Is the tensor product of 2 Hadamard gates entangled?

Question

Assume that you have a system of two qubits in the state $|11 \rangle$. Apply $H \otimes H$, where $H$ is the Hadamard matrix. Is the state $(H \otimes H)|11\rangle$ entangled? I know if we take the tensor product of 2 Hadamard gate we get our initial state, but I'm not quite sure what I'm supposed to do here. I'm more curious how can I represent $H |11\rangle $, I know $H|1\rangle= \dfrac{|0\rangle - |1\rangle}{\sqrt(2)}$ , so is $H|11 \rangle = \dfrac{1}{\sqrt(2) } \big( |00\rangle-|11 \rangle \big)$ or $H|11\rangle= \dfrac{1}{\sqrt(2) } \big(|00\rangle-|10\rangle-|01\rangle+|11\rangle \big) $

Any help would be great. Thanks in advance.

Answer 1

As @Durd2nT pointed out, the answer is no. This is because

$$(H \otimes H) \big(|1\rangle \otimes | 1\rangle\big) = \dfrac{|00\rangle - |01\rangle = |10\rangle + |11\rangle}{2} = \overbrace{\bigg( \dfrac{|0\rangle - |1\rangle }{\sqrt{2} } \bigg)}^{|\psi_A\rangle } \otimes \overbrace{\bigg( \dfrac{|0\rangle - |1\rangle }{\sqrt{2} } \bigg)}^{|\psi_B\rangle} $$

So $(H \otimes H) |1 1\rangle$ produces a product state $|\psi_A \rangle \otimes |\psi_B \rangle$ instead of an entangled state. If a two qubit state is entangled, then you can't write it as a tensor product of two single qubit state like what we just did.

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You can also extend this to more general scenario too.

Suppose you have a quantum state $|\psi_{init} \rangle = |\psi_{init}^A \rangle \otimes |\psi_{init}^B \rangle $ (a product state), then by applying the operator $U = U_A \otimes U_B $ to the state $|\psi_{init} \rangle$, the state of the system will remain as a product state. This is because

$$ U |\psi_{init} \rangle = \big( U_A \otimes U_B \big) \big( |\psi_{init}^A \rangle \otimes |\psi_{init}^B \rangle \big) = \overbrace{U_A |\psi_{init}^A \rangle}^{|\psi_A \rangle} \otimes \overbrace{U_B |\psi_{init}^B \rangle}^{|\psi_B \rangle} = \overbrace{|\psi_A \rangle \otimes |\psi_B \rangle}^{\textrm{product state}}$$

The question you asked, $U_A =H, U_B = H, |\psi_{init}^A \rangle = 1\rangle , |\psi_{init}^B \rangle = 1\rangle $.

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Also note that, it doesn't make sense to write down the operation $H|11\rangle$ as you did in your question. This is because $H$ is a $2 \times 2$ unitary matrix, and $|11\rangle$ is a $4 \times 1$ unit vector. Thus, you can't operate $H$ on $|11\rangle$.

And you said "I know if we take the tensor product of 2 Hadamard gate we get our initial state", I think you are mistaken this with two consecutive Hadamard gate. Two consecutive Hadamard gate applying to a single qubit is different than the tensor product of two Hadamard gate applying to two qubits.

The operation on the right of the figure is the representation of the tensor product of two Hadmard gate $H \otimes H$. It acts on a two qubit system. Whereas the operation on the left is a $H\cdot H$, and it only acts on a single qubit state.

Answer 2

1. You're correct in that $H |1 \rangle = \frac{1}{\sqrt{2}}(|0\rangle - |1 \rangle)$, with $H$ being the Hadamerd gate. However, acting on a product state $|11\rangle$ with $H \otimes \mathbb{1}$ ($H$ on the first qubit), you get \begin{align} H \otimes \mathbb{1} |11 \rangle & = \frac{1}{\sqrt{2}}(|0\rangle - |1 \rangle) |1 \rangle \\ & = \frac{1}{\sqrt{2}}(|01\rangle - |11 \rangle) \end{align} and similarly, \begin{align} H \otimes H |11\rangle & = H|1\rangle \otimes H|1\rangle \\ & = \frac{1}{\sqrt{2}}(|0\rangle - |1 \rangle) \otimes \frac{1}{\sqrt{2}}(|0\rangle - |1 \rangle) \\ & = \frac{1}{2} (|00\rangle - |01\rangle - |10\rangle + |11\rangle) \ . \end{align}

2. As to your question of whether the state $H \otimes H |11\rangle$ produces an entangled state, the answer is no. That is because, by definition, entangled states are non-separable. This means you cannot write an entangled state as a product of two separate states. Take for instance a Bell state, $|\Phi \rangle = \frac{1}{\sqrt{2}}(|00\rangle + |11 \rangle)$. You can NOT find two states $|\phi_1 \rangle, |\phi_2 \rangle$ such that $|\Phi \rangle$ can be written as a product $|\phi_1 \rangle \otimes |\phi_2 \rangle$.

3. You're statement that the tensor product of two Hadamards acted on an initial state results in the initial state is wrong as per the above argument. What you probably got confused with is that the square of the Hadamard gives the identity, $H^2 = \mathbb{1}$. This is because the Hadamard is unitary and Hermitian (or involutory), meaning the matrix is its own inverse.