# Distinguishing between an entangled and non-entangled state (mainly $S(H_A) \otimes S(H_B)$ vs. $S(H_A \otimes H_B)$)

Say I have two quantum systems $A$ and $B$ I can look at the joint (composite) system $AB$ which is given by $H_{AB} \in H_A \otimes H_B$

Measuring a subsystem with respect to a collection of measurement matrices $\textbf{M} = \{M_i\}_{i \in I} \in Meas_{I}(H_A)$ acts as measuring $AB$ with respect to $\textbf{M} \otimes \mathbb{I}_B = \{M_i \otimes \mathbb{I}_B\}_{i \in I}$

Q: Can I do this with entangled states?

I know that making a measurement causes the entangled state of say 2 qubits to decompose in to two states. If I understand correctly, they are not entangled anymore. Now because of this we can seperate the states in to a sum of the product states $|AB> = \sum \alpha_j|j> \otimes |\psi_j>$ over all basis states $|j> \in S(H_A)$. What this says to me is that we can some how distinguish between these qubit states (hence they are not entangled anymore). Do I have this right?

This leads me in to the difference between $S(H_A) \otimes S(H_B)$ vs. $S(H_A \otimes H_B)$. Now the first case is not entangled and we have the "product states" of two wavefunctions/state vectors , but in the second case we have some sort of combination (composite?) of states? I would guess this is when the states are entangled and after measurement they decompose in to say $|\psi> \otimes |\phi> \in S(H_A) \otimes S(H_B)$.

Now if I look at an entangled 2-qubit state that I know is entangled $|\Phi> = |\Phi^+> = \frac{1}{\sqrt{2}}(|0>|0> + |1>|1>) \in H_A \otimes H_B$ (as written out in some notes) I see that we can have an entangled state embedded somehow in the tensor of two state vectors in Hilbert space. I don't know if this is supposed to be $S(H_A \otimes H_B)$ or $S(H_A) \otimes S(H_B)$ or something else completely.

Edit: I removed some tensor math that was incorrect. I had initially thought that if some mixture of states could be decomposed and have just a $\otimes$ and nothing such as an addition or subtraction operator it. I know that is very rudimentary but from my basic understanding of product states: http://en.wikipedia.org/wiki/Product_state I see that if a probability density can be written as the tensor product of two different probability densities, then we have a non-quantum correlation, although in the wiki article above, this is neither quantum or classical in nature. I however also see that a mixed state such as $\rho_{AB}=\frac{1}{2}(|0_A0_B⟩⟨0_A0_B|+|1_A1_B⟩⟨1_A1_B|)$ only has classical correlations. Furthermore I have found that Two states $\rho, \sigma$ are called $\textbf{perfectly distinguishable}$ if there exists a measurement $M \in Meas(H) \text{ with } I = {0,1,...}$ such that $p_0(M,\rho) = 1 = p_1(M,\sigma)$. Now from my reading, in the case of a von Neumann measurement $M = \{M_i\}, p_i$ simplifies to $p_i = tr(M_i \rho)$ and in the case of a complete von Neumann measurement $M = \{ |i \rangle \langle i | \}$, $p_i$ simplifies to $p_i = tr(|i \rangle \langle i | \rho) = \langle i | \rho |i \rangle$

So basically, I just have to take the traces of the respective measurements on probability densities, if they are equal (hence perfectly distinguishable) then the system of, say qubits, are not entangled, otherwise they are? Is this correct?

Thank you,

Brian

-
What do you mean by $S(H)$? The only interpretation I was able to get was $H$ itself and yes, then $S(A)\otimes S(B)$ is the same thing as $S(A\otimes B)$. ;-) Yes, a measurement may be used for entangled and non-entangled states. Yes, the complete measurement of the pure state A brings the state of AB to a product i.e. non-entangled state. –  Luboš Motl Aug 26 '13 at 4:47
The only meaning of your notation $(|0> \otimes |1>)(|0> \otimes |1>)$ would be $|0> \otimes |1>)\otimes(|0> \otimes |1>)$, so it would involve $4$ particles and not $2$, so this notation does not make sense for $2$ particles. –  Trimok Aug 26 '13 at 8:44
" I know that making a measurement causes the entangled state of say 2 qubits to decompose in to two states". No. If you have a initial state $\frac{1}{\sqrt{2}}(|00\rangle +|11\rangle)$, and if you measure the "spin" of the first particle and finding that this first particle is (after the measurement) in a state $|0\rangle$, this means that the global state of the $2$ particles, after the measurement, is $|00\rangle$. So you loose the $|11\rangle$ part. –  Trimok Aug 26 '13 at 8:50
@Lubos, thanks. S is all the norm-1 vectors in $H$. Is there any tie when $S(A)\otimes S(B) \neq S(A \otimes B)$ for some state vector $|\psi \rangle \in S(H)$? I hope this makes sense, I just wonder why there are two notations then for the same thing. I know that $S(A \otimes B)$ would mean that state vectors would be in a product state. I found: If $\rho_A \in D(H_A)$ and $\rho_B \in D(H_B)$ are pure, then so is $\rho_A \otimes \rho_B$ Namely $|\phi_A \rangle \langle \phi_A| \otimes |\phi_B \rangle \langle \phi_B| =$ continued.. –  Relative0 Nov 11 '13 at 2:21
$= (|\phi_A \rangle \otimes |\phi_B \rangle)(\langle \phi_A| \otimes \langle \phi_B|) = |\phi_{AB} \rangle \langle \phi_{AB}|$ for all $|\phi_A \rangle \in S(H_A)$ and $|\phi_B \rangle \in S(H_B)$, where $|\phi_{AB} \rangle = |\phi_A \rangle \otimes |\phi_B \rangle \in S(H_A \otimes H_B)$ How does this relate? This seems to be a product state and thus $S(H_A)\otimes S(H_B) = S(H_A \otimes H_B)$, then again could there possibly be anything else such as in my last comment? Thanks –  Relative0 Nov 11 '13 at 2:34