When constructing the Lagrangian for a two-component left-handed Weyl field $\psi$, in e.g. Srednicki, one rejects the choice of $\partial^\mu \psi \partial_\mu \psi+\partial^\mu \psi^\dagger \partial_\mu \psi^\dagger$ because it will lead to a Hamiltonian that is unbounded below: the Hamiltonian will look like $\dot{\psi} \dot{\psi}+\dot{\psi}^\dagger \dot{\psi}^\dagger+\partial_i \psi \partial_i \psi+\partial_i \psi^\dagger \partial_i \psi^\dagger$. I struggle to see that this is unbounded below. I usually try to understand Lagrangians in terms of classical fields to check for boundedness issues, but any of my attempts to look at complex-number-valued fields are foiled by the fact that we need Grassmann-valued numbers to write a nontrivial kinetic term of the form above, as we're using the 2x2 antisymmetric invariant symbol $\epsilon$ to tie up the indices.

Thus we're led to consider a kinetic term involving both $\psi$ and $\psi^\dagger$. However, for the sake of Lorentz covariance of the Lagrangian density, we can't use the 2x2 antisymmetric invariant symbol $\epsilon$ anymore to couple the fields, since $\psi^\dagger$ is in the complex conjugate representation. This requires the use of $\bar{\sigma}^\mu$ to get a Lorentz covariant Lagrangian. However, the resulting kinetic term of $i\psi^\dagger \bar{\sigma}^\mu \partial_\mu \psi$ is hard for me to understand in terms of whether the Hamiltonian is bounded below.

How can I tell that for a two-component anticommuting Weyl field $\psi$ that a Lagrangian containing $\partial^\mu \psi \partial_\mu \psi+\partial^\mu \psi^\dagger \partial_\mu \psi^\dagger$ gives an unbounded-below Hamiltonian, but a Lagrangian containing $i\psi^\dagger \bar{\sigma}^\mu \partial_\mu \psi$ gives me a Hamiltonian that is bounded below?

  • $\begingroup$ Isn't $\psi\psi$ not Lorentz invariant in the first place since both transform under the same spinor representation? $\endgroup$ Jul 20, 2021 at 17:09
  • 1
    $\begingroup$ @RichardMyers I think the use of the antisymmetric $2$x$2$ $\epsilon$ should make $\psi \psi = \psi_a \psi_b \epsilon^{ab}$ Lorentz invariant. $\endgroup$
    – user196574
    Jul 20, 2021 at 17:43

1 Answer 1


The Weyl field $\psi^{A}$ with $A=1,2$ is an irrep of SL(2,C) which is the double cover of the Lorentz group SO(1,3). The relation between a contravariant 4-vector $x^{\mu}$ (transforming under SO(1,3)) and a Hermitian matrix $X^{\dot{A}}_{\ B}$ (transforming under SL(2,C)) is, $X^{\dot{A}}_{\ B}=[\sigma_{\mu}]^{\dot{A}}_{\ B}x^{\mu}$ where Van der Waerden dotted spinor notation is being used and the $[\sigma_{\mu}]^{\dot{A}}_{\ B}$ are the Pauli matrices. In order to construct an action for the Weyl field, we need a momentum operator. Since the 4-momentum operator is $\hat{p}_{\mu}=i\partial/\partial x^{\mu}$, the $SL(2,C)$ momentum operator must be, $\hat{p}^{\dot{A}}_{\ B}=[\sigma_{\mu}]^{\dot{A}}_{\ B}\eta^{\mu\lambda}i\frac{\partial}{\partial x^{\lambda}}$. The obvious $SL(2,C)$ invariant action is now $S=\int d^{4}x\ \psi^{*}_{\ \dot{A}}\hat{p}^{\dot{A}}_{\ B}\psi^{B}$. I think the first action in the question fails because it does not incorporate a SL(2,C) momentum operator. In order to show that the correct action leads to a positive energy ground state, we bring in the canonically conjugate momentum field $\pi_{A}=\psi^{*}_{\dot{B}}i[\sigma_{0}]^{\dot{B}}_{\ A}$. The Hamiltonian is now $H=\int d^{3}x\ \pi_{C}[\sigma_{0}^{-1}\sigma_{r}]^{C}_{\ B}\partial\psi^{B}/\partial x^{r}$ where lower case Latin indices $r=1,2,3$. The E-L or Hamiltonian EoM is $\hat{p}^{\dot{A}}_{\ B}\psi^{B}=0$. The plane wave solutions are constructed using the ansatz $\psi^{A}(x^{\mu})=\psi^{A}(p^{r})\exp(-ip_{\mu}x^{\mu})$ where the momentum of the wave is $p_{\mu}=(p^{0},-p^{r})$. The EoM implies $\psi^{1}=(p^{1}-ip^{2})c$ and $\psi^{2}=-(p^{0}+p^{3})c$ where $c$ is an arbitrary complex number and also there is a dispersion relation $p^{0}=\pm\sqrt{p^{r}p^{r}}$. The plane wave solutions are eigenstates of the helicity operator $[\sigma_{r}]^{\dot{A}}_{\ B}\hat{p}^{r}/(2|\hat{p}|)$ with helicity eigenvalues $\lambda=\pm1/2$. Let $\omega=+\sqrt{p^{r}p^{r}}$ be the positive frequency. The helicity eigenstates are, \begin{equation} \psi^{A}_{\lambda=-1/2}=c \left[ \begin{array}{c} (p^{1}-ip^{2})\\ -(\omega+p^{3}) \end{array} \right]\exp(-i(\omega t-p^{r}x^{r})); \ \psi^{A}_{\lambda=+1/2}=c \left[ \begin{array}{c} (p^{1}-ip^{2})\\ (\omega-p^{3}) \end{array} \right]\exp(i(\omega t+p^{r}x^{r})) \end{equation} The most general Weyl field is an integral over 3-momentum of these helicity eigenstates. The general Weyl field can be put into the form, \begin{equation} \psi^{A}(x^{\mu})=\int\ \frac{d^{3}p}{(2\pi)^{3/2}}\left\{c_{\lambda=-1/2}(p)\psi^{A}(p)\exp(-ip_{\mu}x^{\mu})-c_{\lambda=+1/2}(-p)\psi^{A}(p)\exp(ip_{\mu}x^{\mu})\right\}|_{p^{0}=\omega} \end{equation} where $\psi^{1}(p)=p^{1}-ip^{2}$ and $\psi^{2}(p)=-(p^{0}+p^{3})$. We substitute this expression for the Weyl field into the Hamiltonian and obtain, \begin{equation} H=\int\ d^{3}p\ 2(p^{0})^{2}(p^{0}+p^{3})\left\{c^{*}_{-1/2}(p)c_{-1/2}(p)-c^{*}_{1/2}(-p)c_{1/2}(-p)\right\}|_{p^{0}=\omega} \end{equation} In order to get this result it was necessary to use the two spin sums, \begin{equation} \psi^{*}_{\dot{A}}(p)p^{r}[\sigma_{r}]^{\dot{A}}_{\ B}\psi^{B}(p)=-2(p^{0})^{2}(p^{0}+p^{3});\ \psi^{*}_{\dot{A}}(-p)p^{r}[\sigma_{r}]^{\dot{A}}_{\ B}\psi^{B}(p)=0 \end{equation} The next step is to quantize the Weyl field. The coefficients $c_{\lambda}(p)$ must be promoted to operators $\hat{c}_{\lambda}(p)$ because emission and absorption operators do not carry spinor indices; they only carry the labels of the single particle states $|p,\lambda\rangle$. We take the FT of the classical field $\psi^{A}(x^{\mu})$ and, using the spin sums, get coefficients $c_{1/2}(p)\exp(i\omega t)$ and $c_{-1/2}(p)\exp(-i\omega t)$ as spatial integrals of the Wey field $\psi^{A}(x^{\mu})$. These formulae are omitted. Now, in the classical Hamiltonian field theory, we have the fundamental Poisson Brackets (PBs) $[\psi^{A}(t,x^{r}),\pi_{B}(t,x'^{r})]_{PB}=\delta^{A}_{B}\delta^{3}(x^{r}-x'^{r})$ etc. Using the expressions for the $c_{\lambda}(p)$ in terms of spatial integrals of $\psi^{a}(t,x^{r})$, a laborious calculation gives the PBs of the $c_{\lambda}(p)$. The two non-zero ones are, \begin{equation} [c_{-1/2}(p)\exp(-i\omega t),c^{*}_{-1/2}(p')\exp(i\omega't)]_{PB}=\frac{-i\delta^{3}(p-p')}{2\omega(\omega+p^{3})}\\ [c_{+1/2}(p)\exp(i\omega t),c^{*}_{+1/2}(p')\exp(-i\omega't)]_{PB}=\frac{-i\delta^{3}(p-p')}{2\omega(\omega-p^{3})} \end{equation} In order to express the coefficients in terms of emission and absorption operators $\hat{\eta}$ and $\hat{\eta}^{\dagger}$ respectively, we'll use the Lorentz invariant (Haar) measure in the resolution of unity for the single particle states. \begin{equation} \sum_{\lambda}\int\ \frac{d^{3}p}{2p^{0}}|p,\lambda\rangle\langle p,\lambda|=1 \end{equation} The commutators or anti-commutators for the emission and absorption operators are now, \begin{equation} [\hat{\eta}^{\dagger}_{p',\lambda '},\hat{\eta}_{p,\lambda}]_{\pm}=\langle p',\lambda '|p,\lambda\rangle=2p^{0}\delta_{\lambda,\lambda'}\delta^{3}(p-p') \end{equation} If we now promote the PBs for the coefficients to commutators or anticommutators according to Dirac's quantization rule, we can infer that the relations between the coefficients and the emission and absorption operators must be, \begin{equation} \eta_{p,-1/2}=c^{*}_{-1/2}(p)\exp(i\omega t)(\omega+p^{3})^{1/2}(2\omega)\\ \eta_{p,+1/2}=c^{*}_{+1/2}(p)\exp(-i\omega t)(\omega-p^{3})^{1/2}(2\omega) \end{equation} Upon substituting for the coefficients, the Hamiltonian for the quantized Weyl field is, \begin{equation} \hat{H}=(1/2)\int\ d^{3}p(\hat{\eta}_{p,-1/2}\hat{\eta}^{\dagger}_{p,-1/2}-\hat{\eta}_{-p,1/2}\hat{\eta}^{\dagger}_{-p,1/2}) \end{equation} Let's assume that the emission and absorption operators obey the commutation rule. The single particle state $\hat{\eta}_{p',1/2}|0\rangle$ is an eigenstate of the Hamiltonian with eigenvalue $-\omega'$. The minus sign in the Hamiltonian means that multiparticle states with helicity 1/2 have negative energy and so the vacuum is unstable. The fix is to assume that the emission and absorption operators obey the anticommutator rule. This removes the minus sign but brings in an infinite constant energy which is ignored. The resulting normally ordered Hamiltonian is, \begin{equation} \hat{H}=(1/2)\int\ d^{3}p(\hat{\eta}_{p,-1/2}\hat{\eta}^{\dagger}_{p,-1/2}+\hat{\eta}^{\dagger}_{-p,1/2}\hat{\eta}_{-p,1/2})-\infty \end{equation} This has positive eigenvalues so the vacuum is stable.

Edit : Why the first action fails

Within the framework of canonical quantization, one first works out the classical Hamiltonian field theory and then promotes the coefficients that don't have spinor labels to emission and absorption operators by working out their Poisson brackets and replacing the PBs by commutators or anti-commutators in order to stabilize the vacuum. This programme does not even start with the first action because the term, \begin{equation} \eta^{\mu\lambda}\frac{\partial \psi_{B}}{\partial x^{\mu}}\frac{\partial \psi^{B}}{\partial x^{\lambda}}= \epsilon_{AB}\eta^{\mu\lambda}\frac{\partial \psi^{A}}{\partial x^{\mu}}\frac{\partial \psi^{B}}{\partial x^{\lambda}} \end{equation} is zero by anti-symmetry of the Levi-Civita tensor. Now, I know that the QFT books get around this by saying that the classical fields are not complex numbers but are classical Grassmann numbers which anti-commute, but this is a level of hand-waving that I don't accept. In support of my view, if one tries to write an action for a Majorana field in terms of Weyl spinors, the mass terms also vanish by antisymmetry. Peskin and Schroeder argue that the fix is to make the classical Majorana field Grassmann odd, however, I tend to the view that the mathematics is trying to tell us that Majorana fields do not exist in nature.

  • $\begingroup$ Thank you for the thorough derivation that the Weyl Hamiltonian is bounded below. Could you please expand a little more on why the first action fails? It appears to be $SL(2,C)$ invariant to me, $\epsilon^{ab} \eta^{\mu \nu} \partial_\mu \psi_a \partial_\nu \psi_b$, and I think if desired the $\partial_\mu$ can be expressed in terms of the $p^\dot{A}_B$ and appropriate invariant symbols. $\endgroup$
    – user196574
    Jul 27, 2021 at 2:36
  • $\begingroup$ @user196574 : I've edited my answer to elaborate on why the first action fails. $\endgroup$
    – user7154
    Jul 27, 2021 at 10:32

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