# What are the conditions under which an unknown quantum state is learnable with arbitrary precision?

Assume that we have an unknown quantum state and we need to learn that unknown state with arbitrary precision.

Under what conditions can we learn the unknown state with arbitrary precision?

One condition that I understand is if I can get an arbitrary number of copies of the unknown states, I can estimate the probabilities and consequently conclude about the amplitudes.

Another condition is the ability to measure in the standard basis. But do we need the ability to measure in all bases other than the standard basis? Put another way, if I restrict myself to measurements in the standard basis only, would I fail to learn the unknown quantum state with arbitrary precision?

I'll assume the setting is that you receive a number of copies of an unknown state $$\rho$$ that you have zero prior knowledge about. Assume what you are able to do is measure this state with some POVM $$\mu\equiv \{\mu(a):a\in\Sigma\}$$, where $$\Sigma$$ is the set of possible outcomes of the measurement. Note that this models the most general kind of measurements you can perform on the state.
1. A number of copies of the state $$\rho$$ increasing with the precision with which you want to estimate it. This is a simple consequence of the statistical nature of measurement outcomes. At each measurement, you will get some outcome $$a\in\Sigma$$ with probability $$\langle \mu(a),\rho\rangle\equiv\operatorname{Tr}(\rho\mu(a))$$. With a finite number of samples, you can never be completely sure what the probability distribution from which you are sampling is (of course, this is a general fact about probabilities, and not specific to quantum mechanics in any way). The precise scaling of the number of copies required with the precision constraints is the subject of quantum estimation/metrology. Some more details about this process are given in Lower bounds on the number of measurements for quantum state tomography.
2. More on point to your question, you need the measurement $$\mu$$ to be informationally complete. Formally, this means that you need the operators $$\mu(a)$$ to be a basis for the underlying operator space. This condition is necessary and sufficient: a given $$d$$-dimensional state $$\rho$$ is fully characterised by the outcome probabilities $$p(a)=\langle\mu(a),\rho\rangle$$ if and only if the set of positive operators $$\mu(a)$$ has rank greater than or equal to $$d^2$$. I'll remind to section 2.3.2 of Watrous' TQI book for the more formal versions of these statements.
For example, if $$\rho$$ is a qubit, then $$d=2$$, and you need a measurement described by four linearly independent operators to fully characterise the state. I should note that this does not contradict the state being characterised by three real numbers in the standard Bloch sphere picture: you need four measurement operators, but these still need to satisfy $$\sum_a \mu(a)=I$$, and thus the probability normalisation constraint means that these still result in three independent real numbers. For explicit examples of such POVMs, and more details about the general formalism, see the answers to this related question.
In particular, a projective measurement is never informationally complete. A projective measurement is bound to contain no more than $$d$$ operators, and thus, at best, characterises the measurement results in a specific measurement basis (more concretely, it only tells you about the diagonal, in some basis, of the given density matrix).