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

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}##.
  • #1
Salmone
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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?
 
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  • #2
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
$$\vec{A}=\frac{1}{2} \vec{B} \times \vec{x}, \quad \phi=0.$$
 
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  • #3
vanhees71 said:
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##.
 
  • #4
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##.
 
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  • #5
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?
 
  • #6
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
$$\vec{A}=\frac{1}{2} \vec{A} \times \vec{x}.$$
 
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  • #7
Okay, thank you so much.
 

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

How is a Hamiltonian obtained in a magnetic field from an EM field?

The Hamiltonian in a magnetic field can be obtained by first writing the Lagrangian for a charged particle moving in an electromagnetic field. The Lagrangian is then used to derive the equations of motion, and the Hamiltonian is obtained by performing a Legendre transformation on the Lagrangian.

What is the significance of obtaining a Hamiltonian in a magnetic field?

The Hamiltonian is a mathematical representation of the total energy of a system, and obtaining it in a magnetic field allows for the study of the dynamics and behavior of charged particles in this specific environment. It is an important tool in understanding and predicting the behavior of particles in magnetic fields.

Can a Hamiltonian be obtained for any type of magnetic field?

Yes, a Hamiltonian can be obtained for any type of magnetic field, as long as the system can be described using the Lagrangian formalism. This includes both static and time-varying magnetic fields.

Is there a specific method for obtaining a Hamiltonian in a magnetic field?

Yes, the most common method for obtaining a Hamiltonian in a magnetic field is through the use of the Lagrangian formalism and performing a Legendre transformation. However, there are other mathematical techniques that can also be used, depending on the specific system and its properties.

How is the Hamiltonian used in the study of particles in a magnetic field?

The Hamiltonian is used to determine the equations of motion for charged particles in a magnetic field, and these equations can then be solved to study the behavior and dynamics of the particles. It is also used to calculate important quantities such as the energy levels and transition probabilities of the particles in the magnetic field.

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