A Quantum linear code/ Dual Code (CSS) proof

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The discussion focuses on the interpretation of symbols in quantum linear codes and dual codes, specifically in the context of CSS codes. It emphasizes the importance of providing additional information or references to clarify the notation used. The mathematical derivations presented illustrate the relationships between elements in the code and their properties, particularly when considering vectors in the dual space. The discussion reiterates that if a vector is in the orthogonal complement, certain conclusions can be drawn easily, while alternative cases require specific conditions to be met. Overall, the conversation highlights the need for clarity in mathematical communication within the topic of quantum coding theory.
steve1763
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What would the proof be for the following identity? I cannot find the proof anywhere
Screenshot 2021-09-05 at 21.52.33.png
 
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How should we know, what your symbols mean? You have to give a bit more information or a link to the paper you are referring too.
 
If ##x\in C^\perp## then it is clear. If not, then there is a ##c_0\in C## such that ##x\cdot c_0 =1##. The you have

##
-1\sum_{c\in C}(-1)^{x\cdot c}=(-1)^{x\cdot c_0}\sum_{c\in C}(-1)^{x\cdot c}=\sum_{c\in C}(-1)^{x\cdot (c-c_0)}=\sum_{c\in C}(-1)^{x\cdot c}
##

The last equality is because ##C## is a subspace.
 
Last edited:
martinbn said:
If ##x\in C^\perp## then it is clear. If not, then there is a ##c_0\in C## such that ##x\cdot c_0 =1##. The you have

##
-1\sum_{c\in C}(-1)^{x\cdot c}=(-1)^{x\cdot c}\sum_{c\in C}(-1)^{x\cdot c}=\sum_{c\in C}(-1)^{x\cdot (c-c_0)}=\sum_{c\in C}(-1)^{x\cdot c}
##

The last equality is because ##C## is a subspace.
Thank you very much
 
For the quantum state ##|l,m\rangle= |2,0\rangle## the z-component of angular momentum is zero and ##|L^2|=6 \hbar^2##. According to uncertainty it is impossible to determine the values of ##L_x, L_y, L_z## simultaneously. However, we know that ##L_x## and ## L_y##, like ##L_z##, get the values ##(-2,-1,0,1,2) \hbar##. In other words, for the state ##|2,0\rangle## we have ##\vec{L}=(L_x, L_y,0)## with ##L_x## and ## L_y## one of the values ##(-2,-1,0,1,2) \hbar##. But none of these...

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