Search Results

When I was doing some work on quantum cloning (so, slightly different applications to the channels you were asking about), I basically ended up setting up the Choi matrix as a description of the actio …

You're missing a bit of algebraic trickery. Remember that $\frac{1}{\sqrt{2}}=\sin(\pi/4)=\cos(\pi/4)$. Thus, $$ \cos(\pi/8)/\sqrt{2}+\sin(\pi/8)/\sqrt{2}=\cos(\pi/8)\cos(\pi/4)+\sin(\pi/8)\sin(\pi/4) …

"Why" is quite a nebulous concept that, ultimately, comes back to your understanding of quantum mechanics. For me, the way that I set up QM is with a set of postulates. The first postulate is that qua …

I guess that what you're after is $ \mathcal{E}^{\otimes 2} $ is defined by the 4 operator elements $$ E_1\otimes E_1,E_1\otimes E_2,E_2\otimes E_1,E_2\otimes E_2. $$ If you apply this to $\rho$, you …

I think it's just notational inconsistency. If you look at the page code, the symbols are generated in two different ways: in the text, someone has just inserted the greek letter symbol (presumably a …

Yes. Generally you should think of filtering as being a measurement. You're only describing the effect of one measurement outcome. There's generally a second one such that the net effect is trace pres …

An alternative way to justify this if to notice that fr any single qubit unitary $U$, acting on the Bell state $|B\rangle$, $$ U\otimes U^\star|B\rangle=|B\rangle. $$ So, I can rewrite $$ U\otimes I|B …

Proofs of the no-cloning theorem vary in quality. It's often something you meet early on in your studies of quantum information, so people want to keep it simple, but those simplifications can miss ou …

Look at the matrix $H_2$: $$ \begin{bmatrix} \alpha+\beta & 0 & 0 & 1 \\ 0 & \alpha-\beta & 1 & 0 \\ 0 & 1 & \beta-\alpha & 0 \\ 1 & 0 & 0 & -(\alpha+\beta) \end{bmatrix}. $$ You can easily divide thi …

If $W=UV^\dagger$ then $U=WV$. This means that you can rewrite $$ U\rho U^\dagger=W(V\rho V^\dagger)W^\dagger. $$ So, let me write $\tilde\rho=V\rho V^\dagger$. Your integral becomes $$ V\mathcal{E}_T …

Imagine you want to make a measurement on a state $|\psi\rangle$ (and we will make the solution work for all possible $|\psi\rangle$). We introduce an ancilla system (Hilbert space dimension at least …

The existing answers have quite an elegant idea behind them. However, my concern is that they don't seem to allow for the introduction of an ancilla. If we introduced an ancilla in a fixed state, and …

The two descriptions are entirely equivalent. It doesn't matter which you use when, it's just a case of using whichever description you personally find to be mathematically the most convenient.

This is a neat idea. However, having individual overlaps being small isn't sufficient in a quantum system. For example, imagine the set of overlaps $$ \langle V_1|V_N\rangle=0,\qquad \langle V_1|V_n\r …

Unitary operations are a subset of CPTP operations. You can think of a CPTP operation as the description of a unitary over a larger system. The advantage of using CPTP maps is that you increase the g …