The use of subscripts in L11 E04 - Classical Mechanics by Leonard Susskind

Question

I am struggling to get my head around the soultion to exercise 4 lecture 11 in The Theoretical Minimum (Classical Mechanics).

The exercise and its solution can be found here: https://tales.mbivert.com/ttm/cm/L11E04.pdf.

I don't fully understand the use of the subscripts i and j in the solution above. I understand they denote spacial dimentions, however, I do not understand why the j is used for Hamilton's equations and the i is used for the Hamiltonion. What is the mathematical reasoning for using seperate subscripts for the starting equations?

Thanks :)

Answer 1

Let the Cartesian coordinates be denotes as $(q_1,q_2,q_3)$ where $q_1=x$, $q_2=y$, and $q_3=z$. Further let the vector potential's components be $(A_1,A_2,A_3)$ where $A_1=A_x(\mathbf{r})$, $A_2=A_y(\mathbf{r})$, $A_3=A_z(\mathbf{r})$. Then the Hamiltonian is:

$$H=\frac{1}{2m}\sum_i\left(p_i-\frac{e}{c}A_i\right)^2$$

Or: $$H=\frac{1}{2m}\left(\left(p_1-\frac{e}{c}A_1\right)^2+\left(p_2-\frac{e}{c}A_2\right)^2+\left(p_3-\frac{e}{c}A_3\right)^2\right)$$

Or if you'd prefer:

$$H=\frac{1}{2m}\left(\left(p_x-\frac{e}{c}A_x\right)^2+\left(p_y-\frac{e}{c}A_y\right)^2+\left(p_z-\frac{e}{c}A_z\right)^2\right)\tag{1}\label{1}$$

Now each of Hamilton's equations is a set of (three in 3D space) equations. For Hamilton's equation: $$\dot{q}_i=\frac{\partial H}{\partial p_i}\tag{2}\label{2}$$

I use $i$ as an index because I expanded $H$ out in Equation \ref{1}.

We actually are referring to the three equations: $$\begin{array}{c} \dot{q}_1=\frac{\partial H}{\partial p_1}\\ \dot{q}_2=\frac{\partial H}{\partial p_2}\\ \dot{q}_3=\frac{\partial H}{\partial p_3} \end{array}$$

Or: $$\begin{array}{c} \dot{q}_x=\frac{\partial H}{\partial p_x}\\ \dot{q}_y=\frac{\partial H}{\partial p_y}\\ \dot{q}_z=\frac{\partial H}{\partial p_z} \end{array}\tag{3}\label{3}$$

Looking at the first line $\dot{q}_x=\frac{\partial H}{\partial p_x}$, we can substitute Equation \ref{1} to get: $$\dot{q}_x=\frac{1}{2m}\frac{\partial}{\partial p_x}\left(\left(p_x-\frac{e}{c}A_x\right)^2+\left(p_y-\frac{e}{c}A_y\right)^2+\left(p_z-\frac{e}{c}A_z\right)^2\right)$$

Now this is a lot to write out, but let me highlight a few things: $$\color{red}{\dot{q}_x}=\frac{1}{2m}\color{red}{\frac{\partial}{\partial p_x}}\left(\color{blue}{\left(p_x-\frac{e}{c}A_x\right)}^2+\color{blue}{\left(p_y-\frac{e}{c}A_y\right)}^2+\color{blue}{\left(p_z-\frac{e}{c}A_z\right)}^2\right)\tag{4}\label{4}$$

The terms in $\color{red}{\text{red}}$ have the same index (i.e., $x$). The terms in $\color{blue}{\text{blue}}$ are all of the same form and span each index (i.e., all $x$, $y$, $z$). Thus for the $\color{blue}{\text{blue}}$ terms, we may want to simplify our notation by summing over all the indices. We can use any symbol, for example, $\boxplus$: $$\color{red}{\dot{q}_x}=\frac{1}{2m}\color{red}{\frac{\partial}{\partial p_x}}\sum_\boxplus\left(p_\boxplus-\frac{e}{c}A_\boxplus\right)^2$$

Now before doing anything with the $\color{red}{red}$ terms let's look at all three Equations contained in \ref{2}, or expanded to Equations \ref{3}: $$\begin{array}{c} \color{red}{\dot{q}_x}=\frac{1}{2m}\color{red}{\frac{\partial}{\partial p_x}}\sum_\boxplus\left(p_\boxplus-\frac{e}{c}A_\boxplus\right)^2\\ \color{red}{\dot{q}_y}=\frac{1}{2m}\color{red}{\frac{\partial}{\partial p_y}}\sum_\boxplus\left(p_\boxplus-\frac{e}{c}A_\boxplus\right)^2\\ \color{red}{\dot{q}_z}=\frac{1}{2m}\color{red}{\frac{\partial}{\partial p_z}}\sum_\boxplus\left(p_\boxplus-\frac{e}{c}A_\boxplus\right)^2 \end{array}$$

Each of these three equations has the two $\color{red}{\text{red}}$ terms with matching indices but are otherwise identical. To express these three equations more compactly, we may represent this symbol with anything EXCEPT $\boxplus$ because that would be confusing since it is in the summation, in Equation \ref{4} there is a difference in what we mean between the different colored terms. Thus let's use $\circledast$ to label the matching indices in each of the three equations. This gives: $$\dot{q}_\circledast=\frac{1}{2m}\frac{\partial}{\partial p_\circledast}\sum_\boxplus\left(p_\boxplus-\frac{e}{c}A_\boxplus\right)^2$$ Or using the definition of the Hamiltonian: $$\dot{q}_\circledast=\frac{\partial H}{\partial p_\circledast},\quad \circledast\in\{x,y,z\}$$

In short, the symbols don't matter, but we use one of them ($j$ in the source or $\circledast$ in mine) to represent which of the three equations we are talking about (i.e., the time evolution of which of the three coordinates) but this time evolution depends on the information from each of the canonical momenta and vector potential coordinates in the same way, so we express each of their contributions as a sum indexed by a different symbol ($i$ in the source or $\boxplus$ in mine).

Answer 2

Imagine you used the same symbol $i$ for both.

You would see $\frac{\partial}{\partial x_i} \sum_{i=1}^n H_i$ where the inner $i$ is a bound variable so not to be conflated with the outer $i$. You could write it this way and then say it is equal to $\frac{\partial}{\partial x_i} \sum_{j=1}^n H_j$ due to the freedom of renaming bound variables.

Or you could just avoid this possible source of typographic mistakes by giving them different names from the start.

Imagine transliterating this into code and you used the same $i$. You would either get a using variable for outer scope warning and it would work as intended but be confusing to read later, or it could mix them up and give you nonsense depending on how you put H_i, H as results of function calls or inlined.

Answer 3

You need two letters - $i$ for the Hamiltonian and $j$ for Hamilton's equations - because you want to bring the differentiation inside the sum. Consider the first term involving the square of the momenta. You can't write

$ \frac{d}{dp_i}\sum_i p_i^2 =\sum_{i}\frac{d}{dp_i}\,p_i^2 $

because you're introducing something inside the sum which depends on the summation index $i$. It would be a little like writing

$ x\int x^2d x = \int x^3 dx $

which is obviously not true. So you need to introduce a new letter $j$, which takes on the same values $\{x,y,z\}$ as $i$. It is OK to write

$ \frac{d}{dp_j}\sum_i p_i^2 = \sum_{i}2p_i\frac{dp_i}{dp_j} = \sum_i 2p_i \delta_{ij} = 2p_j $

since $\frac{d}{dp_j}$ doesn't depend on the summation index $i$. As a general rule, you should avoid having the same index $i,j,...$ appear more than twice in the same term. Hope this helps!

Answer 4

They imply fundamentally different vector derivatives.

$\frac{\partial }{\partial x_a}B_b=(\nabla B_b)_a$, i.e. the $a$ component of the gradient of the $b$ component of vector $\vec{B}$.

$\sum_b\frac{\partial }{\partial x_b}B_b=\nabla \cdot \vec{B}$

$\epsilon_{ijk}\frac{\partial}{\partial x_j}B_k=(\nabla \times \vec{B})_i=$ the $i$ component of the curl of $\vec{B}$.

Consider this proof for the identity for $\nabla \times (\vec {A} \times \vec{B})$

$\nabla \times (\vec {A} \times \vec{B})_a=\epsilon_{abi}\epsilon_{ijk}\frac{\partial}{\partial x_b}(A_jB_k)$

Here the index $b$ can sometimes match $j$ or $k$ or be different. This implies a gradient, whereas if it matched, it would imply a divergence as the next steps show.

$=(\delta_{aj}\delta_{bk}-\delta_{ak}\delta_{bj})\frac{\partial}{\partial x_b}(A_jB_k)$

$=(\delta_{aj}\delta_{bk}-\delta_{ak}\delta_{bj})(B_k\frac{\partial A_j}{\partial x_b}+A_j\frac{\partial B_k}{\partial x_b})$

$=B_b\frac{\partial A_a}{\partial x_b}-B_a\frac{\partial A_b}{\partial x_b}+A_a\frac{\partial B_b}{\partial x_b}-A_b\frac{\partial B_a}{\partial x_b}$

Matching indices imply a dot product or divergence or both, different indices imply any component and so can represent the whole vector. If the index of the partial derivative term matches, you have a divergence. If it is a different index from the index of the vector it implies a gradient of the component in the absence of the Levi-Civita symbol or a curl when it has that symbol.

$\nabla \times (\vec {A} \times \vec{B})= (\vec{B}\cdot \nabla)\vec{A}-\vec{B}(\nabla \cdot \vec{A})+\vec{A}(\nabla \cdot \vec{B})-(\vec{A}\cdot \nabla)\vec{B}$

In the case of the derivative of the Hamiltonian, $j$ and $i$ can be different so a gradient is implied. If they were always the same, it would be a divergence.