# How to measure average value of qubits?

Recently I have been studying quantum mechanics in The Theoretical Minimum by Susskind. In his experiment, when the apparatus rotated by an arbitrary angle within the $$x{–}z$$ plane, the average measurement result is $$\hat{n}\cdot\hat{m}$$. He explains as below:

Classically, if $$\sigma$$ were a vector, we would expect the result of the experiment to be the component of $$\sigma$$ along the $$\hat n$$ axis. If $$\hat n$$ lies at an angle $$\theta$$ with respect to $$z$$, the classical answer would be $$\sigma=\cos\theta$$. But as you might guess, each time we do the experiment we get $$σ = \pm1$$. However, the result is statistically biased so that the average value is $$\cos\theta$$.

My question is: if we get $$\pm1$$ randomly in a qubit, how do we get a statistically biased average value of $$\cos\theta$$?

The way to express an arbitrary Pure Qubit is:

$$|n \rangle = \cos{\frac{\theta}{2}} |0 \rangle + e^{i\phi}\sin{\frac{\theta}{2}} |1 \rangle$$

Where $$n$$ is the Bloch Vector with angle $$\theta$$ with respect to the positive $$z\text{ -axis}$$ and angle $$\phi$$ with respect to the positive $$x\text{ -axis}$$ on the $$xy\text{ -plane}$$

And if we are measuring some $$| \psi \rangle$$ along the axis $$| n \rangle$$, then the probability we obtain $$+\frac{\hbar}{2}$$ is $$|\langle n | \psi \rangle|^2$$ and ofcourse the probability we obtain $$-\frac{\hbar}{2}$$ is $$|\langle -n | \psi \rangle|^2$$, where $$|-n \rangle$$ is simply the vector in the opposite direction of $$|n \rangle$$, because obviously those are the only two eigenstates we can obtain measuring along that axis.

From here you should be able to find expected values and other statistical attributes which would depend on the angles at hand.

Bellow is a Bloch Sphere, which is a very useful visual aid for seeing Rotations of Single Qubits and how that effects observed probabilities.

• Thanks for your nice explanation .But I think the book or lecture is missing the explanation initially. – Amitabha Chakraborty Mar 25 at 4:28