# Non-Relativistic Electron Hamiltonian

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

then by relating the Lagrangian to the Hamiltonian using the identity

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

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

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

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

$$H_I=\frac{g}{2}\frac{e\hbar}{2mc}\overrightarrow{B}\cdot\overrightarrow{\sigma},\tag{4}$$

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

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

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?

• as far as I know, you cannot truly derive it from the classical non-relativistic model. The presence of the $\vec B \cdot \vec \sigma$ term is a quantum effect. You will surely recognize that from the classical point of view you're missing something: how should $\vec s$ be expressed in terms of the canonical coordinates $p,q$ ? To get that term, one has to consider the hamiltonian of the Dirac spinor in presence of an electromagnetic potential. Then by taking the non-relativistic limit one gets a term $\sim \vec B \cdot \vec \sigma$. – tbt Apr 24 '17 at 18:24
• @tbt I think the canonical coordinates here are $(x,p,\sigma) \in T^*\mathbb{R}^3 \times so^*(3) = T^*\mathbb{R}^3 \times su^*(2)$. The Poisson structure is the combination of the canonical symplectic structure on the cotangent bundle $T^*\mathbb{R}^3$ plus the Lie-Poisson structure on the dual of the Lie algebra $so(3) \cong su(3)$. – Futurologist Apr 24 '17 at 21:56
• @Fururologist Sure, you can make up an ad-hoc classical description for the spin. Now where in classical or non-relativistic quantum mechanics are you going to get the corresponding hamiltonian? – Emilio Pisanty Apr 24 '17 at 22:20
• Your formula (3) is wrong: the Hamiltonian is not a sum of a vector and a scalar. Formula (5) can only be valid for very-high-spin (nearly classical) particles; for an electron there may be only two possible projections of the spin on the magnetic field. – Vladimir Kalitvianski Dec 9 '18 at 19:35

## 1 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{d\mu}{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.

• Thanks, this does explain the origin of this term however I am none the wiser as to how a go about deriving $H_I$? – Sam Apr 23 '17 at 17:04
• @Sam I have added some more specific explanations. Check if they help you. – Futurologist Apr 24 '17 at 19:27
• This isn't a derivation; you're just re-postulating $U$ in a form that's transparently identical to what you claim to be deriving. – Emilio Pisanty Apr 24 '17 at 21:22
• @EmilioPisanty Nowhere have I claimed to be deriving $U$. The derivation of $U$ as well as the rest of the terms for that matter was not my intention. According to me, the question is not about the specific derivation of the form of the terms. Instead, what it seems to me, the question is about how to put the whole picture together, as well as what the meaning of the various terms is. Look at the question " is the $H_I$ factor $\phi$?" for example. What I am deriving is the Hamiltonian as a whole, not each of its terms. But if you have better explanations, feel free to pitch in. – Futurologist Apr 24 '17 at 21:51
• The question explicitly asks "how can I derive the factor $H_I$?". As has been noted in the comments to the question, it can only be derived from the Dirac equation. This answer as currently written just feels like a lot of obfuscation and symbol-pushing while strictly avoiding that fact. – Emilio Pisanty Apr 24 '17 at 22:19