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

• I
• Salmone
In summary, after setting ##\vec{E}=0## and solving for ##\vec{\nabla} \phi##, one can find that ##\phi=\text{const}## and that the only possible vector potential is ##\vec{A}=\frac{1}{2} \vec{A} \times \vec{x}##.f

#### 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?

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##.

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?

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.