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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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How does the approximation work? I already tried it with Taylor-series but I failed.

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    $\begingroup$ it should work with Taylor series. $|x'+dx'\rangle = |x'\rangle + dx' \partial_{x'}|x'\rangle + O(dx'^2)$ and then you multiply with $dx'$ $\endgroup$
    – user245141
    Feb 26, 2020 at 16:03
  • $\begingroup$ Does it mean that it is being evaluated at the point $x' - dx'$? Is the ket being regarded as a function? $\endgroup$ Feb 26, 2020 at 16:43
  • $\begingroup$ why at a point $|x'-dx'\rangle$? it is evaluated at $|x'\rangle$. The ket is regarded as a function of $x'$, in the sense that we can define a derivative in a standard manner: $d|x\rangle/dx = \lim_{h\to 0} (|x+h\rangle-|x\rangle)/h$. Note that as $|x'\rangle$ reside in our Hilbert space, we have a well-defined addition and substraction operations on them, and also multiplication by scalars. $\endgroup$
    – user245141
    Feb 26, 2020 at 16:46

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We can define the derivative of a vector in Hilbert space by the usual definition of a derivative: $$\frac{d|x\rangle}{dx}=\lim_{dx\to0}\frac{|x+dx\rangle-|x\rangle}{dx}$$ Similarly we can define higher derivatives. With these in our hand, we can now formally define a Taylor expansion which up to first order looks like: $$|x_0+dx\rangle\approx|x_0\rangle+dx\left(\frac{d|x\rangle}{dx}\right)_{x_0}$$ Now in your case, since the operator itself is in first order, the derivative term will become second order and thus negligible. Finally giving: $$dx’|x’+dx’\rangle\approx dx’|x’\rangle$$

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  • $\begingroup$ stupid question but how do I get the " $dx$ " in the first order? In the Taylor series, the first order consists of the derivation of the function multiplied by $(x-a)$ where in general $x$ represents the variable and $a$ the evaluation point. So I guess $dx$ is coming from $(x-a)$, right? But how exactly? $\endgroup$ Feb 26, 2020 at 18:33
  • $\begingroup$ So that would be $(x-dx)$ but what about the variable $x$? $\endgroup$ Feb 26, 2020 at 18:57
  • $\begingroup$ If you want $dx$ to come out of the difference, set $x\to x+dx$ and $a\to x$. Then you’ll get the right result. We are working with the difference going to zero limit. $\endgroup$ Feb 26, 2020 at 19:27

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