This appears like a question that is rarely addressed in field theory pedagogy (perhaps because the answer is obvious): how does one describe a particle of definite spin in quantum field theory?

For example, given some state in a theory of spinors (say a single particle for simplicity): $|\psi\rangle = \int \psi(p) a^\dagger(p)|p\rangle$, where $a$ is the ladder operator you obtain by canonically quantizing, say a Dirac field. Since this is a Dirac field, it describes a particle of spin 1/2. How does one extract information about the spin of this particle?. For example, what would the creation operator for a single particle of definite spin up look like?

  • $\begingroup$ What do you mean, "Without knowing anything about its position or momentum"? Then you don't know the state! (and thus cannot create it) $\endgroup$ – ACuriousMind Jul 24 '14 at 22:21
  • $\begingroup$ For example, in non-relativistic quantum mechanics, I can write down something like $|0\rangle$, and say this is a state of spin up. This alone doesn't contain any information about its linear momentum, right? $\endgroup$ – zzz Jul 24 '14 at 22:23
  • $\begingroup$ Because, when you write down the spin states as $|0\rangle$ and $|1\rangle$, you implicitly assume that the spin degree of freedom is the only degree of freedom the system/particle has. If you are considering a particle that is not trapped at a spot, but free to move in a line, then "spin-up" is certainly not enough to specify its state. $\endgroup$ – ACuriousMind Jul 24 '14 at 22:25
  • $\begingroup$ Exactly and I'm asking if the same can be done in QFT. $\endgroup$ – zzz Jul 24 '14 at 22:26
  • $\begingroup$ That would mean you have no spatial degrees of freedom, and thus no need for QFT (since it would not differ from QM)! $\endgroup$ – ACuriousMind Jul 24 '14 at 22:29

Fields in QFT are promoted to operators.

The Dirac field operator describes both a particle (electron) and and a anti-particle (positron), and there exist different creation and anihilation operators for the particle ($a^+_s(p), a_s(p))$ and the anti-particle ($b^+_s(p), b_s(p)$). The subscript $s$ indicates that you are creating or destroying a particle of spin $s$ $(\frac{1}{2}$ or $-\frac{1}{2})$ .

For more details, see in Wikipedia Dirac fields.

A particle is described by a state, for instance an electron with definite position and spin, is described by the state: $|p,s\rangle = a^+_s(p)|0\rangle$, that is you apply the creation operator $a^+_s(p)$ on the vacuum state $)|0\rangle$

  • $\begingroup$ You're right, s is the spin degree of freedom, I'll have to go over the derivation again. thanks. $\endgroup$ – zzz Jul 25 '14 at 12:32

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