Lecture 013 - Adding and Deleting Qubits

Adding and Deleting Qubits

Unentangled Qubits

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.

$\begin{align*} &(r|0\rangle + s|1\rangle) \otimes (x|0\rangle + y|1\rangle)\\ =&r|0\rangle \otimes (x|0\rangle + y|1\rangle) + s|1\rangle \otimes (x|0\rangle + y|1\rangle)\\ =&|0\rangle \otimes (rx|0\rangle + ry|1\rangle) + |1\rangle \otimes (sx|0\rangle + sy|1\rangle)\\ =& rx|00\rangle + ry|01\rangle + sx|10\rangle + sy|11\rangle\\ \end{align*}$

Also: given two matrix $F : U \to V, G : W \to Z$ , the following holds:

$(F \otimes G)(|u\rangle \otimes |w\rangle) = F|u\rangle \otimes G|w\rangle$

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

Checking Entanglement

For a two qubit state to be unentangled, there must exist $r, s, x, y$ such that

$\begin{bmatrix} r\\ s\\ \end{bmatrix} \otimes \begin{bmatrix} x\\ y\\ \end{bmatrix} = \begin{bmatrix} rx\\ ry\\ sx\\ sy\\ \end{bmatrix}$

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:

$\begin{align*} |v\rangle =& |a\rangle \otimes |b\rangle\\ =& (\alpha |0\rangle + \beta |1\rangle) \otimes (\gamma |0\rangle + \lambda |1\rangle)\\ =& \alpha \gamma |00\rangle + \alpha \lambda |01\rangle + \beta \gamma |10\rangle + \beta \lambda |11\rangle\\ \end{align*}$

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)

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.

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