## March 8, 2012

### Daya Bay

Congratulations to the Daya Bay Experiment for their announcement of the first measurement of a non-zero value of the neutrino mixing angle, $\theta_{1,3}$ $\sin^2(2 \theta_{1,3}) = 0.092 \pm 0.017.$ which is $5.2\sigma$ away from zero. Since there was no good theoretical reason to expect this mixing angle to vanish, it’s pleasing to see the first definitive experimental results on its value.

One thing, though, that annoys me about the coverage that I’ve seen is the blithe assertion that neutrino masses (and mixing angles) are now “part of the Standard Model.” That is an incredibly dumb thing to say. Yes, it’s true that one can write down a dimension-5 operator, using only Standard Model fields, which gives neutrinos a mass (and which, if non-diagonal in the flavour eigenstate basis, leads to neutrino mixing, too). That operator looks like

(1)$\frac{c_{ij}}{M}\epsilon^{\alpha\beta} (h,\psi^i_\alpha)(h,\psi^j_\beta) +\text{h.c.}$

where $h$ is the Higgs doublet, $\psi^i$ are the lepton doublets, and $(\cdot,\cdot)$ is the skew-symmetric bilinear form on the fundamental representation of $SU(2)$. The complex symmetric $3\times3$ matrix, $c_{ij}$, is dimensionless. If we assume that its entries are $O(1)$, then the mass scale, $M$, suppressing this operator, is enormous, $M\sim 10^{15} \text{GeV}$. This operator is no more a “part of the Standard Model” than is the dimension-6 operator (also expressible, purely in terms of Standard Model fields) which mediates proton decay. Both are suppressed by GUT-scale masses, indicating that have to do with short-distance physics, well beyond the Standard Model’s $M_{EW}\sim 250 \text{GeV}$.

Posted by distler at March 8, 2012 11:41 PM

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### Re: Daya Bay

Nice to see your opinions on this matter.

Posted by: Plato on March 9, 2012 10:03 AM | Permalink | Reply to this

### Re: Daya Bay

It’s probably because my knowledge about physics is limited, but I do not get what you are saying here. From what I understand, assertions about neutrino (Dirac) masses and mixing angles being “now” part of the Standard Model refer to the fact that one can put into the usual SM Lagrangian electron-neutrino-Higgs terms which are exactly analogous to the well-known downquark-upquark-Higgs terms. In particular, they produce the Maki-Nakagawa-Sakata matrix in the same way as the quark terms produce the Cabibbo-Kobayashi-Maskawa matrix. But these terms are of dimension 4 (renormalizable): 2 fermion fields plus one Higgs field. Your highly suppressed dimension-5 term is not what the assertions you criticize are about, or is it? Do you have a reference or a link to a statement you find problematic in this respect? Can you explain the connection to the dimension-5 term in more detail?

Posted by: Cryoweta on March 18, 2012 12:40 AM | Permalink | Reply to this

### Re: Daya Bay

From what I understand, assertions about neutrino (Dirac) masses and mixing angles being “now” part of the Standard Model refer to the fact that one can put into the usual SM Lagrangian electron-neutrino-Higgs terms which are exactly analogous to the well-known downquark-upquark-Higgs terms.

No. That’s precisely the point. The “downquark-upquark-Higgs terms” are renormalizable dimension-4 operators, of the form $\epsilon^{\alpha\beta}\left( \lambda^{(d)}_{i j} (h^\dagger, q^i_\alpha)\tilde{d}^j_\beta +\lambda^{(u)}_{i j} (h,\q^i_\alpha) \tilde{u}^j_\beta \right) + \text{h.c.}$ where $q^i$ is the $SU(2)$ quark doublet and $\tilde{u}^i$, $\tilde{d}^i$ are the $SU(2)$ singlet quarks.

That’s not what gives mass to the neutrinos. Neutrino masses come from the dimension-5 operator, written above. Note that it’s quadratic in the Higgs field, not linear. That’s the lowest dimension operator that you can write down, involving only Standard Model fields, which gives mass to the neutrinos when the Higgs gets a VEV.

There is no dimension-4 operator with that property.

Posted by: Jacques Distler on March 18, 2012 2:19 AM | Permalink | PGP Sig | Reply to this

### Re: Daya Bay

I must be missing something. I meant the following Lagrangian: The fields are a lepton doublet $\ell$ (analogous to $q$), an electron singlet $e$ (analogous to $d$), a neutrino singlet $\nu$ (analogous to $u$), and the Higgs $h$. There are $3\times3$ matrices $\lambda^{(e)}$ and $\lambda^{(\nu)}$, analogous to $\lambda^{(d)}$ and $\lambda^{(u)}$. The Lagrangian is

$\epsilon^{\alpha\beta}\left(\lambda^{(e)}_{ij}(h^\dagger,\ell^i_\alpha)\tilde{e}^j_\beta +\lambda^{(\nu)}_{ij}(h,\ell^i_\alpha)\tilde{\nu}^j_\beta\right) +\text{h.c.}$

Of course the $SU(2)\times\text{U}(1)$ representations of the lepton fields $l,e,\nu$ are different from those of the analogous quark fields, but it is easy to check that the Lagrangian I wrote down is well-defined: the representations fit together with the Higgs representation.

The Lagrangian has dimension 4. It gives rise to masses and mixing angles in exactly the same way as the analogous quark Lagrangian. What’s wrong with it?

The (right-handed) neutrino singlet $\nu$ does not occur in the “old” SM, of course. If you do not allow $\nu$ to occur in the Lagrangian, then we agree at least on the technical side: there is no dimension-4 term with the required properties without $\nu$. In that case I would ask why you don’t allow right-handed neutrinos in the Lagrangian.

Posted by: Cryoweta on March 19, 2012 5:24 AM | Permalink | Reply to this

### Re: Daya Bay

I must be missing something. I meant the following Lagrangian: The fields are a lepton doublet $\ell$ (analogous to $q$), an electron singlet $e$ (analogous to $d$), a neutrino singlet $\nu$ (analogous to $u$) …

There is no such field in the Standard Model.

You can add more fields to the Standard Model — including, perhaps, a sterile neutrino, which is what your field, $\tilde{\nu}$, is conventionally called, except that you’ve decided to add three sterile neutrinos, rather than one (the number of $\tilde{\nu}$ species is logically independent of the number of $\ell$s.

With additional fields, then you are right that there are more operators that you can add to the Lagrangian, including the second term that you added above. But you are still wrong, because that’s not the only renormalizable term, involving the new fermions, that you can add. You can also add the dimension-3 operator

(1)$\tfrac{1}{2}M_{i j}\epsilon^{\alpha\beta}\tilde{\nu}^i_\alpha\tilde{\nu}^j_\beta + \text{h.c.}$

which gives a mass to $\tilde{\nu}$, even without electroweak symmetry-breaking.

The mass matrix for the neutrinos (let’s ignore all mixing, and write the formula for a single generation) looks like: $\begin{pmatrix}0& \lambda^{(\nu)}\langle h\rangle\\ \lambda^{(\nu)}\langle h\rangle & M \end{pmatrix}$ If it should happen that the eigenvalues of $M_{i j}$ are large, then the heavy mass eigenstate is mostly $\tilde{\nu}$. Integrating out the heavy fermion produces the dimension-5 operator that I wrote down. (This is called the “see-saw” mechanism.)

All of that is fine. But it is definitely “beyond the Standard Model.”

Posted by: Jacques Distler on March 19, 2012 8:18 AM | Permalink | PGP Sig | Reply to this

### Re: Daya Bay

You can add more fields to the Standard Model — including, perhaps, a sterile neutrino, which is what your field, $\tilde{\nu}$, is conventionally called,

I am not a physicist and therefore don’t speak the language fluently. I apologize for the inconvenience and appreciate when you point out the correct terminology.

except that you’ve decided to add three sterile neutrinos, rather than one (the number of $\tilde{\nu}$ species is logically independent of the number of $\ell$s.

Sure. It’s just a matter of elegance: the choice I wrote down makes the lepton terms as analogous as possible to the quark terms.

But you are still wrong, because that’s not the only renormalizable term, involving the new fermions, that you can add.

Until now, I hadn’t claimed explicitly that it’s the only such term, but you are right to read this claim between the lines.

You can also add the dimension-3 operator $\tfrac{1}{2}M_{ij}\epsilon^{\alpha\beta}\tilde{\nu}_\alpha^i\tilde{\nu}_\beta^j +\text{h.c.}$ which gives a mass to $\tilde{\nu}$, even without electroweak symmetry-breaking.

Yes, you can add this term, but it is zero (and therefore does not produce a mass). The field $\nu$ lives in a right-handed spinor representation, and a scalar product of two right-handed spinors is $0$. Otherwise already the “old” Standard Model would contain many additional terms like the one you wrote down.

To wrap up this discussion, I wanted to point out that there is a very natural and highly symmetric extension of the “old” Standard Model (by three sterile neutrinos) in which the neutrino masses and lepton mixing angles arise in exactly the same way as their quark analogues, via a renormalizable Lagrangian. I suspect (but might be wrong, not having seen the references) that some of the people you criticized in your original message talk about Lagrangians similar to mine, instead of the dimension-5 Lagrangian you wrote down.

Posted by: Cryoweta on March 19, 2012 9:34 PM | Permalink | Reply to this

### Re: Daya Bay

Yes, you can add this term, but it is zero (and therefore does not produce a mass).

No, that’s incorrect.

a scalar product of two right-handed spinors is 0.

Nonsense. The product is a Lorentz scalar.

All fermion mass terms involve such a skew-bilinear product of two same-chirality Weyl spinors.

Moreover, because the $\tilde{\nu}^a$ are fermions, the mass matrix, $M_{a b}$, is a symmetric matrix.

Otherwise already the “old” Standard Model would contain many additional terms like the one you wrote down.

No, that’s wrong.

$SU(3)\times SU(2)\times U(1)$ gauge invariance forbids all such terms, involving only the fields actually contained in the Standard Model.

To wrap up this discussion, I wanted to point out that there is a very natural and highly symmetric extension of the “old” Standard Model (by three sterile neutrinos)

Not exactly natural. While the Yukawa couplings of the quarks and leptons range from $O(1)$ to $O({10}^{-6})$, your “natural” Yukawa couplings are $O({10}^{-13})$. That’s pretty much the definition of “unnatural”

And hardly innocent. You are also predicting the existence of 3 new, light (nearly massless!) fermion species. Admittedly, they interact only very weakly (via Higgs exchange, with that ridiculously small Yukawa coupling) with ordinary matter, but still …

Posted by: Jacques Distler on March 19, 2012 10:40 PM | Permalink | PGP Sig | Reply to this

### Re: Daya Bay

1) The number of sterile neutrinos that you add to the Standard Model is totally up-for-grabs (modulo some constraints from cosmology). So you should really have written

(1)$\lambda^{(\nu)}_{i a}\epsilon^{\alpha\beta} (h,\ell^i_\alpha) \tilde{\nu}^a_\beta$

to emphasize that these “flavour” indices have a-priori different ranges. Similarly

(2)$\tfrac{1}{2} M_{a b} \epsilon^{\alpha\beta}\tilde{\nu}^a_\alpha \tilde{\nu}^b_\beta$

2) You could impose a $U(1)$ global symmetry to allow (1) but forbid (2). That’s probably the scenario you have in mind. If that’s the case, you still want to consider an arbitrary number, $n$, of sterile neutrino species ($n\geq2$, because we know the $\Delta m^2$ are non-zero).

Hopefully, it’s clear that postulating an arbitrary number of sterile neutrino species is a dramatic extension of the Standard Model.

Posted by: Jacques Distler on March 19, 2012 9:00 PM | Permalink | PGP Sig | Reply to this

### Re: Daya Bay

1) The number of sterile neutrinos that you add to the Standard Model is totally up-for-grabs (modulo some constraints from cosmology).

Of course very few options are ruled out by experiments, because neutrinos are hard to observe. In particular you can take Majorana or Dirac neutrinos, and an arbitrary number of sterile neutrinos. But I think in terms of elegance of the model, one option trumps all the other ones: just put in those right-handed neutrinos (Dirac representation, exactly three generations) that have been, quite artificially, omitted in the construction of the “old” Standard Model.

If all three neutrino generations had mass 0, then these right-handed fields would interact only gravitationally. This is what one gets by taking $\lambda^{(\nu)}_{ij}=0$ in the Lagrangian I wrote down; it is equivalent to the “old” Standard Model (up to nonrenormalizable effects produced by the field $\nu$). But for a general matrix $\lambda^{(\nu)}$, one gets neutrino masses and lepton mixing angles in the same way as for the quarks. The symmetry is striking, in my opinion (very similar to what happens in an $SU(5)$ theory). This is much more natural than an arbitrary “up-for-grabs” choice.

2) You could impose a $\text{U}(1)$ global symmetry to allow (1) but forbid (2). That’s probably the scenario you have in mind.

No, as I said in my other reply, I do not have to forbid (2) because the term vanishes automatically for the $\nu$ field I have in mind: right-handed spinor representation, trivial $\SU(3)\times\SU(2)\times\U(1)$ representation.

Posted by: Cryoweta on March 19, 2012 11:00 PM | Permalink | Reply to this

### Re: Daya Bay

No, as I said in my other reply, I do not have to forbid (2) because the term vanishes automatically for the $\nu$ field I have in mind: right-handed spinor representation, trivial $SU(3)\times SU(2)\times U(1)$ representation.

You can say that any number of times you want; it’s still wrong.

Posted by: Jacques Distler on March 19, 2012 11:07 PM | Permalink | PGP Sig | Reply to this

### Re: Daya Bay

I said that “a scalar product of two right-handed spinors is 0”. You replied: “Nonsense. The product is a Lorentz scalar.”

It’s a Lorentz scalar, namely the one which is zero.

To avoid any possible misunderstanding: The (indefinite) scalar product I am talking about is

$\langle\varphi,\psi\rangle = \bar{\varphi}\psi = \varphi^\dagger\gamma^0\psi$

(as e.g. on p. 43 in Peskin-Schroeder). This is essentially the only way to produce a number from two spinors $\varphi,\psi$ which is linear in $\psi$ and conjugate-linear in $\varphi$ and behaves nicely with respect to Clifford multiplication. It is this scalar product which shows up in all fermion terms of the Standard Model Lagrangian (because there is simply no alternative to this product). I assumed that when you wrote down the Standard Model term

$\epsilon^{\alpha\beta}\left(\lambda^{(d)}_{ij}(h^\dagger,q_\alpha^i)\tilde{d}_\beta^j +\lambda^{(u)}_{ij}(h,q_\alpha^i)\tilde{u}_\beta^j\right) +\text{h.c.},$

your notation $\epsilon^{\alpha\beta}\varphi_\alpha\psi_\beta$ stood for $\langle\varphi,\psi\rangle$ (and that you suppressed the contraction over the indices corresponding to the $\SU(3)$ part). If not, then please explain what it stands for and why you think that what you wrote down is a Standard Model term. Is the $\epsilon$ the same as in your dimension-3 pure-neutrino mass term? Your $\epsilon$ must be different from the one which appears e.g. in formula (20.101) in Peskin-Schroeder, because that one can appear only either in the $u$ or the $d$ term and is probably somehow implicit in your notation. (While we’re at it, why did you put the tildes over $d$ and $u$? I copied that when I wrote down the lepton analogue, but now it’s time to understand your notation in detail.)

I hope we agree that when the left-handed components of two spinors $\varphi,\psi$ are $0$, then $\langle\varphi,\psi\rangle = 0$:

$\begin{pmatrix} 0\\ \varphi_R\end{pmatrix}^\dagger \begin{pmatrix}0 &1 \\ 1&0\end{pmatrix} \begin{pmatrix}0\\ \psi_R\end{pmatrix} = 0 .$

You wrote:

All fermion mass terms involve such a skew-bilinear product of two same-chirality Weyl spinors.

But the quark mass term that you wrote down above contains a left-handed $q$ spinor and a right-handed $d$ resp. $u$ spinor. Please explain why this is not a counterexample to your claim here.

Then you said about terms analogous to your dimension-3 pure-neutrino term (e.g. with quarks instead of neutrinos):

$\SU(3)\times\SU(2)\times\U(1)$ gauge invariance forbids all such terms, involving only the fields actually contained in the Standard Model.

No, gauge invariance is not the reason. The terms we are talking about here are gauge-invariant, namely equal to 0. The reason is again that the chiralities of the two involved spinors fields are the same in each such term.

Consider for instance a hypothetical field $\hat{q}$ whose left- and right-handed spinor parts are both nonzero and which transforms like the Standard Model quark doublet under $\SU(3)\times\SU(2)\times\U(1)$. Then you can write down a dimension-3 mass term like your pure neutrino term (simply by contracting all indices using the standard scalar products), and this term is gauge-invariant; it is also nonzero, because both chiralities are involved. But from the chiral Standard Model fields, such a nonzero term cannot be constructed.

Posted by: Cryoweta on March 20, 2012 2:40 AM | Permalink | Reply to this

### Re: Daya Bay

said that “a scalar product of two right-handed spinors is 0”. You replied: “Nonsense. The product is a Lorentz scalar. It’s a Lorentz scalar, namely the one which is zero.

No. Sorry. You are wrong.

Since you said that you are not a physicist, I’m not sure what to do to correct your misapprehensions.

If you were a physicist, I would point you to the introductory chapters of Wess and Bagger, where they lay out, succinctly, but thoroughly, a set of conventions for 2-component Weyl spinor. There, you would learn that $\psi\psi \equiv \psi^\alpha\psi_\alpha \equiv - \epsilon^{\alpha\beta}\psi_\alpha\psi_\beta$

is

1. a Lorentz scalar
2. most definitely nonzero.

If you were a mathematician, I would point out that:

1. The connected component of $Spin(3,1)$ is isomorphic to $SL(2,\mathbb{C})$.
2. The irreducible left-handed spinor representation is the defining representation, $\mathbf{2}$ and right-handed spinor representation is its complex conjugate, $\overline{\mathbf{2}}$.
3. The anti-symmetric square of the left-handed spinor representation, $\wedge^2(\mathbf{2})$, is the trivial representation (i.e., a Lorentz scalar).

But since I don’t know what your background is, I am at a loss as to how to clear up your misapprehensions.

But the quark mass term that you wrote down above contains a left-handed q spinor and a right-handed d resp. u spinor. Please explain why this is not a counterexample to your claim here.

All of the spinors, in all of the formlæ that I wrote, are left-handed spinors. The terms involving right-handed spinors were only implicitly indicated in the “$\dots+\text{h.c.}$”.

By contrast, the only thing you can build out of a left-handed spinor and a right-handed spinor is a Lorentz-vector (i.e., the $\mathbf{2}\otimes\overline{\mathbf{2}}$ is the vector representation of $Spin(3,1)$).

But, given that you seem to be insistent on repeating flagrantly wrong statements, despite all of my warnings to the contrary, it’s not at all clear that any of the above explanations will do any good whatsoever.

Posted by: Jacques Distler on March 20, 2012 3:20 AM | Permalink | PGP Sig | Reply to this

### Re: Daya Bay

Not exactly natural. While the Yukawa couplings of the quarks and leptons range from $O(1)$ to $O(10^{-6})$, your “natural” Yukawa couplings are $O(10^{-13})$. That’s pretty much the definition of “unnatural”

And hardly innocent. You are also predicting the existence of 3 new, light (nearly massless!) fermion species. Admittedly, they interact only very weakly (via Higgs exchange, with that ridiculously small Yukawa coupling) with ordinary matter, but still …

Okay, the word “naturality” has a technical meaning in particle physics, which is obviously not the one I intended here. Let’s call my colloquial usage “natural$_{cryo}$”.

Your $O(10^{-6})$ vs. $O(10^{-13})$ objection is fair enough. But no one knows why the neutrino masses are as small as we observe them to be. The seesaw mechanism is certainly no complete explanation, it just connects the neutrino-mass hierarchy problem to the Higgs-mass hierarchy problem (sort of).

Of course I don’t have a good answer either. Which is hardly surprising, because, as I said, thinking about particle physics is not even what I do for a living. But now that we are having this discussion, here are my two cents:

I find it natural$_{cryo}$ (much more natural$_{cryo}$ than to postulate seesaw-coupled pairs, which would spoil the quark-lepton symmetry I mentioned before) to expect that when/if we some day understand how the values of the Yukawa couplings arise from short-scale physics, we will quantitatively understand why the coupling of a field which lives in a trivial representation of $\SU(3)\times\SU(2)\times\U(1)$ must be much smaller than the couplings of fields which live in more complicated representations. And why quark fields, which involve a nontrivial $\SU(3)$ representation, are heavier than the corresponding electron-type fields, which involve a trivial $\SU(3)$ representation. (Except possibly for the difference between the strange mass and the muon mass, whose sign is not so clear. And of course up to the caveat that quark masses are notoriously tricky to define.)

So, in a nutshell, my vague would-be explanation for the small neutrino masses is: the simpler the representation, the lighter the corresponding particles. (Yes, I know: $m_u \lt m_d$, but $m_c \gt m_s$ and $m_t \gt m_b$; thus the representation cannot be everything.)

P.S. I just saw your reply to my previous message. Let’s close the whole discussion, apparently some linguistic and/or mental barrier cannot be overcome here.

Posted by: Cryoweta on March 20, 2012 6:40 AM | Permalink | Reply to this

### Re: Daya Bay

So, in a nutshell, my vague would-be explanation for the small neutrino masses is …

I encourage you to pursue these speculations, and see if you can turn them into something less vague and hand-wavy.

However, I think it is abundantly clear that these speculations are not “part of the Standard Model”, as you originally asserted.

P.S. I just saw your reply to my previous message. Let’s close the whole discussion, apparently some linguistic and/or mental barrier cannot be overcome here.

I don’t think that our disagreement, as to what nonzero terms can appear in the Lagrangian, is a merely linguistic one. Hopefully, you now understand the point (an important one, for the subject at hand). But, in any case, I agree that there is probably little profit in continuing the discussion.

Posted by: Jacques Distler on March 21, 2012 11:20 AM | Permalink | PGP Sig | Reply to this

### Re: Daya Bay

I don’t think that our disagreement, as to what nonzero terms can appear in the Lagrangian, is a merely linguistic one.

You are right, it is not merely linguistic. So let’s continue. You said:

1. The connected component of $Spin(3,1)$ is isomorphic to $SL(2,\mathbb{C})$.
2. The irreducible left-handed spinor representation is the defining representation, $\mathbf{2}$ and right-handed spinor representation is its complex conjugate, $\widebar{\mathbf{2}}$.
3. The anti-symmetric square of the left-handed spinor representation, $\bigwedge^2(\mathbf{2})$, is the trivial representation (i.e., a Lorentz scalar).

(You wanted to say “… component of the identity”.) I know all that very well. But it doesn’t help at all to understand how your Lagrangian

(1)$L(\tilde{\nu}) = \tfrac{1}{2}M_{ij}\epsilon^{\alpha\beta}\tilde{\nu}_\alpha^i\tilde{\nu}_\beta^j +\text{h.c.}$

is defined. This Lagrangian is supposed to work for any number of particle generations, right? Then let’s take just one generation, and let’s choose $M_{11}=1$ for simplicity:

(2)$L(\tilde{\nu}) = Re\left( \epsilon^{\alpha\beta}\tilde{\nu}_\alpha\tilde{\nu}_\beta \right) .$

The indices $\alpha,\beta$ run from $1$ to $2$ (or don’t they?), because $\tilde{\nu}$ is just a map from Minkowski spacetime $\mathbb{R}^4$ to the left-handed representation space $\mathbb{C}^2$, with components $\tilde{\nu}_1$ and $\tilde{\nu}_2$ which are functions $\mathbb{R}^4\to\mathbb{C}$. Thus

(3)$L(\tilde{\nu}) = Re\left( \epsilon^{11}\tilde{\nu}_1\tilde{\nu}_1 +(\epsilon^{12} +\epsilon^{21})\tilde{\nu}_1\tilde{\nu}_2 +\epsilon^{22}\tilde{\nu}_2\tilde{\nu}_2 \right)$

(or how else should one interpret your notation?).

Now, what are $\epsilon^{11},\epsilon^{12},\epsilon^{21},\epsilon^{22}$? According to notational customs in physics, which seem to fit vaguely to your “anti-symmetric square” remarks, I would guess that $\epsilon^{11}=\epsilon^{22}=0$ and $\epsilon^{12} = -\epsilon^{21} = 1$. That would imply $L(\tilde{\nu})=0$. But you said:

If you were a physicist, I would point you to the introductory chapters of Wess and Bagger, where they lay out, succinctly, but thoroughly, a set of conventions for 2-component Weyl spinor. There, you would learn that [$\epsilon^{\alpha\beta}\psi_\alpha\psi_\beta$] is […] most definitely nonzero.

Wess and Bagger is not on my bookshelf. But you can easily help me out: just tell me what the four numbers (or functions? or what?) $\epsilon^{11},\epsilon^{12},\epsilon^{21},\epsilon^{22}$ are.

Posted by: Cryoweta on March 27, 2012 2:35 PM | Permalink | Reply to this

### Re: Daya Bay

Your equation (3) is completely wrong. $\epsilon^{1 2} = - \epsilon^{2 1} = 1$, with all other components vanishing. I assume you understand that $\nu_\alpha$ is a fermionic field. Thus $\epsilon^{\alpha\beta} \nu_\alpha\nu_\beta = \nu_1\nu_2 - \nu_2 \nu_1 = 2 \nu_1 \nu_2$ which is, of course, nonzero. The fact that it is also a Lorentz singlet is mere icing on the cake.

Wess and Bagger is not on my bookshelf.

Evidently, it should be.

Posted by: Jacques Distler on March 27, 2012 3:19 PM | Permalink | PGP Sig | Reply to this

### Re: Daya Bay

Your equation (3) is completely wrong. $\epsilon_{12} = −\epsilon_{21} = 1$, with all other components vanishing. I assume you understand that $\nu_\alpha$ is a fermionic field. Thus $\epsilon_{\alpha\beta}\nu_\alpha\nu_\beta = \nu_1\nu_2 -\nu_2\nu_1 = 2\nu_1\nu_2$ which is, of course, nonzero.

I do understand that $\nu\colon\mathbb{R}^4\to\mathbb{C}^2$ is a left-handed spinor field. Its components $\nu_1$ and $\nu_2$ are (as I said explicitly, and you did not object) functions with values in $\mathbb{C}$.

Thus, in my equation (3), $\nu_1$ and $\nu_2$ are complex-valued functions, and the multiplication I meant was clearly the usual commutative pointwise multiplication of complex-valued functions. Now you say that you want to multiply complex-valued functions in an anti-commutative way, that this is how your Lagrangian (2) should be interpreted. Okay, go ahead, tell me which anti-commutative multiplication we should use.

Let’s make it explicit. Our left-handed spinor field $\nu\colon\mathbb{R}^4\to\mathbb{C}^2$ might be given by

$\nu(x_0,x_1,x_2,x_3) = \begin{pmatrix} (3+\text{i})x_1^5 +\sin(x_2+5x_3)\\ 4x_0^3 -\cos(x_0-\text{i}x_3^2) \end{pmatrix}.$

Then how is the product $\cdot$ in

$\begin{split} L(\nu)(x) &= Re\left(2\nu_1(x)\cdot\nu_2(x)\right)\\ &= Re\Big( 2\left( (3+\text{i})x_1^5 +\sin(x_2+5x_3) \right) \cdot \left(4x_0^3 -\cos(x_0-\text{i}x_3^2)\right) \Big) \end{split}$

defined? Which real number is the value of $L(\nu)$ at the spacetime point $(5,6,7,8)$?

Posted by: Cryoweta on March 27, 2012 7:08 PM | Permalink | Reply to this

### Re: Daya Bay

Thus, in my equation (3), $\nu_1$ and $\nu_2$ are complex-valued functions, and the multiplication I meant was clearly the usual commutative pointwise multiplication of complex-valued functions. Now you say that you want to multiply complex-valued functions in an anti-commutative way, that this is how your Lagrangian (2) should be interpreted. Okay, go ahead, tell me which anti-commutative multiplication we should use.

Clearly, when you were citing passages from Peskin and Schroeder above, you didn’t understand a word that you were quoting, as they were using exactly the same anti-commutative multiplication law, for fermionic fields, that I am.

I’m not going to give a satisfactory rendition of the theory of supermanifolds in a blog comment (for that, you can read Manin’s book, or Deligne et al’s “Quantum Fields and Strings”). But, to leave out all the details, we are working in a world where we replace the commutative “ring of functions on $M$” with a $\mathbb{Z}_2$-graded ring.

Pragmatically (i.e., as physicists use them), we have the multiplication rule for fermionic functions, $\psi,\chi$, $\psi(x)\chi(y) = - \chi(y) \psi(x)$ and these form (in the obvious way) a module over the commutative ring of ordinary functions.

Seriously, if you are going to pontificate on extensions to the Standard Model, you really ought to learn what a fermion field is, beforehand.

And, while I like John Baez a lot, appealing to him as some kind of authority on particle physics is downright laughable (as he, himself, would be the first to admit).

Posted by: Jacques Distler on March 27, 2012 8:47 PM | Permalink | PGP Sig | Reply to this

### Re: Daya Bay

However, I think it is abundantly clear that these speculations are not “part of the Standard Model”, as you originally asserted.

Yes, it is clear that these speculations are not “part of the Standard Model”, and no, I did not “originally assert” otherwise. What I said was:

From what I understand, assertions about neutrino (Dirac) masses and mixing angles being “now” part of the Standard Model refer to the fact that one can put into the usual SM Lagrangian electron-neutrino-Higgs terms which are exactly analogous to the well-known downquark-upquark-Higgs terms.

My statement was only about the form of the Lagrangian of the Standard Model. I said nothing about explanations as to why certain parameters in this Lagrangian are large or small.

Here is an extremely close analogue: One can write down the Lagrangian of the “old” Standard Model and use it in computations. One might or might not want to speculate about why the Higgs mass is so far from the GUT scale in that model. Of course such speculations are not “part of the Standard Model”: every possible explanation goes necessarily beyond the Standard Model. The Standard Model works just fine for all practical purposes if we do not care about the Higgs-mass hierarchy “problem”. In exactly the same way one can write down a (renormalizable) Lagrangian including neutrino singlets, use it for theoretical predictions, and call the resulting theory the “new” Standard Model. And here as well one might or might not want to speculate why the neutrino masses are so small in that model, as I did in my previous message. Of course such speculations are not part of any Standard Model (old or new), and I never suggested otherwise.

Several of your comments in this thread sound as if you consider the model I wrote down quite far-fetched. Therefore I want to point out that this model is well-known, certainly not my invention, and that it was not my idea to call it the “new” Standard Model. I do not remember where and when I first learned about it, but a web search turns up several places where it is mentioned. Here is one by John Baez:

There are different theories to explain the masses and oscillations of neutrinos. The most conservative is the “New Standard Model”, a slight variant of the old textbook Standard Model of particle physics. The new version is actually prettier than the old one, because now the masses of leptons work just like the masses of quarks.

Like Baez, I regard the model as a “most conservative” extension of the Old Standard model (and at the same time as the “prettiest” one possible), whereas you think it is “hardly innocent” and belongs into a class of “dramatic extension”s. Has to do with the eye of the beholder, I assume.

Of course the Baez quote cannot be one of the statements you criticized in your original message, because it does not involve a nonrenormalizable Lagrangian. Although I asked you to provide an explicit reference to one of the statements you called “blithe” and “incredibly dumb”, you haven’t done that yet. Therefore I suspect that no suitable reference exists: you just knocked down a straw man.

It’s probably not a good idea of calling any model (in particle physics, cosmology or whatever) “the Standard Model”, because standards usually shift with time, and the resulting change of meaning of a technical term is bound to cause confusion. The Old Standard Model of particle physics is not the standard anymore, but no universally accepted replacement has emerged yet. When someone mentions the “New” Standard Model today (“new standard model” +neutrinos: more than 200000 Google hits), one should always ask which model precisely is meant. I explained in this thread the version I find most natural$_{cryo}$. But I suspect that almost no one who claims today that “neutrino masses (and mixing angles) are now part of the Standard Model” means a model which contains only the fields of the Old Standard Model. The assumption that some (still unidentified) person had made such a claim in the OSM context was the sole premise of your original message about neutrino masses resulting from nonrenormalizable Lagrangians.

By the way, if some reader who stumbles upon this discussion would like to contact me directly for some reason: you can do so at cryoweta “at” hotmail “dot” co “dot” uk.

Posted by: Cryoweta on March 27, 2012 7:40 PM | Permalink | Reply to this

### Re: Daya Bay

Several of your comments in this thread sound as if you consider the model I wrote down quite far-fetched.

Not far-fetched. Ruled out by experiment. The actual model you propose, with no $M_{i j} \epsilon^{\alpha\beta}\tilde{\nu}^i_\alpha\tilde{\nu}^j_\beta$ or $\tfrac{c_{i j}}{M}\epsilon^{\alpha\beta}(h,\ell^i_\alpha)(h,\ell^j_\beta)$ terms, is ruled out by solar neutrino oscillation experiments. Other, more sophisticated, models evade that problem, but are constrained by cosmological observations on BBN, CMB and large-scale structure.

Most of the reviews, I’ve seen, consider the possibility of one or two species of sterile neutrinos; I’m not sure whether 3 species (even with the more general collection of mass terms) is compatible with current observations.

In any case, it is pretty arrogant to call your own pet “theory” (even if it were not already ruled out by experiment), “The New Standard Model.”

There is no “New Standard Model”; nor is the phrase “Old Standard Model” used by any particle physicist I know of. There’s just the “Standard Model”, which has a very specific meaning in our community. And that meaning is not going to change just because you (I refuse to believe John Baez holds the erroneous opinions you ascribe to him) think it should.

Posted by: Jacques Distler on March 27, 2012 11:02 PM | Permalink | PGP Sig | Reply to this

### Re: Daya Bay

Pragmatically (i.e., as physicists use them), we have the multiplication rule for fermionic functions, $\psi,\chi$, $\psi(x)\chi(y) = -\chi(y)\psi(x)$ and these form (in the obvious way) a module over the commutative ring of ordinary functions. Seriously, if you are going to pontificate on extensions to the Standard Model, you really ought to learn what a fermion field is, beforehand.

I shake my head in disbelief and swallow down all the sarcastic replies that come to my mind. Do you understand the difference between classical and quantum fields? (Sadly, I have to ask this now.) Do you understand which of them occur as arguments of Lagrangians in QFT?

Obviously either Jacques Distler or I, one of us has talked blatant nonsense (on a technical question, with no room for interpretations anymore at this point!) for quite a while now. I urge all other readers (if any) of this message thread: Judge for yourself who that person is. Does Distler’s neutrino Lagrangian make sense? Does his anti-commutation formula $\nu_1\nu_2 = -\nu_2\nu_1$ make sense in this context? Don’t take anyone’s word for it, draw your own conclusions.

To Jacques Distler: My challenge stands. I gave you an explicit $\nu\colon\mathbb{R}^4\to\mathbb{C}^2$ and asked what the value of your $L(\nu)$ at the spacetime point $(5,6,7,8)$ is. It’s a real number. Write it down with three digits after the decimal point. Explain in detail how you computed it.

Only you can do that, because until now you are the only person (if there is any at all) who knows the definition of your $L(\nu)$.

The actual model you propose, with no $M_{ij}\epsilon^{\alpha\beta}\tilde{\nu}_ \alpha^i\tilde{\nu}_\beta^j$ or $c_{ij}M\epsilon^{\alpha\beta}(h,\ell_\alpha^i)(h,\ell_\beta^j)$ terms, is ruled out by solar neutrino oscillation experiments.

Interesting. Please provide a reference. Since it would be the first reference you give to back up your claims in this thread, I’m not really expecting anymore that you’ll do. But just in case you’ll surprise me in that respect, let me thank you in advance, and let me make a recommendation to other readers of this blog: check carefully to which extent the reference really supports the claim made above.

There is no “New Standard Model”; nor is the phrase “Old Standard Model” used by any particle physicist I know of.

That might be true: not by anyone you know of. Don’t blame me or the terminology for that. See page 20 of this or page 3 of this (this is from 2004, so the “old” SM has been old for quite some time!), or any other of the several thousand Google hits for “old Standard Model” +neutrino. In addition to more than 200000 Google hits, the term “New Standard Model” has apparently even a standard abbreviation by now: $\nu$SM. (I didn’t know that either until three days ago, but then, I’m not a physicist and don’t have a blog where I accuse other physicists of saying “incredibly dumb” things about neutrinos.)

Now I’m leaving the discussion, this time for good (except possibly if you give a detailed answer to my challenge above, but I’m not holding my breath). Mr Distler, you have the last word.

Posted by: Cryoweta on March 28, 2012 5:57 AM | Permalink | Reply to this

### Re: Daya Bay

Do you understand the difference between classical and quantum fields?

Obviously, I do. And, given that your very next paragraph contains:

To Jacques Distler: My challenge stands. I gave you an explicit $\nu:\mathbb{R}^4 \to \mathbb{C}^2$ and asked what the value of your $L(\nu)$ at the spacetime point $(5,6,7,8)$ is. It’s a real number.

obviously, you don’t.

I’m sorry that you haven’t taken this opportunity to realize that there are some serious gaps in your understanding; gaps that you ought to endeavour to fill, before you embarrass yourself further.

I’ve tried to help, but you’re not going to learn quantum field theory in a blog comments section.

Posted by: Jacques Distler on March 28, 2012 8:09 AM | Permalink | PGP Sig | Reply to this

### fermions are super odd

I am late to this discussion, but I just thought somebody should amplify on Jacques’ response to the puzzlement expressed by Cryoweta in

we have the multiplication rule for fermionic functions, $\psi$, $\chi$, $\psi(x)\chi(y) = - \chi(y)\psi(x)$

[…] Do you understand the difference between classical and quantum fields?

in case Cryoweta or (after his farewell maybe now more likely) some other reader comes back to this, wondering:

anti-commutativity is not a hallmark of quantum operators, as the above question suggests, but of classical, albeit super-geometric observables. Instead, the quantization of a classical superalgebra (a Grassmann algebra) as above is the corresponding Clifford algebra, which is non-anti-commuting (in general), as befits a quantum (super-)algebra.

On the other hand, I think it is fair to say that many standard textbooks do not dicuss this well. A glorious exception is Quantum Fields and Strings where the single page 196 (first volume) – which starts out with the very sentence

A spinor field is a map $\psi : M \to \Pi S$

($M$ is spacetime, $\Pi$ is the shift operator that makes the spinor bundle $S$ super odd)

should serve to entirely clear up the cause of Cryoweta’s (or any other reader’s) puzzlements here.

Maybe one general problem with the textbook situation is that supergeometry is often regarded as a tool relevant only for supersymmetric QFT, while in fact it is the geometry of any world containing fermions.

Posted by: Urs Schreiber on April 13, 2012 1:22 PM | Permalink | Reply to this

### Re: fermions are super odd

A glorious exception is Quantum Fields and Strings where the single page 196 (first volume) – which starts out with the very sentence

A spinor field is a map $\psi: M\to \Pi S$

($M$ is spacetime, $\Pi$ is the shift operator that makes the spinor bundle $S$ super odd).

I agree with the sentiment, but the sentence is inadequate, for a number of reasons.

1. First, it should read $\psi$ is a section of the bundle $\Pi S\to M$.
2. Second, even that statement is inadequate. Locally, such a section is just a map $\mathbb{R}^4\to \Pi (\mathbb{C}^2)$. Unfortunately, interpreted straighforwardly, the space of such maps (itself a supermanifold) has way too few nilpotents. To get a correct statement, you need to work in families $\mathcal{M}\underset{M}{\to}B$, where the base of the family provides an (infinite!) reservoir of nilpotents. Then you can write a sentence like:

$\psi$ is a section of $\Pi \mathcal{S}\to\mathcal{M}$.

But this was not a sentence that Cryoweta (an engineer, I believe, not –as I had first suspected – a mathematician) was going to be able to make heads or tails of.

Anyway, the standard textbooks make perfectly clear what an “anti-commuting field” is, even if they do not provide the proper mathematical context for such.

Posted by: Jacques Distler on April 13, 2012 2:30 PM | Permalink | PGP Sig | Reply to this

### Re: fermions are super odd

it should read $\psi$ is a section of the bundle $\Pi S \to M$.

Where the sentence comes from $S$ is not the bundle, but the fiber.

you need to work in families $\mathcal{M} \to B$, where the base of the family provides an (infinite!) reservoir of nilpotents.

I find it more natural to think in the topos over supermanifolds. There we can talk about an element of the internal hom (at some stage of definition) just as one does naively, and it is still correct.

Anyway, the perfect textbook on these matters probably still needs to be written, but that section I pointed to is at least a serious attempt to tell the full story, and make sure that the reader does not miss the crucial point of supergeometry.

Posted by: Urs Schreiber on April 14, 2012 4:23 PM | Permalink | Reply to this

### Re: fermions are super odd

I find it more natural to think in the topos over supermanifolds. There we can talk about an element of the internal hom (at some stage of definition) just as one does naively, and it is still correct.

My knowledge of topos-theory is pretty much non-existent, but that sounds interesting. What does the phrase “over supermanifolds” (plural) mean?

Posted by: Jacques Distler on April 15, 2012 12:47 AM | Permalink | PGP Sig | Reply to this

### Re: Daya Bay

This is probably the most entertaining argument I’ve ever read on a physics blog, and I’ve seen some of the arguments between you and Smolin (and Motl)!

Posted by: Anon on May 22, 2013 6:59 PM | Permalink | Reply to this

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