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?

  • 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

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