Week 15 - my notes

• Take this state:

\[|q1q0\rangle = \frac{1}{\sqrt{2}}(|00\rangle + |01\rangle)\]

• The number inside the kets tell you the number of qubits
• The number of kets tell you the measurable outcomes for this state

• Remember the Quantum Circuit model? Composed by input states, gates (perform computation), and measurements (obtain the output).
• For a single qubit, measurements result in one of two possible answers. If we measure it using the Z basis, we can only get \(|0\rangle\) or \(|1\rangle\). If we measure it using the X basis, we can only get \(|+\rangle\) or \(|-\rangle\)
• After the measurement, we know the wavefunction colapse happens. Our superposition gets narrowed down to one single state.
• The probability of getting one state is defined by its contribution squared.
• We made normalizations to be able to get correct probabilities. All because of the Born rule. The squares of contribuitions need to add up to 1.
• We can still use the born rule for multi-qubit circuits

• Entanglement is a very defined relationship, where if you know the state of one object, you know the other’s state.
• \[|q1q0\rangle = \frac{1}{\sqrt{2}}(|00\rangle + |11\rangle)\]

• Just because a circit has a CX gate, it doesn’t mean the qubits are automatically entangled.
• To create an entangled circuit, you should always have a superposition in the control qubit.
• Also, the H gate should always come before the CX gate

1. \[|\beta00\rangle = \frac{1}{\sqrt{2}}(|00\rangle + |11\rangle)\]
2. \[|\beta01\rangle = \frac{1}{\sqrt{2}}(|00\rangle - |11\rangle)\]
3. \[|\beta10\rangle = \frac{1}{\sqrt{2}}(|01\rangle + |10\rangle)\]
4. \[|\beta11\rangle = \frac{1}{\sqrt{2}}(|01\rangle - |10\rangle)\]
5. These states are special!
6. They differ in two ways: Parity, Phase