1
$\begingroup$

A bitvector of length N has 2^N different values. How many different values has a quantum bit vector of length N? Is it possible to test two quantum bit vectors for equality?

$\endgroup$
2
$\begingroup$

The state of an $n$-qubit system is defined by $2^n$ complex numbers. Basically the numbers are weightings for each of the classical $n$-bit states, similar to a probability... but square-rooted.

Technically, two quantum states are equal only if all of those weights are equal. But, when it comes to the practical aspects of quantum computing, it's better to focus on continuous concepts like distance rather than discrete concepts like equality because we only get very indirect access to the weights.

For example, quantum mechanics places pretty strict limits on how well you can compare two states. If the vectors defined by the amplitudes of state $a$ and state $b$ are an angle $\theta$ apart, then the chance of any QM process correctly distinguishing between them is upper-bounded by $\frac{1}{2} + \frac{1}{2}|\sin \theta|$ or equivalently $\frac{1}{2} + \frac{1}{2}\sqrt{1 - |\langle a | b\rangle|^2}$ (look up 'Trace Distance'). Distinguishing is strictly easier than equating, and we're already failing probabilistically at distinguishing! That's why people focus more on the distances.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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