The relationship between entanglement of vector states to matrix operations

Question

I don't understand something which is I believe pretty fundamental. It's said that an operation represented by a matrix A is an entanglement if A can't be written as a tensor product of other matrices. On the other hand, I just learned lately that a vector state too can be determined by these criteria. That is, it's entangled if and only if it can't be written as a tensor product of other vector states.

Here comes the confusion. Suppose you take a CNOT for example. This obviously can't be written as a tensor product of other matrices as it's known to be entangled. What're the consequences of applying a CNOT on a vector? I don't understand it, if you take a vector like (1,0,0,0) or (0,0,10) you will get after CNOT (1,0,0,0) or (0,0,0,1) respectively. Which obviously can be written as the tensor product of other vectors. So what's going on here? Is CNOT supposed to entangle the vector state or not?

Why is it that the gate itself can't be decomposed but the vector which it acted upon can? Or if it isn't necessary then, what's the definition or requirements for entanglement which I am missing?

Answer 1

We don’t use the terminology in the way you’re trying to use it.

Entanglement is a property of states. For pure states shared between two parties, if they can be written in the form $$ |\psi\rangle\otimes|\phi\rangle, $$ they are separable. If they are not separable, they are entangled. Something like cnot is said to be “an entangling operation” meaning that when it acts on some separable states (but not an arbitrary separable state), it can produce an entangled state vector. We only say that something is entangled in reference to the state vector.

For example, as you say cnot acting on a state such as $|11\rangle$ simply produces $|10\rangle$, and no entanglement is produced. But, cnot acting on $(|0\rangle+|1\rangle)\otimes|1\rangle$ produces $|01\rangle+|10\rangle$, which is entangled. Hence, cnot is capable of turning a separable state into an entangled state, so it is entangling.

I should also mention that a matrix not being of the form $A\otimes B$ is necessary for it to be entangling but not sufficient. There are operations not of this form which are not entangling, such as swap.

Answer 2

There are 3 different questions you need to ask yourself. First is the operator a tensor product of smaller matrices. Then is your input state a tensor product of smaller vectors. Finally is the result a tensor product of smaller vectors.

Stick with the two qubit example and CNOT that you had.

First question: CNOT is not able to be written as $A \otimes B$ with $A$ and $B$ operating on 1 qubit each. Second question: The examples you gave were $(1,0,0,0)=|0 0 \rangle$ which can be broken up as $|0 \rangle \otimes |0 \rangle$. The other example can as well. It is $|1 \rangle \otimes |0 \rangle$. Third question: The result states were $(1,0,0,0)$ which was not entangled again. The other was $(0,0,0,1)$ which is $|1 \rangle \otimes |1 \rangle$ also not entangled.

The implication goes differently than you seem to think.

If the operator was not entangled and written as $A \otimes B$ and the input state was not entangled and written as $v \otimes w$ with the same dimensions as the operator, then you would get $Av \otimes Bw$ as the output. So not entangled operator AND not entangled input vector with the same dimensionality imply not entangled on the output.

So the example shows if you only have operator entangled but input not, the output can still be not entangled.

As another example, take the identity operator. It can be written as $I \otimes I$ so it is not entangled. Apply it to an entangled input, you get the same state out. That gives the case of unentangled operator and entangled input giving entangled output.

Edit to respond to comment:

CNOT applied to $(H \mid 0 \rangle) \otimes \mid 0 \rangle$ gives $\frac{1}{\sqrt{2}} \mid 0 \rangle \otimes \mid 0 \rangle + \frac{1}{\sqrt{2}} \mid 1 \rangle \otimes \mid 1 \rangle$. So the initial state is not entangled, but the output state is a Bell state.