Bell States from 2-Qubit Computational Basis States via Quantum Circuit (Hadamard and CNOT Gates)
HTML-код
- Опубликовано: 20 янв 2023
- Link to Quantum Playlist:
• Elucidating Quantum Ph...
#quantumcomputing
#quantumphysics
#quantum
Konstantin Lakic 
Наука
The convention used in this video treats qubit wires ordered from top-to-bottom as equivalent to left-to-right ordering in the tensor product, e.g. H tensor product I means H acts on qubit 1 (top) and I acts on qubit 2 (bottom). In other sources, you may also see the bottom-to-top ordering convention, which is the same as turning the quantum circuits in this video upside down.
I had a hard time to understand what a Bell measurement is, because I was confused by shorted explanations. This video makes it completly clear and also links the generation of Bell states and its measurement by linking the adjoint U operator with going backwards in quantum circuit. Thanks a lot.
Thankfully,
I found your channel😃
This was so helpful! Thank you
This is really helpful! Thank you!
Thank you for the video!
Great job!
The CNOT matrix you describe [1,0,0,0],[0,1,0,0],[0,0,0,1],[0,0,1,0]] has the high order bit controlling the low order bit, correct? |00> maps to |00>, |01> maps to |01>, |10> maps to |11>, and |11> maps to |10>. The circuit diagram has the LOB controlling the HOB, correct?
Whether qubit 1 or 2 appear first in a tensor product is a matter of convention. As long as all matrix representations are constructed using the same convention, the matrix multiplication will be correct. To see the convention used in this video, check out ruclips.net/video/0ECbWBBbglw/видео.html
@@Elucyda I think that is my objection. When you make the HxI tensor product the top line is the first operand in the tensor product, When you construct the state |01> = (0,1,0,0)T = |0>x|1> the bottom line is the first operand in the tensor product? Thanks for the link to the video, but that does not show your convention for constructing the state vectors, either. I don't mean to nit pick, but languages like Qiskit, QCEngine, etc. put the low order bit at the top.
@@Elucyda It may be a matter of convention, but just to be clear could you state which convention you are actually using? It your top line the high order bit or the low order bit? Being clear would help other people understand.
Something is wrong with the CNOT gate theory explanation in quantum states which is causing some errors
Could it be the bit order? If the top line is the low order bit and the bottom line is the high order bit then CNOT = [[1,0,0,0],[0,0,0,1],[0,0,1,0],[0,1,0,0]] rather than [[1,0,0,0],[0,1,0,0],[0,0,0,1],[0,0,1,0]] which has the HOB controlling the LOB.
Whether qubit 1 or 2 appear first in a tensor product is a matter of convention. As long as all matrix representations are constructed using the same convention, the matrix multiplication will be correct. To see the convention used in this video, check out ruclips.net/video/0ECbWBBbglw/видео.html
@@Elucyda It may be a matter of convention, but doesn't the difference induce a global phase on the |11> Bell state? With one convention, you get (|01> - |10>)/2^0.5 and with the other one you get (|10> - |01>)/2^0.5