I have seen this question asked a couple of times (here, for example), but I find the answers very lacking in clarity.

Field operators are defined in terms of creation and annihilation operators as $$\phi_A(x)=\sum_\lambda\int\frac{d^3p}{\sqrt{2E_p}}\bigg(u_A(\mathbf{p},\lambda,s)a(\mathbf{p},\lambda,s)e^{ipx}+v_A(\mathbf{p},\lambda,s)a^\dagger(\mathbf{p},\lambda,s)e^{-ipx}\bigg),$$

where $s$ denotes the spin of the particular state $a^\dagger$ creates, and $\lambda=-s,-s+1,...,s-1,s$ denotes the possible helicities of said state (for a massless particle, we do not include $s$). Up to normalization, $$a^{\dagger}(\mathbf{p},\lambda,s)|0\rangle=|\mathbf{p},\lambda,s\rangle.$$

In the expression for the field, the coefficients $u_A$ and $v_A$ are chosen in such a way that the field obeys the following transformation law $$U(\Lambda)\phi_AU^\dagger(\Lambda)=L_A^{\;\;B}(\Lambda)\phi_B,$$

where $U$ is the unitary representation of the Poincaré group acting on the Hilbert space of the theory, and $L$ is a finite dimensional representation of the Lorentz group acting on the fields.

My question is then, what is the explicit mathematical relationship between the spin $s$ of the created states, and the field representation $L$?

I am aware that, for example, the degrees of freedom match in the case of a scalar, spinor or vector with those of a particle of spin $0, 1/2$ or $1$, respectively. But I am asking for an explicit, mathematical explanation of this fact and not the intuitive reasoning behind it. I believe Weinberg does this in eqs. 5.125 - 5.126 of his first volume, but unfortunately I find his notation unreadable.

Could someone clarify this? Thanks.

  • 2
    $\begingroup$ If you feel like you will just get a reprise of earlier answers, then possibly, put up Weinberg equations / analysis as a question, then ask about the bits you don't follow. $\endgroup$
    – user163104
    Aug 2, 2017 at 1:14
  • $\begingroup$ Related question by OP: physics.stackexchange.com/q/349166/2451 $\endgroup$
    – Qmechanic
    Aug 2, 2017 at 6:10

1 Answer 1


I believe I've figured out Weinberg's argument. From now on, the labels $s$ and $m$ on states and creation/annihilation operators will be implicit. I will be dealing with massive particles only, but the analysis for the massless case should be very similar.

Let us first derive the transformation law for creation/annihilation operators. Under a Lorentz transformation, a general state changes as $$U(\Lambda)|p,\lambda\rangle=\sum_{\lambda'}C_{\lambda'\lambda}|\Lambda p,\lambda'\rangle$$

for some matrix $C$. This follows from the fact that the labels $m$ and $s$ come from Lorentz-invariant operators, and $p\rightarrow\Lambda p$ under a Lorentz transformation. On the other hand, $\lambda$ is the label associated to helicity, which is not Lorentz-invariant. States corresponding to a certain $\lambda$ must therefore turn into some linear combination of states corresponding to other $\lambda'$s.

Creation operators satisfy $$a^\dagger(p,\lambda)|0\rangle=\frac{1}{\sqrt{2E_p}}|p,\lambda\rangle,$$

So, under a Lorentz-transformation, we have $$U(\Lambda)a^\dagger(p,\lambda)U(\Lambda)^{-1}U(\Lambda)|0\rangle=\sum_{\lambda'}\sqrt{\frac{E_{\Lambda p}}{E_p}}C_{\lambda'\lambda}a^\dagger(\Lambda p,\lambda')|0\rangle.$$

Since the vacuum is invariant, $U(\Lambda)|0\rangle=|0\rangle$ and we find the transformation law $$U(\Lambda)a^\dagger(p,\lambda)U(\Lambda)^{-1}=\sum_{\lambda'}\sqrt{\frac{E_{\Lambda p}}{E_p}}C_{\lambda'\lambda}a^\dagger(\Lambda p,\lambda').$$

We can express the field in terms of creation/annihilation fields by defining $$\phi_A(x)=\phi_A^+(x)+\phi_A^-(x),$$

where $$\phi_A^+(x)=\sum_\lambda\int\frac{d^3p}{\sqrt{2E_p}}u_A(p,\lambda)a^\dagger(p,\lambda),$$

and similarly for $\phi_A^-(x)$. For simplicity, we will only deal with $\phi_A^+(x)$, since the other one is analogous. By using the transformation law for $a^\dagger$, we find that for the field to transform as $$U(\Lambda)\phi_A(x)U(\Lambda)^{-1}=L_A^{\;\;B}\phi_B(x),$$

the following condition on the coefficient is necessary and sufficient $$\sum_{\lambda'}C_{\lambda'\lambda}^s u_A(p,\lambda')\frac{1}{\sqrt{E_p}}=L_A^{\;\;B}u_B(\Lambda p,\lambda)\frac{1}{\sqrt{E_{\Lambda p}}}.$$

Here, I have added a superscript $s$ to the matrix $C$ to emphasize that, since it rotates the $\lambda$ states into each other, it is a representation of $SU(2)$ that depends on the spin $s$ of the particle we're describing (that is, the particle we have chosen $a^\dagger$ to create).

This last equation gives the relationship between the particular representation under which the field transforms, and the spin $s$ of the particle it describes (I will try to edit my answer again later to give a specific example).


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