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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?

jak
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    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
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    @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
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    @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
  • @JiaYiyang this is exactly the kind of thing I was hoping to hear that I missed. Why do you think that $\partial^{\dot{a}b} \Psi_{c} \partial_{\dot{a}b} \Psi^{c}$ isn't Lorentz invariant? I'm not sure why two spinor fields are needed in order to make the term invariant... The term has no free indices and therefore I thought it is invariant. The term can be rewritten: $ \partial^\mu \sigma_\mu^{a \dot b} \Psi_{c} \partial^\mu (\sigma_\mu)_{a \dot b} \Psi^{c}$, which looks Lorentz invariant to me. – jak Nov 12 '14 at 16:44
  • @JakobH, I'm not entirely sure about your notations, but if you $\sigma_\mu$ has a similar role as Dirac's $\gamma_\mu$, then you can think about why $\bar{\psi}\gamma^{\mu}\partial_{\mu}\psi$ is invariant. Note $\gamma_\mu$ are just coefficients, they don't really transform upon a spacetime transformation, only $\psi$ transforms. What is going on is each $\psi$ , upon lorentz transformation, contribute a matrix transformation $S$ in spinor rep, and because of the algebraic fact $S^{-1}\gamma^\mu S=(\Lambda^{-1})^{\mu}_{\nu}\gamma^\nu$, it cancels the – Jia Yiyang Nov 12 '14 at 18:13
  • vector transformation induced by $\partial_\mu$. As you see, we must be able to sandwich the gamma matrix to induce a vector-like transformation, so for each $\partial_\mu$, we need two spinor fields. – Jia Yiyang Nov 12 '14 at 18:15
  • BTW, this is also the reason why people sometimes say "a spinor is half of a vector." – Jia Yiyang Nov 12 '14 at 18:21
  • Sorry I made a mistake, exactly because of the reasoning I mentioned above, it is not about how many spinors you have in the term, it is about how many $\gamma_\mu$ you can sandwich, e.g. $\bar{\psi}\gamma^{\mu}\gamma^{\nu}\psi$ will transform like a rank-2 tensor. I had the instinctual impression that your term isn't invariant because one of your matrix is put in front, not sandwiched. But again it could be my misunderstanding of your notation, in that case I should retreat to my earlier statement about renormalizability. – Jia Yiyang Nov 13 '14 at 04:46
  • Regarding renormalizability @AndrewMcAddams raised a nontrivial question, but at least let's say such term can't coexist with a conventional Dirac fermion. – Jia Yiyang Nov 13 '14 at 04:51
  • Related to https://physics.stackexchange.com/questions/519110/can-mass-dimension-one-fermion-be-a-dark-matter-candidate. – MadMax Feb 06 '20 at 21:22

3 Answers3

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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

Community
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jak
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  • The pdf link is dead but the same claim is made in Srednicki. A good convincing answer showing the claim is true is still missing though despite there being 3 different duplicate questions on the subject (see also https://physics.stackexchange.com/questions/283810/a-hamiltonian-for-a-left-handed-spinor-field and https://physics.stackexchange.com/questions/584727/why-the-terms-like-partial-mu-psi-partial-mu-psi-h-c-cannot-be-included). – Marten Dec 01 '23 at 16:13
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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.

Andrew McAddams
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  • This is wrong. Anything that would be allowed is within the space of theories we should consider. This theory we cannot consider because it is (apparently) unbounded from below. It is not just that we don't need it to describe the physics that has been realized in nature. There simply isn't any consistent predictive theory that includes this term. – Marten Dec 01 '23 at 16:01
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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.

Alex Nelson
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    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
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    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
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    To me, the crux of the question is why can we not take ${(\partial \psi)}^2$ as the kinetic term instead of the usual one, and have $[\psi] = 1$. With regards to that, it seems that due to the extra derivative, this term is always going to have a larger mass scaling dimension than the usual kinetic term, so might not survive in the IR. I wonder if such a term is common in EFTs closer to the cutoff. – Siva Nov 13 '14 at 00:18
  • JamalS : here are some mistakes: I've meaned $$ L = \partial_{\mu}(\sigma^{\mu}){a\dot{b}}\psi^{\dot{b}}\partial{\alpha}(\tilde {\sigma}^{\alpha})^{\dot{c}a}\psi_{\dot{c}} + a... + h.c. $$ From the requirement that the first summand must have dimension $m^{4}$ we have $[\psi ] = [m]$. – Andrew McAddams Nov 13 '14 at 04:30
  • This doesn't answer the question. The question is why this term is forbidden. Not only would it seem from this analysis that we could use this term instead of the other kinetic term it would also look like we could consider it as an irrelevant term. The claim is much stronger. This term is not allowed at all because it makes the action unbounded from below. So that is what an answer would need to show. – Marten Dec 01 '23 at 15:59