What Is Entanglement? Definition & Examples

[edguy99]

In an effort clarify confusion in my own mind on the definition of entanglement, I looked at wiki and found this:

As an example of entanglement: a subatomic particle decays into an entangled pair of other particles. The decay events obey the various conservation laws, and as a result, the measurement outcomes of one daughter particle must be highly correlated with the measurement outcomes of the other daughter particle (so that the total momenta, angular momenta, energy, and so forth remains roughly the same before and after this process). ​

Which I completely agree with, followed by this:

For instance, a spin-zero particle could decay into a pair of spin-½ particles. Since the total spin before and after this decay must be zero (conservation of angular momentum), whenever the first particle is measured to be spin up on some axis, the other, when measured on the same axis, is always found to be spin down. (This is called the spin anti-correlated case; and if the prior probabilities for measuring each spin are equal, the pair is said to be in the singlet state.)​

Which I feel is not correct. As I understand it, if the spin axis of the original particle was the same as the measurement axis, this would be true (ie, prepare vertical and measure vertical), but if you measure on a different axis, you only have probabilities of seeing this based on cos^2 of the difference between the measurement axis and the original particle axis. Many of the particles prepared in this manner, will not be anti-correlated if measured off of the original particles spin axis.

 

[entropy1]

I think the detections of both sides would correlate following cos²(φ), but this would be a consequence of the spins being opposite. Or at least that view would produce the correlation.

 

[Jilang]

As I understand it, if the spin axis of the original particle was the same as the measurement axis, this would be true (ie, prepare vertical and measure vertical), but if you measure on a different axis, you only have probabilities of seeing this based on cos^2 of the difference between the measurement axis and the original particle axis. Many of the particles prepared in this manner, will not be anti-correlated if measured off of the original particles spin axis.​

But they always are. It's strange but true! However...the measurement of the spin does affect the spin direction so you are not necessarily measuring what you started with. The final direction is function of both the initial spin and the orientation of the detection equipment.​

 

[vanhees71]

In the example you have an original spin state $|s=0,\sigma_z=0 \rangle$. If the particle decays in two spin-1/2 particles, the angular-momentum state must be still such that the total angular momentum is $|J=0,J_z=0 \rangle$, but this is uniquely given in terms of the product states of the two spins (check a QT textbook about addition of angular momenta or Clebsch-Gordan coefficients etc.):
$$|J=0,J_z=0 \rangle=\frac{1}{\sqrt{2}} (|s=1/2,\sigma_z=1/2 \rangle \otimes |s=1/2,\sigma_z=-1/2 \rangle-|s=1/2,\sigma_z=-1/2 \rangle \otimes |s=1/2,\sigma_z=1/2 \rangle.$$
This implies that, if you find when measuring $\sigma_z$ of particle 1 to be $+1/2$ then necessarily particle 2 must have $\sigma_z=-1/2$ etc.

The spin-z component of each single particle is maximally indetermined. The reduced statistical operator for particle 1 is
$$\hat{\rho}_1=\mathrm{Tr}_1 |J=0,J_z=0 \rangle \langle J=0,J_z=0| = \frac{1}{2} \mathbb{1},$$
i.e., the particle is unpolarized. The same is true for particle $2$.

 

[DrChinese]

...

For instance, a spin-zero particle could decay into a pair of spin-½ particles. Since the total spin before and after this decay must be zero (conservation of angular momentum), whenever the first particle is measured to be spin up on some axis, the other, when measured on the same axis, is always found to be spin down. (This is called the spin anti-correlated case; and if the prior probabilities for measuring each spin are equal, the pair is said to be in the singlet state.)​

Which I feel is not correct. As I understand it, if the spin axis of the original particle was the same as the measurement axis, this would be true (ie, prepare vertical and measure vertical), but if you measure on a different axis, you only have probabilities of seeing this based on cos^2 of the difference between the measurement axis and the original particle axis. Many of the particles prepared in this manner, will not be anti-correlated if measured off of the original particles spin axis.

The conservation rule applies here. If the probabilities were as you mention, you would have a Product state and the conservation rule would not hold.

 

[Jilang]

QM can tell you about what happens at detection but not about what happens in between. I would be suspicious that where external fields are involved conservation laws could be relied on.

 

[Nugatory]

QM can tell you about what happens at detection but not about what happens in between. I would be suspicious that where external fields are involved conservation laws could be relied on.

Conservation laws still work; you just have to be careful to include the change in the external field as well. However, there is no external field involved in the decay interaction that @vanhees71 was describing, so this is a bit of a red herring. We started with an isolated quantum system prepared in a state of zero total angular momentum; as long as it remains isolated its angular momentum will remain zero.

 

[edguy99]

In the example you have an original spin state $|s=0,\sigma_z=0 \rangle$. If the particle decays in two spin-1/2 particles, the angular-momentum state must be still such that the total angular momentum is $|J=0,J_z=0 \rangle$, but this is uniquely given in terms of the product states of the two spins (check a QT textbook about addition of angular momenta or Clebsch-Gordan coefficients etc.):
$$|J=0,J_z=0 \rangle=\frac{1}{\sqrt{2}} (|s=1/2,\sigma_z=1/2 \rangle \otimes |s=1/2,\sigma_z=-1/2 \rangle-|s=1/2,\sigma_z=-1/2 \rangle \otimes |s=1/2,\sigma_z=1/2 \rangle.$$
This implies that, if you find when measuring $\sigma_z$ of particle 1 to be $+1/2$ then necessarily particle 2 must have $\sigma_z=-1/2$ etc.

The spin-z component of each single particle is maximally indetermined. The reduced statistical operator for particle 1 is
$$\hat{\rho}_1=\mathrm{Tr}_1 |J=0,J_z=0 \rangle \langle J=0,J_z=0| = \frac{1}{2} \mathbb{1},$$
i.e., the particle is unpolarized. The same is true for particle $2$.

Would this hold true if the basis vectors of the original spin angle or measuring angle were rotated?

 

[Jilang]

@Nugatory , I agree, but the measurement in the SG setup does include a field.

 

[Strilanc]

As I understand it, if the spin axis of the original particle was the same as the measurement axis, this would be true (ie, prepare vertical and measure vertical), but if you measure on a different axis, you only have probabilities of seeing this based on cos^2 of the difference between the measurement axis and the original particle axis. Many of the particles prepared in this manner, will not be anti-correlated if measured off of the original particles spin axis.

The original particle is spin zero, it doesn't have a spin axis.

One of the first things taught about the singlet state is that it doesn't have a preferred axis or orientation. We write it down in some basis, as if it had a preferred orientation, but when you switch to another basis it looks exactly the same.

For example, suppose Alice and Bob share two qubits in the singlet state $|01\rangle - |10\rangle$. If they measured their qubits right now, in the computational basis, which distinguishes between $|0\rangle$ and $|1\rangle$, they will get opposite answers. But instead they are going to rotate their qubits before measuring, so they effectively measure along a different axis.

The rotation operation $U = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ is a 2x2 unitary matrix. Because both Alice and Bob are applying $U$ to their qubit, we end up applying $U \otimes U$ to the two-qubit system. Now turn the crank:

$$\begin{align}
\psi_2
&= (U \otimes U) \cdot \psi_1
\\
&= (U \otimes U) \cdot \big( |01\rangle - |10\rangle \big)
\\
&= (U \otimes U) \cdot |01\rangle - (U \otimes U) \cdot |10\rangle
\\
&= (U \otimes U) \cdot \big( |0\rangle \otimes |1\rangle \big) - (U \otimes U) \cdot \big( |1\rangle \otimes |0\rangle \big)
\\
&= \big( U |0\rangle \big) \otimes \big( U |1\rangle \big) - \big( U |1\rangle \big) \otimes \big( U |0\rangle \big)
\\
&= \big( a |0\rangle + c |1\rangle \big) \otimes \big( b |0\rangle + d |1\rangle \big) - \big( b |0\rangle + d |1\rangle \big) \otimes \big( a |0\rangle + c |1\rangle \big)
\\
&= \big( ab |00\rangle + ad |01\rangle + cb |10\rangle + cd |11\rangle \big) - \big( ba |00\rangle + bc |01\rangle + da |10\rangle + dc |11\rangle \big)
\\
&= (ab-ba) |00\rangle + (ad-bc) |01\rangle + (ca-da) |10\rangle + (cd-dc) |11\rangle
\\
&= (ad-bc) |01\rangle + (cb-da) |10\rangle
\\
&= \det(U) \big( |01\rangle - |10\rangle \big)
\\
&= e^{i \theta} \psi_1
\end{align}$$

As you can see, no matter what rotation Alice and Bob both apply, they end up back in the singlet state. If they limit themselves to operations from SU(2), they don't even pick up a global phase factor (not that it matters, since those have no observable effects).

This proves that the singlet state has no preferred orientation.

Some entangled states do change when the same operation is applied to both sides. For example, if Alice and Bob start in the singlet state and then only Alice flips her qubit over around the Z axis, then X and Y axis measurements would agree whereas the Z axis measurement still disagrees. The Z axis parity "stands out" from the other two. But this is still not the kind of pre-existing orientation that you're talking about. You can still use such states to violate Bell inequalities.

 

[edguy99]

The original particle is spin zero, it doesn't have a spin axis. ...

Absolutely, but the moment it decays we have a spin axis for each spin 1/2 particle on a set of basis vectors.

 

[Strilanc]

Absolutely, but the moment it decays we have a spin axis for each spin 1/2 particle on a set of basis vectors.

Read the rest of my post. The singlet state doesn't have a preferred spin axis.

 

[Jilang]

@strianc, are you saying that there cannot be a state defined for each of the particles individually!?

 

[DrChinese]

@strianc, are you saying that there cannot be a state defined for each of the particles individually!?

If they are entangled on that basis, no.

 

[Strilanc]

@strianc, are you saying that there cannot be a state defined for each of the particles individually!?

Yes. This is a basic well known fact about entanglement. In fact, people often define "entangled" to mean states that can't be factored into parts.

 

[Nugatory]

@Nugatory , I agree, but the measurement in the SG setup does include a field.

indeed it does... But that's part of the measuring apparatus, and the original claim was that we couldn't trust conservation of momentum in an unmeasured system. As long as the quantum system that started as the parent particle remains isolated, it's momentum is constant and momentum is trivially conserved. Eventually that system interacts with the inhomogeneous magnetic field of the SG machines when we measure it; this may change the momentum of the particles, but also changes the momentum of the SG machines in such a way that momentum is still conserved.

 

[Jilang]

Yes. This is a basic well known fact about entanglement. In fact, people often define "entangled" to mean states that can't be factored into parts.

So if one particle for whatever reason ceased to exist, the other particle's state would immediately change?

 

[Jilang]

indeed it does... But that's part of the measuring apparatus, and the original claim was that we couldn't trust conservation of momentum in an unmeasured system. As long as the quantum system that started as the parent particle remains isolated, it's momentum is constant and momentum is trivially conserved. Eventually that system interacts with the inhomogeneous magnetic field of the SG machines when we measure it; this may change the momentum of the particles, but also changes the momentum of the SG machines in such a way that momentum is still conserved.

I didn't see anything in the earlier posts that talked about anything other than measurement.

 

[Strilanc]

So if one particle for whatever reason ceased to exist, the other particle's state would immediately change?

No, that's not how it works at all. Do you expect the color of your left sock to be affected by throwing your right sock into a fire?

The particles are strongly correlated, more correlated than possible with socks, but doing something to one of them still has no observable effect on the other.

 

[Jilang]

No, that's not how it works at all. Do you expect the color of your left sock to be affected by throwing your right sock into a fire?

The particles are strongly correlated, more correlated than possible with socks, but doing something to one of them still has no observable effect on the other.

I agree, although it doesn't explain the mismatches in my sock drawer!

 

[vanhees71]

Read the rest of my post. The singlet state doesn't have a preferred spin axis.

In other words the state $|J=0,J_z=0 \rangle$ doesn't change under rotations; $J=0$ indicates the trivial representation of the rotation group.

 

[Jilang]

I think we can agree on that. When The OP mentions the spin axis of the original particle, I assume it was referring to the spin axis of one of the daughter particles prior to measurement. Is that so?

 

[Nugatory]

When The OP mentions the spin axis of the original particle, I assume it was referring to the spin axis of one of the daughter particles prior to measurement. Is that so?

Perhaps, but that's not a concept that can be applied to a quantum system in the singlet state, which is what we have prior to measurement - so if that's what OP meant, intended, it makes no sense.

 

[Jilang]

Isn't QM silent about what happens before the measurement. Should we even be discussing it on this forum?

 

[Nugatory]

Isn't QM silent about what happens before the measurement. Should we even be discussing it on this forum?

QM is silent about the values of observables that have not been measured, but it not silent about how the wave function evolves in the absence of a measurement - that's what Schrodinger's equation is all about. And this entire thread is basically about the implications of the quantum system having been prepared in the singlet state before it is measured.

 

[Jilang]

OK, but isn't the wavefunction a representation of what we might observe IF we measure it?

 

[mikeyork]

As far as I can see, this whole discussion depends on the orbital state in the CM frame being 0. But we have J = L + S and S = 1/2 + 1/2 (with vector addition not arithmetic), The value of S and the possible individual projections will then depend on L, the parities of all particles involved and whether or not the particles are identical. In all cases there will be a superselection rule of some kind and the resulting particles will always be entangled by that.