The definition of quantum entanglement, found on the internet and the literature is:

On a bipartite system $\mathcal{H}_A \otimes \mathcal{H}_B$, let $\rho$ be a mixed state. It is said to be separable if it is a convex combination of product states $$\rho = \sum_i \lambda_i \rho^A_i \otimes \rho^B_i $$ Here, $\lambda_i\ge0$, and $\rho^A_i,\rho^B_i\ge0$.

If this is not the case, it is said to be entangled.

My question is, how did they come to this definition? Where did it come from and why does it work? Is there any way to start from physical principles and arive to this definition?

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    $\begingroup$ By this definition, no state is entangled. The correct definition is that a state is entangled if it is not a product state. $\endgroup$ – WillO Jan 31 '19 at 0:41
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    $\begingroup$ @WillO no, this is definitely the definition found in the literature, for example in this article definition 3 $\endgroup$ – user353840 Jan 31 '19 at 0:46
  • $\begingroup$ It is nearly trivial to check that the tensor product consists entirely of states of the form you're calling separable. If you've found an article that suggests otherwise, you've found an article with a very elementary error. $\endgroup$ – WillO Jan 31 '19 at 0:56
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    $\begingroup$ @WillO That's true, but in that case you're only talking about pure states (= vector states). This definition deals with mixed states, i.e. density matrices. They are not elements of a Hilbert space, but rather operators acting on a Hilbert space. They are already normalised, so your point does not hold in this context. $\endgroup$ – user353840 Jan 31 '19 at 1:13
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    $\begingroup$ @WillO One could also argue that people are used to think of mixed states when they see $\rho$ rather than $\psi$. $\endgroup$ – Norbert Schuch Jan 31 '19 at 10:12

The answer by PhysicsTeacher is right; I will extend it a little.

The motivation is all about whether or not it is correct to say, of a composite physical system, "It is made of A and B", where we can think of A and B as separate parts of the system, each with their own properties. In most of science it is assumed that this is a valid way of speaking. In quantum physics it is not always correct.

The quantum physics equivalent of saying "A and B each have their own properties" is to assert that the joint quantum state is a product, where contributions of A and B appear in the form: $$ | \mbox{total state} \rangle = |\mbox{state of A}\rangle \otimes |\mbox{state of B}\rangle $$ (and for brevity we often omit the tensor product operator $\otimes$). In this case any interaction with A, such as scattering particles off it, or hitting it with a hammer, or allowing it to fly through a magnetic field gradient, will influence the state of A but not B (assuming B is not nearby). In particular, we can measure A and thus determine its properties and assert "those are the properties of A".

When converted into density matrix notation, the above state becomes $$ \rho = | \psi \rangle \langle \psi | = \rho^A \otimes \rho^B . $$

If we now have a heap of systems like A and B and all of them are in this sort of product state, but we don't know which systems are in which state, then we allow for this by forming the density matrix by adding each one of the above form, multiplied by the probability $p_i$ of that particular one. Thus we get $$ \rho = \sum_i p_i \, \rho^A_i \otimes \rho^B_i . \;\;\;\;\;\;\;\;\;\;\;\;\;\ (1) $$ This is the state you asked about, and I have added an interpretation of the $p_i$ (which you called $\lambda_i$).

The next part of the analysis is to say that the subscript $i$ here indicates a choice of decomposition of the total. There are many ways of making the same $\rho$, by adding up different choices of $\rho_i$. This is like saying one matrix can be expressed as the sum of others in many different ways. Among those different ways, some will have the form (1), some will not. In order to say "the state is not a sum of product states" we have to say "there is no way of forming this sum using product states." That is how entangled states are defined.

Now let me say two more things to motivate the definition. First, for an entangled state defined this way, it will not be possible to say "these are the properties of A" and "these are the properties of B". For example, in the spin-spin singlet state, $$ \frac{1}{\sqrt{2}} \left( | \uparrow \rangle_A | \downarrow \rangle_B - | \downarrow \rangle_A | \uparrow \rangle_B \right) $$ it is not possible to say "the spin of A is up" nor "the spin of A is down" nor "the spin of A is in this direction" nor "the spin of A is in that direction". Nor can one say "the spin of A is in a superposition of up and down"---because if it were correct to say that the spin of A is in a superposition of up and down, then there would exist a direction along which the spin of A is pointing, and one could measure along that direction and get the same result every time. But that is not what is observed. So the way to put the situation into words has to be "when A+B are in the singlet state, then A does not, in and of itself, have a spin direction, and nor does B". One can cash this statement out more fully in terms of local hidden variables, the way John Bell famously did. Here I am giving some general comments on what it means to say that two things are in a quantum entangled state.

Finally, I should add a remark about spatial location and the role of spacetime. The standard definition of entanglement is the one you gave and the one I discussed in the above. It mentions only states and Hilbert space and density operators (or matrices). However, I would like to add that an important added feature, that I would want to look for in order to take an interest in an entangled state, is to ask whether the two entangled subsystems can be physically located in different places, especially spacelike-separated places. If they cannot, then it may be that someone is simply analyzing a given system, such as an electromagnetic field mode, in a way which adopts the language and mathematical tools of entanglement, but does not exhibit the really interesting physical properties of entanglement. The latter are to do with location in spacetime, not just state in Hilbert space.


I believe the motivation is that if there is only a single member in this sum, i.e. the state is separable, then you can think of the system as being comprised of two independent subsystems, each in a definite state, that are combined as per the usual quantum-mechanical rules. You can describe the state of the bi-partite system by describing the state of each part.

This is just like how in a pure entangled state you can't describe the system by describing the state of each part. In the state $|01\rangle+|10\rangle$, we can't do that. In the un-entangled state $|01\rangle$ we can, we can say that the state of the first particle is $|0\rangle$ and the state of the second particle is $|1\rangle$.

EDITED to cut-out mistaken part, so as not to confuse the readers.

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    $\begingroup$ This is not a Schmidt decomposition. $\endgroup$ – Norbert Schuch Jan 31 '19 at 10:12
  • $\begingroup$ Ah. oops. Then... my suggested motivation doesn't really work. It works for the product state, but not the sum of product states. Hmm. $\endgroup$ – PhysicsTeacher Jan 31 '19 at 11:17

The simple definition of entanglement boils down to: any system that is described by a single continuous quantum mechanical wave equation is entangled . In general this affects probability distributions. The electron and proton in the solution of the hydrogen atom are entangled, i.e. the variables and the quantum numbers of its components are interdependent. The concept becomes useful when quantum numbers are involved.

Take the $π^0-> 2γ$ . The $π^0$ is spin zero, this constrains the spins of the two photons, and if the spin of one is measured, the spin of the other because of the entanglement ( the single quantum wavefunction of $π^0$ decay) and conservation of angular momentum, the spin of the other is known.

Take two free hydrogen atoms. Their electrons and protons are not entangled, because the wavefunctions describing them are not solutions of the same boundary conditions as each atom is in a different spacetime region. Once bonded into H2 all particles in the molecule are entangled and the probabilities with their phases have to be taken into account.

The definition above generalizes the simple concept to many body systems that can be described by one wavefunction. A system that is entangled must be modeled with a single wavefunction where individual space time positions of the particles composing the system obey quantum mechanical probabilistic rules. If they are a simple sum of wavefunctions there are no phases in the system and classical physics is the appropriate framework, so there is no entanglement.

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    $\begingroup$ This is either wrong or needs to be extended for clarification, because all quantum systems are described by a quantum wave equation (if you mean Schr. eqn) but not all quantum systems are entangled with one another. e.g. the carbon in this desk in front of me is not entangled with the carbon in the desk where you are, at least in any observable sense. But perhaps you had something else in mind, more field-theory-like? $\endgroup$ – Andrew Steane Jan 31 '19 at 16:08
  • $\begingroup$ "all quantum systems are described by a quantum wave equation" this is wrong, it is not the same equation for disparate quantum systems, the quantum mechanica l solutions have to obey the same boundary conditions. Two hydrogne atoms are not entangled,, nor are their electrons and protons. They are two different quantum mechanical systems described by a similar equation but in different regions of space and time. $\endgroup$ – anna v Jan 31 '19 at 18:45
  • $\begingroup$ @AndrewSteane see my comment above $\endgroup$ – anna v Jan 31 '19 at 18:58
  • $\begingroup$ I roughly see what you mean but I don't regret the comment! $\endgroup$ – Andrew Steane Jan 31 '19 at 20:23
  • $\begingroup$ @AndrewSteane I can understand that. The whole entanglement and qbit mathematical evolution are way over simple basic quantum mechanics. The only reason I answered this question is to stress that it is a very basic concept common to the quantum mechanical framework . Of course in my comment I should have said that two individual hydrogen atoms are not entangles unless in an H2 bond. $\endgroup$ – anna v Feb 1 '19 at 4:29

The reason is that every bipartite classical state can be written in this way. Indeed quantum separable states of that form (if there is more than one non-zero $\lambda_i$) are also called classically correlated.

Thus entanglement captures a feature that does not exist in the classical world.


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