Can we regard a spin-1 particle as combination of two spin-1/2 particles?

Question

As we know, the square of time-reverse operator is -1 acting on fermion and +1 on boson. I can prove it by regarding the time-reverse operator as 2π rotation around $y$-axis multiply the complex conjugation operator,but this require the eigenvectors of $S_y$ to be purely imaginary in any dimensions, which is kind of difficult to prove. My classmate said we only have to prove it with spin-1/2 particle and then regard the higher spin particle as combination of spin-1/2 particles. The conclusion is apparent if his claim is true, but I just wonder if it is the case or not.

Answer 1

The combined representation of two spin-$\frac 1 2$ particles has dimension $4$, since it is the tensor product of two 2-dimensional representations. The spin $1$ representation has dimension $3$. However, it is a well-known fact that $2\cdot 2 = 1 + 3 = 4$. The dimension of the spin $0$ representation is $1$. This hints that, and in fact, fact $\frac 1 2 \otimes \frac 1 2 = 0 \oplus 1$, meaning that two spin-$\frac 1 2$ particles combine to make the direct sum of a spin-$0$ particle and a spin-$1$ particle.

The above is the for the Galilean case. In the Lorentz case, i.e., in Einsteinian relativity, it is somewhat cleaner. There, the product of a left-handed spin-$\frac 1 2$ particle and a right-handed one [there is no distinction between the handednessses in Galilean relativity] makes the 4-vector representation. Let $G$ be the group corresponding to spatial rotations, according to some particular inertial observer's split between space and time. Then under $G$, the three spatial components -- according to the same observer -- do transform like a 3-vector, i.e., like the Galilean spin-$1$ particle, and the time component is of course invariant, i.e., a scalar, i.e., a spin-$0$ particle. Proofs of the above are given and the implications are discussed at length in the two volumes by Penrose and Rindler, Spinors and Space-Time.

Answer 2

(I am not sure to understand what the question is actually.)

Yes it is possible, if ignoring the orbital part of the state describing the relative position of the two fermions (or the relative momentum in an essentially identical relativistic version of my answer).

The state of the system made of the pair of fermions can be described in the Hilbert space $$H = L^2(\mathbb R^3, d^3x)\otimes L^2(\mathbb R^3, d^3X) \otimes \mathbb C^2 \otimes \mathbb C^2$$ Above $\vec{x}$ describes the relative position of the two particles, $\vec{X}$ is the position of the center of mass. Each $\mathbb C^2$ supports the spin observables of the corresponding particle. Now $$C^2 \otimes \mathbb C^2 = \mathbb C^3 \oplus \mathbb C$$ where the first addend on the right-hand side is the triplet subspace spanned by the eigenvectors of total spin with eigenvalue $1$ and the second addend is the singlet subspace spanned by the eigenvectors of total spin with eigenvalue $0$.

The crucial observation is now that the subspace of $H$ $$L^2(\mathbb R^3, d^3x)\otimes L^2(\mathbb R^3, d^3X) \otimes \mathbb C^3$$ is essentially a particle with spin $1$ described by the center of mass of the two particles provided we ignore the first factor $L^2(\mathbb R^3, d^3x)$ referred to the relative degrees of freedom and ending up with $$L^2(\mathbb R^3, d^3X) \otimes \mathbb C^3\:.$$ This procedure (ignore a factor of the Hilbert space) can be obtained by means of a so-called partial trace operation, which generally gives rise to mixed states even if starting from pure states, but it is not very important here.

Summing up, a system of two spin-$1/2$ fermions behaves as a spin-$1$ particle, situated in the center of mass of the system, provided (a) the total spin of the system is defined and is equal to $1$ and (b) we ignore the relative orbital degrees of freedom.

Let us come to the time reversal operation. In view of Wigner theorem it must be represented by an either unitary or anti unitary operator. However, as soon as the Hamiltonian of the system is bounded below but not above, it must be represented by an antiunitary operator. The proof is easy. So here we may assume that the time reversal is antiunitary. Further properties which can be physically justified are that the position observables are invariant under time reversal but the momentum, the angular momentum, and the spin change sing under the action of the time reversal operation.

From these assumptions one may prove that, for a spin-$1/2$ particles described in $$L^2(\mathbb R^3, d^3x)\otimes \mathbb C^2\:,$$ the time reversal operator $U_{1/2}$ has the form $$U_{1/2}= C' \otimes i\sigma_2 C$$ where $C'$ and $C$ are the standard complex conjugations, the former of wavefunctions of $L^2(\mathbb R^3, d^3x)$ and the latter of vectors in $\mathbb C^2$ with respect to the canonical basis (the one of eigenstates of $S_z= \sigma_3/2$).

I stress that we could add a phase $\eta$ in front of $U_{1/2}$ without changing all results I go to to discuss and thus I henceforth omit it.

We have in particular found that the time reversal operation on the spinor part of the spin-$1/2$ particle is the anti unitary operator $$U^{(s)}_{1/2}= i\sigma_2 C\:.$$ This is equivalent to say that $U^{(s)}_{1/2} = e^{i\pi \sigma_2/2} C$ as more or less the OP says (the "rotation" around the $y$ axis is of $\pi$ not $2\pi$!)

Thus we have the known result mentioned by the OP $$U_{1/2}U_{1/2}= C'C'\otimes i\sigma_2 Ci\sigma_2 C= I\otimes\sigma_2 \overline{\sigma_2} CC = - I \otimes \sigma_2\sigma_2 = -I\otimes I = -I\tag{1}$$

When instead considering two such particles, the time reversal symmetry is the tensor product of two time reversal symmetries $$U= U_{1/2}\otimes U_{1/2}\:.$$ Using (1) we have $$UU = (-I)(-I) = I$$

Notice that the triplet subspace (including the orbital degrees of freedom of both the particles) is invariant under $U$ $$U(L^2(\mathbb R^3, d^3x)\otimes L^2(\mathbb R^3, d^3X) \otimes \mathbb C^3) \subset L^2(\mathbb R^3, d^3x)\otimes L^2(\mathbb R^3, d^3X) \otimes \mathbb C^3\:,$$ the proof of this invariance can be obtained by direct inspection, especially exploiting the commutation properties of Pauli matrices, or simply noticing that the total squared spin commute with $U$. If we eventually trace out the relative orbital degrees of freedom $\vec{x}$, we end up with an antiunitary operator $U_1$ describing the time reversal symmetry in the space $$L^2(\mathbb R^3, d^3X) \otimes \mathbb C^3$$ which, as we said above, can be identified with the Hilbert space of a particle with spin $1$ whose position $\vec{X}$ is the one of the center of mass of the system. $$U_1 = C' \otimes (i \sigma_2 C\otimes i\sigma_2 C)|_{\mathbb C^3}$$ The explicit form of $(i \sigma_2 C\otimes i\sigma_2C)|_{\mathbb C^3}$ does not matter (though it can be computed explicitly), since we are only interested in $$U_1U_1 = C'C' \otimes (i \sigma_2 C\otimes i\sigma_2C)|_{\mathbb C^3} (i \sigma_2C \otimes i\sigma_2C)|_{\mathbb C^3}= I \otimes ((i \sigma_2C \otimes i\sigma_2C)\: (i \sigma_2 C\otimes i\sigma_2 C))|_{\mathbb C^3}$$ $$ = I\otimes (-I)\otimes (-I) = I$$ So fake spin-$1$ particles constructed out of a pair of spin-$1/2$ particles must have time reversal operator $T$ satisfying $TT=I$. We can therefore expect that real spin-$1$ particles enjoy the same property.

Answer 3

As the other answers have pointed out, you can consider a spin-1 particle as the projection of two spin-1/2 particles into the triplet subspace. And in fact, this isn't just a math identity, it's actually a useful technique for solving real problems! For example, it is used to solve the spin-1 Heisenberg AKLT model, which has many applications and generalizations, e.g. the Haldane conjecture, the Kitaev chain topological superconductor, matrix product states, etc.

Answer 4

No, because two-spin $1/2$ particles can combine as a singlet (spin $0$) or a triplet (spin $1$).