In the previous article, we looked at single-qubit quantum gates, like the fundamental **Hadamard gate**.

We have detailed the way it was applying to one qubit. But let's see how it could operate on two qubits, by example on |00>.

As we know already from our previous article Expectation value of a product state, the final state will be the tensor product of the two transformed single-qubit states.

which can be developed to:

So by applying the Hadamard gate to two qubits, we generate **the superposition of all four basis states** of the tensor product of the sub-states, with **an equiprobability** of (1/2)^{2} = 0.25 of the computational basis states |0> and |1>.

Below is this quantum circuit visualized in the Quirk browser. Quirk confirms that the probability that the result state collapses to one of the four basis vectors equals to 25%.

**Remark 1**: More generally, starting with a state of n qubits |0...0_{n}⟩, if we apply the Hadamard gate to each qubit it results in the following state:

We end up with 2^{n} observable basis states. Many quantum algorithms use the Hadamard transform as an initial step, since it maps n qubits initialized with |0> to a superposition of all 2^{n} orthogonal states in the |0>,|1> basis with equal weight.

See by example the Deutsch-Jozsa algorithm.

**Remark 2**: When two gates are in parallel, like in the previous case, we can consider **the tensor product of the two gates** and apply that on the 2-qubit state vector, as per the method exposed in Tensor Product Matrices. We'll end up with the same result^{[1]}

Now applying this matrix to the two qbits states, we get:

which is strictly equivalent to:

More generally, it can be proved that the general equation for the Hadamard gate transformation on multiple qubits is:

which can be summarized more succinctly in the very useful equation

where x.z is the bitwise inner product of x and z.

Let's try to verify this result for a few cases.

If n=1, we know that

Which can be formulated this way as well

Checking now for n=2, we get:

[1] This way of applying the tensor product of the gates is even more general that the first way we exposed by applying each gate on each sub-system, since if the two input qubits are entangled, we won't be able to represent the input state as a tensor product of the states of the two qubits.