A problem with some relation to the harmonic oscillator is that of the motion of a charged particle in a constant and uniform magnetic field. In classical mechanics, we know that the Hamiltonian for this system is
where
is the charge of the particle and
is the vector potential. In fact, to obtain the Hamiltonian for any system in the presence of a magnetic field, we simply make the replacement,
In quantum mechanics, we introduce the magnetic field in the same way; this process is referred to as minimal coupling.
Gauge Invariance in Quantum Mechanics
We know from Maxwell's equations that the classical physics of charges in a magnetic field is invariant under a gauge transformation,
We will now show how this is expressed in quantum mechanics.
In the position basis, the Schrödinger equation for a charged particle in a magnetic field is
If we perform a gauge transformation on the magnetic field, then this equation becomes
If we make the substitution,
then we recover the original equation. Therefore, a gauge transformation of the magnetic field effectively introduces a phase factor to the wave function. This does result in a change in the canonical momentum, but it will have no effect on, for example, the probability density for finding the particle at a given position or, as we will see later, on the expectation value of the position or velocity of the particle.
Usually, we use two gauges in magnetic field. One is the Laudau Gauge:
, the other is the Symmetric Gauge:
.
We choose Laudau Gauge in the following calculation.
Motion in electromagnetic field
The Hamiltonian of a particle of charge
and mass
in an external electromagnetic field, which may be time-dependent, is given as follows:

where
is the vector potential and
is the Coulomb potential of the electromagnetic field. In a problem, if there is a momentum operator
, it must be replaced by

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
, we have:
![{\displaystyle {\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1ddca812a6c3243cf7c7fb082b1456fcdbd64b74)
where (
does not depend on
explicitly)
is the equation of motion for the position operator
.
This equation also defines the velocity operator
:

The Hamiltonian can be rewritten as:

Therefore, the Heisenberg equation of motion for the velocity operator is:
![{\displaystyle {\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/162d5b8a0361de72aaeb8bba6ead1846e37042c2)
(Note that
does not depend on
expicitly)
Let's use the following commutator identity:
![{\displaystyle \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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1173fb8102a99df2c9225555f1e3fb094ffdb245)
Substituting, we get:
![{\displaystyle {\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/97092f914e3c58afbccc3c0c28e779f881636a16)
Now let's evaluate
and
:
![{\displaystyle {\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2fc5762d80db95912d3c197c1e123ed47a271767)
![{\displaystyle \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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7da231424ae0291a900a8d6c664f78464e664cba)
and
![{\displaystyle {\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/12bae98980ef693e9f61889b971bc96cca6451df)
Substituting and rearranging, we get:

where

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