For any qubit state a, b is unentangled, iff you can write the state into a \otimes b.

Un-normalized State

If x|00\rangle + y|01\rangle + z|10\rangle + w|11\rangle is un-normalized state, we can normalize it by dividing the state by \sqrt{x^2 + y^2 + z^2 + w^2} // QUESTION: is this true for complex?

A state is normalized iff |x|^2 + |y|^2 + |z|^2 + |w|^2 = 1

Since multiplication by a constant does not change the representation (as long as the normalized version correspond to unit length), we can make the constant (for any c \neq 0) by negative. Therefore, multiplying a state by -1 is the same as the original state.

|v\rangle and -|v\rangle correspond to the same physical state. Just like the same fractions can be written in different ways.

Deleting Qubits by Measureing

In a qubit system, if we measure a qubit x|00\rangle + y|01\rangle + z|10\rangle + w|11\rangle, we simply delete the portion that contradict to our result and normalize the new state: if we measure the first bit A...

With Pr\{A = 0\} = x^2 + y^2, we collapse into x|00\rangle + y|01\rangle (un-normalized)

With Pr\{A = 1\} = z^2 + w^2, we collapse into z|10\rangle + w|11\rangle (un-normalized)

Measuring n qubit together in any order yields the same result.

Measureing a qubit will disentangle the qubit from others. If a qubit is disentangled, we can treat the qubit as if it is not in the system.