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I closed a four-month trade with a +124.5% gain, banking £24k tax-free. Fake crisis, real opportunity On April 2, 2025, Trump announced the most dramatic tariff package in years: a 10% baseline tariff on nearly all imports effective April 5, along with much higher “reciprocal tariffs” on certain countries starting April 9. Within a week, after the market crashed, he declared a 90-day pause on most of the reciprocal tariffs. That pause, which began on April 9, was later extended beyond July 8. ...

Change in notation With the tolerant identity test for a single state $\rho_{AB}$ complete, we now turn to the problem of certifying multiple ($n$) EPR pairs at once. For brevity we write $\Phi = \ket{\text{EPR}}\bra{\text{EPR}}_{AB}$ throughout. The problem statement is as follows: Problem. Given two trace distance tolerances $0 \leq \varepsilon_1 < \varepsilon_2 \leq 1$, a failure probability $\alpha \in (0, 1)$, and $N$ i.i.d. copies of an unknown $2n$-qubit global state $\varrho$ on $A^n B^n$ i.e. the source produces $\varrho^{\otimes N}$, how large must $N$ at least be so that, using only local $Z$/$X$ measurements and classical postprocessing, we can decide with at least confidence $1 - \alpha$ whether ...

Goal. Given $N$ i.i.d. copies of an unknown bipartite state $\rho_{AB}$ held by Alice and Bob, certify that $\rho_{AB}$ held by Alice and Bob is within trace‑distance $\varepsilon$ of $$ \ket{\text{EPR}}_{AB} = \frac{1}{\sqrt2}\left(\ket{00}_{AB} + \ket{11}_{AB}\right), $$ using only sequential* (one qubit at a time), local measurements in the standard ($\{ \ket{0}, \ket{1} \}$) or Hadamard ($\{ \ket{+}, \ket{-} \}$) bases, and classical communication/postprocessing. Importantly, we never perform any joint or Bell‐basis measurement on $AB$. ...

Hello and welcome to my URSS blog! Over the coming weeks, I’ll be sharing my journey into the fascinating field of quantum cryptography. There are two parts to this: “quantum”, the strange and powerful world of quantum physics; and “cryptography”, the science of securing information via encryption and decryption. Let’s start simple. Take this number: $1984909$. Looks random right? Here’s the challenge: can you figure out which two numbers, when multiplied together, give you this result? ...