Charged Particles in an Electromagnetic Field

Gauge

Gauge theory is a type of field theory in which the Lagrangian is invariant under a certain continuous group of local transformations.

Given a distribution of charges and current, and appropriate boundary conditions, the electromagnetic field is unique. However, the electromagnetic potential $A^{\mu }\!$ is not unique. The Maxwell equations can be expressed by electromagnetic field tensor $F^{\mu \nu }\!$, which is defined by

$F^{\mu \nu }=\partial ^{\mu }A^{\nu }-\partial ^{\nu }A^{\mu }$.

If we set

$A'^{\mu }=A^{\mu }+\partial ^{\mu }\chi$,

then

${\begin{aligned}F'^{\mu \nu }&=\partial ^{\mu }A'^{\nu }-\partial ^{\nu }A'^{\mu }\\&=\partial ^{\mu }(A^{\nu }+\partial ^{nu}\chi )-\partial ^{\nu }(A^{\mu }+\partial ^{\mu }\chi )\\&=\partial ^{\mu }A^{\nu }-\partial ^{\nu }A^{\mu }+(\partial ^{\mu }\partial ^{\nu }-\partial ^{\nu }\partial ^{\mu })\chi \\&=F^{\mu \nu }\end{aligned}}$,

eg. the form of Maxwell equations will not change. So, we have a freedom $\partial \chi \!$, which is called Gauge Freedom here.

For the magnetic field case, we can check for gauge invariance:

$\left.\left.{\frac {(p-{\frac {e}{c}}A-\partial \chi )^{2}}{2m}}\right|\varphi \right\rangle =E|\varphi \rangle$,

Let $|\varphi \rangle =e^{{\frac {i}{\hbar }}{\frac {e}{c}}\chi }|\phi \rangle$,

the form ${\frac {(p-{\frac {e}{c}}A)^{2}}{2m}}|\phi \rangle =E|\phi \rangle$ will not change.

Usually, we use two gauges in magnetic field. One is the Laudau Gauge: $A(r)=(-yB,0,0)\!$, the other is the Symmetric Gauge: $A(r)={\frac {1}{2}}(-yB,xB,0)\!$.

We choose Laudau Gauge in the following calculation.

Motion in electromagnetic field

The Hamiltonian of a particle of charge $e\!$ and mass $m\!$ in an external electromagnetic field, which may be time-dependent, is given as follows:

$H={\frac {1}{2m}}\left(\mathbf {p} -{\frac {e}{c}}{\mathbf {A}}({\mathbf {r}},t)\right)^{2}+e\phi ({\mathbf {r}},t)$

where ${\mathbf {A({\mathbf {r}},t)}}\!$ is the vector potential and ${\phi ({\mathbf {r}},t)}\!$ is the Coulomb potential of the electromagnetic field. In a problem, if there is a momentum operator ${\mathbf {p}}\!$, it must be replaced by

$\left({\mathbf {p}}-{\frac {e}{c}}{\mathbf {A}}({\mathbf {r}},t)\right)$

if a particle is under the influence of an electromagnetic field.

Let's find out the Heisenberg equations of motion for the position and velocity operators. For position operator${\mathbf {r}}\!$, we have:

${\begin{aligned}{\frac {d{\mathbf {r}}}{dt}}&={\frac {1}{i\hbar }}\left[{\mathbf {r}},H\right]\\&={\frac {1}{i\hbar }}\left[{\mathbf {r}},{\frac {1}{2m}}\left({\mathbf {p}}-{\frac {e}{c}}{\mathbf {A}}({\mathbf {r}},t)\right)^{2}+e\phi ({\mathbf {r}},t)\right]\\&={\frac {1}{2im\hbar }}\left[{\mathbf {r}},\left({\mathbf {p}}-{\frac {e}{c}}{\mathbf {A}}({\mathbf {r}},t)\right)^{2}\right]\\&={\frac {1}{2im\hbar }}\left[{\mathbf {r}},\left({\mathbf {p}}-{\frac {e}{c}}{\mathbf {A}}({\mathbf {r}},t)\right)\right]\left({\mathbf {p}}-{\frac {e}{c}}{\mathbf {A}}({\mathbf {r}},t)\right)+{\frac {1}{2im\hbar }}\left({\mathbf {p}}-{\frac {e}{c}}{\mathbf {A}}({\mathbf {r}},t)\right)\left[{\mathbf {r}},\left({\mathbf {p}}-{\frac {e}{c}}{\mathbf {A}}({\mathbf {r}},t)\right)\right]\\&={\frac {1}{2im\hbar }}\left[{\mathbf {r}},{\mathbf {p}}\right]\left({\mathbf {p}}-{\frac {e}{c}}{\mathbf {A}}({\mathbf {r}},t)\right)+{\frac {1}{2im\hbar }}\left({\mathbf {p}}-{\frac {e}{c}}{\mathbf {A}}({\mathbf {r}},t)\right)\left[{\mathbf {r}},{\mathbf {p}}\right]\\&={\frac {1}{2im\hbar }}i\hbar \left({\mathbf {p}}-{\frac {e}{c}}{\mathbf {A}}({\mathbf {r}},t)\right)+{\frac {1}{2im\hbar }}\left({\mathbf {p}}-{\frac {e}{c}}{\mathbf {A}}({\mathbf {r}},t)\right)i\hbar \\&={\frac {1}{m}}\left({\mathbf {p}}-{\frac {e}{c}}{\mathbf {A}}({\mathbf {r}},t)\right),\end{aligned}}$

where (${\mathbf {r}}\!$ does not depend on $t\!$ explicitly) is the equation of motion for the position operator ${\mathbf {r}}$. This equation also defines the velocity operator ${\mathbf {v}}$:

${\mathbf {v}}={\frac {1}{m}}\left({\mathbf {p}}-{\frac {e}{c}}{\mathbf {A}}({\mathbf {r}},t)\right)$

The Hamiltonian can be rewritten as:

$H={\frac {m}{2}}{\mathbf {v}}\cdot {\mathbf {v}}+e\phi$

Therefore, the Heisenberg equation of motion for the velocity operator is:

${\begin{aligned}{\frac {d{\mathbf {v}}}{dt}}&={\frac {1}{i\hbar }}\left[{\mathbf {v}},H\right]+{\frac {\partial {\mathbf {v}}}{\partial t}}\\&={\frac {1}{i\hbar }}\left[{\mathbf {v}},{\frac {m}{2}}{\mathbf {v}}\cdot {\mathbf {v}}\right]+{\frac {1}{i\hbar }}\left[{\mathbf {v}},e\phi \right]-{\frac {e}{mc}}{\frac {\partial {\mathbf {A}}}{\partial t}}\end{aligned}}$

(Note that ${\mathbf {p}}\!$ does not depend on $t\!$ expicitly)

Let's use the following commutator identity:

$\left[{\mathbf {v}},{\mathbf {v}}\cdot {\mathbf {v}}\right]={\mathbf {v}}\times \left({\mathbf {v}}\times {\mathbf {v}}\right)-\left({\mathbf {v}}\times {\mathbf {v}}\right)\times {\mathbf {v}}$

Substituting, we get:

${\frac {d{\mathbf {v}}}{dt}}={\frac {1}{i\hbar }}{\frac {m}{2}}\left({\mathbf {v}}\times ({\mathbf {v}}\times {\mathbf {v}})-({\mathbf {v}}\times {\mathbf {v}})\times {\mathbf {v}}\right)+{\frac {1}{i\hbar }}e[{\mathbf {v}},\phi ]-{\frac {e}{mc}}{\frac {\partial {\mathbf {A}}}{\partial t}}$

Now let's evaluate ${\mathbf {v}}\times {\mathbf {v}}\!$ and $[{\mathbf {v}},\phi ]\!$:

${\begin{aligned}({\mathbf {v}}\times {\mathbf {v}})_{i}&=\epsilon _{ijk}v_{j}v_{k}\\&=\epsilon _{ijk}{\frac {1}{m}}\left(p_{j}-{\frac {e}{c}}A_{j}({\mathbf {r}},t)\right){\frac {1}{m}}\left(p_{k}-{\frac {e}{c}}A_{k}({\mathbf {r}},t)\right)\\&=-{\frac {e}{m^{2}c}}\epsilon _{ijk}\left(p_{j}A_{k}({\mathbf {r}},t)+A_{j}({\mathbf {r}},t)p_{k}\right)\\&=-{\frac {e}{m^{2}c}}\epsilon _{ijk}p_{j}A_{k}({\mathbf {r}},t)-{\frac {e}{m^{2}c}}\epsilon _{ijk}A_{j}({\mathbf {r}},t)p_{k}\\&=-{\frac {e}{m^{2}c}}\epsilon _{ijk}p_{j}A_{k}({\mathbf {r}},t)-{\frac {e}{m^{2}c}}\epsilon _{ikj}A_{k}({\mathbf {r}},t)p_{j}{\mbox{(Switching indices in the second terms)}}\\&=-{\frac {e}{m^{2}c}}\epsilon _{ijk}p_{j}A_{k}({\mathbf {r}},t)+{\frac {e}{m^{2}c}}\epsilon _{ijk}A_{k}({\mathbf {r}},t)p_{j}\\&=-{\frac {e}{m^{2}c}}\epsilon _{ijk}\left[p_{j},A_{k}({\mathbf {r}},t)\right]\\&=-{\frac {e}{m^{2}c}}\epsilon _{ijk}{\frac {\hbar }{i}}\nabla _{j}A_{k}({\mathbf {r}},t)\\&=i\hbar {\frac {e}{m^{2}c}}\left(\nabla \times {\mathbf {A}}\right)_{i}\end{aligned}}$

$\rightarrow \left[{\mathbf {v}}\times {\mathbf {v}}\right]=i\hbar {\frac {e}{m^{2}c}}\left(\nabla \times {\mathbf {A}}\right)=i\hbar {\frac {e}{m^{2}c}}{\mathbf {B}}$

and

${\begin{aligned}\left[{\mathbf {v}},\phi \right]&={\frac {1}{m}}\left[{\mathbf {p}}-{\frac {e}{c}}{\mathbf {A}}({\mathbf {r}},t),\phi ({\mathbf {r}},t)\right]\\&={\frac {1}{m}}\left[{\mathbf {p}},\phi ({\mathbf {r}},t)\right]\\&={\frac {1}{m}}{\frac {\hbar }{i}}\nabla \phi \end{aligned}}$

Substituting and rearranging, we get:

$m{\frac {d{\mathbf {v}}}{dt}}={\frac {e}{2c}}\left({\mathbf {v}}\times {\mathbf {B}}-{\mathbf {B}}\times {\mathbf {v}}\right)+e{\mathbf {E}}$

where

${\mathbf {E}}=-\nabla \phi -{\frac {1}{c}}{\frac {\partial {\mathbf {A}}}{\partial t}}$

Above is the quantum mechanical version of the equation for the acceleration of the particle in terms of the Lorentz force.

These results can also be deduced in Hamiltonian dynamics due to the similarity between the Hamiltonian dynamics and quantum mechanics.

Problems about Motion in electromagnetic field

Problem 1

Problem 2