# When does CNOT entangle?

Let $$|\psi_1\rangle$$ and $$|\psi_2\rangle$$ be qubit states such that $$\text{CNOT}|\psi_1\rangle \otimes |\psi_2\rangle$$ is entangled. I'm interested in if there is a simple condition that this imposes on what $$|\psi_1\rangle$$ and $$|\psi_2\rangle$$ can be.

Writing $$|\psi_i\rangle = \alpha_i|0\rangle + \beta_i |1\rangle$$ for each $$i = 1, 2$$, the condition that the final state be entangled is equivalent to saying that the state $$\alpha_1\alpha_2|00\rangle + \alpha_1\beta_2|01\rangle + \beta_1\beta_2|10\rangle + \beta_1\alpha_2|11\rangle$$ be entangled. But now what can I say about the coefficients $$\alpha_i, \beta_i$$ that guarantees this?

• You can compute the density matrix for one of the qubits. Apr 19 at 19:02

A necessary condition is that the control state is not an eigenstate of Z, and the target state is not an eigenstate of X.

A sufficient condition to have a fully entangled state is that the control state is in $$|+\rangle$$ or $$|-\rangle$$, state, and the target is in $$|0\rangle$$ or $$|1\rangle$$.

I'm posting this answer to show how you can get from the two-qubit state which you had already obtained to the conclusions given in the answer of Yaron Jarach.

A general two-qubit state $$\vert\psi\rangle=a\vert00\rangle+b\vert01\rangle+c\vert10\rangle+d\vert11\rangle$$ is not entangled if it can be written as a product of 2 single-qubit states $$\left(u_1\vert0\rangle+v_1\vert1\rangle\right)\left(u_2\vert0\rangle+v_2\vert1\rangle\right) = u_1u_2\vert00\rangle+u_1v_2\vert01\rangle+v_1u_2\vert10\rangle+v_1v_2\vert11\rangle.$$ Equating the above two expressions it follows that a two-qubit state is not entangled if $$ad=bc$$. Applying this to the two-qubit state which you obtained by applying a $$CNOT$$ gate to a product state, we find that your resulting state is entangled if $$\alpha_1\beta_1\alpha_2^2 \neq \alpha_1\beta_1\beta_2^2,$$ which is only the case if the following three conditions are met: \begin{aligned} \alpha_1&\neq0, \\ \beta_1&\neq0, \\ \alpha_2 &\neq \pm\beta_2. \end{aligned} The combination of the first two conditions is equivalent to the condition that the control state is not an eigenstate of the Pauli $$Z$$ matrix, while the third condition is equivalent to the condition that the target state is not an eigenstate of the Pauli $$X$$ matrix.

Might I suggest a slightly clearer method for equating the general statement of entangling to the specific results? Let's write out the unitary of cNOT in terms of Pauli matrices: $$cNOT=I\otimes I+\frac12(I-Z)\otimes(X-I).$$ Notice how the first qubit terms are only $$I$$ and $$Z$$, while the second qubit terms are only $$I$$ and $$X$$. Now, remember that we can choose any orthonormal basis for each qubit. I'm going to choose the orthonormal basis with respect to $$Z$$ (first qubit) and $$X$$ (second qubit). $$|\psi_1\rangle=\alpha_1|0\rangle+\beta_1|1\rangle,\qquad|\psi_2\rangle=\alpha_2|+\rangle+\beta_2|-\rangle.$$ Now, let's evaluate $$|\psi_1\rangle|\psi_2\rangle=\alpha_1\alpha_2|0+\rangle+\alpha_1\beta_2|0-\rangle+\beta_1\alpha_2|1+\rangle+\beta_1\beta_2|1-\rangle,$$ and apply cNOT $$cNOT|\psi_1\rangle|\psi_2\rangle=|\psi_1\rangle|\psi_2\rangle=\alpha_1\alpha_2|0+\rangle+\alpha_1\beta_2|0-\rangle+\beta_1\alpha_2|1+\rangle-\beta_1\beta_2|1-\rangle.$$ See that the only change is the sign on the final term? It should be fairly obvious with a little thought (or maths: try to factor the state) that the only way this final state is not entangled is if $$\beta_1\beta_2=0$$ or $$|\beta_1\beta_2|=1$$, so that the sign change becomes a global phase instead of a local one.