0
$\begingroup$

I'm looking for the mathematical formulation of this trait (from an introductory Quantum Computing course):

enter image description here

How can I generalize this into higher dimensions? Or other observables?

It does not seem provable from the Born rule, which only gives information about the probability amplitudes of $B$.

$\endgroup$
2
  • $\begingroup$ What is $A\downarrow 0$ supposed to denote? And, although one can guess it, what are $\lvert 0\rangle$ and $\lvert 1 \rangle$? Is there an implicit tensor product in the notation $\lvert 0 \rangle\lvert 1 \rangle$? $\endgroup$
    – ACuriousMind
    Oct 28, 2015 at 21:35
  • $\begingroup$ 1) It denotes "A collapses to the spin-down state". 2) They are the computational basis vectors corresponding to the spin-down and spin-up states. 3) Yes. $\endgroup$
    – tba
    Oct 28, 2015 at 22:06

1 Answer 1

1
$\begingroup$

When performing a measurement based on a predicate $P$, such as the first qubit $P(\left|xy\right\rangle) = x$ or the second qubit $P(\left|xy\right\rangle) = x$ or some combination like $P(\left|xy\right\rangle) = x \text{ and } y$, you generally do the same thing:

  1. Let $S$ be the set of states that match P.
  2. Let $Z$ be the set of states that don't match P.
  3. Let $t$ be the sum of the squared magnitudes of states in $S$.
  4. Let $f = 1 - t$ be the sum of the squared magnitudes of states in $Z$.
  5. The predicate measurement tells you if the quantum state is in $S$ or $Z$. It returns "S" with probability $t$, and "Z" with probability $f$.
  6. If the measurement returned "S", then any states in "Z" now have amplitude 0 and any states in "S" had their amplitude scaled by $\frac{1}{t}$.
  7. If the measurement returned "Z", then any states in "S" now have amplitude 0 and any states in "Z" had their amplitude scaled by $\frac{1}{f}$.

In other words: the total squared magnitude of states matching each measurement result tell you how likely that measurement result is. After you know the result, you renormalize the surviving amplitudes so that the total squared magnitude is 100% again.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.