# Measurement of qubit w.r.t basis A qubit can be measured w.r.t to a basis. The result is one of the two basis. The output basis is sampled with probability equal to square norm of the qubit projection on that basis. Let $\ket\psi$ be a qubit and we want to measure it w.r.t basis $\{\ket{b_0},\ket{b_1}\}$, then the measurement result will be $\ket{b_0}$ with probability $\|\langle \ket\psi,\ket{b_0} \rangle\|^2$ . Similarly true for $\ket {b_1}$ Why is the inner product is projection? assume $\ket\psi=\alpha\ket{b_0}+\beta\ket{b_1}$ , then inner product with $\ket{b_0}$ result in $\alpha$ as required. ## Computation basis and Hadamard basis ![[Pasted image 20260526112349.png|331]] ## Breidbart basis ![[Pasted image 20260526112444.png|332]] note: this basis is used in [[CHSH game]] There are many possible basis. The most common are the computational basis and the Hadamard basis. ## Examples - $\ket\psi=\frac{3}{5}\ket0+\frac{4}{5}\ket1$ , measured in the computational basis $\{\ket0,\ket1\}$ . compute $P(0),P(1)$. Solution: $P(0)=(3/5)^2,P(1)=(4/5)^2$. - $\ket\psi=\ket0$, measured in the Hadamard basis $\ket+:=\frac{\ket0+\ket1}{\sqrt 2},\ket-:=\frac{\ket0-\ket1}{\sqrt 2}$. compute $P(\ket+),P(\ket-)$ . Solution: the projection is $\langle \ket0,\ket+\rangle=\langle \ket0,\frac{\ket0+\ket1}{\sqrt 2}\rangle=\frac{1}{\sqrt 2}$ so $P(\ket+)=(\frac{1}{\sqrt 2})^2=0.5$ . - $\ket\psi=\ket0$, measured in the Breidbart basis $\{b_0,b_1\}$ where $b_0=\cos(\frac{\pi}{8})\ket0+\sin(\frac{\pi}{8})\ket1$ , $b_1=-\sin(\frac{\pi}{8})\ket0+\cos(\frac{\pi}{8})\ket 1$ compute $P(b_0),P(b_1)$. Solution: The projection on $b_0$ is $\langle b_0,\psi\rangle=\cos(\pi/8)$ so $P(b_0)=cos^2(\pi/8)\approx0.85$ and so $P(b_1)=1-P(b_0)\approx 0.15$ . - $\ket\psi=\frac{\ket0+\ket1}{\sqrt 2}$, measured in the Breidbart basis $\{b_0,b_1\}$ where $b_0=\cos(\frac{\pi}{8})\ket0+\sin(\frac{\pi}{8})\ket1$ , $b_1=-\sin(\frac{\pi}{8})\ket0+\cos(\frac{\pi}{8})\ket 1$ . compute $P(b_0),P(b_1)$. Solution: The projection on $b_0$ is $\langle b_0,\psi\rangle=\bigg\langle \cos(\frac{\pi}{8})\ket0+\sin(\frac{\pi}{8})\ket1,\frac{\ket0+\ket1}{\sqrt 2}\bigg\rangle=(\cos(\pi/8)+\sin(\pi/8))/\sqrt 2$ so $P(b_0)\approx 0.8536$. Similarly for $P(b_1)$. - quantum-state tomography (the single qubit case) - means reconstructing a classical description of an unknown quantum state by repeatedly preparing and measuring it. - note: trying to compute $\alpha$, $\beta$ by measuring multiple time and estimate the probability $P(0)$,$P(1)$ is not enough because these are only real values in $[0,1]$ and they could have complex values e.g., negative reals which are excluded. - Let's find out the amplitudes of $\ket\psi=\alpha\ket0+\beta\ket1$ . repeating the measurement in the computation basis provides us with the absolute value of the amplitude $P(0)=|\alpha|^2$ and $P(1)=|\beta|^2$. next we want to know their phase difference $\theta_\beta-\theta_\alpha$ because $\ket\psi=|\alpha|e^{i\theta_\alpha}\ket0+|\beta|e^{i\theta_\beta}=e^{i\theta_\alpha}(|\alpha|\ket0+|\beta|e^{i(\theta_\beta-\theta_\alpha)}\ket1)$. For this, we measure in the +/- basis : $\langle \psi, +\rangle=\frac{\alpha+\beta}{\sqrt2}$ and $P(+)=|\frac{\alpha+\beta}{\sqrt2}|^2=\frac{(\alpha+\beta)^*(\alpha+\beta)}{2}=\frac{|\alpha|^2+|\beta|^2+\alpha^*\beta+\beta^*\alpha}{2}= \frac{1}{2}+2Re(\alpha^*\beta)$. Similarly, $P(-)=\frac{1}{2}-2Re(\alpha^*\beta)$ . we get $P(+)-P(-) = 2Re(\alpha^*\beta)$. Also, $\alpha^*\beta=|\alpha||\beta|\cos(\theta_\beta-\theta_\alpha)$ and as we computed approximation for $|\alpha|,|\beta|$ in previous step we can compute the angle. There are two angles that solves the equation and so the right one is selected using another measurement w.r.t Y-basis: $P(+i)-P(-i)=2\text{Im}(\alpha^*\beta)$. - note: the above show that extracting the amplitude to classic bits require multiple measurements in several basis and is exponential in the number of qubits. ## Created 2026-07-12 16:19