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This question already has an answer here:

$$J(d \vec x') = \left(1 - \frac{i \vec p \cdot d \vec x'}{\hbar}\right)$$

This is the infinitesimal translation operator, as defined on p. 46 of Modern Quantum Mechanics by Sakurai & Napolitano.

How is this equation derived?

In particular, I am wondering why it is necessary for this operator to have an imaginary number.

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marked as duplicate by AccidentalFourierTransform, Kyle Kanos, ZeroTheHero, Cosmas Zachos, Jon Custer May 31 '17 at 1:22

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I have derived it in a not so rigorous manner and attached a picture of it. The derivation is different from Sakurai but in the same spirit.

enter image description here

If it has some mistakes, please do rectify.

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The proof starts with the observation (in 1d) that \begin{align} e^{-i x_0\hat p}\hat x e^{i x_0\hat p}&= \left(\hat 1 -i x_0\hat p +\ldots \right)\hat x \left(\hat 1 +i x_0\hat p +\ldots \right)\, ,\\ &=\hat x+ -ix_0\hat p\hat x + i x_0\hat x\hat p+\ldots\, ,\\ &=\hat x-i x_0[\hat p,\hat x]+\ldots \, ,\\ &=\hat x -i x_0 (-i\hbar)=\hat x-x_0 \end{align} which is just a translation by $x_0$. Thus, $$ e^{-i x_0\hat p}\psi(x)=\langle x\vert e^{-i x_0\hat p}\vert \psi\rangle = \langle x+x_0\vert\psi\rangle =\psi(x+x_0) $$ The 3d generalization is immediate.

Note that $U(x_0)=e^{-i x_0\hat p}$ must be unitary, meaning it must satisfy $U^{-1}(x_0)= U^\dagger(x_0)$. Since $\hat p$ is hermitian, $\hat p^\dagger= \hat p$; since $x_0$ is real, $x_0^*=x_0$. Thus, to guarantee unitarity we must have this form so that $$ U^\dagger(x_0)=\left(e^{-i x_0\hat p}\right)^\dagger=e^{+i x_0\hat p} $$ which is obviously $U^{-1}(x_0)$

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