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I'm reading Sakurai's Modern QM right now and in the first chapter he states a number of conditions required for a translation operator: unitarity, $$T(\mathrm{d}\mathbf{x})T(\mathrm{d}\mathbf{x^\prime}) = T(\mathrm{d}\mathbf{x}+\mathrm{d}\mathbf{x^\prime}),$$ $$T(-\mathrm{d}\mathbf{x}) = T^{-1}(\mathrm{d}\mathbf{x}),$$ and going to 1 for small translations.

He then makes the ansatz (1.6.20 in the new version): $$ T(\mathrm{d}\mathbf{x}) = 1-i \mathbf{K} \cdot \mathrm{d}\mathbf{x} $$ and demonstrates that this form of $T$ meets the requirements and is generally the correct choice. Here, $\mathbf K =(\mathrm K_x,\mathrm K_y,\mathrm K_z)$ with each component hermitian.

I was hoping somebody could provide some motivation for this particular choice. Is there any reason why it is to be expected or natural? How would you come up with this form of the operator by yourself?

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  • $\begingroup$ The form comes from the idea that T(dx) is infinitesimally close to the identity. $\endgroup$ – Michael Sep 6 '14 at 20:08
  • $\begingroup$ @Michael this is indeed one of the requirements, but there is still room for a different form, is there not? $\endgroup$ – Danu Sep 6 '14 at 20:11
  • $\begingroup$ Wait... excuse an experimentalist's stupid question... that's the generator for the operator, is it not? Isn't the operator itself the exponential of the generator? $\endgroup$ – CuriousOne Sep 6 '14 at 20:14
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    $\begingroup$ @CuriousOne it's the infinitesimal operator. By summing over infinitesimal translations you get the general translation operator which has the exponential form. This may mean it is the generator, I'll admit that I'm not completely confident with the implications of that term. $\endgroup$ – johnpaton Sep 6 '14 at 20:17
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    $\begingroup$ K is the generator(s) of T. $\endgroup$ – Michael Sep 6 '14 at 20:20
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This may be a bit more abstract than what you're looking for, but this is just a special case of a more general construction; determining the infinitesimal generators of a representation of a Lie Group acting on a vector space. In the case at hand, the Lie group is $G=\mathbb R^3$, the group of spatial translations, and the vector space is the Hilbert space of a quantum system.

The idea is that we want to find operators that act on states of the Hilbert spaces of the theory and which "function in the same way" as the elements of the group of translations do when they act on vectors in three dimensions. The phrase that I just put in quotes mathematically means that we are looking for a group homomorphism, which tells us that the first two conditions you wrote down must be satisfied. Explicitly, then, we are looking for a group homomorphism $\rho_G:G\to \mathrm{GL}(\mathcal H)$, a mapping that assigns an invertible operator on the Hilbert space $\mathcal H$ to each element of the translation group.

Once we accept this, we notice that if the group is a Lie group, then for each element $g$ in (the connected component containing the identity of) $G$, we can write this element as an $g = e^{-ia^iT_i}$, where $T_i$ with $i=1,2,3$ are the elements of the basis of the Lie algebra $\mathfrak g$ of $G$. For example, the $T_i$ could be chosen as the basis which corresponds to translations in the directions of the standard ordered basis vectors $\mathbf e_i$.

Finally, the infinitesimal generators $K_i$ of $\rho_G$ corresponding to $T_i\in\mathfrak g$ are defined as follows: \begin{align} K_i = \frac{d}{d\epsilon}\rho_G(e^{-i\epsilon T_i})\Big|_{\epsilon = 0} \end{align} This corresponds to writing down the ansatz you gave above because the definition of $K_i$ implies that \begin{align} \rho_G(e^{-i\epsilon T_i}) = I + i\epsilon K_i + \mathcal O(\epsilon^2). \end{align}

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