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Fernando Garcia

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Some derivations for in-medium neutrino mixing

January 27th, 2024.

Introduction

The mixing of Neutrinos plays a central role in my current research. In this entry I will derive some equations that aren't used verbatim in my research, but rather set the stage for me to work on.

I assume that the reader is familiar with the distinction between flavor Neutrinos and mass Neutrinos, as well as the general idea behind vacuum neutrino oscillations.

Neutrinos swimming in matter: In-medium mixing angle

Let's consider a model with neutrino flavors: electron and muon. Since the neutral weak current has the same effect on both flavors, we don't expect it to make a difference and thus is ignored. The charged weak interaction, on the other hand, will act on the electron flavor neutrino but not on the muon flavor.

Starting with the weak effective Hamiltonian:

$$H=\frac{1}{\sqrt{2}}{G}_F(J_{\text{c}}^{\mu }{J}_{\text{c}\mu }^{†}+J_{\text{N}}^{\mu }{J}_{\text{N}\mu })$$

The second term (involving the neutral current) vanishes for our purposes. We now recall that the charged current has the form:

$$J_{\text{c}}^{\mu }=\text{Quark terms}+\left(\begin{array}{ccc}\bar{e}& \bar{\mu }& \bar{\tau }\end{array}\right){\gamma }^{\mu }(1-{\gamma }^5)\left(\begin{array}{c}{\nu }_e\\ {\nu }_{\mu }\\ {\nu }_{\tau }\end{array}\right)$$

Performing the contraction on $\mu$ (on the Hamiltonian), while keeping only the electron terms, gives: $$H_{\text{int}}=\frac{1}{\sqrt{2}}{G}_F\bar{e}{\gamma }^{\mu }(1-{\gamma }^5){\nu }_e\overline{{\nu }_e}{\gamma }_{\mu }(1-{\gamma }^5)e$$

(Using Fierz) Let's rearrange the above expression to look like:

$$H_{\text{int}}=\frac{1}{\sqrt{2}}{G}_F\overline{{\nu }_e}{\gamma }^{\mu }(1-{\gamma }^5){\nu }_e\bar{e}{\gamma }_{\mu }(1-{\gamma }^5)e$$

It can then be argued that (having electrons at rest)

$$\bar{e}{\gamma }_{\mu }(1-{\gamma }^5)e={\delta }_{\mu 0}{N}_e$$

Where $N_e$ denotes the number density of electrons. Thus:

$$H_{\text{int}}\to G_F\sqrt{2}{N}_e$$

That is, the medium acts as (through a CC interaction) an additional potential:

$$V=G_F\sqrt{2}{N}_e$$

In the flavor picture, we initially have $H_{\alpha }=U^{†}{H}_{0}U$, but we the added potential, we will now work with the Hamiltonian:

$$H_{\alpha }^{\prime }=U{H}_0{U}^{†}+H_{\text{int}}$$

Where $H_0$ is the Hamiltonian (to describe motion) in the mass picture:

$$i\frac{d}{dt}|{\nu }_i⟩=H_0|{\nu }_i⟩$$

And $U$ (along with $U^{†}$) is the diagonalization matrix used to go from mass states to flavor states (and vice versa).

$$U=\left(\begin{array}{cc}{C}_{\theta }& S_{\theta }\\ -S_{\theta }& C_{\theta }\end{array}\right)U^{†}=\left(\begin{array}{cc}{C}_{\theta }& -S_{\theta }\\ S_{\theta }& C_{\theta }\end{array}\right)$$

This means that in the flavor picture, the equation of motion looks like

$$\begin{array}{rl}-i\frac{d}{dt}\left(\begin{array}{c}{\nu }_e\\ {\nu }_{\mu }\end{array}\right)& =ϵM^2|{\nu }_{\alpha }⟩\\ & =ϵ(U\text{diag}(m_1^2,m_2^2)U^{†}+\left(\begin{array}{cc}A& 0\\ 0& 0\end{array}\right))|{\nu }_{\alpha }⟩\\ & =ϵ\left(\frac{1}{2}[(\mathrm{\Sigma }+A)+\left(\begin{array}{cc}A-\mathrm{\Delta }{C}_{2\theta }& \mathrm{\Delta }{S}_{2\theta }\\ \mathrm{\Delta }{S}_{2\theta }& -A+\mathrm{\Delta }{C}_{2\theta }\end{array}\right)]\right)\left(\begin{array}{c}{\nu }_e\\ {\nu }_{\mu }\end{array}\right)\\ & =ϵ\left[\frac{1}{2}\left(\begin{array}{cc}2A+\mathrm{\Sigma }-\mathrm{\Delta }{C}_{2\theta }& \mathrm{\Delta }{S}_{2\theta }\\ \mathrm{\Delta }{S}_{2\theta }& \mathrm{\Sigma }+\mathrm{\Delta }{C}_{2\theta }\end{array}\right)\right]\left(\begin{array}{c}{\nu }_e\\ {\nu }_{\mu }\end{array}\right)\end{array}$$

Where $ϵ$ is an energy-related constant which won't be too important for us now. Further, the following short-hand notation was used above:

$$\begin{array}{rl}\mathrm{\Sigma }& =m_2^2+m_1^2\\ \mathrm{\Delta }& =m_2^2-m_1^2\end{array}$$

Let's now repeat the familiar diagonalization process. This will move us from the flavor basis to the mass in medium basis. We introduce the diagonalization matrices (and their relation to $M^2$) as follows:

$$U_{\text{m}}=\left(\begin{array}{cc}{C}_{{\theta }_{\text{m}}}& S_{{\theta }_{\text{m}}}\\ -S_{{\theta }_{\text{m}}}& C_{{\theta }_{\text{m}}}\end{array}\right)U_{\text{m}}^{†}=\left(\begin{array}{cc}{C}_{{\theta }_{\text{m}}}& -S_{{\theta }_{\text{m}}}\\ S_{{\theta }_{\text{m}}}& C_{{\theta }_{\text{m}}}\end{array}\right)$$

The eigenvalues of $M^2$, defined to be $M_1^2$ and $M_2^2$ by the diagonalization above, are easily found:

$$\begin{array}{rl}\text{Eigenvalues of }{M}^2& =\frac{1}{2}(A+\mathrm{\Sigma }\mp \sqrt{A^2-2A{C}_{2\theta }\mathrm{\Delta }+C_{2\theta }^2+S_{2\theta }^2{\mathrm{\Delta }}^2})\\ & =\frac{1}{2}(A+\mathrm{\Sigma }\mp \sqrt{(A-\mathrm{\Delta }{C}_{2\theta }{)}^2+(\mathrm{\Delta }{S}_{2\theta }{)}^2})\end{array}$$

In analogy with $\mathrm{\Sigma }$ and $\mathrm{\Delta }$ (defined above as linear combinations of the eigenvalues $m_1^2,m_2^2$) we define

$$\begin{array}{rl}{\mathrm{\Sigma }}^{\text{M}}& =M_2^2+M_1^2\\ {\mathrm{\Delta }}^{\text{M}}& =M_2^2-M_1^2\end{array}$$

Since we know what the eigenvalues $M_1^2,M_2^2$ are (in terms of the mixing angle $\theta$, $A$, $\mathrm{\Sigma }$, and $\mathrm{\Delta }$), we can rewrite these expressions as:

$$\begin{array}{rl}{\mathrm{\Sigma }}^M& =M_2^2+M_1^2\\ & =\frac{1}{2}(A+\mathrm{\Sigma }+\cancel{\sqrt{(A-\mathrm{\Delta }{C}_{2\theta }{)}^2+(\mathrm{\Delta }{S}_{2\theta }{)}^2}})+\frac{1}{2}(A+\mathrm{\Sigma }-\cancel{\sqrt{(A-\mathrm{\Delta }{C}_{2\theta }{)}^2+(\mathrm{\Delta }{S}_{2\theta }{)}^2}})\\ & =A+\mathrm{\Sigma }\end{array}$$

And

$$\begin{array}{rl}{\mathrm{\Delta }}^M& =M_2^2-M_1^2\\ & =\frac{1}{2}(\cancel{A+\mathrm{\Sigma }}+\sqrt{(A-\mathrm{\Delta }{C}_{2\theta }{)}^2+(\mathrm{\Delta }{S}_{2\theta }{)}^2})-\frac{1}{2}(\cancel{A+\mathrm{\Sigma }}-\sqrt{(A-\mathrm{\Delta }{C}_{2\theta }{)}^2+(\mathrm{\Delta }{S}_{2\theta }{)}^2})\\ & =\sqrt{(A-\mathrm{\Delta }{C}_{2\theta }{)}^2+(\mathrm{\Delta }{S}_{2\theta }{)}^2}\\ & =R\end{array}$$

Where we define (to make the manipulations easier to read)

$$R\equiv \sqrt{(A-\mathrm{\Delta }{C}_{2\theta }{)}^2+(\mathrm{\Delta }{S}_{2\theta }{)}^2}$$

We know that (by choice of diagonalization) the in-medium eigenstates are related to flavor eigenstates through the following relation:

Where ${\theta }_m$ is a mixing angle like the one between flavor and mass eigenstates.

The immediate question now is: how can we relate ${\theta }_m$ to the rest of our variables/knowns?

Recall that we know what $M^2$ looks like. Further, notice that this matrix must be equal to $U_{\text{m}}\text{diag}(M_1^2,M_2^2)U_{\text{m}}^{†}$, which is (explicitly):

$$\begin{array}{rl}{U}_{\text{m}}\left(\begin{array}{cc}{M}_1^2& 0\\ 0& M_2^2\end{array}\right)U_{\text{m}}^{†}& =\frac{1}{2}\left(\begin{array}{cc}{M}_2^2+M_1^2-(M_2^2-M_1^2)C_{2{\theta }_{\text{m}}}& (M_2^2-M_1^2)S_{2{\theta }_{\text{m}}}\\ (M_2^2-M_1^2)S_{2{\theta }_{\text{m}}}& M_2^2+M_1^2+(M_2^2-M_1^2)C_{2{\theta }_{\text{m}}}\end{array}\right)\\ & =\frac{1}{2}\left(\begin{array}{cc}{\mathrm{\Sigma }}^M-{\mathrm{\Delta }}^M{C}_{2{\theta }_{\text{m}}}& {\mathrm{\Delta }}^M{S}_{2{\theta }_{\text{m}}}\\ {\mathrm{\Delta }}^M{S}_{2{\theta }_{\text{m}}}& {\mathrm{\Sigma }}^M+{\mathrm{\Delta }}^M{C}_{2{\theta }_{\text{m}}}\end{array}\right)\end{array}$$

So:

$$\begin{array}{rl}{M}^2& =U_{\text{m}}\left(\begin{array}{cc}{M}_1^2& 0\\ 0& M_2^2\end{array}\right)U_{\text{m}}^{†}\\ & \Downarrow \\ \left[\frac{1}{2}\left(\begin{array}{cc}2A+\mathrm{\Sigma }-\mathrm{\Delta }{C}_{2\theta }& \mathrm{\Delta }{S}_{2\theta }\\ \mathrm{\Delta }{S}_{2\theta }& \mathrm{\Sigma }+\mathrm{\Delta }{C}_{2\theta }\end{array}\right)\right]& =\frac{1}{2}\left(\begin{array}{cc}{\mathrm{\Sigma }}^M-{\mathrm{\Delta }}^M{C}_{2{\theta }_{\text{m}}}& {\mathrm{\Delta }}^M{S}_{2{\theta }_{\text{m}}}\\ {\mathrm{\Delta }}^M{S}_{2{\theta }_{\text{m}}}& {\mathrm{\Sigma }}^M+{\mathrm{\Delta }}^M{C}_{2{\theta }_{\text{m}}}\end{array}\right)\end{array}$$

If we compare the top right entries, we see that

$$\frac{1}{2}\mathrm{\Delta }{S}_{2\theta }=\frac{1}{2}{\mathrm{\Delta }}^M{S}_{2{\theta }_{\text{m}}}$$

That is:

$${\sin }^{2}2{\theta }_{\text{m}}=\frac{{\mathrm{\Delta }}^2{\sin }^{2}2\theta }{(A-\mathrm{\Delta }{C}_{2\theta }{)}^2+(\mathrm{\Delta }{S}_{2\theta }{)}^2}$$

Before moving on, we must notice that the in-medium eigenstates are not the same as the mass eigenstates. It is for this reason that we refer to $A$ as the induced mass.

The case of resonance

Now that we know a relation between the in-medium mixing angle and other parameters, we can observe the following (feel free to play with the sliders):

It is clear that there is some value of $A$ for which ${\sin }^2(2{\theta }_{\text{m}})$ obtains its maximum. It is straightforward to see that this maximum occurs at

$$A_{\text{max}}=\mathrm{\Delta }\cos \left(2\theta \right)$$

If our medium is such that this happens, then we will get maximal mixing of neutrinos!

Final remarks

Having developed a foundation to explore neutrino propagation through matter, I will talk about the MSW effect in a later post.

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