# Why are entangled states generic?

1. Jul 22, 2013

### metroplex021

Hi folks --- I was just reading that entangled states are very much the norm in the universe. Can anybody tell me why entanglement is taken to be such a pervasive feature of the world, so that product states are the exception? Has it got something to do with the fact that strictly speaking all particles are interacting with one another, or have their origin in a prior interaction?

Any remarks -- even speculative -- would be gratefully received! Thanks!

2. Jul 22, 2013

### bhobba

You hit it in one.

These days observation is thought of as a kind of entanglement (specifically causing decoherence). It is known that even the cosmic background radiation is enough to cause decoerence and give objects classical properties like position etc.

While it is very difficult to remove this entanglement, requiring, for example, temperatures near absolute zero, its not impossible. And when you do some very very strange things emerge such as the behavior of liquid helium, Bose-Einstein condensates where a macro object literally behaves like on giant single atom, and other wierd things that have been reported eg:
http://physicsworld.com/cws/article/news/2010/mar/18/quantum-effect-spotted-in-a-visible-object

When Einstein said to Bohr do you believe the Moon is there when you are not looking, Born replied, in relation to that, and other questions such as God playing dice with the universe - stop telling God what to do.

The joke however is on them both - they were both wrong - the moon is there when you are not looking because its constantly being observed all the time by being entangled with its environment.

I hasten to add this doesn't resolve all the issues with QM but has been a major advance of recent times.

Thanks
Bill

Last edited: Jul 23, 2013
3. Jul 22, 2013

### metroplex021

Thanks bhobba! So is the idea that since a system is always interacting with its environment, it is always in an entangled state (assuming we're not in one of the special situations you mention, such as being at absolute zero?) Thanks again mate!

4. Jul 23, 2013

### tom.stoer

The reason is that quantum objects are indistinguishable.

Two electrons with momentum and spin do not have additional labels "1" or "2", therefore w/o introducing and additional information their quantum state must be anti-symmetrized (for fermions)

$|p_a,s_a,p_b, s_b\rangle = |p_a,s_a\rangle_1\,\otimes\,|p_b,s_b\rangle - |p_b,s_b\rangle_1\,\otimes\,|p_a,s_a\rangle$

where the indices 1,2 refer to the 1st and 2nd particle and the indices a,b refer to their different quantum numbers.

Note that in ordinary QM using wave functions one always has to write down these symmetrization (for bosons) and anti-symmetrization (for fermions) explicitly, whereas using creation and annihilation operators this is implemented automatically via

$|p_a,s_a,p_b, s_b\rangle = a^\dagger_{p_a,s_a} a^\dagger_{p_b,s_b} |0\rangle$

plus commutation or anti-commutation relations for these operators.

Therefore entanglement arises naturally from symmetrization or antisymmetrization.

One can show that QM w/o this principle does not provide a viable description of n-particle systems; refer e.g. to the so-called Gibbs paradox which is resolved by introducing indistinguishable particles, symmetrization/anti-symmetrization and different state counting.

Last edited: Jul 23, 2013
5. Jul 23, 2013

### bhobba

Yea that's it - basically the classical everyday commonsense world is the way it is because its all entangled. Do take the time to investigate the detail - see Lenny Susskinds Lectures on it:
http://theoreticalminimum.com/courses/quantum-entanglement/2006/fall
'The old Copenhagen interpretation of quantum mechanics associated with Niels Bohr is giving way to a more profound interpretation based on the idea of quantum entanglement. Entanglement not only replaces the obsolete notion of the collapse of the wave function but it is also the basis for Bell's famous theorem, the new paradigm of quantum computing, and finally the widely discussed "many-worlds" interpretation of quantum mechanics originated by Everett.'

I hasten to add, to avoid the ire of those that have gone deeply into the profound interpretational issues of QM, that while Susskind is correct, the notion of collapse has been replaced by decoherence and entanglement it has not solved the measurement problem - but side stepped it to a great extent. However to investigate it at this deep level you really need the technical details and understand one of the key issues - the difference between a proper and improper mixed state. There are others as well such as the so called factoring problem but that is the main one.

Thanks
Bill

Last edited: Jul 23, 2013
6. Jul 23, 2013

### tom.stoer

A few remarks:
- decoherence does not destroy entanglement on the fundamental level
- after decoherence (applied to an electron pair due to thermal photons) entanglement is still present in the electron subspace and the photon subspace
- entanglement has nothing to do with interaction (it applies to free particles)

7. Jul 23, 2013

### Chronos

Just to toss a couple coins in the fountain, all measurements are observer dependent. There is no 'universal' scale of measurement.

8. Jul 23, 2013

### kith

I wouldn't say that this is the reason because also interacting, distinguishable particles are entangled most of the time. If you have two interacting particles, the Hamiltonian doesn't factor (H ≠ H1⊗H2), so an initial product state will evolve into an entangled state. More precisely, it is much more likely to find the system in an entangled state than in a product state at an arbitrary time after the interaction has started. This is because there are much more entangled state vectors than product state vectors in the composite Hilbert space.

Can you elaborate a bit here?

9. Jul 23, 2013

### hilbert2

I'd hypothesize that if you could write down the wavefunction of the whole universe, it would turn out that all particles in the universe are entangled with each other to some extent, as all matter comes from a common source (the Big bang).

Entanglement between two degrees of freedom is generated always when there's a coupling between them is the Hamiltonian of the system.

It's actually hard to quantify the "degree of entanglement" between two particles, but there have been some attempts, see e.g. http://arxiv.org/abs/1304.7058 .

10. Jul 23, 2013

### tom.stoer

No, they aren't.

1) There are no distinguishable particles in QM, except if they are entirely different, e.g. like electrons and photons. But then they can't be entangled b/c the above mentioned formula

$|p_a,s_a,p_b, s_b\rangle = |p_a,s_a\rangle_1\,\otimes\,|p_b,s_b\rangle_2 -|p_b,s_b\rangle_1\,\otimes\,|p_a,s_a\rangle_2$

no longer valid if index 1 and 2 refer to different species. You cannot entangle apples and oranges.

2) Interaction does not cause entanglement; interaction is something you add on top of the fundamental, geometrical structure of the QM Hilbert space.

Two electrons stay indistinguishable even when interacting with a thermal environment. Decoherence does not provide a means to distinguish two identical particles.

They are.

Entanglement is not created via interaction; it is present w/o interaction.

But I know what you mean: if you start with a mixed state w/o interaction no entanglement is created. But in that case we miss the entanglement b/c of the mixed state i.e. b/c of the incomplete knowledge which forces us to describe the system as a mixed state.

We should carefully distinguish between a sub-system which behaves effectively like a mixed state b/c of incomplete knowledge of its and the environment d.o.f., and the fact that indistinguishable particles are always entangled in principle.

Last edited: Jul 23, 2013
11. Jul 23, 2013

### hilbert2

Yes, I meant that if we have a system of two degrees of freedom, $x_{1}$ and $x_{2}$, and the wavefunction of the system is initially of product form: $\psi(x_{1},x_{2})=\phi_{1}(x_{1})\phi_{2}(x_{2})$, then it will stay in product form unless there's a coupling between $x_{1}$ and $x_{2}$ in the hamiltonian. With lack of coupling I mean that $\frac{\partial^{2} H(p_{1},p_{2},x_{1},x_{2})}{\partial x_{1} \partial x_{2}} = 0$ for the classical hamiltonian function H (no "cross terms").

Of course interaction is not necessary for entanglement to exist. Suppose we have two particles that come from infinity, collide and interact for a brief period of time and then fly off to different directions. The entanglement is generated in the interaction but the particles will stay entangled even when they are far away from each other some time after the collision.

I'm not here considering the fact that there exists no real isolated two-particle system without any interaction with an environment.

12. Jul 23, 2013

### tom.stoer

I understand what you mean. But for indistinguishable particles we know explicitly that using this product form is wrong. It leads to wrong results in state counting (entropy) e.g. for ideal, i.e. non-interacting, quantum gases. That's why entanglement in the sense of symmetrization / antisymmetrization is inevitable.

13. Jul 23, 2013

### kith

This is true on the fundamental level but often, we talk about effectively distinguishable systems like the qubits in Quantum Information. How do we reconcile these two notions? For example, if we have two qubits in a Bell state |00>+|11> and Alice performs a measurement and finds 0, the qubits are in state |00> after the measurement. How would we describe this if we want to include the fundamental indistinguishability of the qubits (let's say they are electrons)? This is a quite general question.

What's wrong with the state |a1>⊗|o1>+|a2>⊗|o2> where the |ai> are states of the apple and |oj> are states of the orange?

14. Jul 23, 2013

### tom.stoer

What's wrong with the idea of an entangled state |00> which is zero (!) for fermions and which corresponds to a product state |0>⊗|0> for bosons?

Of course it's more complicated in general b/c we have

$|00\rangle = |\psi_a,0\rangle\otimes|\psi_b,0\rangle + \ldots$

where the psi's refer to all other quantum numbers including position or momentum space. But in the special case when two bosons have identical quantum numbers, the state is a product state.

Try it for an electron and a photon: The first ket lives in the fermionic Hilbertspace and carries charge and spin 1/2; the second ket lives in the bosonic Hilbertspace and carries no charge but spin 1 (helicity 1) and two polarizations. The first ket can never carry the quantum numbers of the second ket and vice versa. There are different spaces, states and operators acting on them.

EDIT: Suppose you have a non-rel. system with two spin 1/2 and one spin 1 particle described in a (2-dim.)^2 and a 3-dim. Hilbert space, respectively. An example for a state is

$|0\rangle = (|+\rangle\otimes|-\rangle - |-\rangle\otimes+\rangle)\otimes|0\rangle$

The spin 1/2 and the spin 1 particles cannot be entangled (symmetrized/antisymmetrized) b/c the fundamental spaces are different; you cannot have a spin 1/2 particle in the state |0>, neither can the spin 1 particle be in the |+1/2> state.

REMARK: these complications are artifacts of old-fashioned QM; they fade away when using creation and annihilation operators.

Last edited: Jul 24, 2013
15. Jul 23, 2013

### hilbert2

Some quotes:
The total electron-photon system can be described in terms of only one Hilbert space.

16. Jul 23, 2013

### tom.stoer

I do not see any state which is not something like "bosonic * fermionic".

17. Jul 24, 2013

### tom.stoer

But I think we should come back the question
All I wanted to indicate is that product states are invalid for indistinguishable particles b/c symmetrization/antisymmetrizations forces them to be entangled on the fundamental level.

18. Jul 24, 2013

### kith

You can't have both: either the state is a product state or it is entangled (this is the very definition of entanglement).

Also a state of zero is not a possible measurement outcome (the probability to get it is zero).

Why does it have to? We only need this to get a symmetrized /antisymmetrized state which isn't necessary for distinguishable particles.

19. Jul 24, 2013

### tom.stoer

Also a state of zero is not a possible measurement outcome (the probability to get it is zero).[/QUOTE]
Please have a look at the elementary algebra

fermions: |00> = |0>⊗|0> - |0>⊗|0> = 0
bosons: |00> = |0>⊗|0> + |0>⊗|0> = |0>⊗|0> (up to normalization)
(but these are two special cases where symmetrization/antisymmetrization is trivial)

It's not "not necessary' but "not possible".

If you want to symmetrize you would have to write something like
|fb> = |f>⊗|b> + |b>⊗|f>
(f = fermionic quantum numbers, b = bosonic quantum numbers)
But this is nonsense algebraically: you can't symmetrize states living in (structurally) different Hilbert spaces.
And it's nonsense physically: |an electron here and a photon over there> is not the same as |an electron over there and a photon here>; two different particles, therefore two different states, therefore no symmetrization.

But I think here we agree.

20. Jul 24, 2013

### hilbert2

Suppose we have a system consisting of a Dirac electron field and a bosonic electromagnetic field.
The degrees of freedom of the system can be taken to be the Fourier modes of the two fields. The fields are not free, there's a coupling between the modes of the two fields in the QED lagrangian. Therefore, any electron is constantly "entangling" itself with the modes of the surrounding EM field. If this would not happen and all states could be represented as products of boson and fermion states, there would actually be no EM interaction between charged particles.

I think the problem is that you're looking at this problem from a mathematician's viewpoint and you can't immediately see that electron-photon entanglement is physically inevitable.

Here's another article about electron-photon entanglement: http://spectrum.ieee.org/tech-talk/semiconductors/devices/a-quantum-dot-first-entanglement .