How many index can you cram into an equation

Index notation, where a summation over some set is indicated by subscripts and superscripts, can be a curse in physics, as things can go crazy pretty fast for some systems. This usually occurs either when the equations require a large amount of them due to the sheer number of terms or just their variety, in the case of the most esoteric physics.

For instance, he's an example from my own master thesis, where some calculation required me to compute Taylor terms up to the 4th order :

\begin{align} \psi (x_2,t + \epsilon ) &=\sqrt{\frac{m}{2\hbar\pi i\varepsilon}} \int_{-\infty}^{\infty} d\eta\ e^{\frac{im}{2\hbar} g_{ab}(x_2) \frac{\eta^a\eta^b}{\varepsilon}}\nonumber\\ &(1 - \frac{im}{2\hbar\varepsilon}(\lambda \eta^\mu \eta^\nu \eta^\rho g_{\mu\nu,\rho})- \frac{m^2}{8\hbar^2\varepsilon^2}(\lambda^2 \eta^\mu \eta^\nu \eta^\rho \eta^\sigma ( g_{\mu\nu,\rho\sigma}+ \eta^\tau \eta^\chi g_{\sigma\tau,\chi}g_{\mu\nu,\rho})) +\mathcal{O}(\eta^7))\nonumber\\ &(1 +\frac{im}{2\hbar\epsilon} g_{ab}(\tilde{x}) C^{ab}- \frac{m^2}{8\hbar^2\epsilon^2} g_{ab}(\tilde{x})g_{pq}(\tilde{x})C^{ab}C^{pq}+\mathcal{O}(C^3)) \nonumber\\ &(\psi(x_2,t)+ \eta^\alpha \psi(x_2,t)_{,\alpha}+ \frac 12 \eta^\alpha \eta^\beta \psi(x_2,t)_{,\alpha\beta} +\mathcal{O}(\eta^3)) \nonumber\\ &(\sqrt{g(x_2)} + \eta^\aleph \sqrt{g(x_2)}_{,\aleph}+ \frac 12 \eta^\aleph \eta^\beth \sqrt{g(x_2)}_{,\aleph\beth}+ \mathcal{O}(\eta^3)) \end{align}

with some hebrew letters added for fun. Index notation is a topic of some humor, for instance from this parody paper :

We use the following notation: Greek letters for vile spinor indices, Roman letters for isospin indices, Cyrillic letters for vector indices, Hebrew letters for 10-dimensional vector indices, Roman numerals for Dirac spinor indices, and radiation therapy for Hodgkin's disease.

Due to this, indexes tend to be dropped, assumed to be implicit by the choice of objects used in the equation. This happens a lot for spinors, in particular.

Let's now see what we could do with the opposite approach, to use as many indexes as possible. Take for example the Dirac equation, in pure unindexed form :

$$(i\hbar \partial\!\!\!/ - mc) \psi = 0$$

The Feynman slash notation $\partial\!\!\!/$ implies here a summation over two spacetime indexes with Dirac's gamma matrices, $\gamma^\mu \partial_\mu$ (this isn't strictly speaking two spacetime indexes since $\gamma$ isn't a tensor but a soldering form, but it is close enough to it). The equation then becomes

$$(i\hbar \gamma^\mu \partial_\mu - mc) \psi = 0$$

For a total of a single index. From there, we'll have to add the actual spinor indices. There are four types of spinor indices, all noted with upper case letters : the basic spinor $\psi^A$, the dual space spinor $\psi_A$, the complex conjugate spinor $\psi^{\dot A}$ and the complex conjugate dual spinor $\psi_{\dot A}$. The spinor of the Dirac equation is actually none of those, but a vector of spinors called the Dirac bispinor.

$$\psi = \begin{pmatrix}\xi^{A}\\\chi^{\dot A}\end{pmatrix}$$

$\xi$ is the left-handed spinor while $\chi$ is the right-handed spinor. In that notation, the gamma matrices become a matrix of the solder forms on those spinors.

$$\gamma^\mu = \begin{pmatrix} 0 && \sigma^\mu_{A\dot A}\\ \bar \sigma^\mu_{A\dot A} && 0 \end{pmatrix}$$

This allows for instance the usual formula to map a spinor to a vector via $\psi^{\dot A} \sigma^\mu_{A\dot A} \psi^A$. The Dirac equation then becomes

\begin{eqnarray} i\hbar (\sigma^\mu_{A\dot A} \partial_\mu - mc I_{A\dot A})\chi^A &=& 0\\ i\hbar (\bar \sigma^\mu_{A\dot A} \partial_\mu - mc I_{A\dot A}) \xi^A &=& 0 \end{eqnarray}

$I$ simply being the identity matrix. We can set the mass to $0$ and end up with the Weyl equation for a single right-handed spinor with no loss of indexes.

$$i\hbar (\sigma^\mu_{A\dot A} \partial_\mu - mc I_{A\dot A})\chi^A = 0$$

Up to three indexes. Now we go onto gauge fields : we replace the partial derivative $\partial_\mu$ with the gauge derivative $\nabla_\mu$. Gauge derivatives are Kozsul connection on a principal bundle and as such are fairly complicated, something of the form (up to some physical factors)

$$\nabla_\mu = \partial_\mu + \tau_\alpha^{ij} A_\mu^\alpha$$

Where $A$ is the gauge field and $\tau$ is the representation of the gauge group. In our case, we'll want to see the gauge derivative of a product of gauge groups, such as $\mathrm{SU}(3) \times \mathrm{SU}(2) \times \mathrm{U}(1)$, for which it will be a sum of the form

$$\nabla_\mu = \mathrm{I}^{ij} \mathrm{I}^{kl} \mathrm{I}^{mn} \partial_\mu + \tau_\alpha^{ij} G_\mu^\alpha \mathrm{I}^{kl} \mathrm{I}^{mn} + \mathrm{I}^{ij} \sigma_\alpha^{kl} W_\mu^\alpha \mathrm{I}^{mn} + \mathrm{I}^{ij} \mathrm{I}^{kl} u^{mn}_\alpha B_\mu^\alpha $$

$G$ is the gluon field, with $\tau$ the Gell-Mann matrices, $W$ is the weak isospin field, with $\sigma$ the Pauli matrices, and $B$ is the weak hypercharge field, with $u$ simply being some one dimensional matrix (we could remove its indices but let's keep it). The spinor field itself is a product of the original spinor space $\mathbb C^2$, $\chi^A_{ikm}$, each index corresponding to a different charge of the standard model. If we drop the mass term (to avoid getting too long), the new equation then becomes something like

$$i\hbar (\sigma^\mu_{A\dot A} (\mathrm{I}^{ij} \mathrm{I}^{kl} \mathrm{I}^{mn} \partial_\mu + \tau_\alpha^{ij} G_\mu^\alpha \mathrm{I}^{kl} \mathrm{I}^{mn} + \mathrm{I}^{ij} \sigma_\alpha^{kl} W_\mu^\alpha \mathrm{I}^{mn} + \mathrm{I}^{ij} \mathrm{I}^{kl} u^{mn}_\alpha B_\mu^\alpha) ) \chi^A_{ikm} = 0$$

Although we could have arbitrarily many indexes here, for a reasonable theory we are up to 10. We can then add the various flavors of particles. Considering every particle as a variation of the triplet, doublet and singlet of the standard model, there are six of them : the three couples of leptons and their neutrinos $(e, \nu_e)$, $(\mu, \nu_\mu)$ and $(\tau, \nu_\tau)$, and the three generations of quarks, $(u,d)$, $(c,s)$ and $(b,t)$. Without the mass term, we can simply add the flavor index (otherwise, the mass term would depend on both the flavor and the charges)

Now we can add curved spacetime to the mix. This changes two things : we replace spacetime indices with frame indices (and add the frame field), and we also need another connection, this time the $\mathrm{SO}(1,3)$ connection of the frame bundle.

$$i\hbar (e^\mu_a \sigma^a_{A\dot A} (\mathrm{I}^{ij} \mathrm{I}^{kl} \mathrm{I}^{mn} \partial_\mu + \tau_\alpha^{ij} G_\mu^\alpha \mathrm{I}^{kl} \mathrm{I}^{mn} + \mathrm{I}^{ij} \sigma_\alpha^{kl} W_\mu^\alpha \mathrm{I}^{mn} + \mathrm{I}^{ij} \mathrm{I}^{kl} u^{mn}_\alpha B_\mu^\alpha) + \sigma^a_{A\dot A} \omega_\mu^\alpha S_{\alpha A}) \chi^A_{ikmf} = 0$$ [...]

While rarely done, it is entirely possible to write quantum states in index notation, since they are vectors themselves. For instance, take the two wavefunctions $\Psi$ and $\Phi$. We can write their product as

$$\langle \Psi, \Phi \rangle = \Psi_{\alpha} \Phi^\alpha$$

We can then add the wavefunction to the equation of motion via the Schwinger-Dyson equation

$$\left\langle \frac{\delta S}{\delta \phi(x)} \right\rangle = \Psi_\alpha \left(\frac{\delta S}{\delta \phi(x)}\right)^\alpha_\beta \Psi^\beta = 0$$

Posted on 2017-10-22 23:21:35

Tags : physics