Motivating the ansatz for the infinitesimal translation operator

Question

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?

Answer 1

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}

Answer 2

In addition to the answers above, I provide a derivation for the finite translations from which the infinitesimal follows. The Lagrange-Boole formula from operational calculus reads \begin{equation} e^{ad_x}f(x)=f(x+a) \end{equation} The action of the translation operator $T_a$ on the eigenkets $\left|x\right>$ is $$T_a\left|x\right>=\left|x+a\right>$$ From above, using completeness, the action of $T_a$ on a state vector $\left|\Psi\right>$ is $$T_a\left|\Psi\right>=\int T_a\left|x\right>\left<x|\Psi\right>dx=\int\left|x+a\right>\left<x|\Psi\right>dx=\int\left|x'\right>\left<x'-a|\Psi\right>dx' $$ Restating the above action in the $X-$basis $$\left<x|T_a|\Psi\right>=\int\left<x|x'\right>\left<x'-a|\Psi\right>dx'=\int\ \delta(x-x')\left<x'-a|\Psi\right>dx'=\left<x-a|\Psi\right> $$ Now using this result and the Lagrange-Boole formula $$e^{-ad_x}\Psi(x)=\Psi(x-a)=\left<x-a|\Psi\right>=\left<x|T_a|\Psi\right> $$ This means that $T_a$ in the $X-$basis is $e^{-ad_x}$. Reverting to the abstract can be done as follows, where $P$ is the momentum operator, $$T_a=e^{-\frac{ai\hbar d_x}{i\hbar}}\rightarrow e^{\frac{aP}{i\hbar}}=e^{-\frac{iaP}{\hbar}} $$ Therefore $T_a=e^{-\frac{iaP}{\hbar}}$ for finite translations. The infinitesimal case, $T_\epsilon$, can be obtained as $$T_\epsilon=e^{-\frac{i\epsilon P}{\hbar}}=1-\frac{i\epsilon P}{\hbar}+O(\epsilon ^2) $$