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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.

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    $\begingroup$ You can compute the probability distribution function of this measurement output exactly. Do you need the details? $\endgroup$ – David Bar Moshe Nov 16 '18 at 9:49
  • $\begingroup$ It will be helpful if you can provide the details @DavidBarMoshe $\endgroup$ – Adex Nov 16 '18 at 10:15
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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 $\pm 0.5$.

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