Daily Archives: Thu, 13 April 2017

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)