The following is a section from Sakurai's book "Modern Quantum mechanics" where he explains the translation operator $J$ commutation with position operator $\hat{x}$ on the subspace $|x' \rangle$:

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On the next page he then states "By choosing $d \vec{x}'$ in the direction of $\hat{x}_j$ and forming the scalar product with $\hat{x}_i$, we obtain $$[x_i, K_j] = i \delta_{ij}"$$ Can anyone see the working that yields that equation?

Thanks for any assistance


1 Answer 1


The equation $-i\mathbf{x}(\mathbf{K}\cdot d\mathbf{x}')+ i(\mathbf{K}\cdot d\mathbf{x}')\mathbf{x}=d\mathbf{x}'$ written in components is: \begin{equation} \sum_j\left(-ix_i K_k dx'_k+ iK_k dx'_k x_i\right)=dx'_i. \end{equation} Now, setting $dx'_k=\delta_{kj}$, we get $-ix_iK_j + iK_j x_i=\delta_{ij}$.

  • $\begingroup$ Thanks for your answer. What substitutions have you made before writing in component form and why are you setting $dx_{k}' = \delta_{kj}$, where does it state that this is done? $\endgroup$
    – Alex
    Commented Mar 14, 2017 at 12:15
  • $\begingroup$ @coconut Do you maybe know why the $\hat{K} \cdot d \vec{x}'$ in $\hat{J}$ is required to be dimensionless? $\endgroup$
    – user100411
    Commented Mar 14, 2017 at 12:39
  • 1
    $\begingroup$ @Alex The substitutions are: the definition of the scalar product $v\cdot w=\sum_i v_iw_i$ and writing the vector $\bf x$ and $d\bf x'$ in their components $x_i$, $dx'_i$. The equations should be valid for any $d\bf x'$, so it can be set to be equal to each $\hat{\bf x}_j$, whose component $k$ is $\delta_{kj}$. $\endgroup$
    – coconut
    Commented Mar 14, 2017 at 15:42
  • $\begingroup$ @JohnDoe I guess that $xJ\left| x\right>=x'\left|x'\right>$ implies that $xJ$ has dimensions of length and therefore $J$ should be dimensionless. $\endgroup$
    – coconut
    Commented Mar 14, 2017 at 17:02
  • $\begingroup$ @coconut I made a mistake, on page 45 of Sakurai he states that "the operator $\hat{K}$ has the dimension of 1/length because $\hat{K} \cdot d \vec{x}'$ must be dimensionless". Do you see the reasoning behind this? $\endgroup$
    – user100411
    Commented Mar 14, 2017 at 19:10

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