When you measure an entangled particle is it obeying a symmetry law?

[Bi-Curious]

Why is it so important to the universe that if you measure the spin of an entangled pair and its up then the other particle must be down.

It seems to have no practical use so why does the universe enforce this rule, how would reality differ if it wasn't true.

And is it a law of symmetry that leads to a conservation law?

 

[ChrisVer]

the conservation of spin?

 

[Nugatory]

Why is it so important to the universe that if you measure the spin of an entangled pair and its up then the other particle must be down.

It seems to have no practical use so why does the universe enforce this rule, how would reality differ if it wasn't true.

And is it a law of symmetry that leads to a conservation law?

It's conservation of angular momentum, a principle that is just as fundamental in classical mechanics - for example, it's needed to solve the problem of planetary motion.

The underlying symmetry is invariance under rotations.

 

[bhobba]

The core is rotational symmetry.

To understand why you need advanced math.

If you have the necessary background Chapter 3 of Ballentine - Quantum Mechanics - A Modern Development will provide the detail.

Even Classical Mechanics can be viewed this way - See Landau - Mechanics:
https://www.amazon.com/dp/0750628960/?tag=pfamazon01-20

When people see this for the first time it can be rather striking - as one review of that book said:
'If physicists could weep, they would weep over this book. The book is devastatingly brief whilst deriving, in its few pages, all the great results of classical mechanics.' and 'The reason for the brevity is that, as pointed out by previous reviewers, Landau derives mechanics from symmetry.'

It's a strange but true fact, possibly the greatest discovery physics has ever made, that symmetry is what underlies much of physics:
http://www.pnas.org/content/93/25/14256.full

At the rock bottom level, when you strip the math bare physics is actually beauty incarnate - but unless you know the technical details its hidden. It was ushered in during the 19th century by a, today, little known program, but at the time very influential, called the Erlangen program:
http://en.wikipedia.org/wiki/Erlangen_program

Thanks
Bill

 

[atyy]

Is the statement true? If the entangled state is |UU>+|DD> then if a measurement finds one spin in the up state, the other spin will also be up.

 

[bhobba]

Is the statement true? If the entangled state is |UU>+|DD> then if a measurement finds one spin in the up state, the other spin will also be up.

If that is the entangled state yes (neglecting the normalisation factor).

It's the wrong way around for the usual photon version though. They have 1/root 2 |u>|d> + 1/root 2 |d>|u>. Its entangled in up and down - they must be opposite from the way that experiment is constructed - the total spin must sum to zero - but we don't know which is which - is the first up and the second down or conversely. It's an entangled superposition of the two. When you observe one it determines the other. But I would use the term strange correlations like Bertlmanns socks:
http://cds.cern.ch/record/142461/files/198009299.pdf

Thanks
Bill

 

[atyy]

If that is the entangled state yes (neglecting the normalisation factor).

It's the wrong way around for the usual photon version though. They have 1/root 2 |u>|d> + 1/root 2 |d>|u>. Its entangled in up and down - they must be opposite from the way that experiment is constructed - the total spin must sum to zero - but we don't know which is which - is the first up and the second down or conversely. It's an entangled superposition of the two. When you observe one it determines the other. But I would use the term strange correlations like Bertlmanns socks:
http://cds.cern.ch/record/142461/files/198009299.pdf

Thanks
Bill

At least theoretically, one can violate a Bell inequality with |UU>+|DD>. An example is given in http://www.theory.caltech.edu/people/preskill/ph229/notes/chap4.pdf (Eq 4.1, 4.41, 4,42).

 

[bhobba]

At least theoretically, one can violate a Bell inequality with |UU>+|DD>. An example is given in http://www.theory.caltech.edu/people/preskill/ph229/notes/chap4.pdf (Eq 4.1, 4.41, 4,42).

I haven't gone through the paper, it doesn't particularly interest me. It may Dr Chinese though - he is really keen on that sort of stuff.

But if it was possible to do that, even theoretically, it would be big news.

Because of that I find it highly unlikely.

Thanks
Bill

 

[atyy]

I haven't gone through the paper, it doesn't particularly interest me. It may Dr Chinese though - he is really keen on that sort of stuff.

But if it was possible to do that, even theoretically, it would be big news.

Because of that I find it highly unlikely.

Thanks
Bill

In the photon case, are you using u and d to refer to helicity or to polarization?

I assume the OP is asking about spin, say spin 1/2, in which case |uu>+|dd> is legitimate.

 

[OQS]

If you have two entangled spin-1/2 particles entangled in the Bell state $|\phi^{+}> = \frac{1}{\sqrt{2}}(|\uparrow\uparrow>+|\downarrow\downarrow>)$ (I am assuming here that you are familiar with bra-ket notation), a measurement on one of the particles yields the state of the second particle via the partial inner product.

For example, if the first particle is measured to be in the spin-up state $|\psi> = |\uparrow>$, then the resulting state of the second particle is given by the partial inner product $<\psi|\phi^{+}> = \frac{1}{\sqrt{2}}|\uparrow>$.

Ignoring the $1/\sqrt{2}$ normalization factor, this tells you that the second particle must be in a spin-up state.

It seems to have no practical use so why does the universe enforce this rule, how would reality differ if it wasn't true.

Based on what I have said previously, all entangled states of particles must exhibit this type of behavior - that is, measurement on one of the particles will affect the physical state of the other. What your question appears to be asking is, what would happen if entanglement was not possible in this universe?

In that case, physics would be restricted to separable states. By definition, separable states are all quantum states which are non-entangled. For the case of a single particle, there would be no discernible difference between standard quantum theory and your theory, as entanglement is only possible between multiple quantum objects (such as two particles).

However, for multiple objects things really break down, and we actually do need these entangled states to correctly describe reality.

 

[stevendaryl]

Why is it so important to the universe that if you measure the spin of an entangled pair and its up then the other particle must be down.

It seems to have no practical use so why does the universe enforce this rule, how would reality differ if it wasn't true.

And is it a law of symmetry that leads to a conservation law?

Entanglement is not always about conservation laws or symmetry. In the general case, entanglement occurs whenever you produce particles whose states are correlated. And particles become correlated whenever they interact. So entanglement is the general case of interacting particles. Correlated spins is just the most stark example.

 

[bhobba]

In the photon case, are you using u and d to refer to helicity or to polarization?

I am considering the spin singlet state:
http://ocw.usu.edu/physics/classical-mechanics/pdf_lectures/29.pdf

Conceptually it doesn't matter as long as they are correlated in a way such when you measure particle 1 you know particle 2.

Thanks
Bill

 

[bhobba]

Hey guys just thinking a bit here - there are 4 bell states:
http://en.wikipedia.org/wiki/Bell_state

I chose the one they call psi+ - others chose phi+

Thanks
Bill

 

[atyy]

So the premise of the question is wrong - measuring one spin in an entangled pair does not mean the other in the pair will be measured to have the opposite spin. I agree with OQS's post #10 and stevendaryl's post #11.

 

[Nugatory]

So the premise of the question is wrong - measuring one spin in an entangled pair does not mean the other in the pair will be measured to have the opposite spin. I agree with OQS's post #10 and stevendaryl's post #11.

I don't think it's so much that the premise of the question is wrong as that it is limited to the particular case that OP was asking about. It seems most likely that he's thinking about the most often discussed case of spin correlations, the one where a single interaction gives rise to an entangled pair with opposite spins.

A bit of a digression: how would an experimenter go about preparing the state |U1>|U2>+|D1>|D2> ?

 

[atyy]

I don't think it's so much that the premise of the question is wrong as that it is limited to the particular case that OP was asking about. It seems most likely that he's thinking about the most often discussed case of spin correlations, the one where a single interaction gives rise to an entangled pair with opposite spins.

But if that's the case, there is no symmetry that prevents |uu>+|dd>.

A bit of a digression: how would an experimenter go about preparing the state |U1>|U2>+|D1>|D2> ?

Off the top of my head, I don't know.

 

[Nugatory]

But if that's the case, there is no symmetry that prevents |uu>+|dd>.

Of course there isn't - but OP's question was why the spins must be opposite in the experiment he's been reading about. I'm inclined to think it's best to nail that answer down before introducing other thought experiments in which the initial premise would be stated differently.

 

[atyy]

Of course there isn't - but OP's question was why the spins must be opposite in the experiment he's been reading about. I'm inclined to think it's best to nail that answer down before introducing other thought experiments in which the initial premise would be stated differently.

So if the state is |ud> + |du>, why do you say that it is angular momentum conservation that makes the measured spins point oppositely?

 

[Zarqon]

A bit of a digression: how would an experimenter go about preparing the state |U1>|U2>+|D1>|D2> ?

Don't have a specific setup in mind, but in general I don't think it's that difficult at least conceptually. What you have to do is make sure there is something else involved in the entanglement scheme that can take up the extra spin. Since you either have two spin up or no spin up, spin is not conserved by the photons themselves, but if the photons are coupled to let's say two atoms, that could also be in either spin state there is no problem with that. One could maybe imagine creating the photons in some cavity QED experiment with strong coupling between atoms and photons, and with the atoms prepared in a clever way, and then not care about the atomic state after the photons have been emitted.

 

[Nugatory]

So if the state is |ud> + |du>, why do you say that it is angular momentum conservation that makes the measured spins point oppositely?

Because the preparation procedure relied on conservation laws to produce that state. That was also the point of my (somewhat rhetorical) question about how you would prepare |uu> + |dd>. Zarqon's reply

Don't have a specific setup in mind, but in general I don't think it's that difficult at least conceptually. What you have to do is make sure there is something else involved in the entanglement scheme that can take up the extra spin. Since you either have two spin up or no spin up, spin is not conserved by the photons themselves, but if the photons are coupled to let's say two atoms, that could also be in either spin state there is no problem with that. One could maybe imagine creating the photons in some cavity QED experiment with strong coupling between atoms and photons, and with the atoms prepared in a clever way, and then not care about the atomic state after the photons have been emitted.

makes this point clear.

 

[atyy]

@Nugatory, yes I agree that angular momentum is important in the state preparation, and also with Zarqon's point that |uu> + |dd> is in principle be doable, though I don't know how easy it is in practice.

Another thought: in the photon case, the analogue of spins is the polarization. Isn't the |+->+|-+> state the same as |hh>+|vv>, which is the analogue of |uu>+|dd>? Eg. http://www.theory.caltech.edu/people/preskill/ph229/notes/chap4.pdf Eq 4.26 and 4.27?

 

[DrChinese]

A bit of a digression: how would an experimenter go about preparing the state |U1>|U2>+|D1>|D2> ?

You mean where the polarization is the same for 2 photons measured at the same angle?

You can do that with Type I PDC (2 crystals). Type II is opposite polarization.