# Quantum optics, 2 qubit gates acting on 2 qubits - cannot be factorized?

1. May 5, 2010

### yaboidjaf

Quantum optics, 2 qubit gates acting on 2 qubits - cannot be factorized??

Hi, I'm struggling to understand why two qubit gates acting on two qubits cannot be factorized, i.e.

G12 $$\neq$$ G1 $$\otimes$$ G2,

where G1 acts on qubit 1 only and G2 acts on qubit 2 only.

2. May 6, 2010

### Tomsk

Re: Quantum optics, 2 qubit gates acting on 2 qubits - cannot be factorized??

This is one of the tricks that you can do with quantum computers that you can't with classical computers. The reason it works is because the two qubits have to be considered as entangled. That means they behave as one system rather than two separate ones. Then the gates that can act on them can act on them together as one. For instance, CNOT flips the 2nd qubit dependent on the state of the 1st, so you can't think of it as one gate acting on the 2nd qubit (with a separate one acting on the first), since you don't know whether to flip it or not by itself.

3. May 6, 2010

### Zarqon

Re: Quantum optics, 2 qubit gates acting on 2 qubits - cannot be factorized??

A two-qubit state that is not factorizable is called an entangle state, and is indeed one of the new resources that are available at the quantum level, but not in classically. If you haven't already done so, it might be a useful exercise to simply for yourself compare an entangled state like a Bell state, to a product of two arbitrary one-qubit states, i.e. compare a Bell state like

$$(\left|0\right>\left|0\right>+\left|1\right>\left|1\right>)$$

to an arbitrary product of single qubit states like

$$(\alpha \left|0\right> + \beta \left|1\right>) \cdot (\gamma \left|0\right> + \delta \left|1\right>)$$.

Doing the multiplication and trying to find any coefficients of the product state that gives you the entangled state, and you easily see that it can't be done.

(note, I omitted the normalization factors, but they don't change the fact that it can't be done)