Independence: Pr\{A = a \cap B = b\} = Pr\{A = a\} \cdot Pr\{B = b\}
Dirty Secret Theorem: if two things are made completely separately, they are independent. Corollary: only CNOT
, CCNOT
or any controlled-x instruction can make two qubits entangled
Tensor (\otimes): an operation to write two quantum states.
Also: given two matrix F : U \to V, G : W \to Z, the following holds:
Intuition:
You can think of |0\rangle, |1\rangle as deterministic state of one (or more) qubit
You can think of \otimes as concating two qubits (look at two qubits together)
You can think of + as "OR" (it is either "this" OR "that")
Note \otimes is not commutative but associative
For a two qubit state to be unentangled, there must exist r, s, x, y such that
For any qubit state a, b is unentangled, iff you can write the state into a \otimes b where a, b are pure state.
To check entanglement efficiently, for two qubits:
So if we multiply coefficients of |00\rangle and |11\rangle, and it is not equal to the multiply coefficients of |01\rangle and |10\rangle, then we know it has to be entangled. (the other direction not proved // TODO)
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.
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.
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