# Evaluating an Equation Using Einstein Summation Notation [closed]

## Problem

As part of a proof for the commutation relation of a vector operator and the angular momentum operator, I need the evaluate the expression

$$(R_\omega)_{ij}V_j = \big([1-\cos(\omega)]\hat{\omega}_i\hat{\omega}_j + \cos(\omega)\delta_{ij} + \sin(\omega)\varepsilon_{ijk}\hat{\omega}_k\big) V_j$$

Here, $$(R_\omega)_{ij}$$ is the representation for the rotation generator I was given and advised to use, and $$V_j$$ is the $$j$$-component of said vector operator. (Eventually, I'm trying to show that $$[L_i,V_j]=i\hbar\varepsilon_{ijk}V_k$$.)

## Attempt

Now, on the LHS, $$j$$-index is repeated, so I should sum over $$j=i,j,k$$. I'm still clunky with this convention, so I computed the sum term-wise, (with $$\hat{\omega}_i\hat{\omega}_j=\delta_{ij}$$ assuming orthonormal unit vectors)

$$(R_\omega)_{ij}V_j = \big(1-\sin(\omega)\hat{\omega}_k\big)V_j$$

Is this correct? I have reservations, because (a) why was I given $$\hat{\omega}_i\hat{\omega}_j$$ instead of another $$\delta$$-function? and (b) this result leaves me with a scalar minus a vector, which seems erroneous to me.

I feel like my execution of the Einstein Summation notation is flawed. I'm familiar with the mechanics of it, but again, I'm clunky with implementation. For instance, am I suppose to sum over $$j$$ on the LHS and then $$i,j$$ on the RHS?

## closed as off-topic by Kyle Kanos, Emilio Pisanty, Jon Custer, user191954, Cosmas ZachosNov 18 '18 at 17:19

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• The convention is to sum over the repeated indices, but not over $i$ which is a free index. So indeed this is a vector equation on both sides. Also, you mustn't just forget the Levi-Civita symbol $\varepsilon_{ijk}$ which, when summed over the two indices $j,k$ will give you the $i$ component of the vector cross product of $V$ and $\hat{\omega}$. Oh, and to avoid confusion, you should say that you are summing over $j=1,2,3$, not $j=i,j,k$. – user197851 Nov 3 '18 at 21:57
• One more thing: $\hat{\omega}_i\hat{\omega}_j$ is not the Kronecker delta. If you sum over $j$, that term will give you the dot product of $\hat{\omega}$ with $V$, multiplied by $\hat{\omega}_i$, multiplied by the $[1-\cos\omega]$ factor. – user197851 Nov 3 '18 at 22:09
• @LonelyProf, thank you for the clarification. Upon review, I see exactly what you’re saying. I was hoping you could clarify one more thing. The expression I’m actually trying to evaluate includes $(R_ω)_{ij}\vec{V}$. Not sure what index I should give $V$ in this case. I see arguments for both $i,k$ and am not sure which is correct, if either. – Grant Cates Nov 5 '18 at 16:21

Hint :

Your notation is a little confusing. Here $$\;\boldsymbol{\hat{\omega}}\;$$ is a unit vector:

$$\begin{equation} \boldsymbol{\hat{\omega}}\boldsymbol{=}\left(\hat{\omega}_1,\hat{\omega}_2,\hat{\omega}_3\right)\,,\quad \Vert\boldsymbol{\hat{\omega}}\Vert^2\boldsymbol{=}\hat{\omega}^2_1\boldsymbol{+}\hat{\omega}^2_2\boldsymbol{+}\hat{\omega}^2_3\boldsymbol{=}1 \tag{01}\label{01} \end{equation}$$ The rotation is around this unit vector through an angle $$\;\omega$$. This is inconvenient notation since anyone would think that $$\;\omega\;$$ is the magnitude of $$\;\boldsymbol{\hat{\omega}}$$. So let replace $$\;\boldsymbol{\hat{\omega}}\;$$ by the unit vector $$\;\boldsymbol{n}\;$$
$$\begin{equation} \boldsymbol{n}\boldsymbol{=}\left(n_1,n_2,n_3\right)\,,\quad \Vert\boldsymbol{n}\Vert^2\boldsymbol{=}n^2_1\boldsymbol{+}n^2_2\boldsymbol{+}n^2_3\boldsymbol{=}1 \tag{02}\label{02} \end{equation}$$ and angle $$\;\omega\;$$ by the angle $$\;\theta$$.

Then your equation is
$$\begin{equation} (R_\theta)_{ij}V_j = \bigl[\left(1\boldsymbol{-}\cos\theta\right)n_i n_j \boldsymbol{+} \delta_{ij}\cos\theta\boldsymbol{+}\sin\theta\varepsilon_{ijk}n_k\bigr] V_j \tag{03}\label{03} \end{equation}$$ in vector form(1) $$\begin{equation} R_\theta\boldsymbol{V}\boldsymbol{=}\cos\theta\cdot\boldsymbol{V}\boldsymbol{+}\left(1\boldsymbol{-}\cos\theta\right)\left(\boldsymbol{n}\boldsymbol{\cdot}\boldsymbol{V}\right)\boldsymbol{n}\boldsymbol{+} \sin\theta\left(\boldsymbol{n}\boldsymbol{\times}\boldsymbol{V}\right) \tag{04}\label{04} \end{equation}$$ Note that the term $$\;\left(\boldsymbol{n}\boldsymbol{\cdot}\boldsymbol{V}\right)\boldsymbol{n}\;$$ is the projection of $$\;\boldsymbol{V}\;$$ on the axis $$\;\boldsymbol{n}\;$$ $$\begin{equation} \rm P_{\boldsymbol{n}}\boldsymbol{V}\boldsymbol{=}\left(\boldsymbol{n}\boldsymbol{\cdot}\boldsymbol{V}\right)\boldsymbol{n}=\left(\boldsymbol{n}\boldsymbol{n}^{\boldsymbol{\top}}\right)\boldsymbol{V}= \begin{pmatrix} \begin{bmatrix} n_1\\ n_2\\ n_3 \end{bmatrix} \begin{bmatrix} n_1 & n_2 & n_3 \vphantom{\dfrac{a}{b}} \end{bmatrix} \end{pmatrix} \begin{bmatrix} V_1\\ V_2\\ V_3 \end{bmatrix} \tag{05}\label{05} \end{equation}$$ that is $$\begin{equation} \rm P_{\boldsymbol{n}}\boldsymbol{=}\boldsymbol{n}\boldsymbol{n}^{\boldsymbol{\top}}= \begin{bmatrix} n_1\\ n_2\\ n_3 \end{bmatrix} \begin{bmatrix} n_1 & n_2 & n_3 \vphantom{\dfrac{a}{b}} \end{bmatrix} = \begin{bmatrix} n^2_1 & n_1n_2 & n_1n_3\\ n_2n_1 & n^2_2 & n_2n_3\\ n_3n_1 & n_3n_2 & n^2_3 \end{bmatrix} \tag{06}\label{06} \end{equation}$$ or $$\begin{equation} \left(\rm P_{\boldsymbol{n}}\right)_{ij}\boldsymbol{=}n_in_j \tag{07}\label{07} \end{equation}$$

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(1) See equation (07) in my answer there :Rotation of a vector