# Are terms with spinors analogous to $( \partial_\mu \Phi )(\partial^\mu \Phi)$ forbidden in the Lagrangian?

For scalar particles, the Lagrangian involves terms of the form $( \partial_\mu \Phi )(\partial^\mu \Phi)$, which is equivalent through integration by parts to $( \partial_\mu \partial^\mu \Phi )\Phi$. I was wondering if analogous terms for spinors are forbidden for some reasons and if not how they are interpreted? For example a term like:

$$\partial^{\dot{a}b} \Psi_{c} \partial_{\dot{a}b} \Psi^{c},$$

Some background:

It's possible to write four-vectors usign the spinor (Van-der-Waerden) notation:

$$v^{a \dot b} = v^\mu \sigma_\mu^{a \dot b} ,$$ where $v^\mu$ can be seen to transform like a four-vector.

Therefore, the usual derivation operator, is in the spinor formalism $$\partial^{a \dot b} = \partial^\mu \sigma_\mu^{a \dot b}$$

and Lorentz invariant terms in the Lagrangian involving first order derivatives are of the form:

$$\Psi_{\dot{a}} \partial^\mu (\sigma_{ \mu})^{\dot{a}b} \Psi_b = (\Psi_L)^{\dagger} \sigma^\mu \partial_\mu \Psi_L$$ and $$\Psi^{\dot{a}} \partial^\mu (\sigma_{ \mu})_{\dot{a}b} \Psi^{b} = (\Psi_R)^{\dagger} \partial^\mu \bar{\sigma}_\mu \Psi_R .$$

I was wondering if terms like

$$\partial^\mu (\sigma_{ \mu})^{\dot{a}b} \Psi_{c} \partial^\nu (\sigma_{ \nu})_{\dot{a}b} \Psi^{c},$$

which would be analogous to the term $( \partial_\mu \Phi )(\partial^\mu \Phi)$ in the scalar case, are forbidden for some reasons, and if not how they are interpreted?

• You had a really good question that was kind of spoiled by the last line, so I removed it for you. – David Z Nov 12 '14 at 15:24
• The term you proposed with spinors will have mass dimension 5, so is not renormalizable. In the light of modern understanding of renormalization, non-renormalizable terms are less of a taboo than they used to be, but if one is only concerned with low energy regime, they won't be very interesting. – Jia Yiyang Nov 12 '14 at 15:38
• @DavidZ: Of course, you make me look it up. – Nikolaj-K Nov 12 '14 at 15:46
• @JiaYiyang : why do you think that it wil have mass dimension 5? We may construct lagrangian with this term and set the dimension of $\psi$ as one. Then terms $\bar{\psi}\gamma^{\mu}\partial_{\mu}\psi$ must be multiplicated by $m$. – Andrew McAddams Nov 12 '14 at 15:51
• @JakobH, Maybe I don't fully understand your notation, but you proposed term doesn't even look lorentz invariant. For every 4-vector $\partial_\mu$ you would need two spinor fields $\Psi$ to make the invariance, while your term has two $\partial_\mu$'s but only two $\Psi$'s – Jia Yiyang Nov 12 '14 at 16:32

Maybe the correct answer is that we don't need to introduce it. Formally this term refers to the free lagrangian, while free lagrangian must produce the equation of motion which corresponds to irreducible representation of Poincare group with mass $m$ and spin $s$. For spinors corresponds to 1/2-spin field, Dirac operator implements irrep of Poincare group.

Well, it'd be non-renormalizable. Observe, the mass dimension of the kinetic term should be...dimensionless. So, for a partial derivative $\partial$, its mass dimension should be $[\partial]=1$. The differential should be the opposite of this, so the 4-volume should have its mass-dimension be $[\mathrm{d}^{4}x]=-4$. Hence the action for a massless fermionic field ("the kinetic part of the action") $$I_{\text{kinetic}}\sim\int\bar{\psi}\gamma^{\mu}\partial_{\mu}\psi\,\mathrm{d}^{4}x$$ should be dimensionless, implying $[\psi^{2}]+1-4=0$ or equivalently $[\psi]=3/2$.

Observe now that the mass dimension for your expression is $$[\int\psi\partial^{2}\psi\,\mathrm{d}^{4}x]=3+2-4=+1$$ which causes renormalizability problems. As to why this causes nonrenormalizability issues, John Baez has a web page dedicated to it.

• Doesn't one usually use $[\psi] = 3/2$? The kinetic energy is taken to be scale-invariant, not the density. – Siva Nov 12 '14 at 18:56
• Assuming $[d^4x] = -4$, we have $[\partial]=1$. We have $[\psi] = 3/2$, therefore $[\psi \partial^2 \psi] = 5$, and hence the coupling $[\lambda] = -1$. To construct a dimensionless parameter in scattering amplitudes, we must multiply it by a $[\Lambda] = 1$, and so as $\Lambda \to \infty$, the term diverges, and the theory is non-renormalizable. – JamalS Nov 12 '14 at 21:44
• @JamalS : we don't have $[\psi] = [m^{\frac{3}{2}}]$ until we don't require the dimensionless of action. But if $$L = \partial^{\mu}\psi^{\dot{a}}(\sigma_{\mu})_{b \dot{a}}\psi^{b} + a \psi^{\dot{a}}(\sigma_{\mu}\partial^{\mu})_{b \dot {a}}\psi^{b} + c...,$$ we can decide that $[\partial_{\mu}\psi^{\dot{a}}(\sigma_{\mu})_{b \dot{a}}\psi^{b}] = 1$, so $[\psi ] = [m]$. – Andrew McAddams Nov 12 '14 at 21:49
• @AndrewMcAddams: Well, of course we demand the action is dimensionless. Since $[\hbar] = [S]$, and we work in natural units, we must have $[S]=0$. – JamalS Nov 12 '14 at 22:03
• @Siva Good catch! – Alex Nelson Nov 12 '14 at 23:46

I stumbled upon a pdf, in another question here, where it is stated that a term of the form $( \partial_\mu \Phi )(\partial^\mu \Phi)$ is forbidden for spinors, because it leads to a hamiltonian that is unbounded from below.

I will update this answer as soon as I have investigated this any further