Everyone familiar with spin angular momentum as Spin is an intrinsic form of angular momentum carried by elementary particles, composite particles (hadrons), and atomic nuclei. To measure spin there are experiments and the most famous one is the Stern-Gerlach experiment.

Now pretend that we don't have two or three-dimensional setups. We allowed having only 1-dimensional systems so

Is there a way with which can determine the spin of a particle given a one-dimensional system?

If Yes How? If not then Can we say that spin only exists for Higher dimensions (2D and 3D)?

Edit: I'm adding a little more details as the answer due to @Qmechanics suggest that there is no sign of spin in $1+1$D space. But the following is seems to mean contradiction to what R. Shankar said :

Let us say we have two identical noninteracting pions and wish to find out if they are bosons or fermions. We put them in $1-$D box and make an energy measurement. Say we find one in the state $n=3$ and the other in the state $ n=4$. The probability distribution in $x$ space would be, depending on their statistics, $$P_{S/A}(x_1,x_2)=2|\psi_{S/A}(x_1,x_2)|^2$$

$$=2|2^{-1/2}[\psi_3(x_1)\psi_4(x_4)\pm \psi_4(x_1)\psi_3(x_2)]|^2$$ $$=|\psi_3(x_1)|^2|\psi_4(x_2)|^2+|\psi_4(x_1)|^2|\psi_3(x_2)|^2\pm[\psi^*_3(x_1)\psi_4(x_2)\psi^*_4(x_2)\psi_3(x_2)+\psi^*_4(x_1)\psi_3(x_1)\psi^*_3(x_2)\psi_4(x_2)]$$

$$\cdots$$ Further, The interference terms tell us if the pions are bosons or fermions. The difference between the two cases is most dramatic as $x_1\rightarrow x_2\rightarrow x$ $$P_A(x_1\rightarrow x_2\rightarrow x)\rightarrow 0 \ (\mathrm{Pauli \ principle \ applied \ to \ state \ } |x\rangle)$$ whereas $$P_S(x_1\rightarrow x_2\rightarrow x)=2\cdots$$ which is twice as big as $P_D(x_1\rightarrow, x_2\rightarrow x)$, the probability density for two distinct label carrying (but otherwise identical ) particles, whose labels are disregarded in the position measurement.

Given the striking difference in the two distributions, we can readily imagine deciding (once and for all) whether pions are bosons or fermions by preparing an the ensemble of systems (with particles in $n= 3$ and $4$) and measuring $P(x_1 , x_2)$.

Note that the details in the edit are not complete. Please refer to


Principle of Quantum Mechanics R. Shankar

Chapter 10 : SYSTEMS WITH $N$ DEGREES OF FREEDOM: Determination of Particle Statistics


1 Answer 1


In a relativistic 1+1D theory a (massive or massless) particle has no internal spin/helicity (because the stabilizer subgroup/isotropy group/little group of the Lorentz group $O(1,1)$ is trivial). The Lorentz algebra $o(1,1)$ has only 1 generator, a boost, which is conserved in collisions.

  • $\begingroup$ I'm not getting your answer. Can you talk a bit in simple language? I don't have much knowledge of symmetry groups etc. $\endgroup$ Commented Dec 5, 2020 at 11:42
  • $\begingroup$ It means spin is not a relevant or even well-defined quantity in 1+1D. $\endgroup$
    – NDewolf
    Commented Dec 6, 2020 at 9:03
  • $\begingroup$ By 1+1D, Do you mean 1-dimensional space and time? $\endgroup$ Commented Dec 6, 2020 at 12:19
  • $\begingroup$ Well, 1 space and 1 time dimension. $\endgroup$
    – Qmechanic
    Commented Dec 6, 2020 at 12:21
  • $\begingroup$ @Qmechanic This is not right to say that spin is not relevant or even a well-defined quantity for $1+1$D. If you put two fermions in the 1D box then their wave function must follow the Pauli exclusion principle and thus different properties than bosons. $\endgroup$ Commented Dec 8, 2020 at 6:21

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