2 votes

Investigating the scaling of the error of a Trotter-Suzuki-approximation

You can use the Baker–Campbell–Hausdorff formula that states that for $e^Ae^B = e^C$ (assuming $e^A,e^B \approx I$) $C$ is given by: $$C = A + B + \frac12[A, B] + \frac1{12}[A, [A, B]] + \frac1{12}[B, ...
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2 votes

Homeomorphism or stereographic projection corresponding to the set of mixed states within the Bloch sphere

I'm late to the party, but here's my take: Pure qubit states As you said, the space of pure states of a single qubit can be described as a complex projective line $\mathbb{C}P^1$, which is ...
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1 vote

Do unitary matrices acting on entangled states always give a quantum state?

Yes, you can absolutely simplify down. The problem you're having isn't one of simplification, rather one of counting all the terms in the normalisation, although you appear to have picked up a stray $\...
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1 vote

Investigating the scaling of the error of a Trotter-Suzuki-approximation

For this particular calculation, you can keep your results exact for quite a long time. To see this, start with the exact thing $$ H_0=e^{i\pi/2(X+Z)/\sqrt{2}}=i\frac{X+Z}{\sqrt{2}}. $$ Now for the ...
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