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 $[L_i,V_j]=i\hbar\varepsilon_{ijk}V_k$.)


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 Zachos Nov 18 '18 at 17:19

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – Kyle Kanos, Emilio Pisanty, Jon Custer, Community, Cosmas Zachos
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 1
    $\begingroup$ 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 $j=i,j,k$. $\endgroup$ – user197851 Nov 3 '18 at 21:57
  • $\begingroup$ 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. $\endgroup$ – user197851 Nov 3 '18 at 22:09
  • $\begingroup$ @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. $\endgroup$ – 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}


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


Not the answer you're looking for? Browse other questions tagged or ask your own question.