Can I find the states of individual qubits in a quantum register using only linear algebra?

Question

Say I have a quantum register consisting of two qubits like this $\left| -,0\right>$ which as a vector would be $\frac{1}{\sqrt{2}}(1, 0, -1, 0)$. If I only started with this vector, would it possible using only the vector formulation to find the two qubits that make it up ie. $\left|-\right> = \frac{1}{\sqrt{2}}(1, -1)$ and $\left|0\right> = (1, 0)$? How would that be done in a general case where we only have a vector $(v_{1},v_{2}, ....v_{n})$ and we know it consists of $log_{2}(n)$ qubits?

Answer 1

Here's a simple method that will work on any state that is not entangled with other qubits. It's also pretty efficient; it's the method used by the amplitude displays in Quirk.

1. Find the index $k$ of any one of the largest magnitude entries in the vector. In your case this would be index 0 or index 2. Technically any non-zero entry will do, but if you aren't being perfectly precise then picking the largest will reduce numerical error.

2. Split $k$ into the "included in my subsystem" part and the "excluded from my subsystem" part. For example the index $k=2$ splits into the index pair $(k_i, k_e) = (1, 0)$ whereas the index 0 would split into (0, 0).

3. Let $w_{r}$ be the vector formed by holding $k_i=r$ fixed while you vary $k_e$ over all possible exterior values and tale the corresponding amplitude in the state vector. For example, you hold the 1 in (1, 0) constant while iterating the other. So you need to look up (1, 0) which is $-1$ and (1, 1) which is $0$. So $w_{k_i}$ would be $(-1, 0)$.

4. Let $v = w_{k_i} / |w_{k_i}|$ be the renormalized vector. This is the state vector of the subsystem, up to global phase. Note that in our case it equals $-|0\rangle$, which is in fact one of the qubit states (up to global phase).

5. [Optional] Verify. If $v$ is actually the subsystem vector, then all $w_r$ should be parallel to $v$. You can check this by checking that $|\sum_r \langle v | w_r\rangle | = 1$. If it's not, the internal and external states are entangled.