# How does equality works for qubit-vectors?

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?

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.
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.