These are the best lectures ever made by Qiskit/IBM by far. Please assign more people to the team to get the whole series completed as soon as possible. Thanks for your effort.
At 46:35 you mentioned using density matrices in the "third unit of this series". This confuses me a bit, as this is lesson 3, and I don't see any mention of density matrices in any of the other courses either. Was this cut content and the lessons were renumbered? Do you intent to release lessons on the general description using density matrices in the near future? I've been really enjoying these videos and getting a lot out of them. Thank you for your hard work in producing them.
@36:00 - I don't know why, but I didn't make the mental connection between projections and subspaces until they were mentioned in the same breath here. I guess it finally took hearing the topic explained from multiple angles... (no pun intended)
An important distinction is that matrix multiplication is not the same as the "tensor product". Specifically it helped me to see the definitions in Lecture 1.2 (pp71-72) for: (1) controlled-NOT operation where 1st qubit is the control, and (2) controlled-NOT operation where the 2nd qubit is the control. Once I considered this, the matrix representation at 12:23 came out in the wash.
What does it mean when you say that the two mesurements are the same. Suppose i measure a 1 on the measured qubit. What does it mean for the first two?
Hello, I have a question: regarding what was said around minute 54, what does it mean to initialize a quantum gate? and how do it? Thanks for the excellent lectures Prof. Watrous!
Hello. First, thank you very much for these precious lectures. Second, I am curious about the example of representing the projective measurement by quantum circuit at 42:15 . I tried to check it my self and for the projective measurement \pi_1 acting on |00>, to inner products of two qubits and finally get: 1/2{|01> - |01> - |10> +|10> } . and I stopped . how these inner products are evaluated. And I reached by tracking the circuit and the measurement of the lower qubit gives ket zero with probability 1 entering |00?> for the upper qubits. please I would be grateful if you help me on this . Thank you very much
It's difficult to communicate mathematically here in the comment section of RU-vid, but I'll try to at least offer you some clues. This circuit is implementing what's known as the swap test, and if you look it up you should be able to find out more about it. Try setting the two input qubits to each of the Bell states to see what happens. The key is that the first three Bell states are symmetric - meaning that swapping the qubits does nothing - while the fourth Bell state is anti-symmetric - meaning that swapping the qubits injects a -1 phase. Think about how this affects the bottom qubit.
It will give same probability for both states |01> and |10> , resulted minus will do nothing (as it will be squared) to the measurement result. Am I right?@@John.Watrous
@@asmaa.ali6 If you start with phi^+, phi^-, psi^+ as inputs to the top two wires, the controlled-swap (C-SWAP) gate does not change any of these states. For example, |0 psi^+> goes to |+ psi^+> via H, then C-SWAP doesn't do anything so you still have |+ psi^+>. Then you apply H again turning it back to |0 psi^+>. Finally, the bottom wire outputs 0 with probability 1 and the top two wires give psi^+. Hence psi^+ is unchanged. Same story for phi^- and phi^+. As for |psi^->, we get something a bit different (you will not get an overall minus sign that squares to 1 as your intuition might suggest). First, applying H gives |+ psi^->. Applying C-SWAP now gives |- psi^->. Notice that the bottom qubit is now in the - state! You should check this step by expanding out |+ psi^-> in the 0,1 basis and see what the C-SWAP gate does to the latter two qubits. That minus sign "travels" to the first qubit by changing it from the + state to the - state. Then applying H gives |1 psi^->. So now you get the other outcome with probability 1.
Thanks - but I'm not a professor anymore! I have officially resigned my faculty position to work for IBM Quantum, so that I can devote myself to creating educational content like this. I believe education should be free, and that people who want to learn should not be turned away and denied that opportunity. I'm grateful for the opportunity to do this.
Very engaging! BTW, I have a question. Can you give me an example of experimental measurements corresponding to measure a state in basis of |0> and |1> versus a state of |+> and |->? Like, when we represent spin up and down to be |0> and |1>, what would correspond to |+> or |->?
Thanks for the content, you're doing a lucrative job, but I've got a question. This video uses a CNOT gate that is quite different from what I usually encounter, and I even tried to take the opposite of the matrix, but it didn't work, so if you can explain this to me I would appreciate it. Thanks
This is the usual CNOT gate. If it's the matrix representation that's causing concern, then I suspect that Qiskit's qubit ordering convention is to blame. It's difficult to explain further through comments, though, so this might be a good question to bring over to Quantum Computing Stack Exchange.
Happy to discover the "circuit" approach, even when it is strongly emphasized these circuits shouldn't be understood as electric circuits. From an engineering perspective (my background) this aids a lot. Even one dreams with an FPGA-like deployment where complex applications are built on top of mature IP and one does not to start from the scratch when creating end user applications. Just out of personal curiosity, time is assumed "from left to right" but I'm not able to see the mathematical resource representing the advancement of time to synchronized status changes. Maybe is it there, and I just didn't get it, or maybe time evolution is conceived different for quantum information processing.
Time goes from left to right in a quantum circuit diagram, so for the circuit shown in this part of the video the first gate is H, then S is second, then H again, and finally T is last. This means that if we run the circuit starting from the initialized |0> state for instance, it's first transformed into H |0> because we apply H first, then into SH |0> because we apply S second, then into HSH |0> when we apply H again, and then finally into THSH |0>. In short, this reversal happens because we're multiplying vectors by matrices from the left, so the matrices on the right get multiplied to the vector first while the ones on the left get multiplied last.
@29:00 - ah, that's a mild let-down; one of the things I was kind of hoping for was a decent explanation of Graham-Schmidt (without having to go through an entire linear algebra course); I started going through Linear Algebra Done Right (Axler), which gets to it later, but I've stalled... That's not to say I'm ungrateful. I'm very much enjoying this (free) course so-far.
I have to draw the line somewhere! At a very high level, we can always add a new vector by first picking a vector that's linearly independent of the existing ones. (There will always be at least one standard basis vector like this, so it's not hard to find one.) Then, project this vector onto the subspace orthogonal to all of the existing vectors and normalize. You can recurse as long as you have dimensions remaining. I'll leave it to Sal Khan to explain the Gram-Schmidt procedure itself: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-rHonltF77zI.html
Thanks for the great video. Probably I am asking a question that can be answered later, but here it is: say I know the general description of the quantum circuit i.e. how qubits can interact with each other, but I haven't decided on the basis yet i.e. instead of working on {|0>, |1>}, I would like to work with the basis {|00>, |01>, |10>, |11>} and do the measurements based on this basis using inner product. Is the circuit model general enough for such application?
One of those is a multi-qubit system, the other is a single qubit system. It doesn't make much sense to talk about a basis of multiple qubits when your system has only one. Change of basis is simply a rotation (i.e. Hadamard changes from Z to X basis and back) so you can change the basis in your circuit at will but you can not magically alter the dimensions of the system. There is also the fact that every gate you introduce to change basis will introduce noise into the system so you need to be careful. X and Z basis measurements are universal so often there is no reason to work with a basis other than these two
No, this is correct - but it's a common point of confusion that comes from the Qiskit ordering convention, which I discuss just prior to this moment in the video.
Hi, on 50:53 is there a typo on the last equation by linearity shouldn't the result be 1/√2∣0⟩ ⊗ ∣0⟩ + 1√2 ∣1⟩ ⊗ ∣0⟩ so the last ket equals to |0⟩ instead of |1⟩ , also thank you for the amazing lectures
I'm not sure I fully understand the question, please feel free to clarify. The measurements function in the usual way, as described in the previous lessons - so they do cause a collapse of the state. What the diagram depicts are qubits going into standard basis measurements with the classical measurement outcomes coming out. The state of the qubits has collapsed, and we're ignoring them in the diagram. Their states are determined by the measurement results.
Lesson 4 has been recorded and we're now in post-production - it will be released sometime this month (February). Subscribe and ring the bell if you wish, and you'll be notified when it's released.
Is the matrix representation of the C-NOT gate really correct (in the video at 12:30)? Given the (for me somewhat unfamiliar) |XY> ordering, shouldn't the state |00> transform into |00>, |10> into |10>, |01> into |11> and |11> into |01>? The corresponding matrix would then be different (and can be obtained by cyclically shifting the given matrix by 1 element up and to the left). I also wonder if the Qiskit textbook has the same error?
You have the transformation right, but this is consistent with the matrix. The rows and columns are ordered in the natural way: 00, 01, 10, 11. So, for example, because |10> maps to |10>, we have a 1 in the third row and third column, because 10 comes third in our list of strings. If we cyclically shift the entries of the matrix up and to the left, we obtain a transformation where |00> maps to |10>, |01> maps to |01>, |10> maps to |00>, and |11> maps to |11>, which is a different transformation. We're not talking about changing the ordering of strings, here - this is as I described in Lesson 2 - this is about how we order the qubits in a circuit diagram. (It is also connected with the way we number/index the qubits, using the so-called little endian convention, but this is not so important here because we've chosen our own names, X and Y, for the qubits.) This is a *very* common source of confusion.
thank you for the explanation, the ordering from top/left (|00>) to bottom/right (|11>) explains my confusion. I should probably have had again a look to the previous lectures, which I had seen a month ago. Thank you for setting up this lecture.
