Can one ever find the elements of a superposition state?

Question

Given some set of basis states $\{\vert 0\rangle, \vert 1\rangle, \vert 2\rangle...\vert N\rangle\}$ and an unknown superposition of the form $\frac{1}{\sqrt{2}}(\vert i \rangle + \vert j \rangle)$, what exactly forbids us from computing $i$ and $j$? I know about the no cloning theorem but why can I not see it from a single copy of the state?

Answer 1

Given only one copy of such a state, it is not possible to determine it with any good probability. Reason being there is no way, in principle, to extract information from the system without making a measurement on the system. And when we go for a measurement, we project it to a basis element $i,j$ if we turn out to pick these by chance from the set $\{1,2,...,N\}$, in otherwise cases, we can never make out what is $i$ or $j$ from the given set of degrees of freedom. To answer a more general question, if we don't even know what the set $\{1,2,...,N\}$ is; Suppose someone prepares a quantum state and gives to you without telling what kind of experimental apparatus may be the basis, there is no way in principle to figure out anything (if someone gives $n$ isolated levels of a harmonic oscillator and you are looking for two states of polarization in it, for instance). Hence, in all literature, we assume that the type of the experimental apparatus is always known, which means that the basis $\{1,2,...,N\}$ is known but a general state $|\psi\rangle=\frac{1}{\sqrt{N}}\sum_i c_i |i\rangle$ is unknown in this Basis, where we can evaluate the coefficients $c_i$s by repeated measurements. One such technique is the State Tomography. Otherwise, the most ordinary and default way is to use the projectors $|i\rangle \langle i|$ formed from the basis and see what we get as the final state after the measurement.

Answer 2

Denote $| \psi_{i,j} \rangle = \frac{1}{\sqrt{2}} ( | i \rangle + | j \rangle) $ for fixed $i$ and $j$. Alice gives a randomly drawn one of this form.

That is for each $0 \leq i<j \leq N$ she assigns a probability $p_{ij}$ and sends the corresponding state $| \psi_{ij} \rangle$ with that probability. When you say unknown of that form, I'll assume $p_{ij}=\frac{2}{N(N+1)}$ uniformly.

Then Bob receives a $\rho=\sum_{0 \leq i < j \leq N} p_{ij} | \psi_{ij} \rangle \langle \psi_{ij} |$. Bob can do any transformation followed by measurement he wants to get a variable $Y$.

How much information about $p_{ij}$ is contained therein? $\leq S(\rho)$. The second term is 0 in this case. I haven't calculated this entropy out though.

Holevo's Theorem

Answer 3

To determine the state, you’ll have to measure it. As you've said, the no-cloning theorem forbids us from making copies, so we just have to work with the single copy we're give. If we restrict to projective measurements, there are no more than $N+1$ outcomes to the measurement. However, you want to distinguish $\binom{N+1}{2}$ different outcomes. Hence, it’s not possible with just one copy of the state.

Of course, one must mention the exception of $N=1$ but that is a triviality because you already know what state it is! This still leaves the case of $N=2$ not covered by my argument, which one should look at explicitly.

An alternative approach is as follows: imagine you could determine from a single copy what the values of $i$ and $j$ are. In that case, you could use that knowledge to produce as many copies as you like. In other words, you can perform quantum cloning. As you state in the question, this is impossible (except when all the states are orthogonal to each other, i.e. $N=1$).

For a fully rigorous answer, one strategy is to ask about how well one can distinguish $(|0\rangle+|1\rangle)/\sqrt{2}$ from $(|0\rangle+|2\rangle)/\sqrt{2}$. If one is to complete the specified task, one must certainly be able to do this sub-task. However, the optimal measurement to distinguish these two states is known as the Helstrom measurement. See my previous answer here for further details. Since the two states are not orthogonal, you derive that it is impossible to distinguish them perfectly.