# Are all bound states entangled?

In QM, a bound state is a special state of a particle subject to ta potential such that the particle tends to remain localized in space.

The potential may be external or it may be the result of the presence of another particle; in the latter case, one can equivalently define a bound state as a state representing two or more particles whose interaction energy exceeds the total energy of each separate particle.

https://en.wikipedia.org/wiki/Bound_state

So basically all bound states of particles need energy to separate the parts.

Quantum entanglement is a physical phenomenon that occurs when pairs or groups of particles are generated, interact, or share spatial proximity in ways such that the quantum state of each particle cannot be described independently of the state of the others, even when the particles are separated by a large distance.

https://en.wikipedia.org/wiki/Quantum_entanglement

Now entanglement is defined by sharing spatial proximity in some cases (in ways such that the quantum state of each particle cannot be described independently of the state of the others).

Based on the definitions, entangled particles might not create a bound system, because for example two entangled photons that fly apart, do not need energy to be separated.

But a bound system of particles might be entangled, but I am not sure if it is always the case, so that all bound systems are made up of entangled parts. For example, the constituents of an atom, the quarks that make up the proton and neutrons, and the electrons are creating a bound system. But are all the parts (quarks and electrons) entangled too?

Question:

1. Are all bound systems (QM) entangled too?
• Two entangled photons can be made to be arbitrarily far away from each other. Entanglement is non-local and not necessarily related to space. You can entangle things by getting them close to each other, but that is not a necessary condition. Commented Oct 9, 2019 at 23:53
• Since photons can be entangled but cannot be bound, the answer to the second half of the question in the title is No: all entangled systems are not bound. Commented Oct 9, 2019 at 23:53
• @KFGauss correct, agree, that is what I wrote. Commented Oct 10, 2019 at 2:08
• I think you should change the title to "Are all bound states entangled?". It seems like you already know that not all entangled states are bound. Commented Oct 10, 2019 at 3:53
• @LewisMiller different particles can be entangled too. physics.stackexchange.com/questions/224317/… Commented Oct 10, 2019 at 15:51

Are all bound states entangled?

No. At least not if you allow that one of the particles can be treated classically. I suppose this answer might inappropriately avoid the question, but nevertheless this is a common way to treat the hydrogen atom and a common way to use the term "bound state."

In treatments of the hydrogen atom, the problem is reduced from two-body to one-body and then the relative coordinate is treated quantum mechanically. This resulting in the usual $$\phi_{n\ell m}(\vec r)$$ states of the hydrogen atom. (Here $$\vec r$$ is the relative coordinate.)

There are an infinite number of such bound states. None of those bound states are entangled, since there is only one effective quantum particle (that described by the reduced coordinate).

Perhaps the above discussion seems like a cheat. So, consider more generally a quantum mechanical system with two positions of two particles treated quantum mechanically. Suppose particle 1 can be found to have eigenstates $$\phi_a(\vec r_1)$$ of some observable operator $$\hat A$$. Then the entire quantum mechanical wave function can be written as: $$\Psi(\vec r_1, \vec r_2) = \sum_a \chi_a(\vec r_2)\phi_a(\vec r_1)\;,\tag{1}$$ where the $$\chi_a(\vec r_2)$$ can be though of as parameters in the expansion in terms of the $$\phi_a(\vec r_1)$$, but also the $$\chi_a$$ are the wave functions of particle 2. Such a state in Eq. (1) is generally entangled, but not in the case when there is only one term in the series like: $$\Psi_1(\vec r_1, \vec r_2) = \chi_1(\vec r_2)\phi_1(\vec r_2)\;. \tag{2}$$ A state like that in Eq. (2) is a direct product and is not entangled.

Now, is the state given in Eq. (2) a bound state? Well, we do now know because we have not said what the Hamiltonian is. But in the case where we take the Hamiltonian to be $$\frac{-\hbar^2}{2m_2}\nabla^2_{\vec r_2} + \frac{-\hbar^2}{2m_2}\nabla^2_{\vec r_1} - \frac{|e|^2}{4\pi\epsilon_0 r_1}$$ then we are right back to the hydrogen atom and we see that such a direct product state (non-entangled state) can describe a bound state.

Regarding the converse question, a typical textbook/pedagogical entangled state of two spins like $$\frac{1}{\sqrt{2}}(|\uparrow\rangle \uparrow\rangle +|\downarrow\rangle|\downarrow\rangle)$$ is entangled, but not necessarily "bound" in any sense.

• Regarding your last, two spin example, that state can only be an eigenstate of what is normally called an interacting Hamiltonian. I can define interacting Hamiltonian (for two parties) as an operator that cannot be written as $A\otimes B$.
– lcv
Commented May 3 at 5:27
• So in the sense above a bound state would be an eigenstate of an interacting Hamiltonian and it would be necessarily entangled (some care is needed to treat the degenerate case).
– lcv
Commented May 3 at 5:31
• Anyway I'm not saying your answer is not accurate rather the definition of bound states as per the Wikipedia page seems to be. That page is a bit shaky. At one point it is stated that bound states is equivalent to belonging to the Hilbert's space, but in other references to interacting (and free) state is made.
– lcv
Commented May 3 at 5:35

I intentionally didn't read the comments below to see if I could figure it out myself. So I'll give it a try.

Entanglement surely doesn't apply to bound states. In bound states, each state can be described independently of the other states. Of course, you have to take their interactions in account, which means they depend on each other but only in the sense that they interact with each other.

That means that if you could measure the state in which, for example, a certain electron, in a "heavy" atom finds itself (even if they are indistinguishable in each shell, which doesn't imply entanglement either) none of the other constituents will instantaneously jump into a certain state, because of the simple fact that all the constituents already find themselves in a certain state because they continuously (which is quantum-mechanically wrongly put, but that's of little importance in this context) interact with each other.

So in whatever bound states one is interested, it are exactly these interactions which are the cause of non-entanglement.

Two electrons, for example, can't find themselves in a superimposed state, insofar their position is concerned. If you measure the position of one of them when they are far apart from each other then the other electron will not instantaneously (even though this is a somewhat vague notion, but I'm sure you know what I mean) jump in a correlated position, because their positions can't be correlated because of the distance between them. It's different for their spins though. They stay correlated (entangled) all the way because the spins don't depend on the distance between the electrons.

This may seem very strange at first sight (or spooky, as Einstein called it) but it is actually quite logical when you think a bit (or a lot) deeper about it.