: Bob's later measurement reveals statistical consistency with Alice's measurement per a.

Understand that QM does not assert step b occurs. This step is inserted simply to supply one possible explanation of the mechanics. If b. were true, you would essentially have instantaneous action at a distance. Some interpretations of QM include this feature.

But note that the experimental evidence supports alternate explanations (i.e. other interpretations), since there is no difference in the observed outcomes if the ordering is reversed (Alice measures at T=2, and Bob at T-1). In this scenario, causal order is reversed. Again, QM does not assert that causal order is reversed. It is simply silent on the point.

 

[PeroK]

if I have two particles in an entangled state, I make them travel in different directions, and I measure the state of only one of them then I know the outcome of the measurement of the other.
But when I take a measurement on the first particle, what happens to the second? Does it undergo a measurement? Does the measurement of one particle force the measurement of the other?

If you have an entangled state, then any measurement is a measurement of that state, not a measurement of a single particle state. That's the point about entanglement.

 

[vanhees71]

As many in this forum know, I'm of a different opinion, because I think it's very clear that there is no instantaneous collapse, because the most comprehensive quantum theory we have is relativistic quantum field theory, and it is from the very beginning constructed in a way that there are no faster-than-light causal connections between events possible.

If you have an entangled pair of particles (or photons, which are most easy to operate with in the lab when it comes to entanglement) and you measure, e.g., the momentum of one of the particles, nothing happens to the other particle, at least not for the time it takes to send a signal from you to the other particle, i.e., you need at least the time light needs to travel from your place to the other particle's place.

That you know immediately the value of the other particle's momentum is due to the preparation of the two particles in the entangled state, i.e., although the single-particle momenta of both particles are indetermined before the measurement, they are strongly correlated due to the preparation in the entangled state in the very beginning (e.g., by decay of a heavier particle at rest into two ligher particles, which then have strictly opposite momentum because of momentum conservation, but you don't know in which direction both fly), i.e., you know when measuring the momentum that this particle now has momentum $\vec{p}$, then you know that somebody measuring the other particle's momentum must bei $-\vec{p}$. It doesn't matter, who measures his particle's momentum first or if you do the measurement at the same time (in your common rest frames), you'll always find a random distribution of momenta (or directions of momenta in our decay example) but you find 100% correlation, i.e., the outcomes of the particle-momenta measurements always give opposite momenta.

I'm a proponent of the minimal statistical interpretation, i.e., there is nothing to be known about particles than the probabilities for the outcome of measurements, quantum (field) theory provides. What happens to the particles in the process of measurement depends of course on how specifically your measurement apparatus works, i.e., whether or not after the measurement you have a particle with the measured value of the observable you measured, depends on the measurement device. For sure, there's no instantaneous influence of far distant local measurements on two particles, no matter whether they were prepared in an entangled or an unentangled state.

I'm also agnostic about the question, how it comes that we always measure (within the accuracy of the measurement device) one specific value for an observable that was undetermined by the preparation of the measured quantum system. It's just the construction of the apparatus, and all we know before the measurement is the probability for the outcome of this measurement. At the present state of our knowledge that's really all we can know about quantum systems.

The apparent classical behavior of macroscopic objects, including measurement devices, is well understood by quantum-statistical physics, i.e., considering only the relevant macroscopic observables (e.g., the position of the center of momentum of a planet moving around the Sun), are spatio-temporal averages over many microscopic degrees of freedom, and those tend to behave in almost all cases classical.

 

[antonantal]

If we represent the two particles as qubits, an entangled state would be for example:$$|\psi\rangle=\frac{|00\rangle+|11\rangle}{\sqrt2}$$At measurement, you randomly and equally likely get one of only two outcomes: both qubits 0 or both 1.

Mathematically, it is because the state vector doesn't contain any other components than $|00\rangle$ and $|11\rangle$.

Physically, one way to interpret this is that there are many parallel universes: half of them have both qubits 0 and the other half have both qubits 1. When you measure e.g. first qubit as 0 you only find out in which universe you are located: that is in one from the first half. And in that universe the other qubit has been 0 all along, so there is no faster-than-light communication involved.
This is the many-worlds interpretation of QM, and there is no proof that it is either true or false as far as I know, but one can still use it as a framework for thought experiments.

 

[vanhees71]

NB: The state of the first particle is given by "tracing out" the second particle. This socalled "reduced" state for the first particle is given by the corresponding partial trace of the statistical operator. For the two-particle state the statistical operator is
$$\hat{\rho}=|\psi \rangle \langle \psi|,$$
and the partial trace is given by
$$\hat{\rho}_1=\mathrm{Tr}_{2} \hat{\rho} = \sum_{j_1,j_2,k=0}^1 |j_1 \rangle \langle j_1,k|\hat{\rho}|j_2,k \rangle \langle j_2| = \frac{1}{2} (|0 \rangle \langle 0| + |1 \rangle \langle1|)=\frac{1}{2} \hat{1},$$
i.e., the outcome of the measurement on particle 1 is totally indetermined ("maximum-entropy state"), and hyou get with 50% probability 0 and with 50% probality 1. The same holds of course for particle 2.

 

[antonantal]

...i.e., the outcome of the measurement on particle 1 is totally indetermined ("maximum-entropy state"), and hyou get with 50% probability 0 and with 50% probality 1. The same holds of course for particle 2.

I just want to expand a bit on this statement because not long ago I was struggling to understand how this is different from measuring a single qubit in state $\frac{|0\rangle+|1\rangle}{\sqrt2}$ which also gives with 50% probability 0 and with 50% probability 1.

The difference is that in the case of a qubit from a pair of entangled qubits, you have ambiguity on the state, i.e. the state itself can be either $|0\rangle$ OR $|1\rangle$, each with 50% probability (this is called a mixed state).
In the case of a single-qubit state $\frac{|0\rangle+|1\rangle}{\sqrt2}$ we have a pure stare which is a superposition of $|0\rangle$ AND $|1\rangle$. So the state is perfectly known, we have ambiguity on the measurement only. In fact, if you choose to measure in a basis that contains the $\frac{|0\rangle+|1\rangle}{\sqrt2}$ vector, the measurement result will always be 1. In the other case (a qubit from an entangled pair), the ambiguity still remains regardless what measurement basis you choose.

To further emphasize the difference, note that you cannot consider each qubit from the state $|\psi\rangle=\frac{|00\rangle+|11\rangle}{\sqrt2}$ to be in the pure superposition state $\frac{|0\rangle+|1\rangle}{\sqrt2}$, because in this case the two-qubit state would have to be:
$$\frac{|0\rangle+|1\rangle}{\sqrt2}\otimes\frac{|0\rangle+|1\rangle}{\sqrt2}=\frac{|00\rangle+|01\rangle+|10\rangle+|11\rangle}{2}$$

 

[accdd]

I understood that measurement is a physically vague concept (although mathematically defined as the collapse of the wave function), it happens when a particle hits an instrument, i.e., a macroscopic object. is that right? Now, if we have two particles in an entagled state and I measure one particle with an instrument I collapse the wave function of the composite system. Why should the other one undergo a measurement? After all, it didn't interact with anything, it didn't hit a macroscopic object.
vanhees71, I know that probability in the quantum domain is different from classical probability because of Bell's inequalities. I am wrong?
I'm using a translator, sorry for mistakes.

 

[Nugatory]

Now, if we have two particles in an entagled state and I measure one particle with an instrument I collapse the wave function of the composite system.

One way or another, you must purge yourself of this misconception.

You do not have a system composed of two particles. You are not measuring one or the other particle.

You have a single quantum system described by a single wave function, and there are two measurements that we can perform on this single system: Result at detector A and result at detector B. Yes, the two detectors are in different places, but you must not allow this to tempt you into thinking that you're measuring something at position A or at position B - that's letting your classical intuition mislead you. You are performing a measurement on a single quantum system whether you start with detector A or detector B.

It turns out that either measurement leaves the system in a state such that with further mathematical massaging (see posts above) we can start to describe the post measurement state as two independent particles. The details will open up an interpretational can of worms so for now I’ll settle for the B-level summary: the measurement breaks the entanglement.