How to obtain Hamiltonian in a magnetic field from EM field?

[Salmone]

To calculate the Hamiltonian of a charged particle immersed in an electromagnetic field, one calculates the Lagrangian with Euler's equation obtaining $L=\frac{1}{2}mv^2-e\phi+e\vec{v}\cdot\vec{A}$ where $\phi$ is the scalar potential and $\vec{A}$ the vector potential, and then we go to the Hamiltonian by calculating the conjugate momentum which is $\vec{p}=m\vec{v}+e\vec{A}$ obtaining $H=\frac{1}{2m}(\vec{p}-e\vec{A})^2 +e\phi$. In the case of a particle immersed in a constant magnetic field $\vec{B}=(0,0,B)$ the Hamiltonian is $H=\frac{1}{2m}(\vec{p}-e\vec{A})^2$, but how is this obtained? Do you go directly from the $H$ in the EM field or do you compute the Lagrangian from zero?

 

[vanhees71]

I don't know, what you question is. If it's about, how the Lagrangian is derived, it's simply that via the Euler-Lagrange equations you get the correct (non-relativistic) equation of motion for a charged particle in the em. field, i.e.,
$$m \ddot{\vec{x}}=e (\vec{E}+\vec{v} \times \vec{B}),$$
where
$$\vec{E}=-\partial_t \vec{A} -\vec{\nabla} \phi, \quad \vec{B}=\vec{\nabla} \times \vec{A}.$$
For the homogeneous magnetic field you can use
$$\vec{A}=\frac{1}{2} \vec{B} \times \vec{x}, \quad \phi=0.$$

 

[Salmone]

I don't know, what you question is. If it's about, how the Lagrangian is derived, it's simply that via the Euler-Lagrange equations you get the correct (non-relativistic) equation of motion for a charged particle in the em. field, i.e.,
$$m \ddot{\vec{x}}=e (\vec{E}+\vec{v} \times \vec{B}),$$
where
$$\vec{E}=-\partial_t \vec{A} -\vec{\nabla} \phi, \quad \vec{B}=\vec{\nabla} \times \vec{A}.$$
For the homogeneous magnetic field you can use
$$\vec{A}=\frac{1}{2} \vec{B} \times \vec{x}, \quad \phi=0.$$

I know how Lagrangian and Hamiltonian are derived in the case of a charged particle inside an electromagnetic field, I don't know how to derive the Hamiltonian in the case of a charged particle in a magnetic field of type $\vec{B}=(0,0,B)$. Why do I have to use that Gauge transformation? If $\vec{E}=0$, then what I obtain is $\vec{\nabla} \phi=0$.

 

[vanhees71]

For the homogeneous magnetic field, for the Hamiltonian you need just some vector potential, which is of course determined only up to a gauge transformation, i.e., any other vector potential
$$\vec{A}'=\vec{A}-\vec{\nabla} \chi$$
describes the same homogeneous $\vec{B}$-field, for any arbitrary scalar field, $\chi$.

 

[Salmone]

Yes but how do we obtain that $\phi=0$ from the consideration that $\vec{E}=0$ and that $\vec{B}=(0,0,B)$? We have an EM Hamiltonian $H=\frac{1}{2m}(\vec{p}-e\vec{A})^2 +e\phi$, now we set $\vec{E}=0$ and $\vec{B}=(0,0,B)$, what are logical consequencies that lead us to the final Hamiltonian?

 

[vanhees71]

You just need a vector and a scalar potential such that
$$\vec{E}=-\partial_t \vec{A}-\vec{\nabla} \phi=0, \quad \vec{\nabla} \times \vec{A}=\vec{B}=\text{const}.$$
It's obvious that you can make the potentials time independent. Then from the first equation you have $\phi=\text{const}$. Then it's easy to check that a possible choice for the vector potential is
$$\vec{A}=\frac{1}{2} \vec{A} \times \vec{x}.$$

 

[Salmone]

Okay, thank you so much.