MSW Hamiltonian

1. Nov 21, 2014

Matterwave

Hi guys,

I was wondering if anyone knows of a good source/paper that I can find online that details how to derive the MSW Hamiltonian from a field-theoretic approach, or from an effective field theory approach or anything like that? Every time I study the MSW effect, the Hamiltonian $\sqrt{2}G_F n_e \left|\nu_e\right>\left<\nu_e\right|$ is just given. I know it comes from the electron neutrinos being the only flavor of neutrinos which have charged current interactions with electrons, while the neutral current interactions are flavor blind, but I don't know any more details other than that. Thanks.

2. Nov 21, 2014

ShayanJ

There is something in this which looks like a derivation!

3. Nov 21, 2014

Matterwave

That paper goes into a little bit more detail. But what I'm mostly confused on is how did you get $V_{CC}=\sqrt{2}G_F N_e$ etc., from the Feynman diagrams. In fact, I am confused on how to produce an "effective potential" $V_\alpha$ at all by using a Feynman diagram since I am only familiar with using Feynman diagrams to get cross-sections. I do not recall ever using a Feynman diagram to get $V$ in the Schroedinger equation! Ostensibly, the Feynman diagrams give you the interactions with the other background particles and can then be used to find a $V$, but I don't know how to do this. To be clear, I am not really looking to understand how to calculate the Feynman diagrams so much as how the process goes from "Feynman diagram -> cross section(?) -> V"?? I might just be missing something really simple...

I will take a look at the reference in that paper. Thanks. :)

Last edited: Nov 21, 2014
4. Nov 21, 2014

ShayanJ

Last edited by a moderator: May 7, 2017
5. Nov 21, 2014

Orodruin

Staff Emeritus
The MSW potential is due to coherent forward scattering so there is no cross section involved. The effect appears at amplitude level, which is why it is proportional to $G_F$ and not $G_F^2$ as a cross section would be. There are several ways of arguing here, and for me the most consistent one is to use finite temperature and density propagators (see Nötzold, Raffelt, Nucl. Phys. B 307 (1988) 924-936, Georg Raffelt has a pdf on his homepage http://wwwth.mpp.mpg.de/members/raffelt/). In that case, it is simply the FTD contribution to a loop diagram, which alters the dispersion relation for the neutrinos.

Another way of doing it is to work in Fermi theory (which is completely justified as you would need extreme neutrino energies to even be close to the W and Z resonances) and take an expectation value over the background electrons/quarks (although the quark contribution will cancel out with the NC contribution from electrons for electrically neutral matter). The result of this is an operator with the form $\bar\nu_e \gamma^0 \nu_e$, which is second order in the neutrino fields and thus alters the neutrino dispersion (of course, it is the same contribution as you get in FTD field theory). The $\gamma^0$ appears as the background violates Lorentz invariance (it really is $V_\mu \gamma^\mu$, where $V$ is the 4-velocity of the background.

Then there is of course also Wolfenstein's original paper (Phys.Rev. D17 (1978) 2369-2374), but I do not remember how much detail he really goes into.

6. Nov 22, 2014

Matterwave

Hmmm, this seems to be a bit above my understanding of QFT. I have only ever used Feynman diagrams to obtain cross sections...what do you mean with "the effect appears at amplitude level"? I would have thought the effect would be something akin to Coulomb scattering of two electrons in QED, except now you have one neutrino and one electron which scatter again into one neutrino and one electron...?

Neglecting that part of the puzzle for right now, let me confirm if I understand at least how to go from $V_\alpha$ as defined by $H_I\left|\nu_\alpha\right>=V_\alpha\left|\nu_\alpha\right>$ to obtaining the matrix representation of $H_I$ (c.f. Shyan's first posted paper, section 3.2). To get the matrix elements I just do $H_{I,\alpha\beta}=\left<\nu_\beta\right|H_I\left|\nu_\alpha\right>=\left<\nu_\beta\right|V_\alpha\left|\nu_\alpha\right>$ right? So this gives me the condition that since $H_I$ is Hermitian that $\left<\nu_\beta\right|V_\alpha\left|\nu_\alpha\right>^*=\left<\nu_\alpha\right|V_\beta\left|\nu_\beta\right>$ right? Lastly since $V_\alpha = \sqrt{2}G_F n_e \delta_{\alpha e}-\frac{\sqrt{2}}{2}G_F n_n$, we immediately get $H_{I,ee}=\sqrt{2}G_F n_e-\frac{\sqrt{2}}{2}G_F n_n$ and $H_{I,xx}=-\frac{\sqrt{2}}{2}G_F n_n$ where "x" represents either of the other 2 neutrino species. All of the off-diagonal terms are zero since $\left<\nu_e|\nu_x\right>=0$ and the $V_\alpha$ is simply a number and not an operator?

Where I'm trying to get with this is how to figure out when there will be off-diagonal terms in $H_I$. It looks like if the resulting $V_\alpha$ is not an operator which links different flavor states, then I won't have off-diagonal elements...? In other words, I need $V_\alpha$ to contain a term like $\left|\nu_e\right>\left<\nu_x\right|$? What kind of a Feynman diagram would give a $V_\alpha$ that looks like that? A diagram where an electron neutrino enters the vertex and then leaves as an "x" neutrino?

Really, I am asking these questions because I work with simulating neutrino coherent forward scattering on a background matter density + a neutrino background. However, up till now, I've only dealt with "given the Hamiltonian is this, figure out the evolution of the neutrino flavors", which is a problem I'm very familiar with. But at some point I will need to know "Given the interactions are this, figure out the effective Hamiltonian governing the evolution of neutrino flavors". In the coherent forward scattering case, the Hamiltonian for neutrino-neutrino interactions is given by a sum over all the neutrinos (neglecting anti-neutrinos): $H_I=\sqrt{2}G_F\sum_a (1-\cos\theta_a)n_{a}\left|\psi_a\right>\left<\psi_a\right|$ where $\left|\psi_a\right>$ is the flavor state of the neutrino $\nu_a$ and $\theta_a$ is the angle neutrino a makes with our test neutrino. In this case, there are definitely off-diagonal terms since the flavor state of neutrino a may not be a flavor eigen-state. I'm wondering how this came about. I was hoping that the MSW case would be a simple case from which I could build some intuition on how to go from "Feynman diagrams -> V(alpha) -> H_I", and then I could tackle this (what I assume to be) much more complicated case of neutrino-neutrino scattering.

Last edited: Nov 22, 2014
7. Nov 22, 2014

Orodruin

Staff Emeritus
This is the point, it is not a scatter and the neutrino does not change momentum. The effect is in the propagator and is more akin to the refractive index of light propagating in a medium. In an effective background approach, you essentially replace the electron fields $\bar e \gamma^\mu P_L e \to n_e V^\mu$ (perhaps some factor of 2 or so as well). This gives you an effective Lagrangian with an extra term which is second order in the neutrino fields. Just varying the resulting action to obtain the equations of motion should then net you an additional term proportional to $G_F n_e \gamma^0 \nu_e$ in the Dirac equation.

Alternatively, it is essentially a background contribution to the self-energy of the neutrinos computed to one-loop order.

Correct, the $V_\alpha$ are simply the eigenvalues of the interaction Hamiltonian.

Yes. Such phenomena can occur if you have a state with a coherent superposition of electrons and muons, which is difficult since they have very different masses and so normally these terms do not appear. Alternatively, they could appear due to flavour changing neutral currents or other non-standard interactions (there has been a significant amount of research on this topic). In addition, you could have such things in neutrino dense environments such as the interior of a supernova, where neutrinos could undergo coherent scattering on a neutrino background (and the neutrino background can very well be in a coherent superposition of $\nu_e$ and $\nu_x$). In fact, this makes the flavour propagation non-linear and this was a very popular field of study a few years back (it still is, but the hype has gone down as it is very difficult to make predictions).

Ok, I guess I should read the full message before starting to write a reply ...
The basic principle still holds, the neutrino background can be in a coherent superposition of different flavours, which leads to off-diagonal terms.

The Nötzold-Raffelt paper, references therein, and there-to should be very relevant to what you want to do.

8. Nov 22, 2014

Orodruin

Staff Emeritus
For a longer general treatment, the paper Raffelt, Sigl, General kinetic description of relativistic mixed neutrinos,
Nucl. Phys. B 406 (1993) 423-451, would also be relevant. It is also in Georg's homepage.

9. Nov 22, 2014

Matterwave

Hmmm, this scheme seems more complicated than I at first thought. I will look into the references and hopefully be able to figure out some details. Thanks!

10. Nov 22, 2014

Orodruin

Staff Emeritus
Well, the basic reason for why this happens is not that convoluted. In order to have coherent forward scattering, the in and out states of the spectator must be the same (including the same momentum). Thus, in order to have something off-diagonal in the effective potential, you need to have something in the background that is off-diagonal (as long as we stay in the SM). Off diagonal in flavour space means that the density matrix consists of a coherent superposition of different flavours. For charged leptons, this does not really happen due to the different masses (also, muons tend to be scarce and taus even more so). However, for neutrinos, it happens and the effective potential essentially becomes proportional to the flavour state of the background.

11. Nov 22, 2014

Matterwave

Don't you mean that in and out states have to have the same momentum, but can have different flavors? I thought that was how we got these off-diagonal terms...? I thought that's what it meant to have a diagram with a e-neutrino entering and a x-neutrino exiting.

12. Nov 22, 2014

Orodruin

Staff Emeritus
In order for the scattering to be coherent, the in and out states of the background must be the same (not only the momentum, but also the flavour state). This does not mean that the background must be flavour diagonal. The only thing is that for normal matter, the background is flavour diagonal, since you will not find many coherent superpositions of electrons and muons. If you have a neutrino background, it can be flavour off-diagonal and thus lead to an off-diagonal potential. If you have non-standard flavour changing interactions, this could also happen even if the background is flavour diagonal. The question you have to ask is: Can my interaction term change the neutrino flavour state without changing the background.

13. Nov 22, 2014

Matterwave

I was guessing that the Feynman diagram that does this is $\nu_e\nu_x\rightarrow_Z\nu_x\nu_e$. So you have a $\nu_x$ in the background of some arbitrary flavor-state (non-eigenstate) and your $\nu_e$ enters the vertex. Your $\nu_e$ takes on the flavor information of the background neutrino and the background neutrino takes on the flavor information of your $\nu_e$, this would be the "exchange diagram" between just $\nu_e\nu_x\rightarrow_Z \nu_e\nu_x$? But it seems the background neutrino has to become a flavor eigenstate for this to happen, so you are saying this is not a correct "coherent forward scattering" event?

14. Nov 22, 2014

Orodruin

Staff Emeritus
Exactly, it is not coherent, since you are changing the background state. If the background neutrino is not in a flavour eigenstate, you will get off-diagonal terms.

15. Nov 22, 2014

Matterwave

But then what does that Feynman diagram look like? The coherent forward scattering event that leads to off-diagonal terms in the Hamiltonian I mean. Or is there not a Feynman diagram for this process of coherent forward scattering?

16. Nov 22, 2014

Orodruin

Staff Emeritus
Just like that of any neutrino-neutrino scatter, you just have to sum coherently over all that contribute. If it helps, you can project out the off diagonal elements by assuming that the incoming propagating neutrino is of one flavour and the outgoing propagating neutrino of another.

17. Nov 23, 2014

Matterwave

I will definitely have to study this in more detail. Thanks for the sources and the info!