# How does the position operator change in the Heisenberg picture as depicted in Modern Quantum Mechanics by JJ Sakurai?

In the quote shown below, they use the second approach, i.e the operator changes with time and not the state kets, making use of the unitary operator for infinitesimal translation. I understand that you can get the second to last step by ignoring the higher order terms of $$d\mathbf x$$ (if I'm not wrong), but I am unable to figure out how the commutation bracket leads to the final equation $$(2.2.7)$$.

In contrast, if we follow approach 2, we obtain

\begin{aligned}[b]|\alpha\rangle&\rightarrow|\alpha\rangle\\\mathbf x&\rightarrow\left(1+\frac{i\mathbf p\cdot d\mathbf x'}{\hbar}\right)\mathbf x\left(1-\frac{i\mathbf p\cdot d\mathbf x'}{\hbar}\right)\\&=\mathbf x+\left(\frac i{\hbar}\right)[\mathbf p\cdot d\mathbf x',\mathbf x]\\&=\mathbf x+d\mathbf x'\end{aligned}\tag{2.2.7}

We leave it as an exercise for the reader to show that both approaches lead to the same result for the expectation value of $$\mathbf x$$:

$$\langle\mathbf x\rangle\rightarrow\langle\mathbf x\rangle+\langle d\mathbf x'\rangle\tag{2.2.8}$$

• Please note that images of text and mathematics are very strongly discouraged on the site. Please use text and Mathjax. Commented Jul 15, 2021 at 10:09
• My bad.Did not know that that was the case. Will keep it mind. Thanks! Commented Jul 15, 2021 at 10:31

You're asking for a proof that $$\left[\mathbf{p}\cdot d\mathbf{x}^\prime,\,\mathbf{x}\right]=-i\hbar d\mathbf{x}^\prime$$, i.e. $$\left[p_i dx^\prime_i,\,x_j\right]=-i\hbar dx^\prime_j$$ in Einstein notation (for convenience I'm keeping indices downstairs). Since $$dx^\prime_i$$ is a c-number, this follows from the CCR $$[p_i,\,x_j]=-i\hbar\delta_{ij}$$.

• Thank you! I was too stuck up on the operators to have thought about it this way. Also, its safe to assume that the higher order terms have been ignored right? Commented Jul 15, 2021 at 7:08
• @NiranjanHaridasMenon They have, yes. That's why we don't have $dx^\prime_jdx^\prime_k$ terms.
– J.G.
Commented Jul 15, 2021 at 7:09

It's because momentum is the generator for translation. In classical mechanics , infinitesimal canonical transformation $$\eta\to \eta+\delta \eta$$ generated by the function G, will satisfy $$\delta \eta=\epsilon[\eta, G]$$ with $$|\epsilon|<<1$$. In this case we have $$\epsilon G=p.dx$$

Here is the way I maneged the problem: first rewrite $$\mathbf{p \cdot dx'}$$ in terms of the translation operator $$\mathscr{T}(\mathbf{dx'})$$ $$\mathscr{T}(\mathbf{dx'}) = 1 - \frac{i \mathbf{p \cdot dx'}}{\hbar} \rightarrow \mathbf{p \cdot dx'} = i\hbar\mathscr{T}(\mathbf{dx'}) - i\hbar$$

Now open the commutator $$[\mathbf{p \cdot dx' , x}] = \mathbf{p \cdot dx' x} - \mathbf{xp \cdot dx'}$$ and substitute $$\mathbf{p \cdot dx'}$$ by the relation above. You'll find: $$[\mathbf{p \cdot dx' , x}] = i\hbar \mathscr{T}(\mathbf{dx'}) \mathbf{x} - i\hbar \mathbf{x} \mathscr{T}(\mathbf{dx'}) = i\hbar[\mathscr{T}(\mathbf{dx'}) ,\mathbf{x}] = -i\hbar[\mathbf{x} , \mathscr{T}(\mathbf{dx'})]$$

Finally, the book has already proved that $$[\mathbf{x} , \mathscr{T}(\mathbf{dx'})] = \mathbf{dx'}$$, so $$[\mathbf{p \cdot dx' , x}] = -i\hbar \mathbf{dx'}$$