You can add location information to your Tweets, such as your city or precise location, from the web and via third-party applications. You always have the option to delete your Tweet location history. Learn more
No claim that this is new! But it was new & pleasing to me, a slightly new way of seeing an old friend.
Update: In fact, it is possible to analyze the two-qubit circuit with almost no algebra. Ignoring the final Z^z gate, the main thing to analyze is this circuit:pic.twitter.com/WcJOsjJCrN
We run the state through the CNOT, and then Hadamard + measurement is equivalent to measuring in the <+| or <-| basis:pic.twitter.com/oWK4nI63El
The resulting conditional states are as shown, and we can simply cancel the Z^z term:pic.twitter.com/Pvuyw9FSVQ
I really like the first part of the proof: it's just a quantum one-time pad. I don't quite know how to think about the second part of the proof. Why does measuring in the |+>, |-> basis work? It's easy to prove it does, but I don't have a really good explanation for why.
You can reach this by pulling the Z^z back through the CNOT to get Z^z on qubits 1 and 2, and since Z^z on the qubit 2 now acts on |0>, it can be removed. The Z^z on qubit 1 can be pulled back through the CNOT and Hadamard, then the measurement output is always |0>.
Twitter may be over capacity or experiencing a momentary hiccup. Try again or visit Twitter Status for more information.
