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Suppose we have a quantum state $|\psi \rangle$ of $n$ qubits, where $|\psi\rangle=\sum_{x∈\{0,1\}^n}\alpha_x |x\rangle$,and we measure the first qubit of $|\psi\rangle$ in the computational basis. What is the probability that the measurement outcome is $1$, in terms of the $\alpha_x $ coefficients?

I'm not quite sure how to approach this. Usually the computational basis is $\{|0\rangle,|1\rangle\}$ and I'm not sure what ket I am meant to apply to the $|\psi\rangle$.

I'm also not sure what matrix I need to use to do the measurement.

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$x \in \{0, 1\}^n$ means that in the sum we have all possible bitstrings with length $n$. Let's take $n = 3$:

$$|\psi\rangle = \sum_{x \in \{0, 1\}^n} \alpha_x |x\rangle= \alpha_{000} |000\rangle + \alpha_{001} |001\rangle + \alpha_{010} |010\rangle + \\ + \alpha_{011} |011\rangle + \alpha_{100} |100\rangle + \alpha_{101} |101\rangle + \alpha_{110} |110\rangle + \alpha_{111} |111\rangle$$

If we use projecter $P = |1\rangle \langle 1 | \otimes I \otimes I$ then the probability of measuring first qubit $|1\rangle$ will be equal to:

$$p = \langle \psi | P | \psi \rangle = |a_{100}|^2 + |a_{101}|^2 + |a_{110}|^2 + |a_{111}|^2$$

More about projective measurements can be found in M. Nielsen and I. Chuang's textbook pages 87-88 (for $M$ in the textbook one can take $Z \otimes I \otimes I$ operator).

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The probability that the measurement outcome is 1 in the first qubit is $\sum|\alpha_x|^2$ for all $x$ whose first bit is 1.

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