Notation for n-qubits.

Question

I have been asked to, given a n-qubit state and measuring in the computational basis on the first qubit, to calculate the probability of obtaining the outcome 1 and writing the state of the system after the measurement.

The problem is I do not understand the notation of it. Could someone help? This is the system:

$|ψ⟩ = \sum_{x∈\{0,1\}^n} a_x|x⟩ ∈ C^{2^n}$

I know how to do it for 1-2-3...qubits, but do not know how to unroll this formula.

Thanks in advance!

Answer 1

We may represent the state as follows: $$ |\psi\rangle = \sum_{b_1,b_2,\dots,b_n\in\{0,1\}} \alpha_{b_1b_2\cdots b_n}|b_1b_2\dots b_n\rangle $$ where $\alpha_{b_1b_2\dots b_n}\in\mathbb{C}$ are the coefficients in the computation basis. Assuming this state is normalized, we have $$ \langle \psi|\psi\rangle = \sum_{b_1b_2,\dots,b_n\in\{0,1\}} |\alpha_{b_1b_2\cdots b_n}|^2 = 1. $$

If we were to measure the first qubit and obtain the outcome that the first qubit is in the $|1\rangle$ state, this collapses the state $|\psi\rangle $ into superpositions of states whose first bit is $|1\rangle$ (i.e., $b_1=1$). That is, the resulting (unnormalized) state is $$ |\psi'\rangle = \sum_{b_2,\dots,b_n\in\{0,1\}} \alpha_{1b_2\cdots b_n}|1b_2\dotsb_n\rangle. $$ The probability of having obtained this result is the norm $\langle \psi'|\psi'\rangle = \sum_{b_2,\dots,b_n\in\{0,1\}} |\alpha_{1b_2\cdots b_n}|^2$.

Does this help?