Hi everybody,
I found some issues in the codebook, but I cannot open a new issue in the GitHub repo because the repository has been archived.
I also forked the codebook repository, but unfortunately, I’m unable to pull my revisions due to the same problem.
How can I report those issues?
Thank you all.
Hi @mbzadegan !
Here in the PennyLane forum is a good place to report issues! What issue have you encountered.
Best wishes,
David
Hi @dren !
In proof (4) of Xanadu-Quantum-Codebook/nodes/I.1/Lesson.md at main · XanaduAI/Xanadu-Quantum-Codebook · GitHub , we proved that <0| and |1> are orthogonal. But you can see in the attached photo that the main texts mentioned |0> and |1> are orthogonal, and, as far as we know, <0| is not equal to |0>. So, the first issue in your text is that you should replace |0> with <0| in all locations on this page if you want to refer to this (4) mathematics proof.
Please let me know if my reason is not clear to you.
Thank you all.
Thanks @mbzadegan !
You raise an excellent and subtle point! Indeed, \langle 0| \neq |0\rangle. However, finding that \langle 0| 1\rangle = 0 does not mean that |1\rangle is orthogonal to \langle 0 |. Rather, it is correct to state: “Because \langle 0| 1\rangle = 0, |0\rangle is orthogonal to |1\rangle and vice versa.”
We must juggle two notations simultaneously: the bra-ket notation and standard linear algebra vector notation.
For standard linear algebra, a column vector of N dimensions can be written as \vec{c} = (c_1, c_2, \dots c_N)^T. The row vector \vec{r} = (r_1, r_2, \dots, r_N). We can define the inner product of two complex vectors \vec{v} and \vec{w} to be <\vec{v},\vec{w}> = \sum_{k=1}^{N} v_k^* w_k , where the superscript ^{*} represents the complex conjugate and k is a subscript. That is the same as performing \vec{v}^{T^*}\vec{w}.
Vectors \vec{v} and \vec{w} are orthogonal to each other if and only if <\vec{v}, \vec{w}>=0. To see this, recall the dot product picture of two vectors are 90^\circ to other other. The dot product of those two vectors is 0.
For the bra-ket notation, |0\rangle = (1,0)^T = \vec{v} and |1\rangle=(0,1)^T=\vec{w} are each column vectors. To prove that they are orthogonal, we need to perform the bra-ket version of \vec{v}^{T^*}\vec{w}, which is \langle v | w\rangle. Equation 4 exactly checks this case and finds \langle 1 | 0 \rangle = 0, so |0\rangle and |1\rangle are orthogonal vectors (i.e., no part of either vector overlaps with the other).
I hope that helps! Your question was excellent.
Best wishes,
David
Thank you for your comprehensive explanation,
Although I still believe that mathematical proof (4) is irrelevant to the main text because mathematical proof (4) uses “bra zero”, which is not the same as “ket zero” that used in main texts.
I hope I have expressed my meaning correctly.
Sincerely.
Hi @mbzadegan !
Great point! If I understand your meaning correctly, you think that Equation 4 does not prove that |0\rangle and |1\rangle are orthogonal. Rather, Equation 4 proves that \langle 0| and |1\rangle are orthogonal. Did I understand correctly?
If so, keep in mind that you can prove two vectors are orthogonal by taking the dot product \vec{v} \cdot \vec{w} and seeing if the result is 0. For example \begin{pmatrix} 1 \\ 0 \end{pmatrix} \cdot \begin{pmatrix} 0 \\ 1 \end{pmatrix} = 1\times 0 + 0\times 1 = 0.
That dot product is exactly the same as if I wrote \vec{v}^T \vec{w} = \begin{pmatrix} 1& 0 \end{pmatrix} \begin{pmatrix} 0 \\ 1 \end{pmatrix} = 1\times 0 + 0\times 1=0. This is the inner product form, where you use the rules of matrix multiplication.
Translating the vectors into Dirac bra-ket notation, the inner product becomes \langle 0| |1 \rangle, which brings you to Equation 4.
So, if the question is: Are |0\rangle and |1\rangle orthogonal? We can jump directly to checking their inner product in bra-ket notation in Equation 4 i.e., \langle 0 | 1\rangle = 0 \rightarrow Yes!
I hope that helps!
Yes, equation (4) only proved that Bra-0 and Ket-1 are orthogonal. I knew that |0\rangle and |1\rangle are orthogonal, but not through Equation (4), and we need to clarify this for readers with your last details.
Thank you for your consideration.
Thanks for your feedback @mbzadegan.
