AM final project - Problem 1

문제: AM Finalterm problem / 답안 모음: AM finalterm answer

Problem 1

Based on the following effective action:

where is a momentum-space Berry connection, derive the Hamiltonian equation of motion and solve them to find the below expression

L = (\mathbf{p} + e\mathbf{A}) \cdot \dot{\mathbf{x}} - (|\mathbf{p}| + e\phi) - \mathbf{a} \cdot \dot{\mathbf{p}}

\frac{d}{dt}\frac{\partial L}{\partial \dot{x}^j} - \frac{\partial L}{\partial x^j} = 0

\frac{\partial L}{\partial \dot{x}^j} = p_j + eA_j

\frac{d}{dt}\frac{\partial L}{\partial \dot{x}^j} = \frac{d}{dt}(p_j + eA_j) = \dot{p}j + e,\dot{x}^i\partial{x^i} A_j,

\frac{\partial L}{\partial x^j} = e,\partial_{x^j} A_i,\dot{x}^i - e,\partial_{x^j}\phi

\dot{p}j + e,\partial{x^i} A_j,\dot{x}^i - e,\partial_{x^j} A_i,\dot{x}^i + e,\partial_{x^j}\phi = 0

\dot{p}j = -e,\partial{x^j}\phi + e(\partial_{x^j} A_i - \partial_{x^i} A_j)\dot{x}^i = eE_j + e(\dot{x}\times B)_j

\boxed{\dot{\mathbf{p}} = e\mathbf{E} + e\dot{\mathbf{x}}\times \mathbf{B}}
\tag{I}

\frac{d}{dt}\frac{\partial L}{\partial \dot{p}^j} - \frac{\partial L}{\partial p^j} = 0

\frac{\partial L}{\partial \dot{p}^j} = -a_j \quad \Rightarrow \quad \frac{d}{dt}(-a_j) = -\dot{p}^i\partial_{p^i}a_j,

\frac{\partial L}{\partial p^j} = \dot{x}^j - \partial_{p^j}a_i,\dot{p}^i - \frac{p^j}{|p|}

-\partial_{p^i}a_j,\dot{p}^i - \dot{x}^j + \partial_{p^j}a_i,\dot{p}^i + \hat{p}^j = 0

\dot{x}^j = \hat{p}^j + (\partial_{p^j}a_i - \partial_{p^i}a_j)\dot{p}^i = \hat{p}^j - \epsilon_{ijk}\Theta_{k},\dot{p}^i = \hat{p}^j - [\boldsymbol{\Theta} \times \dot{\mathbf{p}}]

\boxed{\dot{\mathbf{x}} = \hat{\mathbf{p}} - \boldsymbol{\Theta} \times \dot{\mathbf{p}}}
\tag{II}

\dot{\mathbf{x}} = \hat{\mathbf{p}} - \boldsymbol{\Theta} \times (e\mathbf{E} + e\dot{\mathbf{x}}\times \mathbf{B})

= \hat{\mathbf{p}} - e(\boldsymbol{\Theta} \times \mathbf{E}) - e,\boldsymbol{\Theta} \times (\dot{\mathbf{x}} \times \mathbf{B})

\dot{\mathbf{x}} = \hat{\mathbf{p}} - e(\boldsymbol{\Theta} \times \mathbf{E}) - e(\boldsymbol{\Theta} \cdot \mathbf{B})\dot{\mathbf{x}} + e(\boldsymbol{\Theta} \cdot \dot{\mathbf{x}})\mathbf{B}

(1 + e,\boldsymbol{\Theta} \cdot \mathbf{B})\dot{\mathbf{x}} - e(\boldsymbol{\Theta} \cdot \dot{\mathbf{x}})\mathbf{B} = \hat{\mathbf{p}} - e(\boldsymbol{\Theta} \times \mathbf{E})
\tag{III}

(\boldsymbol{\Theta} \cdot \dot{\mathbf{x}})(1 + e,\boldsymbol{\Theta} \cdot \mathbf{B}) - e(\boldsymbol{\Theta} \cdot \dot{\mathbf{x}})(\boldsymbol{\Theta} \cdot \mathbf{B}) = \boldsymbol{\Theta} \cdot \hat{\mathbf{p}} - e,\boldsymbol{\Theta} \cdot (\boldsymbol{\Theta} \times \mathbf{E})

(\boldsymbol{\Theta} \cdot \dot{\mathbf{x}})\Big[(1 + e,\boldsymbol{\Theta} \cdot \mathbf{B}) - e(\boldsymbol{\Theta} \cdot \mathbf{B})\Big] = \boldsymbol{\Theta} \cdot \hat{\mathbf{p}}

\boldsymbol{\Theta} \cdot \dot{\mathbf{x}} = \boldsymbol{\Theta} \cdot \hat{\mathbf{p}}
\tag{IV}

(1 + e,\boldsymbol{\Theta} \cdot \mathbf{B})\dot{\mathbf{x}} - e(\boldsymbol{\Theta} \cdot \hat{\mathbf{p}})\mathbf{B} = \hat{\mathbf{p}} - e(\boldsymbol{\Theta} \times \mathbf{E})

\boxed{\mathfrak{m},\frac{d\mathbf{x}}{dt} = \hat{\mathbf{p}} + e(\mathbf{E} \times \boldsymbol{\Theta}) + e(\boldsymbol{\Theta} \cdot \hat{\mathbf{p}})\mathbf{B}} \tag{2.1-1}

\dot{\mathbf{p}} = e\mathbf{E} + e(\hat{\mathbf{p}} - \boldsymbol{\Theta} \times \dot{\mathbf{p}}) \times \mathbf{B}

= e\mathbf{E} + e\hat{\mathbf{p}} \times \mathbf{B} - e(\boldsymbol{\Theta} \times \dot{\mathbf{p}}) \times \mathbf{B}

\dot{\mathbf{p}} + e(\boldsymbol{\Theta} \cdot \mathbf{B})\dot{\mathbf{p}} = e\mathbf{E} + e\hat{\mathbf{p}} \times \mathbf{B} + e,\boldsymbol{\Theta}(\dot{\mathbf{p}} \cdot \mathbf{B}) \tag{V}

(1 + e,\boldsymbol{\Theta} \cdot \mathbf{B})(\dot{\mathbf{p}} \cdot \mathbf{B}) = e(\mathbf{E}\cdot \mathbf{B}) + e(\hat{\mathbf{p}} \times \mathbf{B})\cdot \mathbf{B} + e,(\boldsymbol{\Theta}\cdot \mathbf{B})(\dot{\mathbf{p}} \cdot \mathbf{B})

(1 + e,\boldsymbol{\Theta} \cdot \mathbf{B})(\dot{\mathbf{p}} \cdot \mathbf{B}) - e(\boldsymbol{\Theta} \cdot \mathbf{B})(\dot{\mathbf{p}} \cdot \mathbf{B}) = e(\mathbf{E} \cdot \mathbf{B})

\dot{\mathbf{p}} \cdot \mathbf{B} = e(\mathbf{E} \cdot \mathbf{B})
\tag{VI}

\boxed{\mathfrak{m},\frac{d\mathbf{p}}{dt} = e\mathbf{E} + e\hat{\mathbf{p}} \times \mathbf{B} + e^2(\mathbf{E} \cdot \mathbf{B})\boldsymbol{\Theta}} \tag{2.1-2}

\mathfrak{m} = 1 + e,\boldsymbol{\Theta} \cdot \mathbf{B}

\begin{cases}
\mathfrak{m},\dfrac{d\mathbf{x}}{dt} = \hat{\mathbf{p}} + e\mathbf{E} \times \boldsymbol{\Theta} + (\boldsymbol{\Theta} \cdot \hat{\mathbf{p}}),e\mathbf{B} \[8pt]
\mathfrak{m},\dfrac{d\mathbf{p}}{dt} = e\mathbf{E} + e\hat{\mathbf{p}} \times \mathbf{B} + e^{2}(\mathbf{E} \cdot \mathbf{B})\boldsymbol{\Theta}
\end{cases}