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I am trying to understand the gaps in my knowledge that prevents me from completely understanding quantum field theory. Sometimes I ask pretty basic questions, but please excuse me if I make a blunder.

Suppose we have a part of a generic Lagragian: $$\begin{eqnarray} \mathcal{L} &=\ ...\ - \frac{1}{2}\phi_a M^2_{ab}\phi_b\ + \ ...\ , \end{eqnarray}$$

where $M^2_{ab}$ is Real, and symmetric, and $\phi_a$ is a collection of Real scalar fields.

By linear algebra we know that a Real and a symmetric matrix $\mathrm{A}$ can be diagonalized by the similarity transformation $R^\mathrm{T}AR = A_{diagonal}$. $R$ is an orthogonal matrix. This transformation will change the basis of the initial matrix. Therefore, if I redefine the fields as $\varphi = R\phi$, I can obtain the following: $$\begin{eqnarray} \mathcal{L} &=\ ...\ - \frac{1}{2}\varphi_a m^2_{a}\varphi_a\ + \ ...\ . \end{eqnarray}$$

In this example we used an orthogonal matrix to change the basis to write the Lagrangian in terms of mass eigenstates. My questions here:

  1. What was the basis of the initial fields $\phi_a$? I know it is generic, but why do we need to define such a generic matrix and work with generic basis in the first place? Why do we bother about diagonalization, but not start with already diagonalized matrix and mass eigenstates? I know that I gave you this example, so I must explain it to you, but I ask for examples in QFT why this is important? What are the use cases of similarity transformations in various examples?

  2. We usually say symmetry transformations are active transformations. Here, there is no symmetry transformation but rather a change of basis (similarity transformation). I think this is a passive transformation. Could you tell me if there is nothing wrong with my understanding here?

  3. Even though I know that similarity transformations and symmetry transformations are two different things, I cannot stop noticing the following connection: Here, I told you that $\phi_a$ is a collection of Real scalar fields, and we know that if I write them as a column vector I want to talk about a symmetry, and this symmetry is an $SO$ symmetry. Its group elements are orthogonal matrices. Just as the change of basis matrices that we used to diagonalize $M^2_{ab}$. Is there a connection, or is this just a coincidence because I also told you that $M^2_{ab}$ is Real and symmetric? If it is just a coincidence can you give me some examples that falsifies this, and also where they are used?

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    $\begingroup$ In a free theory, there is no reason not to work with mass eigenstates from the very beginning. But with interacting theories there can be other useful bases e.g. flavour eigenstates. $\endgroup$ Commented Sep 11, 2022 at 15:12
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    $\begingroup$ For the future viewers of this post: I also found this discussion to be helpful. $\endgroup$ Commented Sep 11, 2022 at 18:27

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I would like to give you a very important example related with the comment posted by Connor Behan. In the standard model Lagragian we find out that diagonalization is not possible for all terms in the mass basis. As a consequence we learn about quark mixing through the Cabibbo–Kobayashi–Maskawa matrix.

I will try to go straight to the points since I dont know how much you are familiar with the mathematics of the Standard Model.

The Standard Model is a non-abelian quantum field theory combining quantum chromodynamics, which has a $SU(3)$ gauge group, with the electroweak theory, which has a $SU(2)\times U(1)$ gauge group. The fermionic particles are grouped into $SU(2)$ doublets and singlets

\begin{equation} SU(2) \quad doublets: \quad q_{i\alpha L}= \begin{pmatrix}u_{i\alpha} \\ d_{i\alpha}\end{pmatrix}_L, \quad l_{iL}= \begin{pmatrix} \nu_i \\ e_i\end{pmatrix}_L \end{equation} \begin{equation} SU(2) \quad singlets: \quad u_{i\alpha R}, \quad d_{i\alpha R}, \quad e_{iR} \end{equation}

where the $i$-index refers to the respective up-type or down-type quark fields (the up-type are u,c and t while the down-type are d, s and b) and $\alpha$ refers to the three possible colours of a quark, "red" (R), "blue" (B) or "green" (G). The mathematical formulation of the SM involves a Lagrangian (density), $L_{SM}$, which, in addition to Lorentz invariance, has an internal symmetry given by the gauge group $SU(3)\times SU(2)\times U(1)$. It is usually divided into four sectors, $ L_{SM}= L_{gauge} + L_{f} +L_{\phi} + L_{Yuk}$, namely the gauge, fermion, scalar and Yukawa sectors of the theory.

Through the Higgs mechanism, when the vacuum is $\langle\phi\rangle_{VAC}=(0\quad \tfrac{(v+ H(x))}{\sqrt{2}})^T$ the Yukawa sector (ignoring the color indices since these will not be important for this example),

\begin{equation} L_{Yuk}= -\Big[ \overline{q}_{iL}\Gamma^u_{ij}\tilde{\phi}u_{jR} +\overline{q}_{iL}\Gamma^d_{ij}\phi d_{jR} +\overline{l}_{iL}\Gamma^e_{ij}\phi e_{jR} +h.c. \Big] \end{equation}

where $\tilde{\phi}=i\sigma^2\phi^*$, gives rise to fermion mass terms after expanding around $\langle\phi\rangle_{VAC}$

\begin{equation} L_{Yuk}\rightarrow \overline{u}_{iL}\Gamma^u_{ij}\frac{(\nu+H)}{\sqrt{2}}u_{jR}+ (d,e \quad terms) + h.c. =\\ = \overline{u}_{L}M^u(1+\frac{H}{\nu})u_{R}+ (d,e \quad terms) + h.c. \qquad (1) \end{equation}

where in the second equality $u_{L,R}, d_{L,R}$ and $e_{L,R}$ are family column vectors, $u_{L,R}= (u_{L,R} \quad c_{L,R}\quad t_{L,R})^T$ etc, and $M^{u,d,e}=\tfrac{\nu}{\sqrt{2}}\Gamma^{u,d,e}$ are mass matrices. These matrices, in the most general case, need not to be diagonal. To identify the masses of the physical particles it is necessary to diagonalize them by two separate unitary transformations, $A_L$ and $A_R$ such that $A^{u\dagger}_LM^{u}A^u_R=D^u$. Doing so, the physical fields are rotated from the weak basis to the mass basis,

\begin{equation} (1)\rightarrow \overline{u}_{L} A^u_LA^{u\dagger}_LM^{u}A^u_RA^{u\dagger}_Ru_{R}+...+ (d,e \quad terms) + h.c. =\\= \overline{u}^{'}_{L}D^uu^{'}_{R} + \overline{d}^{'}_{L}D^dd^{'}_{R} + \overline{e}^{'}_{L}D^ee^{'}_{R} + ... + h.c. \nonumber \end{equation}

When generalizing this rotation to the rest of $L_{SM}$ every term remains diagonal in the new basis except for the weak charged currents of the quarks, $J^{\mu}_{W}$, where a unitary quark mixing matrix, the Cabibbo-Kobayashi-Maskawa matrix or, $V_{CKM}$ matrix, arises \begin{equation} J^{\mu\dagger}_{W}= 2\overline{u}_{L}\gamma^{\mu}d_L \rightarrow 2\overline{u}_{L}A^u_LA^{u\dagger}_L\gamma^{\mu}A^d_LA^{d\dagger}_Ld_L = 2\overline{u}^{'}_{L}\gamma^{\mu}V_{CKM}d^{'}_L \end{equation} since $A^{u\dagger}_LA^{d}_L = V_{CKM}$ is not equal to unit in the most general case. The $V_{CKM}$ matrix entries are proportional to the probability of a transition from one flavour $i$-quark to another flavour $j$-quark by $|V_{ji}|^2$. Currently, the best determination of the magnitudes of the $V_{CKM}$ matrix elements is \begin{equation} V_{CKM}= \begin{bmatrix} |V_{ud}| & |V_{us}| & |V_{ub}|\\ |V_{cd}| & |V_{cs}| & |V_{cb}|\\ |V_{td}| & |V_{ts}| & |V_{tb}| \end{bmatrix} = \begin{bmatrix} 0.97370\pm0.00014 & 0.2245\pm0.0008 & 0.00382\pm0.00024\\ 0.221\pm0.004 & 0.987\pm0.011 & 0.0410\pm0.0014\\ 0.0080\pm0.0003 & 0.0388\pm0.0011 & 1.013\pm0.030 \end{bmatrix}\label{VCKMM} \end{equation} telling that each quark has a large probability to undergo a flavour transition into a quark of the same generation (diagonal elements are close to unity) but a considerably smaller probability to transit to a quark of other generation (off-diagonal elements are very small).

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    $\begingroup$ Thank you for this detailed answer. I am familiar with the SM, however I often have hard time fully comprehending it, but your answer is totaly useful. Can you, however make a comment on my 3rd question? I notice that here is $SU(2)_L \times SU(2)_R$ symmetry, which also showed itself in the change of basis. So I think they are connected, or one occurs due to the other. Is that right? $\endgroup$ Commented Sep 11, 2022 at 17:41
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    $\begingroup$ The symmetry you mention in the comment is (an accidental) global symmetry that follows from the Higgs sector before SSB. You need to write Higgs bi-doublets and consider the coupling of the U(1) gauge group to be zero in order to find such symmetry. It is an important symmetry as one can conclude about the mass of the gauge bosons when SSB takes place. Regarding your 3rd question, if the fields are not real then the matrix does not belong to a representation of SO(N) groups but instead to a representation of SU(N) groups $\endgroup$
    – Legoinha
    Commented Sep 11, 2022 at 18:01

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