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I know that the quantum circuit $\text{CNOT}\; (H \otimes I)$, where $\text{CNOT}$ is the controlled-not gate and $H$ the Hadamard gate, takes the computational basis of two qubits $|00\rangle,|01\rangle,|10\rangle,|11\rangle$ to the Bell states, which are maximally entangled.

Would $\text{CNOT}\;(H \otimes I)$ give us an entangled state even if the qubits are initially not in a standard basis state, i.e. $|0\rangle$ or $|1\rangle$?

I was interested if, when you say that some qubit A is maximally entangled with another qubit B, is there a similar circuit that (a) produces that state, and (b) from which you can easily show that the qubits are maximally entangled?

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1. Not all states produced by $\text{CNOT} \; (H \otimes I)$ are entangled.

For the first part of your question: no, not all states which arise as the output of $\text{CNOT} \; (H \otimes I)$ are entangled. Specifically, you can consider $$\begin{align*} |\psi\rangle \;&=\; (H \otimes I) \; \text{CNOT} \;\Bigl[ \tfrac{1}{\sqrt 2}\bigl( |00\rangle + |10\rangle \bigr) \Bigr] \\ \;&=\; (H \otimes I) \Bigl[ \tfrac{1}{\sqrt 2}\bigl( |00\rangle + |11\rangle \bigr) \Bigr] \\ \;&=\; \tfrac{1}{2} \bigl( |00\rangle + |01\rangle + |10\rangle - |11\rangle\bigr), \end{align*}$$ which is a maximally entangled state (notice from the second line above that it differs from a Bell state by a local unitary). However, by construction, if you apply $\text{CNOT} \; (H \otimes I)$ to $|\psi\rangle$, you will get back out the state $$ \text{CNOT} \; (H \otimes I) \;|\psi\rangle \;=\; \tfrac{1}{\sqrt 2}\bigl( |00\rangle + |10\rangle \bigr) \;=\; |+\rangle\otimes|0\rangle\;, $$ which has no entanglement at all. Even if you're interested only in inputs which are product states, we can see that the circuit maps $|+\rangle \otimes |0\rangle$ to another product state — specifically, $|0\rangle|0\rangle$.


2. All maximally entangled two-qubit states can be easily described by a similar circuit to $\text{CNOT} \; (H \otimes I)$ acting on the standard basis.

For the second part of your question: if you allow yourself arbitrary single-qubit unitaries, then for any maximally entangled two-qubit state $|\psi\rangle$, you can certainly construct a circuit which constructs $|\psi\rangle$ from standard basis states, and which allows you easily to see that the state is maximally entangled. Every two-qubit state has a Schmidt decomposition, $$ |\psi\rangle \;=\; u_0|\alpha_0\rangle|\beta_0\rangle \;+\; u_1|\alpha_1\rangle|\beta_1\rangle \;,$$ where $u_0 \geqslant u_1 \geqslant 0$ — in particular, $u_1 = 0$ for $|\psi\rangle$ a product state — and where $\{ |\alpha_0\rangle, |\alpha_1\rangle \}$ and $\{ |\beta_0\rangle, |\beta_1\rangle \}$ are each orthonormal bases for $\mathbb C^2$. In order for $|\psi\rangle$ to be maximally entangled, we need $u_0 = u_1 = \tfrac{1}{\sqrt 2}$. Consider single-qubit unitary matrices $A$ and $B$, such that $$\begin{alignat*}{4} A |x\rangle \;&=\; |\alpha_x\rangle &\quad& \text{for $x \in \{0,1\}$}, \\[1ex] B |y\rangle \;&=\; |\beta_y\rangle && \text{for $y \in \{0,1\}$}; \end{alignat*}$$ then we can describe $|\psi\rangle = (A \otimes B)\;\text{CNOT}\;(H \otimes I)\;|0\rangle|0\rangle$, that is, the effect of applying $(A \otimes B)$ to the Bell state $|\Phi^+\rangle = \tfrac{1}{\sqrt 2}\bigl( |00\rangle + |11\rangle )$. Furthermore, it's not hard to show that any standard basis state is mapped by that circuit to some maximally entangled state similar to $|\psi\rangle$, and that any circuit of this form (whatever $A$ and $B$ might be) maps any standard basis state to a maximally entangled two-qubit state.

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