The commutation of translation operator with Hamiltonian under electromagnetic field

Question

The book "Introduction to Solid-state" by Madelung gives the commutation of translation operator with Hamiltonian under an electromagnetic field in section 2.2.9 . The Hamiltonian is

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

It then gives the commutator without proof:

$$\left[T_{R}, H\right]=\left(\xi \cdot R+\frac{e^{2} B^{2}}{2 m} R_{x}^{2}\right) T_{R} \quad \mathrm{w} \mathrm{i} \mathrm{th} \quad \xi=e \boldsymbol{v} \times \boldsymbol{B}+e \boldsymbol{E}+e \dot{\boldsymbol{A}}$$

I failed to work it out. Could anyone please help me?

My derivation is as follows

Inserting $\mathbf{A}=(0,Bx,0)$ into the Hamiltonian, we have: $$ \begin{array}{ll} H &=\frac{1}{2 m}(\mathbf{p}+e \mathbf{A})^{2}+V(r)+e \mathbf{E} \cdot \mathbf{r} \\ &=\frac{1}{2 m}(p^2+e\mathbf{p} \cdot \mathbf{A} +e \mathbf{A}\cdot \mathbf{p} +e^2A^2)+e \mathbf{E} \cdot \mathbf{r}+V(r)\\ &=\frac{1}{2 m}(p^2+ep_yBx + eBxp_y+e^2B^2x^2)+e \mathbf{E} \cdot \mathbf{r}+V(r). \end{array} $$ Now we consider the commutation of this Hamiltonian with a translation operator $T(\mathbf{R})$, where $\mathbf{R}=(R_x,R_y,R_z)$. We first work out the commutation of $T(\mathbf{R})$ with the term related to the electric field:

$$ \begin{array}{ll} [T(\mathbf{R}),e \mathbf{E} \cdot \mathbf{r}]&=e[T(\mathbf{R}),\mathbf{E}]\cdot\mathbf{r}+e\mathbf{E}\cdot[T(\mathbf{R}),\mathbf{r}]\\ &=e[\mathbf{E}(\mathbf{r}+\mathbf{R})-\mathbf{E}(\mathbf{r})]T(\mathbf{R})+e\mathbf{E}\cdot\mathbf{R}T(\mathbf{R})\\ &=e\mathbf{E}\cdot\mathbf{R}T(\mathbf{R}). \end{array} $$ Then we go to the terms related to the magnetic field, we first pay attention to the term linearly dependent to the magnetic field: $$ \begin{array}{ll} [T(\mathbf{R}),e p_yBx]&=ep_yB[T(\mathbf{R}),x]\\ &=ep_yBR_xT(\mathbf{R}). \end{array} $$ We then pay attention to the terms quadratically dependent to the magnetic field:

$$ \begin{array}{ll} [T(\mathbf{R}),e^2 B^2x^2]&=e B^2[T(\mathbf{R}),x^2]\\ &=e^2B^2[T(\mathbf{R}),x^2]\\ &=e^2B^2{x[T(\mathbf{R}),x]+[T(\mathbf{R}),x]x}\\ &=e^2B^2[xR_xT(\mathbf{R})+R_xT(\mathbf{R})x]\\ &=e^2B^2[xR_xT(\mathbf{R})+R_xxT(\mathbf{R})++R^2_xT(\mathbf{R})]\\ &=2e^2B^2xR_xT(\mathbf{R})+e^2B^2R^2_xT(\mathbf{R}). \end{array} $$ In summary, the commutation is

$$ \begin{array}{ll} [T(\mathbf{R}),H]&=e\mathbf{E}\cdot\mathbf{R}T(\mathbf{R})+\frac{1}{2m}\{2ep_yBR_xT(\mathbf{R})+2e B^2xR_xT(\mathbf{R})+e B^2R^2_xT(\mathbf{R})\}\\ &=\{e\mathbf{E}\cdot\mathbf{R}+\frac{1}{m}[ep_yBR_x+ e B^2xR_x]+\frac{1}{2m}e B^2R^2_x\}T(\mathbf{R})\\ &=\{e\mathbf{E}\cdot\mathbf{R}+ev_yBR_x+\frac{1}{m} (e^2B^2xR_x)+\frac{1}{2m}e^2B^2R^2_x\}T(\mathbf{R}) \end{array} $$ The final results given in the book is :

$$ \begin{array}{} \left[T(\mathbf{R}), H\right]&=\left(\xi \cdot \mathbf{R}+\frac{e^{2} B^{2}}{2 m} R_{x}^{2}\right) T(\mathbf{R}) \quad \mathrm{w} \mathrm{i} \mathrm{th} \quad \xi=e \boldsymbol{v} \times \boldsymbol{B}+e \boldsymbol{E}+e \dot{\boldsymbol{A}}\\ &=(eBv_yR_x-eBv_xR_y+\frac{e^{2} B^{2}}{2 m} R_{x}^{2}+e\mathbf{E}\cdot\mathbf{R}+e\dot{\boldsymbol{A}}\cdot\mathbf{R})T(\mathbf{R})\\ &=(eBv_yR_x+\frac{e^{2} B^{2}}{2 m} R_{x}^{2}+e\mathbf{E}\cdot\mathbf{R})T(\mathbf{R}) \end{array} $$

Comparing the two results, we find an extral term $\frac{1}{m} (e^2B^2xR_x)$. Is this term left out by the book?

Answer 1

1. 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 ${\bf A} = (0, x B,0)$.

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

3. 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 $[T_{\bf R},{\bf r}] = {\bf R} T_{\bf R}$.

4. 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

     $$v_n R_m \partial_m A_n = v_n R_m\epsilon_{mnk} B_k + v_n R_m\partial_n A_m. \tag{6} $$

     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 $$\frac{1}{2m} 2{\bf P} \cdot {\bf W}T_{\bf R} = e ( {\bf v} \times {\bf B} + \dot{\bf A}) \cdot {\bf R}T_{\bf R}. \tag{8}$$

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

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