# Hamiltonian for a charged particle in a magnetic field

• zhangnous
In summary, the author finds that the Hamiltonian of a charged particle in a magnetic field is given by summing the products of two vectors, one pointing in the direction of the magnetic field and the other perpendicular to it, and solving for the vector that points in the direction of the particle's motion.
zhangnous
I find a exercise in Leonard Susskind's book Classical Mechanics

the Hamiltonian of a charged particle in a magnetic field(ignore the electric field) is $$H=\sum_{i} \left\{ \frac{1}{2m} \left[ p_{i}-\frac{e}{c}A_{i}(x) \right]\left[ p_{i}-\frac{e}{c}A_{i}(x) \right] \right\},$$and$$p_{i}=m\dot x_{i}+\frac{e}{c}A_{i}(x)$$and Hamilton's equations of motion is $$\dot q_{i} = \frac{\partial H}{\partial p_{i}}$$$$\dot p_{i} = -\frac{\partial H}{\partial q_{i}}$$then I solve these equations in x. $$\dot x = \frac{\partial H}{\partial p_{x}} = \frac{p_{x}-\frac{e}{c}A_{x}}{m}$$and$$\dot p_{x}=m\ddot x + \frac{e}{c}\left( \frac{\partial A_{x}}{\partial x}\dot x+\frac{\partial A_{x}}{\partial y}\dot y+\frac{\partial A_{x}}{\partial z}\dot z \right)$$$$-\frac{\partial H}{\partial x}=-\left[ \frac{p_{x}-\frac{e}{c}A_{x}}{m}(-\frac{e}{c}\frac{\partial A_{x}}{\partial x})+\frac{p_{y}-\frac{e}{c}A_{y}}{m}(-\frac{e}{c}\frac{\partial A_{y}}{\partial x})+\frac{p_{z}-\frac{e}{c}A_{z}}{m}(-\frac{e}{c}\frac{\partial A_{z}}{\partial x}) \right]=\frac{e}{c}\left( \frac{\partial A_{x}}{\partial x}\dot x+\frac{\partial A_{y}}{\partial x}\dot y+\frac{\partial A_{z}}{\partial x}\dot z \right)$$then$$m\ddot x + \frac{e}{c}\left( \frac{\partial A_{x}}{\partial x}\dot x+\frac{\partial A_{x}}{\partial y}\dot y+\frac{\partial A_{x}}{\partial z}\dot z \right)=\frac{e}{c}\left( \frac{\partial A_{x}}{\partial x}\dot x+\frac{\partial A_{y}}{\partial x}\dot y+\frac{\partial A_{z}}{\partial x}\dot z \right)$$and get $$ma_{x}=\frac{e}{c}(B_{z}\dot y-B_{y}\dot z)$$

Am I a right way to work it out? I am not sure if it is right, because it seems like a math structure game.

vanhees71
Looks absolutely fine! The first Hamilton equation gives ##\dot{x}_i = \frac{1}{m} \left( p_i - \frac{e}{c}A_i \right)## and the second Hamilton equation gives\begin{align*}\dot{p}_i = m\ddot{x}_i + \frac{e}{c} \sum_j \left[ \frac{\partial A_i}{\partial x_j} \dot{x}_j \right] &=\frac{e}{mc} \sum_j \left[ \left( p_j - \frac{e}{c}A_j \right) \frac{\partial A_j}{\partial x_i} \right]\\

&= \frac{e}{c} \sum_j \left[ \frac{\partial A_j}{\partial x_i} \dot{x}_j \right]\end{align*}so you end up with\begin{align*}m\ddot{x}_i = \frac{e}{c} \sum_j \left[ \left( \frac{\partial A_j}{\partial x_i} - \frac{\partial A_i}{\partial x_j} \right) \dot{x}_j \right] = \frac{e}{c} \sum_j \left[ F_{ij} \dot{x}_j \right] = \frac{e}{c} \left(\mathbf{v} \times \mathbf{H} \right)_i
\end{align*}Note that in this last equation the indices of the Maxwell tensor ##F_{ij}## run over the set ##\{ 1,2,3 \}##, so here we are only dealing with magnetic components. To obtain the entire Lorentz force, you need to modify the original Hamiltonian to include an additional term ##e\phi##.

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vanhees71, Twigg and zhangnous
etotheipi said:
Looks absolutely fine! The first Hamilton equation gives ##\dot{x}_i = \frac{1}{m} \left( p_i - \frac{e}{c}A_i \right)## and the second Hamilton equation gives\begin{align*}\dot{p}_i = m\ddot{x}_i + \frac{e}{c} \sum_j \left[ \frac{\partial A_i}{\partial x_j} \dot{x}_j \right] &=\frac{e}{mc} \sum_j \left[ \left( p_j - \frac{e}{c}A_j \right) \frac{\partial A_j}{\partial x_i} \right]\\

&= \frac{e}{c} \sum_j \left[ \frac{\partial A_j}{\partial x_i} \dot{x}_j \right]\end{align*}
So the first Hamilton equation gives ##\dot{x}_i = \frac{1}{m} \left( p_i - \frac{e}{c}A_i \right)##, which is just the canonical momentum ##p_{i}=m\dot x_{i}+\frac{e}{c}A_{i}## in this special situation, and can also be used into the second Hamilton equation. It seems strange for me and looks like these two equation play with themselves.

But I remember the situation of one free particle. Hamiltonian is ##H = \frac{p^2}{2m} + V(x)##. First gives##\dot x = \frac{\partial H}{\partial p}=\frac{p}{m}## and ##\dot p = m\ddot x## then second gives ##\dot p = -\frac{\partial H}{\partial x}=-\frac{\partial V(x)}{\partial x}=F##. Finally have Newton's equation ##F=m\ddot x##. Which is also the two equation play with each other. This seems I just playing a joke with my own mind.

Thanks for you comprehensive explanation

vanhees71, Twigg and etotheipi
Hamilton's equations should be equivalent to Newton's laws in these cases. You can derive Lagrange's equations either through variational principles or by d'Alembert's principle. With Einstein summation convention (all following formulae assume summation over repeated indices), the Lagrangian itself satisfies ##d\mathcal{L} = \frac{\partial \mathcal{L}}{\partial q^i} dq^i + \frac{\partial \mathcal{}}{\partial \dot{q}^i} d\dot{q}^i = \dot{p}_i dq^i + p_i d\dot{q}^i##. You can do a Legendre transform ##\mathcal{H} := p_i \dot{q}^i - \mathcal{L}## from which it follows that ##d\mathcal{H} = -\dot{p}_i dq^i + \dot{q}^i dp_i## and by comparison to the expression for the total derivative of ##\mathcal{H}## you identify ##\dot{q}^i = \partial \mathcal{H} / \partial p_i## and ##\dot{p}_i = - \partial \mathcal{H} / \partial q^i##.

It all follows over to classical field theory, e.g. for a particle interacting with an electromagnetic field you have a Lagrangian $$\mathcal{L} = -mc^2 \sqrt{1-\frac{v^2}{c^2}} + \frac{e}{c} A_i v^i - e\phi$$Defining ##P^i = mv^i/\sqrt{1-\frac{v^2}{c^2}} + \frac{e}{c}A^i## one can check that this gives $$\mathcal{H} = e\phi + \sqrt{m^2c^4 + \left(P_i - \frac{e}{c} A_i \right) \left(P^i - \frac{e}{c} A^i \right) }$$which in the limit ##v/c \rightarrow 0## reduces to ##\mathcal{H} = e\phi + \frac{1}{2m} \left(P_i - \frac{e}{c} A_i \right) \left(P^i - \frac{e}{c} A^i \right)##, i.e. Susskind's expression just with the ##e\phi## term included. Also notice that ##\lim_{v/c \rightarrow 0} P^i = p^i = mv^i + \frac{e}{c}A^i##.

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vanhees71, zhangnous and Twigg

## 1. What is a Hamiltonian for a charged particle in a magnetic field?

The Hamiltonian for a charged particle in a magnetic field is a mathematical expression that describes the energy of a charged particle moving in a magnetic field. It takes into account the particle's kinetic energy, potential energy, and the effects of the magnetic field on its motion.

## 2. How is the Hamiltonian for a charged particle in a magnetic field derived?

The Hamiltonian for a charged particle in a magnetic field is derived from the classical mechanics equations of motion, specifically the Lorentz force equation. It is also derived from the quantum mechanical Schrödinger equation.

## 3. What is the significance of the Hamiltonian for a charged particle in a magnetic field?

The Hamiltonian for a charged particle in a magnetic field is significant because it allows us to understand and predict the behavior of charged particles in the presence of a magnetic field. It is used in many fields of physics, including electromagnetism, quantum mechanics, and solid state physics.

## 4. How does the Hamiltonian for a charged particle in a magnetic field affect the particle's motion?

The Hamiltonian for a charged particle in a magnetic field affects the particle's motion by introducing a force perpendicular to the particle's velocity, known as the Lorentz force. This force causes the particle to move in a circular or helical path, depending on the strength and direction of the magnetic field.

## 5. Are there any simplifications or approximations made in the Hamiltonian for a charged particle in a magnetic field?

Yes, there are simplifications and approximations made in the Hamiltonian for a charged particle in a magnetic field. In classical mechanics, the Hamiltonian assumes that the particle is moving at non-relativistic speeds and that the magnetic field is constant. In quantum mechanics, the Hamiltonian assumes that the particle's spin is negligible and that the magnetic field is uniform.

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