Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

A well know result is that Clifford group preserve the Pauli group under conjugation or, in other words:

  • $C(P_{1} \otimes P_{2})C^{\dagger} = P_{3} \otimes P_{4}$, with $C \in$ Clifford group and $P_{n} \in$ Pauli group.

How we can prove this?

Thank's...

share|improve this question
    
Can you say some more about what these groups are? –  Ryan Thorngren Sep 12 '12 at 4:22
    
This actually seems like more of a pure math question to me, and that suggests it may get a better response on Mathematics... I can migrate it if people think that would be appropriate. –  David Z Sep 12 '12 at 5:00
    
Also, it seems like a relatively trivial problem of multiplying finitely many matrices, but please define these to get a proper answer. You should know that "preserving a group" usually means preserving the algebraic relations, and then all conjugations do that. In this case, I assume you mean multiplying and product of $\sigma_x,\sigma_y,\sigma_z$ by some finite set of matrices and their inverse keeps you in the set of these matrices and their products –  Ron Maimon Sep 12 '12 at 6:33
    
More about this relation in home.lu.lv/~sd20008/papers/essays/…. –  user901366 Sep 12 '12 at 12:54
add comment

1 Answer

up vote 6 down vote accepted

Usually the Clifford group is defined to be the group of unitaries that preserve the Pauli group under conjugation, so no proof is needed.

If instead you are asking, how can we prove that a certain unitary (such as the controlled-NOT) is in the Clifford group, the usual straightforward way to do this is just to calculate. Conjugation is a group homomorphism, so it is sufficient to check a generating set of the Pauli group. For instance, single-qubit X and Z operators are enough, so in the 2-qubit case, you should check the action of conjugation for X_1, X_2, Z_1, and Z_2.

See quant-ph/9807006 for more about the Clifford group.

share|improve this answer
    
Hi, thanks by your answer. My question is really about the proof this definition, my intention is know the technique applied in this proof to study the construction of groups that show same structure (preservation about conjugation). –  user901366 Sep 25 '12 at 17:57
1  
I don't understand what you are asking. How does one prove a definition? There are some standard generalizations to qudit Clifford groups, which share many of the properties of the qubit version. The general group theory construction of preserving a subgroup under conjugation is called the "normalizer". –  Daniel Gottesman Sep 27 '12 at 14:49
    
Ok, now I understand, my question is really about find group normalizer's. –  user901366 Sep 28 '12 at 13:46
    
Gottesman, I make other question about a more general relation in Matrix separability preservation under conjugation?. Do you know any work in the group theory about this? –  user901366 Sep 28 '12 at 13:52
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.