4.1.2 Electromagnetic Coupling#

Prompts

What is the kinetic momentum \(\hat{\boldsymbol{\pi}} = \hat{\boldsymbol{p}} - q\boldsymbol{A}\), and why does it — not the canonical \(\hat{\boldsymbol{p}}\) — set the velocity? Why is this answer forced once the minimally coupled Hamiltonian is fixed?
Why does the Heisenberg equation for \(\hat{\boldsymbol{\pi}}\) produce something with the form of the Lorentz force? Where do \(\boldsymbol{E}\) and \(\boldsymbol{B}\) enter the calculation, and why must the magnetic term be symmetrized at the operator level?
Why are the gauge-invariant combinations \(-\nabla\phi - \partial_t\boldsymbol{A}\) and \(\nabla\times\boldsymbol{A}\) the physical fields? What goes wrong if one tries to treat \(\phi\) or \(\boldsymbol{A}\) themselves as observables?
What is the field strength tensor \(F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu\), and why is it gauge-invariant? How does it package \(\boldsymbol{E}\) and \(\boldsymbol{B}\) into one object?

Lecture Notes#

Overview#

The gauge connection \((\phi, \boldsymbol{A})\) was forced in §4.1.1 by the local phase redundancy of \(\psi\). This section turns to its dynamical content: the Heisenberg equation for the minimally coupled \(\hat{H}\) identifies the kinetic momentum \(\hat{\boldsymbol{\pi}} = \hat{\boldsymbol{p}} - q\boldsymbol{A}\) as the velocity operator, produces the quantum Lorentz force, and reads off the physical electric and magnetic fields \(\boldsymbol{E}, \boldsymbol{B}\) from the potentials. The gauge connection is not bookkeeping — its dynamics is electromagnetism.

Kinetic Momentum#

From §4.1.1, local U(1) covariance gives the minimally coupled Hamiltonian

\[ \hat{H}=\frac{(\hat{\boldsymbol{p}}-q\boldsymbol{A})^{2}}{2m}+q\phi. \]

We now ask what this Hamiltonian means dynamically. In the Heisenberg picture, operators carry the time dependence, so we study equations of motion directly. The position equation is

\[ \frac{\mathrm{d}\hat{\boldsymbol{x}}}{\mathrm{d}t} =\frac{1}{\mathrm{i}\hbar}[\hat{\boldsymbol{x}},\hat{H}] =\frac{\hat{\boldsymbol{p}}-q\boldsymbol{A}}{m}. \]

Derivation: Velocity from the Hamiltonian

Use the minimally coupled Hamiltonian

\[ \hat{H}=\frac{1}{2m}(\hat{p}_j-qA_j)(\hat{p}_j-qA_j)+q\phi, \]

where repeated spatial indices are summed. Since \(\hat{x}_i\) commutes with any function of position,

\[\begin{split} \begin{split} [\hat{x}_i,A_j(\hat{\boldsymbol{x}},t)] &= 0, \\ [\hat{x}_i,\phi(\hat{\boldsymbol{x}},t)] &= 0. \end{split} \end{split}\]

Before assembling \([\hat{x}_i,\hat{H}]\), evaluate the building-block commutator with the kinetic momentum. Linearity of the commutator and the fact that \(\hat{x}_i\) commutes with any function of \(\hat{\boldsymbol{x}}\) give

\[\begin{split} \begin{split} [\hat{x}_i,\hat{p}_j-qA_j] &= [\hat{x}_i,\hat{p}_j] - q\,[\hat{x}_i,A_j(\hat{\boldsymbol{x}},t)] \\ &= \mathrm{i}\hbar\delta_{ij} - 0 \\ &= \mathrm{i}\hbar\delta_{ij}, \end{split} \end{split}\]

where the second line uses the canonical commutation \([\hat{x}_i,\hat{p}_j]=\mathrm{i}\hbar\delta_{ij}\) and \([\hat{x}_i,A_j(\hat{\boldsymbol{x}},t)]=0\) from above. The \(q\phi\) piece of \(\hat{H}\) also commutes with \(\hat{x}_i\), so only the kinetic term contributes. Applying the product rule \([A,BC]=[A,B]C+B[A,C]\) to \(\hat{\pi}_j\hat{\pi}_j\) with \(\hat{\pi}_j = \hat{p}_j - qA_j\),

\[\begin{split} \begin{split} [\hat{x}_i,\hat{H}] &=\frac{1}{2m}\left([\hat{x}_i,\hat{p}_j-qA_j](\hat{p}_j-qA_j)+(\hat{p}_j-qA_j)[\hat{x}_i,\hat{p}_j-qA_j]\right)\\ &=\frac{1}{2m}\left(\mathrm{i}\hbar\delta_{ij}(\hat{p}_j-qA_j)+(\hat{p}_j-qA_j)\mathrm{i}\hbar\delta_{ij}\right)\\ &=\frac{1}{2m}\left(\mathrm{i}\hbar(\hat{p}_i-qA_i)+(\hat{p}_i-qA_i)\mathrm{i}\hbar\right)\\ &=\frac{\mathrm{i}\hbar}{m}(\hat{p}_i-qA_i). \end{split} \end{split}\]

Therefore

\[ \frac{\mathrm{d}\hat{x}_i}{\mathrm{d}t} =\frac{1}{\mathrm{i}\hbar}[\hat{x}_i,\hat{H}] =\frac{\hat{p}_i-qA_i}{m}, \]

or, in vector form,

\[ \frac{\mathrm{d}\hat{\boldsymbol{x}}}{\mathrm{d}t} =\frac{\hat{\boldsymbol{p}}-q\boldsymbol{A}}{m}. \]

This identifies the velocity operator and kinetic momentum operator:

Kinetic Momentum

\[\begin{split} \begin{split} \hat{\boldsymbol{\pi}} &= \hat{\boldsymbol{p}} - q\boldsymbol{A}, \\ \hat{\boldsymbol{v}} &= \frac{\hat{\boldsymbol{\pi}}}{m}. \end{split} \end{split}\]

The kinetic momentum \(\hat{\boldsymbol{\pi}}\) determines motion. The canonical momentum \(\hat{\boldsymbol{p}}=-\mathrm{i}\hbar\nabla\) is gauge-dependent; \(\hat{\boldsymbol{\pi}}\) is the mechanical momentum that appears in the velocity.

Taking expectation values of this operator equation gives the measurable average velocity, but the dynamics itself is already valid at the operator level.

Caution: Kinetic Momentum Does Not Commute

Unlike canonical momentum (\([\hat{p}_i, \hat{p}_j] = 0\)), kinetic momentum components satisfy \([\hat{\pi}_i, \hat{\pi}_j] = \mathrm{i}q\hbar\epsilon_{ijk}B_k\) in a magnetic field. This noncommutativity is central to Landau quantization (§4.3).

The Lorentz Force Law#

Having identified \(\hat{\boldsymbol{v}} = \hat{\boldsymbol{\pi}}/m = (\hat{\boldsymbol{p}} - q\boldsymbol{A})/m\), we now ask how this velocity changes in time. The Heisenberg equation \(\mathrm{d}\hat{O}/\mathrm{d}t = [\hat{O},\hat{H}]/(\mathrm{i}\hbar)\) applies to operators with no explicit time dependence. But \(\hat{\boldsymbol{v}}\) depends on time in two ways:

Implicitly, through operator dynamics of \(\hat{\boldsymbol{x}}(t)\) and \(\hat{\boldsymbol{p}}(t)\) — captured by the commutator \([\hat{\boldsymbol{v}},\hat{H}]/(\mathrm{i}\hbar)\).
Explicitly, through \(t\)-dependence in \(\boldsymbol{A}(\hat{\boldsymbol{x}},t)\) — contributing \(\partial_t\hat{\boldsymbol{v}} = -(q/m)\partial_t\boldsymbol{A}\).

The full equation of motion is therefore

\[ \frac{\mathrm{d}\hat{\boldsymbol{v}}}{\mathrm{d}t} =\frac{1}{\mathrm{i}\hbar}[\hat{\boldsymbol{v}},\hat{H}] +\partial_t\hat{\boldsymbol{v}}, \]

where \(\partial_t\hat{\boldsymbol{v}} = -(q/m)\partial_t\boldsymbol{A}\) is the explicit time derivative. Evaluating the commutator gives the quantum Lorentz force:

Lorentz Force from Gauge Theory

\[ m\frac{\mathrm{d}\hat{\boldsymbol{v}}}{\mathrm{d}t} =q\left[\boldsymbol{E}+\frac{1}{2}(\hat{\boldsymbol{v}}\times\boldsymbol{B}-\boldsymbol{B}\times\hat{\boldsymbol{v}})\right]. \]

Here \(\boldsymbol{E}= -\nabla\phi-\partial_t\boldsymbol{A}\) and \(\boldsymbol{B}=\nabla\times\boldsymbol{A}\), evaluated at the position operator \(\hat{\boldsymbol{x}}\). The magnetic term is symmetrized because \(\hat{\boldsymbol{v}}\) need not commute with \(\boldsymbol{B}(\hat{\boldsymbol{x}},t)\).

For uniform fields, or in the classical limit where ordering is irrelevant, this reduces to the familiar Lorentz force \(m\boldsymbol{a}=q(\boldsymbol{E}+\boldsymbol{v}\times\boldsymbol{B})\).

Derivation: Operator Equation for Kinetic Momentum

We work in components throughout; repeated spatial indices are summed. Recall \(\hat{v}_i = \hat{\pi}_i/m\) with \(\hat{\pi}_i = \hat{p}_i - qA_i(\hat{\boldsymbol{x}},t)\).

Step 1 — Fundamental commutator: \([\hat{p}_i, f(\hat{\boldsymbol{x}})]\).

We establish the identity \([\hat{p}_i, f(\hat{\boldsymbol{x}})] = -\mathrm{i}\hbar\,(\partial_i f)(\hat{\boldsymbol{x}})\) by acting on an arbitrary state \(\psi(\boldsymbol{x})\). In the position representation \(\hat{p}_i = -\mathrm{i}\hbar\,\partial_i\):

\[\begin{split} \begin{split} [\hat{p}_i, f]\,\psi &= \hat{p}_i\,(f\,\psi) - f\,\hat{p}_i\,\psi \\ &= -\mathrm{i}\hbar\,\partial_i(f\,\psi) + \mathrm{i}\hbar\,f\,\partial_i\psi \\ &= -\mathrm{i}\hbar\,(\partial_i f)\,\psi - \mathrm{i}\hbar\,f\,\partial_i\psi + \mathrm{i}\hbar\,f\,\partial_i\psi \\ &= -\mathrm{i}\hbar\,(\partial_i f)\,\psi. \end{split} \end{split}\]

Since this holds for all \(\psi\), we obtain the operator identity

\[ [\hat{p}_i, f(\hat{\boldsymbol{x}})] = -\mathrm{i}\hbar\,(\partial_i f)(\hat{\boldsymbol{x}}). \]

The kinetic momentum \(\hat{\pi}_i = \hat{p}_i - qA_i\) inherits this identity because \(A_i\) is a function of \(\hat{\boldsymbol{x}}\) and therefore commutes with any other function of \(\hat{\boldsymbol{x}}\):

\[ [\hat{\pi}_i, f(\hat{\boldsymbol{x}})] = [\hat{p}_i - qA_i,\, f(\hat{\boldsymbol{x}})] = [\hat{p}_i, f(\hat{\boldsymbol{x}})] = -\mathrm{i}\hbar\,(\partial_i f)(\hat{\boldsymbol{x}}). \]

Step 2 — Kinetic-momentum commutator: \([\hat{\pi}_i, \hat{\pi}_j]\).

\[\begin{split} \begin{split} [\hat{\pi}_i, \hat{\pi}_j] &= [\hat{p}_i - qA_i,\; \hat{p}_j - qA_j] \\ &= [\hat{p}_i, \hat{p}_j] - q[\hat{p}_i, A_j] - q[A_i, \hat{p}_j] + q^2[A_i, A_j] \\ &= 0 - q(-\mathrm{i}\hbar\,\partial_i A_j) - q(\mathrm{i}\hbar\,\partial_j A_i) + 0 \\ &= \mathrm{i}\hbar\,q\,(\partial_i A_j - \partial_j A_i) \\ &= \mathrm{i}\hbar\,q\,\epsilon_{ijk}\,\epsilon_{klm}\,\partial_l A_m \\ &= \mathrm{i}\hbar\,q\,\epsilon_{ijk}\,B_k, \end{split} \end{split}\]

where the third line uses \([\hat{p}_i, A_j] = -\mathrm{i}\hbar\,\partial_i A_j\) (Step 1) and \([A_i, \hat{p}_j] = -[\hat{p}_j, A_i] = \mathrm{i}\hbar\,\partial_j A_i\). The last two lines insert the contracted-epsilon identity \(\epsilon_{ijk}\,\epsilon_{klm} = \delta_{il}\delta_{jm} - \delta_{im}\delta_{jl}\) in reverse to write \(\partial_i A_j - \partial_j A_i = \epsilon_{ijk}\,\epsilon_{klm}\,\partial_l A_m\), then recognise the curl in components \(B_k = \epsilon_{klm}\,\partial_l A_m\), i.e. \(\boldsymbol{B} = \nabla\times\boldsymbol{A}\).

Step 3 — Scalar-potential commutator: \([\hat{\pi}_i, q\phi]\).

Since \(\boldsymbol{A}\) commutes with \(\phi\) (both are functions of \(\hat{\boldsymbol{x}}\)), only the \(\hat{p}_i\) part contributes. Applying the Step 1 identity \([\hat{p}_i, f(\hat{\boldsymbol{x}})] = -\mathrm{i}\hbar\,(\partial_i f)(\hat{\boldsymbol{x}})\) with \(f = \phi\):

\[\begin{split} \begin{split} [\hat{\pi}_i, q\phi] &= [\hat{p}_i - qA_i,\; q\phi] \\ &= q\,[\hat{p}_i, \phi] \\ &= -\mathrm{i}\hbar\,q\,\partial_i\phi. \end{split} \end{split}\]

Step 4 — Kinetic-energy commutator: \([\hat{\pi}_i, \hat{\pi}_j\hat{\pi}_j]\).

Using the product rule \([A, BC] = [A,B]\,C + B\,[A,C]\):

\[\begin{split} \begin{split} [\hat{\pi}_i, \hat{\pi}_j\hat{\pi}_j] &= [\hat{\pi}_i, \hat{\pi}_j]\,\hat{\pi}_j + \hat{\pi}_j\,[\hat{\pi}_i, \hat{\pi}_j] \\ &= \mathrm{i}\hbar\,q\,\epsilon_{ijk}\,B_k\,\hat{\pi}_j + \hat{\pi}_j\,\mathrm{i}\hbar\,q\,\epsilon_{ijk}\,B_k \\ &= \mathrm{i}\hbar\,q\,\epsilon_{ijk}\,(B_k\,\hat{\pi}_j + \hat{\pi}_j\,B_k). \end{split} \end{split}\]

Note the symmetrized form: \(\hat{\pi}_j\) and \(B_k\) do not commute in general (since \(B_k\) depends on \(\hat{\boldsymbol{x}}\)), so both orderings must be kept.

Step 5 — Evaluate \([\hat{\pi}_i, \hat{H}]\).

Write the Hamiltonian as \(\hat{H} = \hat{\pi}_j\hat{\pi}_j/(2m) + q\phi\). Combining Steps 3 and 4:

\[\begin{split} \begin{split} [\hat{\pi}_i, \hat{H}] &= \frac{1}{2m}\,[\hat{\pi}_i, \hat{\pi}_j\hat{\pi}_j] + [\hat{\pi}_i, q\phi] \\ &= \frac{\mathrm{i}\hbar\,q}{2m}\,\epsilon_{ijk}\,(B_k\,\hat{\pi}_j + \hat{\pi}_j\,B_k) - \mathrm{i}\hbar\,q\,\partial_i\phi. \end{split} \end{split}\]

Therefore

\[ \frac{1}{\mathrm{i}\hbar}\,[\hat{\pi}_i, \hat{H}] = \frac{q}{2m}\,\epsilon_{ijk}\,(B_k\,\hat{\pi}_j + \hat{\pi}_j\,B_k) - q\,\partial_i\phi. \]

Step 6 — Assemble the full equation of motion.

The velocity operator \(\hat{v}_i = \hat{\pi}_i/m\) has explicit time dependence through \(\boldsymbol{A}(\hat{\boldsymbol{x}},t)\): \(\partial_t\hat{v}_i = -(q/m)\,\partial_t A_i\). The Heisenberg equation with explicit time dependence gives:

\[\begin{split} \begin{split} m\,\frac{\mathrm{d}\hat{v}_i}{\mathrm{d}t} &= \frac{1}{\mathrm{i}\hbar}\,[\hat{\pi}_i, \hat{H}] + m\,\partial_t\hat{v}_i \\ &= \frac{q}{2m}\,\epsilon_{ijk}\,(B_k\,\hat{\pi}_j + \hat{\pi}_j\,B_k) - q\,\partial_i\phi - q\,\partial_t A_i \\ &= q\,(-\partial_i\phi - \partial_t A_i) + \frac{q}{2m}\,\epsilon_{ijk}\,(B_k\,\hat{\pi}_j + \hat{\pi}_j\,B_k) \\ &= q\,E_i + \frac{q}{2}\,\epsilon_{ijk}\,(\hat{v}_j\,B_k + B_k\,\hat{v}_j), \end{split} \end{split}\]

where \(E_i = -\partial_i\phi - \partial_t A_i\) and we used \(\hat{\pi}_j/m = \hat{v}_j\).

The cross-product combination \(\hat{\boldsymbol{v}}\times\boldsymbol{B} - \boldsymbol{B}\times\hat{\boldsymbol{v}}\) matches the symmetrized form in the result above. Writing \((\hat{\boldsymbol{v}}\times\boldsymbol{B})_i = \epsilon_{ijk}\,\hat{v}_j B_k\) and \((\boldsymbol{B}\times\hat{\boldsymbol{v}})_i = \epsilon_{ijk}\,B_j \hat{v}_k\) in components,

\[\begin{split} \begin{split} (\hat{\boldsymbol{v}}\times\boldsymbol{B} - \boldsymbol{B}\times\hat{\boldsymbol{v}})_i &= \epsilon_{ijk}\,\hat{v}_j B_k - \epsilon_{ijk}\,B_j \hat{v}_k \\ &= \epsilon_{ijk}\,\hat{v}_j B_k - \epsilon_{ikj}\,B_k \hat{v}_j \\ &= \epsilon_{ijk}\,\hat{v}_j B_k + \epsilon_{ijk}\,B_k \hat{v}_j \\ &= \epsilon_{ijk}\,(\hat{v}_j B_k + B_k \hat{v}_j), \end{split} \end{split}\]

where line 2 relabels the dummy indices \(j \leftrightarrow k\) in the second term and line 3 uses \(\epsilon_{ikj} = -\epsilon_{ijk}\). The minus sign in the cross-product difference becomes a plus sign in the symmetrized form precisely because of this antisymmetry. Substituting back:

\[ m\,\frac{\mathrm{d}\hat{\boldsymbol{v}}}{\mathrm{d}t} = q\left[\boldsymbol{E} + \frac{1}{2}\,(\hat{\boldsymbol{v}}\times\boldsymbol{B} - \boldsymbol{B}\times\hat{\boldsymbol{v}})\right]. \]

Discussion: Lorentz force from gauge invariance

In classical mechanics, the Lorentz force \(m\boldsymbol{a} = q(\boldsymbol{E} + \boldsymbol{v} \times \boldsymbol{B})\) is an empirical law. In quantum mechanics, we derived it from the gauge principle alone. Does this mean quantum mechanics is “deeper” than classical mechanics? Could a different gauge group produce a different force law?

Electric and Magnetic Fields from Potentials#

The gauge potentials \((\phi, \boldsymbol{A})\) are gauge-dependent and not directly observable. The physical fields are:

\[\begin{split} \begin{split} \boldsymbol{E} &= -\nabla\phi - \frac{\partial\boldsymbol{A}}{\partial t}, \\ \boldsymbol{B} &= \nabla \times \boldsymbol{A}. \end{split} \end{split}\]

In spacetime notation with \(A_\mu = (\phi, \boldsymbol{A})\), both fields are encoded in the field strength tensor:

Field Strength Tensor

In natural unit \(c=1\),

\[ F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu. \]

\(F_{\mu\nu}\) is gauge-invariant: under \(A_\mu \to A_\mu + \partial_\mu\alpha\), the antisymmetry gives \(F'_{\mu\nu} = F_{\mu\nu}\) because \(\partial_\mu\partial_\nu\alpha = \partial_\nu\partial_\mu\alpha\).

This tensor notation makes the core dynamical equations compact:

Maxwell equations (covariant form)

\[ \partial_\mu F^{\mu\nu}=J^\nu, \]

\[ \partial_\lambda F_{\mu\nu}+\partial_\mu F_{\nu\lambda}+\partial_\nu F_{\lambda\mu}=0. \]

Covariant Lorentz-force equation

\[ m\frac{\mathrm{d}u^\mu}{\mathrm{d}\tau} = q\,F^{\mu\nu}u_\nu. \]

Spacetime Notation: A Gentle Introduction

The key idea is simple: space and time are two aspects of the same thing, and the index \(\mu = 0, 1, 2, 3\) lets us treat them on equal footing.

We set \(c = 1\) to keep the 4-vector notation compact (so \(x^0 = t\), \(A^0 = \phi\), \(p^0 = E\), \(F_{0i} = E_i\), etc.). Four-vectors. Each physical quantity has a time component (\(\mu = 0\)) and three spatial components (\(\mu = 1, 2, 3\)):

Four-vector	\(\mu = 0\) (time)	\(\mu = 1,2,3\) (space)
Coordinates \(x^\mu\)	\(x^0 = t\)	\(\boldsymbol{x} = (x^1, x^2, x^3)\)
Proper velocity \(u^\mu=\mathrm{d}x^\mu/\mathrm{d}\tau\)	\(u^0 = \gamma\)	\(\boldsymbol{u}=\gamma\boldsymbol{v}\)
Derivatives \(\partial_\mu\)	\(\partial_0 = \partial_t\)	\(\nabla = (\partial_1, \partial_2, \partial_3)\)
Energy-momentum \(p^\mu\)	\(p^0 = E\)	\(\boldsymbol{p} = (p^1, p^2, p^3)\)
Gauge field \(A^\mu\)	\(A^0 = \phi\)	\(\boldsymbol{A} = (A^1, A^2, A^3)\)
Covariant derivative \(D_\mu\)	\(D_0 = \partial_t + \mathrm{i}(q/\hbar)\phi\)	\(\boldsymbol{D} = \nabla - \mathrm{i}(q/\hbar)\boldsymbol{A}\)

Notice the pattern: every familiar pair (energy and momentum, scalar and vector potential, time and space derivatives) is unified into a single four-vector.

Index conventions. Upper and lower indices are related by the Lorentz metric \(g_{\mu\nu} = \mathrm{diag}(+1, -1, -1, -1)\):

\[\begin{split} \begin{split} a_\mu &= g_{\mu\nu}\, a^\nu, \\ a_0 &= a^0, \\ a_i &= -a^i \quad (i = 1,2,3). \end{split} \end{split}\]

Repeated indices (one upper, one lower) are summed over: \(a^\mu b_\mu = a^0 b^0 - \boldsymbol{a}\cdot\boldsymbol{b}\). This is the Einstein summation convention.

The field strength tensor. With this notation, \(F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu\) is a \(4 \times 4\) antisymmetric matrix whose components are exactly \(\boldsymbol{E}\) and \(\boldsymbol{B}\):

\[\begin{split} F_{\mu\nu} = \begin{pmatrix} 0 & E_x & E_y & E_z \\ -E_x & 0 & -B_z & B_y \\ -E_y & B_z & 0 & -B_x \\ -E_z & -B_y & B_x & 0 \end{pmatrix} \end{split}\]

\[\begin{split} F^{\mu\nu} = \begin{pmatrix} 0 & -E_x & -E_y & -E_z \\ E_x & 0 & -B_z & B_y \\ E_y & B_z & 0 & -B_x \\ E_z & -B_y & B_x & 0 \end{pmatrix} \end{split}\]

For example, with \(A^{\mu} = (\phi, \boldsymbol{A})\) and lowered components \(A_{\mu} = (\phi, -\boldsymbol{A})\), \(F_{01} = \partial_{t}(-A_{x}) - \partial_{x}\phi = -\partial_{t}A_{x} - \partial_{x}\phi = E_{x}\) (using \(E_{x} = -\partial_{x}\phi - \partial_{t}A_{x}\)), and \(F_{12} = \partial_{x}(-A_{y}) - \partial_{y}(-A_{x}) = -(\partial_{x}A_{y} - \partial_{y}A_{x}) = -B_{z}\), in agreement with the matrix above.

The beauty of \(F_{\mu\nu}\) is that it packages six quantities (\(\boldsymbol{E}\) and \(\boldsymbol{B}\)) into one geometric object that transforms simply under changes of reference frame.

Summary#

Kinetic momentum \(\hat{\boldsymbol{\pi}} = \hat{\boldsymbol{p}} - q\boldsymbol{A}\) determines velocity; its components do not commute in a magnetic field.
The Lorentz force \(m\boldsymbol{a} = q(\boldsymbol{E} + \boldsymbol{v}\times\boldsymbol{B})\) is derived from the gauge-invariant Hamiltonian, not assumed.
Electric and magnetic fields arise from gauge potentials: \(\boldsymbol{E} = -\nabla\phi - \partial_t\boldsymbol{A}\), \(\boldsymbol{B} = \nabla\times\boldsymbol{A}\).
The field strength tensor \(F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu\) is gauge-invariant.
The gauge connection of §4.1.1 carries dynamics: redundancy + minimally coupled \(\hat{H}\) ⇒ electromagnetism, with \(F_{\mu\nu}\) as the gauge-invariant content of \((\phi,\boldsymbol{A})\).

Homework#

1. Velocity for the Pauli Hamiltonian. Spin-1/2 motion in electromagnetic fields is governed by the Pauli Hamiltonian

\[ \hat{H}_{\mathrm{P}} = \frac{(\hat{\boldsymbol{p}} - q\boldsymbol{A}(\hat{\boldsymbol{x}}))^{2}}{2m} - \frac{q\hbar}{2m}\,\hat{\boldsymbol{\sigma}}\cdot\boldsymbol{B}(\hat{\boldsymbol{x}}) + q\phi(\hat{\boldsymbol{x}}), \]

where \(\hat{\boldsymbol{\sigma}} = (\hat{\sigma}^x, \hat{\sigma}^y, \hat{\sigma}^z)\) are Pauli operators acting on the spin index and \(\boldsymbol{B} = \nabla\times\boldsymbol{A}\).

(a) Compute \([\hat{x}_i, \hat{H}_{\mathrm{P}}]\) and show that the spin-magnetic term contributes nothing. Conclude that \(\mathrm{d}\hat{\boldsymbol{x}}/\mathrm{d}t = \hat{\boldsymbol{\pi}}/m\), identical to the spinless result.

(b) Generalize: argue that any term built from \(\hat{\boldsymbol{x}}\) and \(\hat{\boldsymbol{\sigma}}\) alone (without an accompanying \(\hat{\boldsymbol{p}}\)) cannot contribute to the velocity operator, regardless of the spin algebra.

(c) Where does the spin then influence motion? Compute the contribution of the spin-magnetic term to \(m\,\mathrm{d}\hat{\boldsymbol{v}}/\mathrm{d}t\) and show that it produces a Stern-Gerlach-type force \((q\hbar/2m)\,\nabla(\hat{\boldsymbol{\sigma}}\cdot\boldsymbol{B})\) — the spin couples to motion only through the gradient of \(\boldsymbol{B}\), not through \(\boldsymbol{B}\) itself.

2. Velocity uncertainty in B. For an electron in a uniform magnetic field \(\boldsymbol{B} = B\hat{z}\), the velocity components \(\hat{v}_i = \hat{\pi}_i/m\) obey \([\hat{v}_x, \hat{v}_y] = \mathrm{i}\hbar qB/m^2\).

(a) Use the commutator to derive a Heisenberg uncertainty relation for \(\Delta v_x\,\Delta v_y\).

(b) Translate this into an uncertainty for the cyclotron orbit area \(\pi r_c^2\) (with \(r_c = mv_\perp/(qB)\)). Show that the minimum orbit area is of order \(\pi\ell_B^2\), where \(\ell_B^2 = \hbar/(qB)\).

3. Cyclotron motion from Heisenberg. A particle of charge \(q\) moves in a uniform \(\boldsymbol{B} = B\hat{z}\) with \(\hat{H} = \hat{\boldsymbol{\pi}}^2/(2m)\) and \(\boldsymbol{E} = 0\).

(a) Use the Heisenberg equation \(\mathrm{d}\hat{\pi}_i/\mathrm{d}t = [\hat{\pi}_i, \hat{H}]/(\mathrm{i}\hbar)\) together with \([\hat{\pi}_x, \hat{\pi}_y] = \mathrm{i}q\hbar B\) to derive coupled equations of motion for \(\hat{\pi}_x\) and \(\hat{\pi}_y\).

(b) Solve them and show that \(\hat{\pi}_x(t), \hat{\pi}_y(t)\) rotate at frequency \(\omega_c = qB/m\), recovering classical cyclotron motion at the operator level.

(c) Verify that \(\hat{\boldsymbol{\pi}}^2\) is conserved, and connect this to energy conservation in the absence of an electric field.

4. Symmetric gauge verification. In a uniform magnetic field \(\boldsymbol{B} = B\hat{z}\), use the symmetric gauge \(\boldsymbol{A} = \frac{B}{2}(-y, x, 0)\).

(a) Verify that \(\boldsymbol{B} = \nabla \times \boldsymbol{A}\) gives \(B\hat{z}\).

(b) Compute \([\hat{\pi}_x, \hat{\pi}_y]\) explicitly using this gauge and confirm \([\hat{\pi}_x, \hat{\pi}_y] = \mathrm{i}\hbar qB\).

(c) Interpret this non-zero commutator physically: what does it say about simultaneous measurability of velocity components?

5. Field strength tensor components. Working in natural units (so the speed of light is set to one), write out all components of \(F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu\) as a \(4\times 4\) matrix (rows and columns labeled \(0, x, y, z\)). Verify that \(F_{0i} = E_i\) and \(F_{ij} = -\epsilon_{ijk}B_k\) by computing a few entries explicitly from the definitions \(\boldsymbol{E} = -\nabla\phi - \partial_t\boldsymbol{A}\) and \(\boldsymbol{B} = \nabla\times\boldsymbol{A}\).