1. What is ZX applied to |0> ?

2. What is ZX applied to H|0> ?

Answer in form of: a|0> + b|1>

To understand the effect of the ZX gate, we need to know the Pauli-X (X) gate and the Pauli-Z (Z) gate.

1. The Pauli-X gate, represented as X, flips the state of a qubit. Specifically, X|0⟩ = |1⟩ and X|1⟩ = |0⟩.

2. The Pauli-Z gate, represented as Z, applies a phase flip to the qubit. It leaves the state |0⟩ unchanged, but changes the sign of the state |1⟩. Specifically, Z|0⟩ = |0⟩ and Z|1⟩ = -|1⟩.

Now, let's answer the questions by using the ZX gate.

1. ZX applied to |0⟩:
To apply the ZX gate to |0⟩, we first apply the X gate, and then apply the Z gate.
X|0⟩ = |1⟩ (Applying X)
Z|1⟩ = -|1⟩ (Applying Z)
Therefore, ZX|0⟩ = -|1⟩ = -|1⟩|0⟩ + 0|1⟩.

2. ZX applied to H|0⟩:
To apply the ZX gate to H|0⟩, we need to know the effect of the Hadamard (H) gate first. The H gate transforms the basis states as follows:
H|0⟩ = (|0⟩ + |1⟩)/√2
H|1⟩ = (|0⟩ - |1⟩)/√2

Now, let's compute ZX applied to H|0⟩:
H|0⟩ = (|0⟩ + |1⟩)/√2 (Applying H)
X((|0⟩ + |1⟩)/√2) = (X|0⟩ + X|1⟩)/√2 = (|1⟩ + |0⟩)/√2 = (|0⟩ + |1⟩)/√2 (Applying X)
Z((|0⟩ + |1⟩)/√2) = (Z|0⟩ + Z|1⟩)/√2 = (|0⟩ - |1⟩)/√2 (Applying Z)

Therefore, ZX(H|0⟩) = (|0⟩ - |1⟩)/√2 = -(|1⟩ - |0⟩)/√2 = (-|1⟩ + |0⟩)/√2 = -|1⟩|0⟩ + |0⟩|1⟩.

The final answers are:
1. ZX applied to |0⟩ = -|1⟩|0⟩ + 0|1⟩
2. ZX applied to H|0⟩ = -|1⟩|0⟩ + |0⟩|1⟩