# Proof that entanglement is independent of distance

I've already read this quite often but never seen a proof—maybe it's just so clear to physicists, but I'm not really sure how to prove it.

Currently I'm pretty confused so the following might be total nonsense:

As far as I understood an entangled state is always in some kind of superposition, e.g. $$|\Phi\rangle = \frac{1}{\sqrt 2}(|1\rangle_A|1\rangle_B + |0\rangle_A|0\rangle_B)$$ because otherwise it would be a product state, e.g. $$|\Phi\rangle = |1\rangle_A|1\rangle_B.$$

Now my intuition would be that a difference in distance would be caused by some kind of movement (in space) of the state in $A$ or $B$. For this kind of evolution one should have some unitary operators $U_A$ and $U_B$ that act on the respective subspaces, such that $U:=U_A\otimes U_B$ acts on $|\Phi\rangle$.

Considering the special case of the second system not being affected at all (i.e. $U_B = I_B$, the identity), what guarantees that $U_A|1\rangle$ won't turn into $|0\rangle$? This in turn would yield a "different entanglement" like $$\frac{1}{\sqrt 2}(|0\rangle_A|1\rangle_B + |1\rangle_A|0\rangle_B).$$

• In the interesting cases $U_A$ and $U_B$ are some sort of time evolution operators. In this case you have to look at your Hamiltonian. You could achieve exactly the operator you describe by passing spin $A$ through a magnetic field. Normally, however, you have set up you experiment such that any spin terms are negligible, so it is trivial that the spins do not change. – By Symmetry Jul 22 '15 at 11:58
• Are you asking why entanglement does not change if you act with a unitary on one part of the system? – Norbert Schuch Jul 22 '15 at 12:55
• @Peter The state you are considering is a pure state (in the sense that pure states can be described by wave-functions, impure states must be described by a density matrix)! – Sebastian Riese Jul 22 '15 at 13:23
• Product state is a common term (as it can be represented as a direct product of elements of the two single particle Hilbert spaces). – Sebastian Riese Jul 22 '15 at 15:45
• @Peter But for that your question sounds far too complicated. I think it would be more clear if you would ask "Why is entanglement invariant under if I act on a state with local unitaries, i.e., $|\psi\rangle \mapsto U_A\otimes U_B|\psi\rangle$." -- if this is what you are looking for. – Norbert Schuch Jul 22 '15 at 16:29

Firstly, the notation $|\Phi\rangle = \frac{1}{\sqrt 2}(|1\rangle_A|1\rangle_B + |0\rangle_A|0\rangle_B)$ already assumes away anything spatial because you are only writing the spin degrees of freedom.

Secondly you mention superposition as if it avoids being a pure state. A superposition is still a pure state. Superposition is like a sum. You don't look at the number 5 and say it is a sum, you look at at 2+3=5 and say that the sum of 2 and 3 is 5. Similarly you can look at the equation $|\Phi\rangle = \frac{1}{\sqrt 2}(|1\rangle_A|1\rangle_B + |0\rangle_A|0\rangle_B)$ and say that $|\Phi\rangle$ is a superposition (sum) of $|1\rangle_A|1\rangle_B$ and $|0\rangle_A|0\rangle_B.$ more properly to say something is a superposition of some other things you mean it is a (possibly complex) linear combination of the other things. The result of a linear combination of some vectors is just another vector, no different than any other vector.

Being pure is the same as having your state be represented by a vector in your Hilbert Space. So an entangled state is just as pure as a factorizable state or any other vector in your Hilbert Space. Mixed states involve taking linear combinations of density matrixes of pure states, real linear combinations, nonnegative linear combinations, combinations where the terms add up to one and are all between zero and one. Basically where your system has a classical probability to be in each of some pure states. And these are combinations of density matrices, not combinations of Hilbert Space vectors.

So a pure state could be factorizable like $|\Phi 11 \rangle =|1\rangle_A|1\rangle_B$ or it could be not factorizable (entangled) like $|\Phi\rangle = \frac{1}{\sqrt 2}(|1\rangle_A|1\rangle_B + |0\rangle_A|0\rangle_B).$ But those are both pure because a complex linear combination of Hilbert Space vectors is still a Hilbert Space vector.

And if you made density matrices out of either of those they would be pure, but a real linear combination of density matrices could be another density matrix and that density matrix might be mixed (or might be pure again).

So complex linear combinations of factorizable states might be factorizable or might be entangled, but they are still pure. A real linear combination of density matrices of pure states might be pure or it might be mixed but it is still a density matrix.

OK, so let's not mention pure versus mixed again since it is irrelevant. So you have (pure) states like $|1\rangle_A|1\rangle_B$ and $|0\rangle_A|0\rangle_B$ both of which are factorizable since each is a product of states of a single particle. But the set of factorizable states is not a subspace, in particular it is not closed under linear combinations. So the (pure) state $|\Phi\rangle = \frac{1}{\sqrt 2}(|1\rangle_A|1\rangle_B + |0\rangle_A|0\rangle_B)$ might not (and indeed is not) factorizable. The reason we know it is not factorizable is because it behaves differently than factorizable states behave. Since we know it is not factorizable (therefore entangled, since that is what entangled means) the closest we can get to describing it in terms of factorizable states is to wrote is as a linear combination (superposition) of factorizable states. This is possible because the factorizable states form a basis so any (pure) state whatsoever can be written as a complex linear combination (superposition) of factorizable states.

So the terminology is hopefully cleared up. But there is still the notation. Look at the hydrogen atom. If you just described the electron you could use $|nlm\rangle$ to describe the spatial part but there are still two spin degrees of freedom so you could use $|nlms\rangle$ to fully describe that electron. So a general (pure) state is written like $\sum \alpha_{nlms}|nlms\rangle,$ where the sum is over all nonnegative $n,$ and for each $n$ there is a finite range of $l$ and for each $l$ there is a finite range of $m$ and the term $s$ also has two possibilities.

This is all for just one particle. But it is possible to have a single particle have its spin degrees of freedom "entangled" with its spatial degrees of freedom. In fact, that's usually how a spin measurement works you interact with the particle so that it's spin degree of freedom becomes "entangled" with its spatial degrees of freedom. Something similar to $|200\frac{1}{2}\rangle+|100\frac{-1}{2}\rangle$ rather than something like $|200\frac{1}{2}\rangle.$ However you want to have something you could write like $|1\rangle_A|1\rangle_B$ and you wanted it to be factorizable so you wanted the state $|1\rangle_A$ to have a well defined spin state so it can't be something like to to $|200\frac{1}{2}\rangle+|100\frac{-1}{2}\rangle$ and has to be something like $|200\frac{1}{2}\rangle.$ So in fact it can be something like $\sum \alpha_{nlms}|nlm\frac{1}{2}\rangle,$ where the sum is over all nonnegative $n,$ and for each $n$ there is a finite range of $l$ and for each $l$ there is a finite range of $m$ and but the term $s$ has to be the possibility $\frac{1}{2}$ regardless of what $n, l,$ or $m$ is. Nature allowed it to be anything, but unless $\alpha_{nlm\frac{-1}{2}}=0$ for every $n,l,$ and $m$ then the state $\sum \alpha_{nlms}|nlm\frac{1}{2}\rangle$ can't be summarized as $|1\rangle.$

Note that I put quotes around the word entangled since the math is similar but technically they are the same particle just different degrees of freedom. I wanted you to see the similarities mathematically, but not get confused about the actual definition.

OK, do spatial degrees of freedom are real and are just being suppressed by the notation. And this can't be done for an arbitrary state only for certain states one where half the $\alpha$ linear combination coefficients of the single particle state are zero so the single particle state has a definite spin eigenvalue.

So we designed the spatial states to be unrelated to the spin state in order to be able to use the notation that suppresses the spatial degrees of freedom. In reality for a hydrogen atom there would be some spin-orbit coupling so a state like $|nlms\rangle$ wouldn't be an energy eigenstate. But if you wrote something as a true energy eigenstate for the actual Hamiltonian in reality, then nothing would really happen.

So this might have answered your actual question. However you talked about unitary evolutions on the single particle state. If your two particles are identical (e.g. two electrons) then it isn't actually possible to act on one and not the other, but we can assume they are distinguishable if that's what you want.

If you send $|1\rangle_A$ to $|0\rangle_A$ by a unitary transformation, then you must also send $|0\rangle_A$ to $e^{i\theta}|0\rangle_A$ since it has to send an orthonormal set to an orthonormal set. But a unitary transformation can do almost anything as long as it preserves the inner product. It can even send a factorizable state into an entangled state and vice versa. Though not if it only acts on one particle.

Though this last fact about unitary transformations has nothing to do with spatial degrees of freedom or distance.

So let's get to how spatial degrees usually transition from not mattering to mattering in actual normal practice.

You could have a beam travelling forwards, left, or right. So the spatial degrees of freedom could be $|F\rangle_A,$ $|L\rangle_A,$ and $|R\rangle_A.$ And a Stern-Gerlach device. Would be designed to send $|F\rangle_A|1\rangle_A$ to $|L\rangle_A|1\rangle_A$ and to send $|F\rangle_A|0\rangle_A$ to $|R\rangle_A|0\rangle_A$ and so by linearity we know how it works on any forward spatial degree of freedom. So imagine that the spatial degree of freedom of particle A (the part your notation suppressed) was going in the forward direction. Then now it evolves so that the spatial degree of freedom is "entangled" with the spin degree of freedom. Which means

$|\Phi\rangle$ = $\frac{1}{\sqrt 2}(|1\rangle_A|1\rangle_B + |0\rangle_A|0\rangle_B)$ = $\frac{1}{\sqrt 2}(|F\rangle_A|1\rangle_A|1\rangle_B + |F\rangle_A|0\rangle_A|0\rangle_B)$ evolves into $\frac{1}{\sqrt 2}(|L\rangle_A|1\rangle_A|1\rangle_B + |R\rangle_A|0\rangle_A|0\rangle_B)$ and this means that the spin of $B$ is entangled with the spatial parts of $A.$ So even if we couldn't before interact with just one particle, now we can clearly interact with just $|1\rangle_A|1\rangle_B$ or just $|0\rangle_A|0\rangle_B$ by putting something in the oath of the left going bean or the right going beam respectively.

So in particular a detector can detect $|1\rangle_A|1\rangle_B$ or $|0\rangle_A|0\rangle_B$ and that is the real lesson of entanglement, the measurement found $|1\rangle_A|1\rangle_B$ or $|0\rangle_A|0\rangle_B$ it didn't just find $|1\rangle_A$ or $|0\rangle_A.$

But we can make things become entangled and we can make them become unentangled. In fact that measurement, if it gives you the result $|1\rangle_A|1\rangle_B$ or the result $|0\rangle_A|0\rangle_B$ is giving you unentangled spins either way. That's life. Entanglement isn't magic, it's just where some things have correlations more than having their own properties. You can interact with them to realize that correlation (and then they aren't entangled anymore since the properties are realized completely). Or you can interact with them to change which things are correlated (for instance to have the same spins rather than opposite spins).

Or you can even interact with them to undo the entanglement without realizing the properties. They are all just vectors in a Hilbert Space. There is no magic.

• this is too long, i'm afraid. can you precis it? – innisfree Jul 22 '15 at 18:31
• What is the reason for the "finitenesses" and dependences on each other in this part: "the sum is over all nonnegative n, and for each n there is a finite range of l and for each l there is a finite range of m"? – Peter Jul 26 '15 at 11:19
• @Peter $|m| < l + 1 < n + 1$ and its basically labelled so $n$ is like the row on the periodic table and all the parts with different $l$ can be grouped together on the periodic table. It isn't important, we could instead use $Z=1,2, ...$ (for position on the periodic table) and it would work just as well. It also relates to a convention about labeling a state based on how it interacts with an operator so there are three operators $H, L^2,$ and $L_z$ and if you study them you'll learn why they have those ranges. – Timaeus Jul 26 '15 at 15:39
• @innisfree I have to clear up some misconceptions about the four terms factorizable, entangled, pure, and mixed. So I have to explain all four. Then I have to set up how to describe spatial degrees of freedom since I didn't know if the OP knows how to do them but they asked about them. Then I have to address unitary transformations and how they can and usually relate to those five things. That is a lot of things. I'm sure I could do a better job, but really a textbook is a better resource when considering so many topics. – Timaeus Jul 26 '15 at 15:45

An entangled state has to such that it can't be written in the form $|i\rangle_A|j\rangle_B$. If the state can't be written in that form, then neither system alone has a pure reduced density matrix, i.e. - a density matrix of the form $|i\rangle\langle i|$ as opposed to a density matrix of the form $\sum_ip_i|i\rangle\langle i|$, which is called a mixed state.

The rest of what you have written is very confused. Suppose that you have two atoms five metres apart. The state of the electrons in one of those atoms may not be much affected by those of the other unless you deliberately couple them, e.g. - shine a laser on one atom or not depending on the outcome of a measurement on the other. If you bring them close enough for their electron wave functions to overlap with relatively high probability (i.e. a distance of the order one angstrom), then they may affect one another. The same will remain true if they are one metre apart or ten metres apart. If you excite the electron of one of the atoms into a higher energy level, then what matters is how the electron interacts with the nucleus of the atom of which it is a part, not what is happening to some distant atom.

So there is no particular reason to expect the atoms' state to depend on their distance from one another if they are far apart.

• Because we need to bring them close together in order to entangle them I wanted to consider movements of them (which is needed to increase their distance), but I don't get why those movements (=unitary evolution of the state) leaves the entanglement intact. The "The same will remain true if they are one metre apart" and "there is no particular reason to expect the atoms' state to depend on their distance" is just what I would like to prove. – Peter Jul 22 '15 at 12:43