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The commutator of position operator $\mathbf x$ with the translation operator defined as $\mathscr{J}\left(d \mathbf{x}^{\prime}\right)\left|\mathbf{x}^{\prime}\right\rangle=\left|\mathbf{x}^{\prime}+d \mathbf{x}^{\prime}\right\rangle$ is $\mathbf {dx'}$ where sakurai says "$\mathbf {dx'}$ is a number multiplied by a identity operator"

So: $[\mathbf x, \mathscr{J( d\mathbf x'}]= d\mathbf x' \mathbf 1 $

Then $[\mathbf x, \mathscr{J( \mathbf dx'}] |u \rangle = \mathbf{dx'}\mathbf 1|u \rangle =\mathbf {dx'}$ $| u\rangle$

Isn't this number $\mathbf {dx'}$ actually a euclidean vector? And we are multiplying a euclidean vector with a ket vector, which I've not seen how that's defined.

What does $\mathbf {dx'}$ $|u\rangle$ mean then?

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    $\begingroup$ There are some problems with your latex, what do you want to indicate with these dots? Also you switch between primed and unprimed and bold and non bold d and x a lot, is that on purpose? $\endgroup$
    – Kuhlambo
    Aug 23, 2022 at 17:23
  • $\begingroup$ This chapter of Sakurai is why I started hating QM at some point, he is saying $\mathbf{x}^{\prime}+d \mathbf{x}^{\prime}$ is an eigenvalue, what this means is that he is using a product space of the eigen spaces of the x, y and z position operators , this means $\hat x_i |\mathbf{x}\rangle =x_i |\mathbf{x}\rangle $, but what an eigenvalue that is itself a vector would be, i really don't know. And in the chapter Sakurai himself refers to (1.4 ) he also defines it like above, for each operator separately, what this vector eigenvalue business is I don't know. I just really hate this book. $\endgroup$
    – Kuhlambo
    Aug 23, 2022 at 19:03
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    $\begingroup$ Maybe this answer helps but I think it's just a really horrible book to learn from... a vector can't be an eigenvalue, maybe in some fancy notation, but not at this stage in a beginner book and not without comment. $\endgroup$
    – Kuhlambo
    Aug 23, 2022 at 19:17
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    $\begingroup$ The point is that one must say what $|\mathbf{x}\rangle$ means. In general, it should mean something like $|x\rangle \otimes |y\rangle \otimes |z\rangle$. The $X$ operator on this new Hilbert space (the three-fold tensor product of 1D spaces) is then lifted to $X\otimes \mathbb I \otimes \mathbb I$ and similarly for $Y$ and $Z$. $\endgroup$ Aug 23, 2022 at 19:18
  • $\begingroup$ @JasonFunderberker I am not confident enough with this math to write an answer but I think this is what kashmirs confusion stems from, as did mine long ago. So maybe that would really help. Sakurai is so "student friendly" that tensor products are never brought up... $\endgroup$
    – Kuhlambo
    Aug 23, 2022 at 19:26

1 Answer 1

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The expression $$[\hat{\mathbf x}, \mathscr T(\mathrm d\mathbf x)] = \mathrm d\mathbf x\ \mathbb I$$ is shorthand for the three equations $$[\hat x,\mathscr T(\mathrm d\mathbf x) ] = \mathrm dx \ \mathbb I\qquad [\hat y,\mathscr T(\mathrm d\mathbf x)] = \mathrm dy\ \mathbb I\qquad [\hat z,\mathscr T(\mathrm d\mathbf x)] = \mathrm dz \ \mathbb I$$

or, in index form, $[\hat x_i ,\mathscr T(\mathrm d\mathbf x)] = \mathrm dx_i$. This can be proven straightforwardly:

$$[\hat x_i ,\mathscr T(\mathrm d\mathbf x)] |\mathbf x\rangle = \hat x_i \mathscr T(\mathrm d\mathbf x)|\mathbf x\rangle - \mathscr T(\mathrm d\mathbf x)\hat x_i |\mathbf x\rangle = (x_i + \mathrm dx_i)|\mathbf x+\mathrm d\mathbf x\rangle - x_i |\mathbf x + \mathrm d\mathbf x\rangle$$ $$= \mathrm dx_i |\mathbf x+\mathrm d\mathbf x\rangle = \mathrm dx_i |\mathbf x\rangle + \mathscr O(|\mathrm d\mathbf x|^2)$$ $$\implies [\hat x_i, \mathscr T(\mathrm d\mathbf x)] = \mathrm dx_i \ \mathbb I + \mathscr O(|\mathrm d\mathbf x|^2)$$


Typically we don't define the multiplication of kets by vectors, so when you see an expression like this it should be interpreted as shorthand (though we should also remember that vector operators have well-defined transformation properties under rotations).

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    $\begingroup$ It's making a lot of sense now. Thank you :) $\endgroup$
    – Kashmiri
    Aug 24, 2022 at 6:29

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