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2 votes
Accepted

Does every code have transversal Pauli group?

I don't think so - consider e.g. the 'diagonal' representation $$\phi:SU(2) \rightarrow GL((\mathbb{C}^2)^{\otimes 4})$$ $$U \mapsto U^{\otimes 4}. $$ The Clebsch-Gordan series tells us that, for spin ...
  • 502
2 votes

Does every code have transversal Pauli group?

Every code that can be implemented by a stabilizer circuit (this includes stabilizer codes, gauge codes, floquet codes, etc) has this type of subset-transversal Pauli gate. In such a code, the X, Y, ...
  • 28.7k
0 votes
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Find min of a quantum state L2 norm

For unit vectors $|\psi\rangle, |\phi\rangle \in \mathbb{C^n}$ we have: $$ ||e^{i\alpha}|\psi\rangle-|\phi\rangle ||^2 = \left(e^{-\alpha i}\langle\psi|-\langle \phi| \right) \left(e^{i \alpha} |\psi\...
  • 2,033
0 votes

Weight enumerators for Hermitian operator

This is just fleshing out some of the themes from Adam Zalcman's answer, which I have already accepted. The standard proof that $ B_j\geq A_j $ for a projection $ H $ uses the spectral theorem to ...
2 votes
Accepted

Weight enumerators for Hermitian operator

TL;DR: No. We can actually blow $A_j$ up to infinity while simultaneously sinking $B_j$ negative. Sneaky plan Coefficients $A_j$ cannot be negative, but if $H$ squares to identity and anticommutes ...
  • 18.2k
2 votes
Accepted

Weight enumerators for Hermitian operator (wrong $ B_j $ definition)

TL;DR: No. Suppose that $H=\alpha H'$ for some scale factor $\alpha\in\mathbb{R}$. The key observation is that $A_j$ is independent of $\alpha$, but $B_j$ is linear in $\alpha$. We can use this to ...
  • 18.2k
3 votes

Why can all quantum circuits be converted into circuits that use only real matrices?

To give a more general perspective on this: in linear algebra in general you can replace complex numbers with real numbers by enlarging the space and suitably redefining operations. A simple way is to ...
  • 21.2k
13 votes

Why can all quantum circuits be converted into circuits that use only real matrices?

The fact that "we need work only with quantum Turing machines (QTMs) with real-valued transitions" is proved by Bernstein and Vazirani in their paper Quantum complexity theory (1993). ...
4 votes

Why can all quantum circuits be converted into circuits that use only real matrices?

As Mark notes, the result is quite old, possibly dating back to the 90s. However, a quick search yields this paper, which briefly discusses this result. Then it cites this thesis for proofs. Technical ...
3 votes
Accepted

Where am I going wrong in my understanding of qubit associativity?

Looks like a simple error in the left-associative calculation: $$ \left( \begin{pmatrix} 0\\ 1\\ \end{pmatrix} \otimes \begin{pmatrix} 1\\ 0\\ \end{pmatrix} \right) \otimes \begin{pmatrix} 0\\ 1\\ \...
4 votes
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Building universal gate set for $SU(d^n)$ from universal gate set for $SU(d)$

Let $ G $ be a finite set of gates, of size $ |G| $. Let $ <G> $ be the words of $ G $. That is, $ <G> $ is the group generated by $ G $. Suppose that $ <G> $ is dense in $ SU(d) $. ...
3 votes
Accepted

What is the difference between Gate.power() and Gate.repeat()?

The method Gate.repeat() accepts positive integers only. While Gate.power() accepts real values. So, you can use it, for eaxmple,...
2 votes
Accepted

Bounding operator norm by total variation distance

No you cannot, here's a counterexample. Let $U=I$ be the identity matrix and let $S = \sum_{i} (-1)^{\delta_{0,i}} |i\rangle \langle i|$ where $\delta_{i,j}$ is the Kronecker delta. That is, $S$ is ...
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1 vote

Modular Addition general explanation

To add to hft's great answer, addition modulo 2 is often used as a synonym for the bit-wise XOR function, as you can see from a truth table: ...
  • 823
1 vote
Accepted

Modular Addition general explanation

What exactly does a modulo 2 addition stand for? It means that numbers are considered congruent (often stated as "equal" or "equal... mod 2") when they have the same remainder ...
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