# Quantum spin measurement

The state of a spin $$\frac{1}{2}$$ particle is $$|0\rangle$$ which is eigenstate of $$\sigma_z$$. What is the most generalized way to show that the results of any spin measurement along any direction in x-y plane is completely random.

• You can compute the probability distribution function of this measurement output exactly. Do you need the details? – David Bar Moshe Nov 16 '18 at 9:49
• It will be helpful if you can provide the details @DavidBarMoshe – Adex Nov 16 '18 at 10:15

Any set of commuting observables in any quantum state can be characterized by a joint classical distribution function describing the probabilities of its measurement outputs in that quantum state. Since you need a single observable and it is of course self commuting, the above is valid in your case.

The obsevable in your case is:

$$\sigma = \cos \phi \sigma_x + \sin \phi \sigma_y$$

In a state having a densiyy matrix $$\rho$$, the characteristic function of the probability density of this observable is given by: (this is the most important thing to remember here)

$$g(t) = tr(\rho e^{it\sigma})$$

In our case $$\rho$$ is the projector on the spin up state, and it is not difficult to show that the exponent is:

$$\cos (t/2 )+i \sin( t/2 )\sigma$$

Thus:

$$g(t) = \cos (t/2)$$

The probability density function is the Fourier transform of the characteristic function:

$$f(s) = \frac{1}{2\pi}\int_{-\infty}^{\infty} g(t) e^{ist} dt$$

Where $$s$$ is the measurement outcome of $$\sigma$$. We get:

$$f(s) = 0.5 \delta(s-0.5) + 0.5 \delta(s+0.5)$$

This is the probability distribution function of a Bernoullian random variable, uniformly distrinuted at $$\pm 0.5$$.