# Non-Relativistic Electron Hamiltonian

Physics Asked by Sam on August 24, 2020

I have determined a Hamiltonian for an electron using an appropriate Lagrangian of the form
$$L=frac{1}{2m}(moverrightarrow{v}+frac{q}{c}overrightarrow{A})^2-frac{q^2}{2mc^2}overrightarrow{A}cdotoverrightarrow{A}+qphi.tag{1}$$

then by relating the Lagrangian to the Hamiltonian using the identity

$$H=overrightarrow{v}cdotoverrightarrow{p}-L=overrightarrow{v}(moverrightarrow{v}+frac{q}{c}overrightarrow{A})-frac{1}{2m}((moverrightarrow{v}+frac{q}{c}overrightarrow{A})^2-frac{q^2}{2mc^2}overrightarrow{A}cdotoverrightarrow{A}-qphi.tag{2}$$

I’ve then simplified this and made $q=-e$ so that is describing an electron.

$$H=frac{1}{2m}(overrightarrow{p}+frac{e}{c}overrightarrow{A})-ephi.tag{3}$$

This is as far as I have managed to go however I have read that, if we have an electron in a purely magnetic field there is an additional interaction such that

$$H_I=frac{g}{2}frac{ehbar}{2mc}overrightarrow{B}cdotoverrightarrow{sigma},tag{4}$$

where $overrightarrow{sigma}=2overrightarrow{s}$ this makes our Hamiltonian the following

$$H=frac{1}{2m}(overrightarrow{p}+frac{e}{c}overrightarrow{A})^2+frac{g}{2}frac{ehbar}{2mc}overrightarrow{B}cdotoverrightarrow{sigma}.tag{5}$$

Note: $g=2$

How do I derive the factor $H_I$? is the $H_I$ factor $phi$ when the electron is in the magnetic field?

## One Answer

I assume that the analogous situation in classical mechanics is the following. If you are describing the motion of an object, its Hamiltonian looks one way if you treat your object as a mass-point. However, if you start considering the object as a rigid body and not a point, your model starts to include description of the time-evolution of the body's center of mass and the evolution of it's angular momentum. The situation you are encountering seems to be analogous. Your first Hamiltonian describes the motion of a charged particle treated as a charged point. However, the addition of the second term includes description of its magnetic moment, i.e. your charged particle is considered as a dipole (or something that has a magnetic moment, like a rotating charged sphere) rather than a point. So the second, extended Hamiltonian describes the evolution of both the position of the electron and the evolution of its magnetic moment.

I hope this explanation makes sense.

The factor $H_I$ has nothing to do with the electric potential $phi$. It has to do with the dynamics of the magnetic moment.

Here is how I see the "derivation" you seek. When describing classically your charged particle, say the electron, you think of it as follows: it is a little magnetized segment, not a point. The center of the segment is its center of mass. When describing the dynamics of this magnetized segment, you would like to know where it is in space and how it is oriented in space. Thus, to describe fully the configuration of the segment, you want to know the position $x in mathbb{R}^3$ of its center, where $x = (x_1, x_2, x_3)$ is a vector in three space, and the orientation of its axis (which is aligned with the segment and is oriented according to the magnetic field of the segment) determined by a vector $mu = (mu_1, mu_2, mu_3) in mathbb{R}^3$ in three space, so that the vector $mu$ is aligned with the segment. The vector's orientation and length are determined in accordance with the segment's own magnetic field.

Now, you want to know how $x = x(t)$ and $mu = mu(t)$ evolve in time under external static (i.e. time independent) magnetic field, given by a magnetic vector potential $$A = A(x) = big(A_1(x), A_2(x), A_3(x)big)$$ Observe that the magnetic field doesn't change with time (hence static), so $phi equiv 0$. However, the magnetic field could be inhomogeneous in space, i.e. it may vary from point to point. The vector field of the external magnetic field is determined by $B = B(x) = nabla times A(x)$. As it is usually the case in classical dynamics, in order to track the time evolution of your object's configuration $x(t), , mu(t)$, you have to track, in the Lagrangian picture, the time evolution of the quantities $x(t),, v(t), , mu(t)$ where $v(t) = frac{dx}{dt}(t)$ is the velocity of $x(t)$. In the Hamiltonian picture, you have to track $x(t),, p(t), , mu(t)$ where $p(t)$ is the conjugate momentum of $x$ and carries more or less the same information as the velocity $v$. The Lagrangian and the Hamiltonian functions consist of the terms that determine the dynamics of $x, p$ plus the terms that determine the dynamics of the magnetic moment $mu$. Thus, the Lagraingian for $x, v$ is $$L_0(x,dot{x})=frac{1}{2} , m {v}^2+ {q}, {A}(x) cdot v = frac{1}{2} , m {dot{x}}^2+ {q}, {A}(x) cdot dot{x}$$ where $v = dot{x} = frac{dx}{dt}$. The Lagrangian term responsible for the time evolution of the magnetic moment $mu$ is the magnetic potential energy $$U = U(x, mu) = - , B(x) cdot mu$$ In these notations $, cdot ,$ is the dot product of the three dimensional space $mathbb{R}^3$ and $v^2 = v cdot v$. To form the total Lagrangian, we need to subtract the potential energy $U$ from the rest of the lagrangian $L_0$, obtaining the full lagrangian that determines the time evolution of the position $x$, the velocity $dot{x}$ and the magnetic moment $mu$: $$L = L_0 - U$$ $$L(x, dot{x}, mu) = L_0(x, dot{x}) - U(x, mu)$$ $$L(x, dot{x}, mu) =frac{1}{2} , m {v}^2+ {q}, {A}(x) cdot v + B(x)cdot mu$$ Analogously, in the Hamiltonian picture, the Hamiltonian for $x, p$ is $$H_0(x,p)=frac{1}{2m} big( p - {q}, {A}(x) big)^2$$ The Hamiltonian term responsible for the time evolution of the magnetic moment $mu$ is the magnetic potential energy $$H_I = U = U(x, mu) = - , B(x) cdot mu$$ To form the total Hamiltonian, we add the potential energy $U$, i.e. the magnetic moment potential Hamiltonian $H_I$, to the rest of the Hamiltonian $H_0$, obtaining the full Hamiltonian $H$ that determines the time evolution of the position $x$, the momentum $p$ and the magnetic moment $mu$: $$H = H_0 + U$$ $$H(x, p, mu) = H_0(x, p) + H_I(x, mu) = H_0(x, p) + U(x, mu)$$ $$H(x, p, mu) =frac{1}{2m} big( p - {q}, {A}(x) big)^2 - B(x)cdot mu$$ Consequently, the evolution equations for $x, p, mu$ are begin{align*} frac{dx}{dt} &= frac{partial H_0}{partial p}(x,p)\ frac{dp}{dt} &= - frac{partial H_0}{partial x}(x,p) - frac{partial U}{partial x}(x,mu)\ frac{dmu}{dt} &= frac{partial U}{partial mu}(x,mu) times mu = - , B(x) times mu end{align*} Observe the form of the third equation. It looks like that because $mu$ is a type of quantity very closely related to angular momentum and its dynamics is driven by torques not by forces. Double check all the signs (pluses and minuses), because I did not check any of these derivations carefully, I simply derived them off the top of my head, relying on general principle and philosophy.

Finally, if you think about the magnetic potential $U = - , B(x) cdot mu$, it makes sense. Whenever the magnetic moment $mu$ is aligned with the magnetic vector field $B(x)$ the potential $U$ attains its minimum value, and if you look at the third equation, you see that $B(x) times mu = 0$. Thus, no precession of $mu$ will occur as long as $mu$ is aligned with the magnetic field. However, if $mu$ is not aligned, the potential $U$ increases, and precession starts to emerge because $B(x) times mu neq 0$. In the special case when $mu$ is perpendicular to the magnetic field, the potential $U=0$ which is its largest value, so the precession of $mu$ is the strongest.

Answered by Futurologist on August 24, 2020

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