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I've been told that electrons can have $+{1\over 2}$ or $-{1\over 2}$ spin. And that because of the Pauli exclusion principle this is how they can occupy the same shell. But when I look online I only ever see that particles with mass have only positive spin of ${1\over 2}$ (especially when I look at the table of elementary particles). So I thought that maybe that particles always have a spin of ${1\over 2}$, and it's just the direction of spin that makes it + or - because it's a handy way to explain it. Is that the case? Can all particles have the ability to have a negative spin?

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    $\begingroup$ Not all massive particles have spin = $\frac{1}{2}$ (eg. Higgs boson has spin $0$). For a particle with spin $s$, there are $2s+1$ ways to project the spin on an axis (conventionally, the z axis). It just turns out that some projections are on the positive side, and some on the negative. $\endgroup$ – Avantgarde Jun 19 '17 at 20:06
  • $\begingroup$ Quick comment: note that the table of elementary particles also lists electron with spin $\frac12$. $\endgroup$ – The Vee Jun 22 '17 at 19:36
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In quantum mechanics, spin is a kind of angular momentum that is intrinsic to a particle. Like normal angular momentum, when we go to the quantum world there are two numbers that describe the spin of a body, which we label $s$ and $m_s$. $s$ can be integers or half integers, depending on what you're trying to describe, and $m_s$ is an integer or half integer that takes on values from $-s$ to $s$, differing by at least $1$.

The spin value quoted in these particle lists you are looking at is the $s$ value, which gives you the magnitude of a particle's spin vector through the relationship $\sqrt{s(s+1)}$, which can be derived from quantum mechanics. This is the value always quoted for the spin of a particle.

The $+\frac{1}{2}$ and $-\frac{1}{2}$ are the values of $m_s$ and they are the components of the spin along some axis, usually the z-axis.

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  • $\begingroup$ "when we go to the quantum world there are two numbers that describe the spin of a body" that is not entirely correct. The spin is always described by the value of $s$: the reason why you need two, namely $(s,m_s)$ is in order to uniquely identify the quantum state in a degenerate subspace of the Hilbert space of spin $s$. $\endgroup$ – gented Jun 20 '17 at 14:25
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This is a case where our use of the word "spin" is a little sloppy. We use it to describe both the spin quantum number which is 1/2 for electrons, and also to describe the component of the spin vector along the z-axis which for electrons can be either +1/2 or -1/2. It is usually obvious by the context which of these two usages is meant, but it can be confusing for a beginner. In one case we are referring to the eigenvalue of the spin operator squared ($S^2|>=s(s+1)|>$) and in the other case to the eigenvalue of the z-component of the spin operator ($S_z|>=s_z|>$). For electrons $s=1/2$ while $s_z=1/2, -1/2$.

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Your third sentence is correct. The number quoted in most references is the magnitude of the spin, whereas the "direction" can be positive or negative.

The reason I put direction in quotes above is because it only beomes an actual spatial direction relative to an external field. Spin is an internal magnetic moment that many (but not all) particles possess, and the sign refers to its reaction in an external field: positive spin means it is pushed in the direction of an external field, whereas negative spin means it is pushed in the opposite direction.

In addition, spin need not necessarily be $\frac{1}{2}$; in fact, any integer or half-integer can work. For example, mesons have spin 0 or 1, so they either don't react to a magnetic field or react twice as strongly. Delta particles have spin $\frac{3}{2}$, so they sometimes experience three times the force as an electron.

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  • $\begingroup$ Do antiparticles have the same spin as their counterparts? Will a positron always have a spin of ${1\over 2}$ just like an electron? $\endgroup$ – matryoshka Jun 19 '17 at 20:14
  • $\begingroup$ If the two particles are created from the same decay, then yes. Otherwise, either particle can have spin of either sign. The only intrinsic property is the maximum magnitude. $\endgroup$ – probably_someone Jun 19 '17 at 20:22

protected by Qmechanic Jun 20 '17 at 13:22

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