Which of these statements about tensor products is incorrect?

[nomadreid]

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\otimesB
(b) An entangled state C is a pure state, and hence there does not exist A and B such that A\otimesB = C
(c) By appropriate interactions one can produce entangled states.

Thanks.

 

[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\otimesH_B

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

(C) is correct.

 

[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.)

 

[nomadreid]

Thank you very much, Ravi Mohan and conana. That clears up a lot of confusion on my part. Very helpful.