# Which of these statements about tensor products is incorrect?

1. Jun 23, 2013

I have read the following three simplifications in various places, but together they give a contradiction, so at least one of them must be an oversimplification. Which one?
(a) Interaction between two systems A and B is described by A$\otimes$B
(b) An entangled state C is a pure state, and hence there does not exist A and B such that A$\otimes$B = C
(c) By appropriate interactions one can produce entangled states.

Thanks.

2. Jun 23, 2013

### Ravi Mohan

(A) is not correct. The correct statement should be
If $H_A$ is the hilbertspace of system A and $H_B$ is space of system B, then the hilbertspace of composite system is $H_A$$\otimes$$H_B$

(B) is not correct. The correct statement should be
If $|C\rangle$$\in$$H_A$$\otimes$$H_B$ represents entangled state then it can not be written in the form $|A\rangle$$\otimes$$|B\rangle$ where $|A\rangle$$\in$$H_A$ and $|B\rangle$$\in$$H_B$

(C) is correct.

3. Jun 23, 2013

### conana

(a) interaction between two systems whose individual Hilbert spaces are $H_A$ and $H_B$ can be describe by states $|\psi\rangle$ belonging to the direct product of these Hilbert spaces $H=H_A\otimes H_B$.

(b) an entangled state $|\psi\rangle_C$ belongs to this direct product Hilbert space, however, it cannot be expressed simply as a tensor product of states belonging to $H_A$ and $H_B$. That is, for $|\psi\rangle_A\in H_A$ and $|\psi\rangle_B\in H_B$, one cannot write $|\psi\rangle_C=|\psi\rangle_A\otimes|\psi\rangle_B$, rather $|\psi\rangle_C=\sum_{i,j}\alpha_{ij}|\psi_i\rangle_A\otimes|\psi_j \rangle_B$ where $\{|\psi_i\rangle_A\}\subset H_A$, $\{|\psi_i\rangle_B\}\subset H_B$ and there MUST be more than one non-vanishing terms in the sum for an entangled state $|\psi\rangle_C$.

(c) this is correct. (Edit: Actually, we can get an entangled state simply by appropriately partitioning our total system into subsystems A and B.)

Last edited: Jun 23, 2013
4. Jun 23, 2013