The commutation of translation operator with Hamiltonian under electromagnetic field

The Hamiltonian (2.131) is

$$ H = frac{1}{2m}({bf p} + e{bf A})^2 + V({bf r}) + e {bf E} cdot {bf r}. tag{1}$$

The electric and magnetic fields are constant (homogeneous, so independent of position), and furthermore from section 2.1.6 we take the particular magnetic field ${bf B} = (0,0,B)$ with choice of vector potential ${bf A} = (0, x B,0)$.

The potential $V({bf r})$ is considered to be invariant under lattice translations $T_{bf R}$, so $[T_{bf R}, V({bf r})] = 0$.

We calculate

$$ [T_{bf R},e {bf E} cdot {bf r}] = e {bf E} cdot [T_{bf R},{bf r}] = e {bf E} cdot {bf R} T_{bf R}, tag{2}$$ where in the first equality we used the fact that the electric field ${bf E}({bf r})$ is independent of position, and in the last equality $[T_{bf R},{bf r}] = {bf R} T_{bf R}$.

So the only difficult part is the first term of the Hamiltonian. Using the notation ${bf P} = {bf p} + e {bf A}$, that term is equal to ${bf P}^2/2m$.

• We first calculate

  $$ [T_{bf R}, {bf P}] = [T_{bf R}, {bf p} + e {bf A}] = e [T_{bf R}, {bf A}] = e ({bf R} cdot nabla ) {bf A} T_{bf R} tag{3}$$

  In the second equality we used that the momentum operator ${bf p}$ is the generator of translations and therefore commutes with the translation operator. In the last equality we used the expansion

  $$ T_{bf R} f({bf r}) = f({bf r} + {bf R}) = f({bf r}) + {bf R} cdot nabla f({bf r}) + ldots$$ for any function $f({bf r})$.

• We write $[T_{bf R}, {bf P}] equiv {bf W}T_{bf R}$ with ${bf W} equiv e ({bf R} cdot nabla ) {bf A}$. Then

  $$begin{eqnarray} [T_{bf R}, {bf P}^2] &=& {bf P} cdot [T_{bf R},{bf P}] + [T_{bf R},{bf P}]cdot{bf P} = {bf P} cdot{bf W}T_{bf R} + {bf W}cdot T_{bf R}{bf P} nonumber &=& 2{bf P} cdot{bf W}T_{bf R} + {bf W}cdot {bf W} T_{bf R}. tag{4} end{eqnarray}$$ The first equality is just a commutator identity; we used $ T_{bf R}{bf P} = {bf P}T_{bf R} + [T_{bf R},{bf P}]$ in the last step, as well as ${bf P} cdot {bf W} = {bf W} cdot {bf P}$ (since $nabla cdot {bf A} = 0$).

• Since ${bf A}$ only depends on $x$ and not on $y$ and $z$, we have $({bf R} cdot nabla ) {bf A} = R_x partial_x {bf A}$. Using the explicit choice of ${bf A} = (0,xB,0)$, we see $partial_x {bf A} = (0,B,0)$ (remember $B$ is the constant magnetic field amplitude). This leads to the term

$$frac{1}{2m} {bf W}cdot {bf W} T_{bf R} = frac{e^2 B^2}{2m}R_x^2T_{bf R}. tag{5}$$

• For the final term we need a trick that is also employed in (2.48)-(2.51) in section 2.1.6. First note that ${bf v} equiv dot{bf r} = nabla_{bf p} H = {bf P}/m$ from (2.48). This is also true for our Hamiltonian (1).

  Then we have $frac{1}{2m} 2{bf P} cdot {bf W} = e{bf v} cdot ({bf R} cdot nabla) {bf A}$. The vector notation is getting messy, so let's use index notation: $e v_n R_m partial_m A_n$.

  We know that ${bf B} = nabla times {bf A}$, or $B_k = epsilon_{kmn} partial_m A_n$. Using an $epsilon$-identity, one can then obtain $partial_m A_n = epsilon_{mnk} B_k + partial_n A_m$. So we have

  $$v_n R_m partial_m A_n = v_n R_mepsilon_{mnk} B_k + v_n R_mpartial_n A_m. tag{6} $$

  The first term is in the desired form $({bf v} times {bf B})cdot {bf R}$. We need 'the trick' for the second term. We again use that ${bf A}$ only depends on $x$, so that only $v_x partial_x$ does not vanish in $v_m partial_m$. Also, $partial_x A_m = (0,B,0)_m$. Combining this, we have $$v_n partial_n A_m = (0,v_x B,0)_m = (0,dot{x} B,0)_m = partial_t(0,xB,0)_m = dot{A}_m. tag{7}$$ Finally we see that $$frac{1}{2m} 2{bf P} cdot {bf W}T_{bf R} = e ( {bf v} times {bf B} + dot{bf A}) cdot {bf R}T_{bf R}. tag{8}$$

Together, (2), (5) and (8) lead to the commutator

$$[T_{bf R},H] = e ( {bf v} times {bf B} + {bf E} + dot{bf A}) cdot {bf R}T_{bf R} + frac{e^2 B^2}{2m}R_x^2T_{bf R}. tag{9}$$