Lecture 9 - Multiple-Qubit Systems

We looked at:

![[Physics CPE 345 Quantum Computing Lecture slides Week 4 Multi qubit systems 240424 2.pdf]]

Note:

If we have $n$ qubits, we can have $2^{n}$ total basis states
$| 1011 ⟩ = {| 1 ⟩}_{1} \otimes {| 0 ⟩}_{2} \otimes {| 1 ⟩}_{3} \otimes {| 1 ⟩}_{4}$ .

Example

Show that the two single qubits description is consistent with the one system description by finding the mathematical relationships between $a_{1, 2}$ and $b_{1, 2}$ and $a, b, c, d$ .

Proof
We know that:

| a_{1} |^{2} + | b_{1} |^{2} = | a_{2} |^{2} + | b_{2} |^{2} = 1, | a |^{2} + | b |^{2} + | c |^{2} + | d |^{2} = 1

by their probabilities. Notice:

(a_{1} | 0 ⟩ + b_{1} | 1 ⟩) (a_{2} | 0 ⟩ + b_{2} | 1 ⟩) = a_{1} a_{2} | 00 ⟩ + b_{1} b_{2} | 11 ⟩ + a_{1} b_{2} | 01 ⟩ + a_{2} b_{1} | 10 ⟩

Thus:

a = a_{1} a_{2}, b = a_{1} b_{2}, c = a_{2} b_{1}, d = b_{1} b_{2}

☐
Thus doing the conversions from:

![[Physics CPE 345 Quantum Computing Lecture slides Week 4 Multi qubit systems 240424 2.pdf#page=7]]

In general if we have $a | 00 ⟩ + \dots$ for each of these:

a = a_{1} a_{2}, b = a_{1} b_{2}, c = a_{2} b_{1}, d = b_{1} b_{2}

Thus for example:
(a) Using symmetry have all $a_{i}, b_{i} = \frac{1}{\sqrt{2}}$ .
Note (b) and (d) are impossible.