4.1.2 Electromagnetic Coupling#

Worked solutions for the homework problems in the 4.1.2 Electromagnetic Coupling lecture. Each problem statement is reproduced verbatim, followed by a complete worked solution.

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.

Solution.

(a) Write the Pauli Hamiltonian as a sum of three pieces,

\[ \hat{H}_{\mathrm{P}} = \underbrace{\frac{\hat{\pi}_j\hat{\pi}_j}{2m}}_{\text{kinetic}} \;\underbrace{-\;\frac{q\hbar}{2m}\,\hat{\boldsymbol{\sigma}}\cdot\boldsymbol{B}(\hat{\boldsymbol{x}})}_{\text{spin-magnetic}} \;+\;\underbrace{q\phi(\hat{\boldsymbol{x}})}_{\text{electric}}, \]

with \(\hat{\pi}_j = \hat{p}_j - qA_j(\hat{\boldsymbol{x}})\) and a sum over the repeated index \(j\). The position operator \(\hat{x}_i\) commutes with any function of \(\hat{\boldsymbol{x}}\), and it acts on a different tensor factor than the Pauli operators, so \([\hat{x}_i,\hat{\sigma}^a]=0\). Both the spin-magnetic term and the electric term are built only from \(\hat{\boldsymbol{x}}\) and \(\hat{\boldsymbol{\sigma}}\):

\[\begin{split} \begin{split} \left[\hat{x}_i,\; -\frac{q\hbar}{2m}\,\hat{\boldsymbol{\sigma}}\cdot\boldsymbol{B}(\hat{\boldsymbol{x}})\right] &= -\frac{q\hbar}{2m}\,\hat{\sigma}^a\,[\hat{x}_i,\, B_a(\hat{\boldsymbol{x}})] = 0, \\ [\hat{x}_i,\; q\phi(\hat{\boldsymbol{x}})] &= 0. \end{split} \end{split}\]

Only the kinetic term survives. With \([\hat{x}_i,\hat{\pi}_j] = [\hat{x}_i,\hat{p}_j] = \mathrm{i}\hbar\delta_{ij}\) and the product rule \([A,BC]=[A,B]C+B[A,C]\),

\[\begin{split} \begin{split} [\hat{x}_i,\hat{H}_{\mathrm{P}}] &= \frac{1}{2m}\,[\hat{x}_i,\hat{\pi}_j\hat{\pi}_j] \\ &= \frac{1}{2m}\big([\hat{x}_i,\hat{\pi}_j]\hat{\pi}_j + \hat{\pi}_j[\hat{x}_i,\hat{\pi}_j]\big) \\ &= \frac{1}{2m}\big(\mathrm{i}\hbar\delta_{ij}\hat{\pi}_j + \hat{\pi}_j\,\mathrm{i}\hbar\delta_{ij}\big) \\ &= \frac{\mathrm{i}\hbar}{m}\,\hat{\pi}_i. \end{split} \end{split}\]

Therefore

\[ \frac{\mathrm{d}\hat{x}_i}{\mathrm{d}t} = \frac{1}{\mathrm{i}\hbar}[\hat{x}_i,\hat{H}_{\mathrm{P}}] = \frac{\hat{\pi}_i}{m}, \qquad\text{i.e.}\qquad \frac{\mathrm{d}\hat{\boldsymbol{x}}}{\mathrm{d}t} = \frac{\hat{\boldsymbol{\pi}}}{m}, \]

exactly the spinless result. The Stern-Gerlach term changes the dynamics of the velocity (part c), but it leaves the definition of the velocity operator untouched.

(b) The velocity operator is \(\mathrm{d}\hat{x}_i/\mathrm{d}t = [\hat{x}_i,\hat{H}]/(\mathrm{i}\hbar)\), so a Hamiltonian term contributes to the velocity only if it fails to commute with \(\hat{x}_i\). Consider any term \(\hat{T}(\hat{\boldsymbol{x}},\hat{\boldsymbol{\sigma}})\) assembled from position operators and Pauli operators alone. The position operators commute among themselves, \([\hat{x}_i,\hat{x}_j]=0\), and they commute with every Pauli operator, \([\hat{x}_i,\hat{\sigma}^a]=0\), because the two act on different tensor factors of the Hilbert space (orbital \(\otimes\) spin). Hence \(\hat{x}_i\) commutes with every monomial in \(\hat{\boldsymbol{x}}\) and \(\hat{\boldsymbol{\sigma}}\), and therefore with any function of them:

\[ [\hat{x}_i,\hat{T}(\hat{\boldsymbol{x}},\hat{\boldsymbol{\sigma}})] = 0. \]

The spin algebra \(\hat{\sigma}^a\hat{\sigma}^b = \delta^{ab} + \mathrm{i}\epsilon^{abc}\hat{\sigma}^c\) never enters this conclusion: it is an algebra internal to the spin factor and cannot generate an \(\hat{x}_i\)-noncommuting object. Position operators fail to commute only with the canonical momentum, \([\hat{x}_i,\hat{p}_j]=\mathrm{i}\hbar\delta_{ij}\), so only Hamiltonian terms containing \(\hat{\boldsymbol{p}}\) contribute to the velocity operator. This is why the velocity is always \(\hat{\boldsymbol{\pi}}/m\) for a Hamiltonian quadratic in \(\hat{\boldsymbol{\pi}}\) — the precise form of any spin- or potential-energy term is irrelevant to it.

(c) Spin enters through the acceleration, i.e. through the equation of motion for \(\hat{\boldsymbol{v}}\). Splitting \(\hat{H}_{\mathrm{P}} = \hat{H}_{\text{orb}} + \hat{H}_{\text{spin}}\) with \(\hat{H}_{\text{orb}} = \hat{\boldsymbol{\pi}}^2/2m + q\phi\) and \(\hat{H}_{\text{spin}} = -(q\hbar/2m)\,\hat{\boldsymbol{\sigma}}\cdot\boldsymbol{B}\), the velocity obeys

\[ m\frac{\mathrm{d}\hat{v}_i}{\mathrm{d}t} = \frac{1}{\mathrm{i}\hbar}[\hat{\pi}_i,\hat{H}_{\text{orb}}] + \frac{1}{\mathrm{i}\hbar}[\hat{\pi}_i,\hat{H}_{\text{spin}}] + m\,\partial_t\hat{v}_i. \]

The orbital and explicit-time pieces give the ordinary Lorentz force (lecture derivation). The new contribution is the spin commutator. Since \(\hat{H}_{\text{spin}}\) is a function of \(\hat{\boldsymbol{x}}\) (with Pauli operators riding along as constants for the \(\hat{\boldsymbol{x}}\)-commutator), apply \([\hat{\pi}_i, f(\hat{\boldsymbol{x}})] = -\mathrm{i}\hbar\,\partial_i f\) with \(f = \hat{H}_{\text{spin}}\):

\[\begin{split} \begin{split} [\hat{\pi}_i,\hat{H}_{\text{spin}}] &= \left[\hat{\pi}_i,\; -\frac{q\hbar}{2m}\,\hat{\sigma}^a B_a(\hat{\boldsymbol{x}})\right] \\ &= -\frac{q\hbar}{2m}\,\hat{\sigma}^a\,[\hat{\pi}_i, B_a(\hat{\boldsymbol{x}})] \\ &= -\frac{q\hbar}{2m}\,\hat{\sigma}^a\,(-\mathrm{i}\hbar\,\partial_i B_a) \\ &= \mathrm{i}\hbar\,\frac{q\hbar}{2m}\,\partial_i(\hat{\boldsymbol{\sigma}}\cdot\boldsymbol{B}). \end{split} \end{split}\]

Dividing by \(\mathrm{i}\hbar\), the spin-magnetic term adds

\[ \left(m\frac{\mathrm{d}\hat{v}_i}{\mathrm{d}t}\right)_{\!\text{spin}} = \frac{1}{\mathrm{i}\hbar}[\hat{\pi}_i,\hat{H}_{\text{spin}}] = \frac{q\hbar}{2m}\,\partial_i(\hat{\boldsymbol{\sigma}}\cdot\boldsymbol{B}), \qquad\text{i.e.}\qquad \boldsymbol{F}_{\text{spin}} = \frac{q\hbar}{2m}\,\nabla(\hat{\boldsymbol{\sigma}}\cdot\boldsymbol{B}). \]

This is a Stern-Gerlach force. Introducing the spin magnetic moment operator \(\hat{\boldsymbol{\mu}} = (q\hbar/2m)\,\hat{\boldsymbol{\sigma}}\), the spin term in the Hamiltonian is \(\hat{H}_{\text{spin}} = -\hat{\boldsymbol{\mu}}\cdot\boldsymbol{B}\) and the force is \(\boldsymbol{F}_{\text{spin}} = \nabla(\hat{\boldsymbol{\mu}}\cdot\boldsymbol{B})\) — the gradient of the magnetic potential energy. Because only the gradient of \(\boldsymbol{B}\) appears, a uniform field exerts no translational force on the spin (it only makes the moment precess); a force on the center of mass requires a field gradient. This is exactly the mechanism that splits an atomic beam in the Stern-Gerlach experiment.

The full equation of motion for the Pauli particle is therefore

\[ m\frac{\mathrm{d}\hat{\boldsymbol{v}}}{\mathrm{d}t} = q\boldsymbol{E} + \frac{q}{2}(\hat{\boldsymbol{v}}\times\boldsymbol{B} - \boldsymbol{B}\times\hat{\boldsymbol{v}}) + \frac{q\hbar}{2m}\,\nabla(\hat{\boldsymbol{\sigma}}\cdot\boldsymbol{B}), \]

the orbital Lorentz force plus the spin-gradient force.

2. Velocity uncertainty in B. For an electron in a uniform magnetic field \(\boldsymbol{B} = B\boldsymbol{e}_z\) (with \(\boldsymbol{e}_z\) the unit vector along the \(z\)-axis), 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 \(\ell_B^2 = \hbar/(qB)\).

Solution.

Take \(q>0\) and \(B>0\) for definiteness (otherwise replace \(qB\) by \(|qB|\) below). The commutator follows directly from the kinetic-momentum algebra: \([\hat{\pi}_x,\hat{\pi}_y] = \mathrm{i}\hbar q\,\epsilon_{xyz}B_z = \mathrm{i}\hbar qB\), so

\[ [\hat{v}_x,\hat{v}_y] = \frac{1}{m^2}[\hat{\pi}_x,\hat{\pi}_y] = \frac{\mathrm{i}\hbar qB}{m^2}. \]

(a) The Robertson uncertainty relation for any two observables is \(\Delta A\,\Delta B \ge \tfrac12\,|\langle[\hat{A},\hat{B}]\rangle|\). Here the commutator is a pure c-number, so its expectation value is itself, independent of the state:

\[ \Delta v_x\,\Delta v_y \;\ge\; \frac{1}{2}\,\bigl|\langle[\hat{v}_x,\hat{v}_y]\rangle\bigr| \;=\; \frac{1}{2}\,\frac{\hbar qB}{m^2} \;=\; \frac{\hbar qB}{2m^2}. \]

The two velocity components are jointly uncertain in a magnetic field: sharpening \(v_x\) necessarily blurs \(v_y\). In zero field the commutator vanishes and the two are simultaneously sharp.

(b) Consider a stationary (cyclotron / Landau) state, for which the orbit is centered, \(\langle\hat{v}_x\rangle = \langle\hat{v}_y\rangle = 0\). Then \(\langle\hat{v}_x^2\rangle = (\Delta v_x)^2\) and likewise for \(y\), so the mean-square transverse speed is

\[ \langle v_\perp^2\rangle = \langle\hat{v}_x^2\rangle + \langle\hat{v}_y^2\rangle = (\Delta v_x)^2 + (\Delta v_y)^2. \]

By the arithmetic-geometric-mean inequality and then the result of (a),

\[ (\Delta v_x)^2 + (\Delta v_y)^2 \;\ge\; 2\,\Delta v_x\,\Delta v_y \;\ge\; 2\cdot\frac{\hbar qB}{2m^2} = \frac{\hbar qB}{m^2}, \]

so \(\langle v_\perp^2\rangle \ge \hbar qB/m^2\). The cyclotron radius is \(r_c = m v_\perp/(qB)\), hence the orbit area is \(\pi r_c^2 = \pi m^2 v_\perp^2/(qB)^2\). Taking expectation values,

\[ \langle \pi r_c^2\rangle = \frac{\pi m^2}{(qB)^2}\,\langle v_\perp^2\rangle \;\ge\; \frac{\pi m^2}{(qB)^2}\cdot\frac{\hbar qB}{m^2} = \frac{\pi\hbar}{qB} = \pi\ell_B^2, \]

with the magnetic length \(\ell_B^2 = \hbar/(qB)\). The cyclotron orbit therefore cannot enclose an area smaller than \(\sim\pi\ell_B^2\). The bound is saturated by the lowest Landau level: there \(\langle v_\perp^2\rangle = \hbar\omega_c/m = \hbar qB/m^2\) (transverse kinetic energy \(\tfrac12 m\langle v_\perp^2\rangle = \tfrac12\hbar\omega_c\)), giving exactly \(\pi r_c^2 = \pi\ell_B^2\).

(c) A small orbit area means a small \(r_c\), hence a small \(v_\perp\) — both velocity components simultaneously small and sharply defined. But in a magnetic field \(\hat{v}_x\) and \(\hat{v}_y\) do not commute; they are conjugate variables, exactly like position and momentum. The uncertainty relation of (a) forbids them from being jointly localized near zero. Fixing the kinetic energy fixes \(\langle v_\perp^2\rangle = \langle v_x^2\rangle + \langle v_y^2\rangle\), but it does not allow that sum to drop below \(\hbar qB/m^2\): even the ground state retains an irreducible zero-point cyclotron motion with energy \(\tfrac12\hbar\omega_c\). Physically, the magnetic field turns the transverse \((v_x,v_y)\) plane into a quantum phase space with a minimum cell of area \(\sim\pi\ell_B^2\); the orbit cannot be squeezed below that cell without violating the velocity uncertainty relation. This is the kinematic origin of Landau quantization (§4.3).

★ 3. Cyclotron motion from Heisenberg. A particle of charge \(q\) moves in a uniform \(\boldsymbol{B} = B\boldsymbol{e}_z\) 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.

Solution.

For a uniform static field the vector potential can be chosen time-independent, so \(\hat{\pi}_i\) has no explicit time dependence and the plain Heisenberg equation \(\mathrm{d}\hat{\pi}_i/\mathrm{d}t = [\hat{\pi}_i,\hat{H}]/(\mathrm{i}\hbar)\) applies. The relevant commutators are \([\hat{\pi}_x,\hat{\pi}_y] = \mathrm{i}q\hbar B\) and \([\hat{\pi}_x,\hat{\pi}_z] = [\hat{\pi}_y,\hat{\pi}_z] = 0\) (since \(\boldsymbol{B}\) points along the \(z\)-axis, \(\epsilon_{xzk}B_k = \epsilon_{yzk}B_k = 0\)).

(a) With \(\hat{H} = (\hat{\pi}_x^2 + \hat{\pi}_y^2 + \hat{\pi}_z^2)/2m\), evaluate \([\hat{\pi}_x,\hat{H}]\) using the product rule. The \(\hat{\pi}_x^2\) and \(\hat{\pi}_z^2\) pieces commute with \(\hat{\pi}_x\), so only \(\hat{\pi}_y^2\) contributes:

\[\begin{split} \begin{split} [\hat{\pi}_x,\hat{H}] &= \frac{1}{2m}\,[\hat{\pi}_x,\hat{\pi}_y^2] \\ &= \frac{1}{2m}\big([\hat{\pi}_x,\hat{\pi}_y]\hat{\pi}_y + \hat{\pi}_y[\hat{\pi}_x,\hat{\pi}_y]\big) \\ &= \frac{1}{2m}\big(\mathrm{i}q\hbar B\,\hat{\pi}_y + \hat{\pi}_y\,\mathrm{i}q\hbar B\big) \\ &= \frac{\mathrm{i}q\hbar B}{m}\,\hat{\pi}_y, \end{split} \end{split}\]

where the c-number \(\mathrm{i}q\hbar B\) was pulled out freely. Likewise, only \(\hat{\pi}_x^2\) contributes to \([\hat{\pi}_y,\hat{H}]\), and with \([\hat{\pi}_y,\hat{\pi}_x] = -\mathrm{i}q\hbar B\),

\[ [\hat{\pi}_y,\hat{H}] = \frac{1}{2m}\,[\hat{\pi}_y,\hat{\pi}_x^2] = \frac{1}{2m}\big({-2\mathrm{i}q\hbar B}\,\hat{\pi}_x\big) = -\frac{\mathrm{i}q\hbar B}{m}\,\hat{\pi}_x. \]

Dividing by \(\mathrm{i}\hbar\) and writing \(\omega_c = qB/m\) gives the coupled operator equations of motion

\[ \frac{\mathrm{d}\hat{\pi}_x}{\mathrm{d}t} = \omega_c\,\hat{\pi}_y, \qquad \frac{\mathrm{d}\hat{\pi}_y}{\mathrm{d}t} = -\omega_c\,\hat{\pi}_x. \]

(The \(z\)-component is constant, \(\mathrm{d}\hat{\pi}_z/\mathrm{d}t = [\hat{\pi}_z,\hat{H}]/(\mathrm{i}\hbar) = 0\), so the motion along the field is free.)

(b) Form the complex combination \(\hat{\pi}_\pm = \hat{\pi}_x \pm \mathrm{i}\hat{\pi}_y\). Then

\[ \frac{\mathrm{d}\hat{\pi}_+}{\mathrm{d}t} = \frac{\mathrm{d}\hat{\pi}_x}{\mathrm{d}t} + \mathrm{i}\frac{\mathrm{d}\hat{\pi}_y}{\mathrm{d}t} = \omega_c\hat{\pi}_y - \mathrm{i}\omega_c\hat{\pi}_x = -\mathrm{i}\omega_c(\hat{\pi}_x + \mathrm{i}\hat{\pi}_y) = -\mathrm{i}\omega_c\,\hat{\pi}_+, \]

a single decoupled equation with solution \(\hat{\pi}_+(t) = \mathrm{e}^{-\mathrm{i}\omega_c t}\,\hat{\pi}_+(0)\). Taking real and imaginary parts,

\[\begin{split} \begin{split} \hat{\pi}_x(t) &= \hphantom{-}\cos(\omega_c t)\,\hat{\pi}_x(0) + \sin(\omega_c t)\,\hat{\pi}_y(0), \\ \hat{\pi}_y(t) &= -\sin(\omega_c t)\,\hat{\pi}_x(0) + \cos(\omega_c t)\,\hat{\pi}_y(0). \end{split} \end{split}\]

The pair \((\hat{\pi}_x,\hat{\pi}_y)\) rotates rigidly in the velocity plane at the cyclotron frequency \(\omega_c = qB/m\). Because the rotation matrix has c-number entries, this is the operator-level statement of classical cyclotron motion: every expectation value \(\langle\hat{\pi}_x\rangle(t), \langle\hat{\pi}_y\rangle(t)\) traces a circle at frequency \(\omega_c\), exactly the classical orbit. Quantum mechanics adds nothing to the frequency — it only constrains the amplitude through the velocity uncertainty relation of Problem 2.

(c) The kinetic energy operator \(\hat{\boldsymbol{\pi}}^2 = 2m\hat{H}\) commutes with the Hamiltonian, since any operator commutes with itself:

\[ \frac{\mathrm{d}\hat{\boldsymbol{\pi}}^2}{\mathrm{d}t} = \frac{1}{\mathrm{i}\hbar}[\hat{\boldsymbol{\pi}}^2,\hat{H}] = \frac{1}{\mathrm{i}\hbar}[2m\hat{H},\hat{H}] = 0. \]

One can also see it from the explicit solution: using \(\cos^2 + \sin^2 = 1\) and that the cross terms cancel,

\[ \hat{\pi}_x(t)^2 + \hat{\pi}_y(t)^2 = \hat{\pi}_x(0)^2 + \hat{\pi}_y(0)^2, \]

while \(\hat{\pi}_z\) is separately constant, so \(\hat{\boldsymbol{\pi}}^2 = \hat{\pi}_x^2 + \hat{\pi}_y^2 + \hat{\pi}_z^2\) is conserved. Physically, the magnetic force \(q\hat{\boldsymbol{v}}\times\boldsymbol{B}\) is always perpendicular to the velocity and therefore does no work; with \(\boldsymbol{E} = 0\) there is no other force, so the kinetic energy \(\hat{H} = \hat{\boldsymbol{\pi}}^2/2m\) is a constant of motion. The field bends the trajectory into a circle but never changes its speed — energy conservation at the operator level.

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

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

(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?

Solution.

(a) With \(\boldsymbol{A} = \tfrac{B}{2}(-y,\,x,\,0)\), the components are \(A_x = -By/2\), \(A_y = Bx/2\), \(A_z = 0\). The curl is

\[\begin{split} \begin{split} (\nabla\times\boldsymbol{A})_x &= \partial_y A_z - \partial_z A_y = 0 - 0 = 0, \\ (\nabla\times\boldsymbol{A})_y &= \partial_z A_x - \partial_x A_z = 0 - 0 = 0, \\ (\nabla\times\boldsymbol{A})_z &= \partial_x A_y - \partial_y A_x = \partial_x\!\left(\tfrac{Bx}{2}\right) - \partial_y\!\left(-\tfrac{By}{2}\right) = \tfrac{B}{2} + \tfrac{B}{2} = B. \end{split} \end{split}\]

Hence \(\nabla\times\boldsymbol{A} = B\boldsymbol{e}_z = \boldsymbol{B}\), as required. (The “symmetric” name refers to the equal \(B/2\) split between the \(x\) and \(y\) contributions, in contrast with the asymmetric Landau gauge \(\boldsymbol{A} = B(-y,0,0)\) or \(B(0,x,0)\).)

(b) The kinetic-momentum components in this gauge are

\[ \hat{\pi}_x = \hat{p}_x - qA_x = \hat{p}_x + \frac{qB}{2}\,\hat{y}, \qquad \hat{\pi}_y = \hat{p}_y - qA_y = \hat{p}_y - \frac{qB}{2}\,\hat{x}. \]

Expand the commutator by bilinearity. The terms \([\hat{p}_x,\hat{p}_y]\) and \([\hat{y},\hat{x}]\) vanish, leaving two canonical commutators:

\[\begin{split} \begin{split} [\hat{\pi}_x,\hat{\pi}_y] &= \left[\hat{p}_x + \tfrac{qB}{2}\hat{y},\;\; \hat{p}_y - \tfrac{qB}{2}\hat{x}\right] \\ &= [\hat{p}_x,\hat{p}_y] - \frac{qB}{2}[\hat{p}_x,\hat{x}] + \frac{qB}{2}[\hat{y},\hat{p}_y] - \left(\tfrac{qB}{2}\right)^2[\hat{y},\hat{x}] \\ &= 0 - \frac{qB}{2}(-\mathrm{i}\hbar) + \frac{qB}{2}(+\mathrm{i}\hbar) - 0 \\ &= \frac{\mathrm{i}\hbar qB}{2} + \frac{\mathrm{i}\hbar qB}{2} \\ &= \mathrm{i}\hbar qB, \end{split} \end{split}\]

using \([\hat{p}_x,\hat{x}] = -\mathrm{i}\hbar\) and \([\hat{y},\hat{p}_y] = +\mathrm{i}\hbar\). The result \([\hat{\pi}_x,\hat{\pi}_y] = \mathrm{i}\hbar qB\) matches the gauge-independent formula \([\hat{\pi}_i,\hat{\pi}_j] = \mathrm{i}\hbar q\,\epsilon_{ijk}B_k\) — the two halves \(\mathrm{i}\hbar qB/2\) come symmetrically from the \(\hat{y}\)-term of \(\hat{\pi}_x\) and the \(\hat{x}\)-term of \(\hat{\pi}_y\), whereas in the Landau gauge one half would carry the whole result. The commutator is the same because it depends only on the physical field \(\boldsymbol{B} = \nabla\times\boldsymbol{A}\), not on the gauge.

(c) A nonzero commutator \([\hat{\pi}_x,\hat{\pi}_y] = \mathrm{i}\hbar qB \neq 0\) means the velocity components \(\hat{v}_x = \hat{\pi}_x/m\) and \(\hat{v}_y = \hat{\pi}_y/m\) are incompatible observables: they have no common eigenbasis and cannot be assigned simultaneously sharp values. A measurement of \(v_x\) disturbs \(v_y\), and they obey the uncertainty relation \(\Delta v_x\,\Delta v_y \ge \hbar qB/(2m^2)\) derived in Problem 2. In a magnetic field the two transverse velocity components behave like a conjugate pair — analogous to position and momentum — which is what makes the cyclotron motion quantize into Landau levels. With \(B \to 0\) the commutator vanishes and the velocity components become jointly measurable again, recovering the familiar field-free situation where the full velocity vector is a sharp observable.

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 \(\boldsymbol{B} = \nabla\times\boldsymbol{A}\).

Solution.

Set \(c = 1\). Use the metric \(g_{\mu\nu} = \mathrm{diag}(+1,-1,-1,-1)\), so the gauge four-potential has upper components \(A^\mu = (\phi,\boldsymbol{A})\) and lowered components

\[ A_\mu = g_{\mu\nu}A^\nu = (\phi,\,-\boldsymbol{A}), \qquad\text{i.e.}\qquad A_0 = \phi,\quad A_i = -A^i. \]

The derivative is \(\partial_\mu = \partial/\partial x^\mu = (\partial_t,\,\nabla)\), with \(\partial_0 = \partial_t\) and \(\partial_i\) the ordinary spatial gradient component. The field strength tensor \(F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu\) is antisymmetric, so its diagonal vanishes and only six independent entries need evaluation.

Time-space entries \(F_{0i}\). For \(i = x,y,z\),

\[\begin{split} \begin{split} F_{0i} &= \partial_0 A_i - \partial_i A_0 \\ &= \partial_t(-A^i) - \partial_i\phi \\ &= -\partial_t A^i - \partial_i\phi \\ &= E_i, \end{split} \end{split}\]

using the definition \(\boldsymbol{E} = -\nabla\phi - \partial_t\boldsymbol{A}\), component-wise \(E_i = -\partial_i\phi - \partial_t A^i\). Explicitly for \(i=x\): \(F_{0x} = \partial_t(-A_x) - \partial_x\phi = -\partial_t A_x - \partial_x\phi = E_x\).

Space-space entries \(F_{ij}\). For spatial \(i,j\),

\[\begin{split} \begin{split} F_{ij} &= \partial_i A_j - \partial_j A_i \\ &= \partial_i(-A^j) - \partial_j(-A^i) \\ &= -(\partial_i A^j - \partial_j A^i) \\ &= -\epsilon_{ijk}B_k, \end{split} \end{split}\]

where the last step uses that the curl \(B_k = \epsilon_{klm}\partial_l A^m\) is equivalent to \(\partial_i A^j - \partial_j A^i = \epsilon_{ijk}B_k\) (contract \(\epsilon_{ijk}B_k = \epsilon_{ijk}\epsilon_{klm}\partial_l A^m = (\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl})\partial_l A^m = \partial_i A^j - \partial_j A^i\)). A sample entry: \(F_{xy} = \partial_x(-A_y) - \partial_y(-A_x) = -(\partial_x A_y - \partial_y A_x) = -B_z\), and indeed \(-\epsilon_{xyk}B_k = -\epsilon_{xyz}B_z = -B_z\). Similarly \(F_{xz} = -(\partial_x A_z - \partial_z A_x) = -\epsilon_{xzk}B_k = -\epsilon_{xzy}B_y = +B_y\) and \(F_{yz} = -\epsilon_{yzk}B_k = -\epsilon_{yzx}B_x = -B_x\).

The matrix. Collecting all entries with rows/columns ordered \((0,x,y,z)\) and imposing antisymmetry \(F_{\nu\mu} = -F_{\mu\nu}\):

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

The first row/column carries the electric field (\(F_{0i} = E_i\)); the lower-right \(3\times3\) block carries the magnetic field through \(F_{ij} = -\epsilon_{ijk}B_k\). A single antisymmetric tensor thus packages all six components of \(\boldsymbol{E}\) and \(\boldsymbol{B}\).

For completeness, raising both indices with the metric, \(F^{\mu\nu} = g^{\mu\alpha}g^{\nu\beta}F_{\alpha\beta}\), flips the sign of each entry that carries exactly one spatial index, i.e. the electric (time-space) block:

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

The magnetic block is unchanged (two spatial indices, two sign flips that cancel). Gauge invariance of \(F_{\mu\nu}\) is immediate: under \(A_\mu \to A_\mu + \partial_\mu\alpha\), the extra piece is \(\partial_\mu\partial_\nu\alpha - \partial_\nu\partial_\mu\alpha = 0\) since partial derivatives commute — consistent with \(\boldsymbol{E}\) and \(\boldsymbol{B}\) being the gauge-invariant physical fields.