0
$\begingroup$

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 $|x' \rangle$:

enter image description here

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

$\endgroup$

1 Answer 1

1
$\begingroup$

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 $-ix_iK_j + iK_j x_i=\delta_{ij}$.

$\endgroup$
8
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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