What is ZX applied to H|0>?

To determine what happens when the ZX gate is applied to the state |H⟩ = (1/√2)(|0⟩ + |1⟩), we need to understand the effect of the ZX gate on qubits in the basis states |0⟩ and |1⟩.

The ZX gate is a two-qubit gate, which means it acts on two qubits simultaneously. It is commonly used for entangling qubits and creating superposition states.

The ZX gate can be represented by the following matrix:
```
[ 1 0 0 0 ]
[ 0 1 0 0 ]
[ 0 0 0 -1 ]
[ 0 0 -1 0 ]
```

To find the result of applying the ZX gate to the state |H⟩ = (1/√2)(|0⟩ + |1⟩), we can apply the matrix representation of the ZX gate to the state vector:

```
[ 1 0 0 0 ] [ 1 0 0 0 ] [ 1 x 1 + 0 x 0 + 0 x 0 + 0 x 0 ]
[ 0 1 0 0 ] . [ 1 0 0 0 ] = [ 0 x 1 + 1 x 0 + 0 x 0 + 0 x 0 ]
[ 0 0 0 -1 ] [ 1 0 0 0 ] [ 0 x 1 + 0 x 0 + 0 x 0 + -1 x 0 ]
[ 0 0 -1 0 ] [ 1 0 0 0 ] [ 0 x 1 + 0 x 0 + -1 x 0 + 0 x 0 ]
```

Simplifying the multiplication:

```
[ 1 0 0 0 ] [ 1 ] [ 1 ]
[ 0 1 0 0 ] . [ 1 ] = [ 0 ]
[ 0 0 0 -1 ] [ 0 ] [ 0 ]
[ 0 0 -1 0 ] [ 0 ] [ 0 ]
```

So, applying the ZX gate to the state |H⟩ = (1/√2)(|0⟩ + |1⟩) results in the state |Ψ⟩ = |0⟩. This means that the ZX gate collapses the superposition state |H⟩ to the computational basis state |0⟩.