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This question already has an answer here:

i have seen some analogies of spin using playing cards but i am struggling to grasp the concept due to this making no sense in terms of playing cards

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marked as duplicate by ACuriousMind, Prahar, Kyle Kanos, John Rennie, Ryan Unger Apr 14 '15 at 20:01

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    $\begingroup$ For color, see this question. $\endgroup$ – ACuriousMind Apr 14 '15 at 11:18
  • $\begingroup$ this still leaves how a particle can have a spin of 2 $\endgroup$ – ziggy Apr 14 '15 at 11:19
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    $\begingroup$ What do you mean, "how"? From Noldorin's answer: "Specifically, the allowed values of the spin quantum number s are non-negative multiples of 1/2." Don't cling to that card analogy, it's just an analogy that breaks down if you think about it too hard. Spin is a technical term with a very precise technical meaning. $\endgroup$ – ACuriousMind Apr 14 '15 at 11:23
  • $\begingroup$ ok, so the card idea is not a good one that does not work too well. i'll just accept that it can happen considering that it is not literally referring to spin as something like a card. having a spin of 2 does fit the explanation you directed me to. $\endgroup$ – ziggy Apr 14 '15 at 11:29
  • $\begingroup$ You can help us by describing the analogy with playing cards that you refer to. At least one of us has never seen it. $\endgroup$ – garyp Apr 14 '15 at 12:22
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I hear that analogy too. Spin 0: any rotation left the "object" invariant, like a circle who rotates. Spin 1/2: half rotation to het the initial state of the object, and here we are: any figure of the playing card "has spin 1/2".

Spin 1: any non figure card, like who knows, the ace of clubs. One integer rotation to get is as it was initially.

Spin 2: no example in playing cards. But you may think about the Moebius strip. You have to make 2 rounds to get back to the initial state/position. This is the best analogy for spin 2.

Now we should have to find an analogy for spin 3/2 :D

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  • $\begingroup$ what about a hyper cube? $\endgroup$ – ziggy Apr 14 '15 at 11:44
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    $\begingroup$ Don't you have it backwards? I'm pretty sure the playing card is spin $2$ and the spin $\frac{1}{2}$ is where it starts to get tricky (because fermions are represented by spinors not vectors). $\endgroup$ – or1426 Apr 14 '15 at 12:01

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