# Translation invariance in two dimensions

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 12, Exercise 12.1.1.

In preparation for an examination of rotation invariance, we’ll have a look at translational invariance in two dimensions. We can apply much of what we did with translation in one dimension, where we showed that the momentum ${P}$ is the generator of translations. In particular, the translation operator ${T\left(\varepsilon\right)}$ for an infinitesimal translation ${\varepsilon}$ is

$\displaystyle T\left(\varepsilon\right)=I-\frac{i\varepsilon}{\hbar}P \ \ \ \ \ (1)$

In two dimensions, we can write an infinitesimal translation as ${\boldsymbol{\delta}a}$ where

$\displaystyle \boldsymbol{\delta}a=\delta a_{x}\hat{\mathbf{x}}+\delta a_{y}\hat{\mathbf{y}} \ \ \ \ \ (2)$

In one dimension, we showed earlier that

$\displaystyle \left\langle x\left|T\left(\varepsilon\right)\right|\psi\right\rangle =\psi\left(x-\varepsilon\right) \ \ \ \ \ (3)$

The analogous relation in two dimensions is

$\displaystyle \left\langle x,y\left|T\left(\boldsymbol{\delta}a\right)\right|\psi\right\rangle =\psi\left(x-\delta a_{x},y-\delta a_{y}\right) \ \ \ \ \ (4)$

We can verify that the correct form for ${T\left(\boldsymbol{\delta}a\right)}$ is

 $\displaystyle T\left(\boldsymbol{\delta}a\right)$ $\displaystyle =$ $\displaystyle I-\frac{i}{\hbar}\boldsymbol{\delta}a\cdot\mathbf{P}\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle I-\frac{i}{\hbar}\left(\delta a_{x}P_{x}+\delta a_{y}P_{y}\right) \ \ \ \ \ (6)$

Using the representation of momentum in the position basis, which is

 $\displaystyle P_{x}$ $\displaystyle =$ $\displaystyle -i\hbar\frac{\partial}{\partial x}\ \ \ \ \ (7)$ $\displaystyle P_{y}$ $\displaystyle =$ $\displaystyle -i\hbar\frac{\partial}{\partial y} \ \ \ \ \ (8)$

the LHS of 4 is, using ${\left\langle x,y\left|\psi\right.\right\rangle =\psi\left(x,y\right)}$:

 $\displaystyle \left\langle x,y\left|T\left(\boldsymbol{\delta}a\right)\right|\psi\right\rangle$ $\displaystyle =$ $\displaystyle \left\langle x,y\left|I-\frac{i}{\hbar}\left(\delta a_{x}P_{x}+\delta a_{y}P_{y}\right)\right|\psi\right\rangle \ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \psi\left(x,y\right)-\delta a_{x}\frac{\partial\psi}{\partial x}-\delta a_{y}\frac{\partial\psi}{\partial y} \ \ \ \ \ (10)$

The last line is also what we get if we expand the RHS of 4 to first order in ${\boldsymbol{\delta}a}$, which verifies that 5 is correct, so that the two-dimensional momentum ${\mathbf{P}}$ is the generator of two-dimensional translations.

We can apply the exponentiation technique we used in the one-dimensional case to obtain the translation operator for a finite translation in two dimensions. We need to be careful that we don’t run into problems with non-commuting operators, but in view of 7 and 8 and the fact that derivatives with respect to different independent variables commute, we see that

$\displaystyle \left[P_{x},P_{y}\right]=0 \ \ \ \ \ (11)$

We can divide a finite translation ${\mathbf{a}}$ into ${N}$ small steps, each of size ${\frac{\mathbf{a}}{N}}$, so that the translation is

$\displaystyle T\left(\mathbf{a}\right)=\left(I-\frac{i}{\hbar N}\mathbf{a}\cdot\mathbf{P}\right)^{N} \ \ \ \ \ (12)$

Because the two components of momentum commute, we can take the limit of this expression to get the exponential form:

$\displaystyle T\left(\mathbf{a}\right)=\lim_{N\rightarrow\infty}\left(I-\frac{i}{\hbar N}\mathbf{a}\cdot\mathbf{P}\right)^{N}=e^{-i\mathbf{a}\cdot\mathbf{P}/\hbar} \ \ \ \ \ (13)$

Again, because the two components of momentum commute, we can combine two translations, by ${\mathbf{a}}$ and then by ${\mathbf{b}}$, to get

$\displaystyle T\left(\mathbf{b}\right)T\left(\mathbf{a}\right)=e^{-i\mathbf{b}\cdot\mathbf{P}/\hbar}e^{-i\mathbf{a}\cdot\mathbf{P}/\hbar}=e^{-i\left(\mathbf{a}+\mathbf{b}\right)\cdot\mathbf{P}/\hbar}=T\left(\mathbf{b}+\mathbf{a}\right) \ \ \ \ \ (14)$