Random notes about physicists’ obsession with Einstein notations

3 minute read

I took a quantum physics class on Coursera this year and I found that the Mathematical language spoken by the two communities, math vs physics, are quite different. Here are two examples.

Proof of a vector calculus formula using index notation

Proposition: For any smooth vector $\boldsymbol{a}$ and $\boldsymbol{b}$ \(\nabla \times ( \boldsymbol{a}\times \boldsymbol{b}) = \boldsymbol{a}\nabla \cdot \boldsymbol{b} + \boldsymbol{b} \cdot \nabla \boldsymbol{a} - \boldsymbol{b} \nabla \cdot \boldsymbol{a} - \boldsymbol{a} \cdot \nabla \boldsymbol{b}.\)

Proof: Using the following Levi-Civita definition of cross product

\[\boldsymbol{a} \times \boldsymbol{b} =\boldsymbol{e}_i \epsilon_{ijk}a_ib_j\]

where \(\epsilon_{ijk} =\begin{cases} +1 & \text{if } i,j,k \text{ are in clockwise permutation}, \\ -1 & \text{if } i,j,k \text{ are in counterclockwise permutation, and} \\ \;\;\,0 & \text{if }i=j \text{ or } j=k \text{ or } k=i. \end{cases}\)

Then we have

\[\begin{aligned} \nabla \times (\boldsymbol{a} \times \boldsymbol{b}) &=\partial_l \hat{e}_l \times (a_i b_j \hat{e}_k \epsilon_{ijk}) \\[5pt] &=\partial_l a_i b_j \epsilon_{ijk} \underbrace{ (\hat{e}_l \times \hat{e}_k)}_{(\hat{e}_l \times \hat{e}_k) = \hat{e}_m \epsilon_{lkm} } \\[5pt] &=\partial_l a_i b_j \hat{e}_m \underbrace{\epsilon_{ijk} \epsilon_{mlk}}_{\text{contracted epsilon identity}} \\[5pt] &=\partial_l a_i b_j \hat{e}_m \underbrace{(\delta_{im} \delta_{jl} - \delta_{il} \delta_{jm})}_{\text{They sift other subscripts}} \\[5pt] &=\partial_j (a_i b_j \hat{e}_i)- \partial_i (a_i b_j \hat{e}_j) \\[5pt] &=\color{red}{a_i \partial_j b_j \hat{e}_i + b_j \partial_j a_i \hat{e}_i} - (\color{cyan}{a_i \partial_i b_j \hat{e}_j + b_j \partial_i a_i \hat{e}_j}) \\ &= \boldsymbol{a}(\nabla \cdot \boldsymbol{b}) + (\boldsymbol{b} \cdot \nabla)\boldsymbol{a} - (\boldsymbol{a}\cdot \nabla)\boldsymbol{b} - \boldsymbol{b}(\nabla \cdot \boldsymbol{a}) \end{aligned}\]

Question: Why did the deltas vanish?

Due to Kronecker $\delta$’s sifting property. Recall the definition of Kronecker delta:

\[\delta_{ij}=\begin{cases} 0,\quad \text{if } i\ne j, \\ 1,\quad \text{if } i=j. \end{cases}\]

Thus, for $j\in\mathrm Z$:

\[\sum\limits_{-\infty}^{\infty}a_i\delta_{ij} = a_j\]

This is just like filtering (or sifting), because only when $i=j$ does $\delta_{ij} = 1$. Other terms are zeroes. This also works for partial derivatives.

For example,

\[\partial_l a_i b_j \hat{e}_m \delta_{im} \delta_{jl} = \partial_l a_i b_j \hat{e}_i \delta_{jl}\]

If $l\ne j$ then
\(\partial_l a_i b_j \hat{e}_i \delta_{jl} = \partial_l a_i b_j \hat{e}_i(0) =0;\)

If $l=j$ then \(\partial_l a_i b_j \hat{e}_i \delta_{jl} = \partial_j a_i b_j \hat{e}_i (1) = \partial_j a_i b_j \hat{e}_i,\)

thus, \(\partial_l a_i b_j \hat{e}_m \delta_{im} \delta_{jl} =\partial_j (a_i b_j \hat{e}_i).\)

Computing the angular momentum squared using mathematcian’s notation

First explicitly we can write down the defintion of the angular momentum \(\mathbf{L}\psi = k\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \partial_x & \partial_y & \partial_z \\ \phi \partial_x\psi & \phi \partial_y\psi & \phi \partial_z\psi \end{vmatrix} \\[5pt] =k\left(\begin{vmatrix}\partial_y & \partial_z \\ \phi \partial_y\psi & \phi \partial_z\psi \end{vmatrix}, -\begin{vmatrix} \partial_x & \partial_z \\ \phi \partial_x\psi & \phi \partial_z\psi \end{vmatrix}, \begin{vmatrix} \partial_x & \partial_y \\ \phi \partial_x\psi & \phi \partial_y\psi \end{vmatrix}\right) \\[10pt] =k\begin{pmatrix}\partial_y\phi \partial_z\psi - \partial_z\phi \partial_y\psi \\ \partial_z\phi \partial_x\psi - \partial_x\phi \partial_z\psi \\ \partial_x\phi \partial_y\psi - \partial_y\phi \partial_x\psi\end{pmatrix} := \begin{pmatrix}\mathbf{L}_x\psi \\ \mathbf{L}_y\psi \\ \mathbf{L}_z\psi\end{pmatrix}.\)


\[\begin{aligned} \mathbf{L}\cdot \mathbf{L}\psi &= \mathbf{L}_x \mathbf{L}_x \psi+ \mathbf{L}_y\mathbf{L}_y \psi + \mathbf{L}_z \mathbf{L}_z \psi \\ &= k\mathbf{L}_x(\partial_y\phi \partial_z\psi - \partial_z\phi \partial_y\psi) \\ &\quad + k\mathbf{L}_y(\partial_z\phi \partial_x\psi - \partial_x\phi \partial_z\psi) \\ &\quad \;+k\mathbf{L}_z (\partial_x\phi \partial_y\psi - \partial_y\phi \partial_x\psi) \\ &=k^2\Big[ \partial_y\phi \partial_z(\partial_y\phi \partial_z\psi - \partial_z\phi \partial_y\psi) - \partial_z\phi \partial_y(\partial_y\phi \partial_z\psi - \partial_z\phi \partial_y\psi) \\ &\; +\partial_z\phi \partial_x(\partial_z\phi \partial_x\psi - \partial_x\phi \partial_z\psi) - \partial_x\phi \partial_z(\partial_z\phi \partial_x\psi - \partial_x\phi \partial_z\psi) \\ &\; + \partial_x\phi \partial_y(\partial_x\phi \partial_y\psi - \partial_y\phi \partial_x\psi) - \partial_y\phi \partial_x(\partial_x\phi \partial_y\psi - \partial_y\phi \partial_x\psi)\Big]. \end{aligned}\]

Making use of $\phi = (x^2+y^2+z^2)/2$ and simplifying this you will have:

\[\begin{aligned} \mathbf{L}\cdot \mathbf{L}\psi &= k^2\Big( (y^2+z^2)\partial_{xx}\psi + (z^2+x^2)\partial_{yy}\psi + (x^2+y^2)\partial_{zz}\psi \\ &\quad \;-2xy\partial_{xy}\psi - 2yz\partial_{yz}\psi - 2zx\partial_{zx} \psi -2x\partial_x\psi - 2y\partial_y\psi -2z\partial_z\psi\Big). \end{aligned}\]

Using some algebra trick yields the traditional multivariate calculus formula:

\[\mathbf{L}\cdot \mathbf{L}\psi = k^2\Big(\Delta \psi |\mathbf{r}|^2 - (\mathbf{r}\cdot \nabla)^2 \psi- \mathbf{r}\cdot \nabla \psi \Big), \tag{1}\]


\[(\mathbf{r}\cdot \nabla)^2 = (x\partial_x+y\partial_y+z\partial_z)^2.\]

So $(1)$ is our ultimate formula for angular momentum squared. In quantum mechanics.

$\mathbf{L}^2$ has $\hbar^2 l(l+1)$ as its eigenvalue, so let’s check if the result above agrees with the deduction in index notation, when $\psi = \chi_{0,0} = x$: \(\mathbf{L}^2 \chi_{0,0} = i^2\hbar^2(\Delta x |\mathbf{r}|^2 - (\mathbf{r}\cdot \nabla)^2 x - \mathbf{r}\cdot \nabla x ) = 2\hbar^2 x= \hbar^2 l(l+1)\chi_{0,0}\)

where $l=1$ so we are happy. Let’s check one more $\psi =\chi_{1,-1} = (x-iy)/\sqrt{2}$:

\[\mathbf{L}^2 \chi_{1,-1} = i^2\hbar^2\Big(0 |\mathbf{r}|^2 - (\mathbf{r}\cdot \nabla) (x-iy)/\sqrt{2} - \mathbf{r}\cdot (1/\sqrt{2},-i/\sqrt{2},0) \Big) = 2\hbar^2 \chi_{1,-1}.\]

All is good.