Book a Demo!
CoCalc Logo Icon
StoreFeaturesDocsShareSupportNewsAboutPoliciesSign UpSign In
quantum-kittens
GitHub Repository: quantum-kittens/platypus
 Path: blob/main/notebooks/summer-school/2021/lec1.2.ipynb
3855 views
Kernel: Python 3

Introduction to Quantum Circuits

In this lecture Elisa talks about quantum circuits, ‘circuit model’, introducing and explaining what are ‘Single Qubit gates’, and ‘Two-Qubit gates’ through the example of ‘Classical’ and ‘Quantum’ gates. This lecture also discusses some basic gates such as the Pauli and Hadamard gates. The lecture continues with the explanation of the multiple states operations (‘Multipartite quantum state’) and explains the difference between correlated (entangled) vs uncorrelated states. Elisa discussed how we can represent some states using tensor products, but entangled states cannot be represented in this way.

FAQ

Is a CNOT gate sort of like converting quantum data to classical data? No, the operation that would be most analogous to converting to classical data would be measurement. Measurement can convert a quantum state to either a 0 or a 1, with the probabilities of each measurement dependent on the quantum state before measurement. The CNOT operation can operate on quantum states (i.e. with a superposition state). So, a qubit in a superposition can be used as the control qubit and another as the target qubit, and finally the qubit after the operation can also be in a superposition (depending on what states the C and T-qubits were in)
Is it clockwise rotation or anticlockwise rotation in bit flip gate? (e.g. when we apply an X gate to evolve state ∣0⟩|0\rangle into state ∣1⟩|1\rangle) There are many different ways you could do this. It not only can be clock-wise or counter-clockwise, you can also do it about the X-axis, the Y-axis, or any arbitrary axis in the X/Y plane for that matter. In the case of IBM-Q's hardware specifically, I believe it is a counter-clockwise rotation about the Y axis.
What kind of correlation makes entanglement? Is it a chemical link, strong link or something else? If we don’t use quantum gate, how else can we create quantum entanglement? Nature can generate entangled particles, through the decay of a particle. For example, in the decay of a particle into two photons, the two photons can be entangled.
What is the difference between correlated state and entangled state? For a pure state, is correlated equal to entangled? While for mix state, why can a state be correlated but not entangled? For a pure 2-qubit state: if it can be written as ∣v⟩∣w⟩|v\rangle|w\rangle (that is, the tensor product of 1-qubit states ∣v⟩|v\rangle and ∣w⟩|w\rangle) then it is uncorrelated; if it cannot be written this way, then it is correlated and entangled. For a mixed state, the situation is more complicated; there is a long discussion about this in https://en.wikipedia.org/wiki/Quantum_entanglement.
Why is the CNOT gate matrix produced from these combinations specifically? A CNOT is a particularly simple and useful 2-qubit gate; it turns out that any n-qubit gate can be built of a few types of 1-qubit gates and CNOT gates.
Are the Bell state and entangled state the same? Because Alice and Bob shared same initial entangled state ∣psi⟩00|psi\rangle 00. And this is also one of the same state from the 4 bell state. Because Alice and Bob shared same initial entangled state ∣psi⟩00|psi\rangle 00. And this is also one of the same state from the 4 bell state Alice needs to do a CNOT as part of the algorithm. CNOT is a 2-qubit gate. Alice has her own state, but needs another. What can we use for the other qubit? We give Alice (and Bob) some entangled state to use. For the example today, we give Alice (and Bob) the ∣00⟩|00\rangle Bell state because it is one example of an entangled state. There are other options (any entangled state will do). But, the Bell states are great choices because the circuit used to make them is relatively simple.
What are unitary operators? Unitary operators preserve the length and the angle between vectors. For example, you can think of rotation in R2. Rotations do not change the angle or the length between two vectors.
What are the examples of tensor states? Every state generated by only single qubit gates can be an example. For instance, the state 12(∣00⟩+∣01⟩+∣10⟩+∣11⟩)=∣++⟩=∣+⟩⊗∣+⟩\tfrac{1}{2}(|00\rangle + |01\rangle + |10\rangle + |11\rangle) = |++\rangle = |+\rangle \otimes|+\rangle. It's like a factorization.
we have complete information of a pure state but not of a mixed state". For example, if we consider a superposition state ∣+⟩|+\rangle, it is a pure state, but we do not have complete information right? We are still taking in terms of probability right?. In fact, the superposition state ∣+⟩|+\rangle represents its complete information. For instance, we can measure it by applying Hadamard gate and measurement, to find it gives 100% probability to give ∣0⟩|0\rangle state. You can compare it to the single qubit in Bell state: 12(∣00⟩+∣11⟩)\tfrac{1}{\sqrt{2}}(|00\rangle + |11\rangle). If you measure the first qubit, you get 50% of ∣0⟩|0\rangle and ∣1⟩|1\rangle state, regardless of any single qubit gate you applied before measurement.
How do we simulate qubits and quantum algorithms on a classical computer systems?, What are the limitations of this vs working on actual quantum computer? Actually the simulation of the quantum system is a major issue itself, but you can simply understand that it is just doing quantum mechanics numerically. Since simulating the quantum system requires resources exponentially growing with the volume of the problem, it has no benefit on running the system by a quantum algorithm. If we use the actual quantum computer, each operation may need more time, but the total volume of the algorithm reduces significantly.
Are the mixed states just entangled states? More specifically, part of the entangled state; the information of the other parts is not in touch or just ignored

Other resources