{tag=Tensor product}

We don't need to understand a super generalized version of <tensor products> to know what they mean in basic <quantum computing>!

Intuitively, taking a <tensor product> of two <qubits> simply means putting them together on the same quantum system/computer.

When we write the <bra-ket notation>: $\ket{00}$ that is the same as $\ket{0} \otimes \ket{0}$.

The tensor product is called a "product" because it distributes over addition.

E.g. consider:
$$
(\frac{\ket{0} + \ket{1}}{\sqrt{2}}) \otimes \ket{0} =
\frac{\ket{0} \otimes \ket{0} + \ket{1} \otimes \ket{0}}{\sqrt{2}} =
\frac{\ket{00} + \ket{10}}{\sqrt{2}}
$$

Intuitively, in this operation we just put a <Hadamard gate> qubit together with a second pure $\ket{0}$ qubit.

And the outcome still has the second qubit as always 0, because we haven't made them interact.

The <quantum state> $\frac{\ket{00} + \ket{10}}{\sqrt{2}}$ is called a <separable state>, because it can be written as a single product of two different qubits. We have simply brought two qubits together, without making them interact.

If we then add a <CNOT gate> to make a <Bell state>:
$$
\frac{\ket{00} + \ket{11}}{\sqrt{2}} =
\frac{\ket{0} \otimes \ket{0} + \ket{1} \otimes \ket{1}}{\sqrt{2}}
$$
we can now see that the <Bell state> is <non-separable>: we've made the two qubits interact, and there is no way to write this state with a single <tensor product>. The qubits are fundamentally <entangled>.


Ciro Santilli @cirosantilli 34

 Incoming links: Separable state