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I would like to know how uncertainty principle plays in measuring particle spin on different axis, specially in a language that compares it against decomposing a signal into its frequencies using Fourier transform.

For example my understanding is that since a particle has been measured in z direction and we have a very definite answer of spin $|u\rangle$ or $|d\rangle$ then if we want to calculate the spin in lets say x direction it would give us the most vague answer of 50% $|u\rangle$ and 50% $|d\rangle$. That is $$|r\rangle=\frac{1}{\sqrt{2}}|u\rangle+\frac{1}{\sqrt{2}}|d\rangle$$ or $$|l\rangle=\frac{1}{\sqrt{2}}|u\rangle-\frac{1}{\sqrt{2}}|d\rangle \, ,$$

but I am not sure how I can make the analogy between this and frequency decomposition. For example what is the complete range for spin or what is the increment or decrement of standard deviation(uncertainty) in spin example exactly?

Also if in signal decomposition we have a case that the signal spans the whole range, hence letting us make sharp calculations of constituent frequencies, then how we can translate such into an example in spins?

I would appreciate if the particle spin part be described in bra-ket notation.

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    $\begingroup$ Time and frequency domain functions are infinite dimensional while $x$ and $z$ spins are two dimensional. So I don't know how good an analogy there can be. $\endgroup$ Jul 23, 2021 at 18:02
  • $\begingroup$ But regardless of that difference I still feel that there can be analogies because the main principle behind both, that is uncertainty principle, is the same! $\endgroup$
    – al pal
    Jul 23, 2021 at 18:58
  • $\begingroup$ no analogies since the commutators are different. $\endgroup$ Jul 23, 2021 at 22:08
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    $\begingroup$ The uncertainty relation for spins is (for instance) $\Delta S_x\Delta S_y\ge \frac{1}{2}\vert \langle S_z\rangle\vert$. The average value $\langle S_z\rangle$ is state dependent and in fact can be $0$, something not possible in a Fourier transform. $\endgroup$ Jul 26, 2021 at 22:17
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    $\begingroup$ yes but the right hand side cannot be $0$, and does not depend on the problem. In fact, the right hand side is $\approx 1$. Point being: this analogy between FT and uncertainty relation is not a very good analogy, especially as in the time domain $\Delta E\Delta t$ is subject to tricky interpretation since time in QM is not an operator. see physics.stackexchange.com/q/53802/36194 $\endgroup$ Jul 26, 2021 at 23:07

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