4.1.3 Gauge Invariance#

Worked solutions for the homework problems in the 4.1.3 Gauge Invariance lecture. Each problem statement is reproduced verbatim, followed by a complete worked solution.

1. Gauge-transforming an observable. Consider the expectation value \(\langle\hat{O}\rangle = \langle\psi\vert\hat{O}\vert\psi\rangle\).

(a) Under the gauge transformation \(\vert\psi\rangle \to \mathrm{e}^{\mathrm{i}q\alpha/\hbar}\vert\psi\rangle\), show that \(\langle\hat{O}\rangle\) is invariant if and only if \([\hat{O}, \mathrm{e}^{\mathrm{i}q\alpha/\hbar}] = 0\). Which familiar operators satisfy this, and which do not?

(b) One claims that \(\langle\hat{\boldsymbol{p}}\rangle\) is gauge-invariant because “momentum is physical.” Find the flaw by computing \(\langle\hat{\boldsymbol{p}}\rangle\) in two gauges for a particle in a uniform field \(\boldsymbol{B} = B\boldsymbol{e}_z\) (with \(\boldsymbol{e}_z\) the unit vector along the \(z\)-axis).

Solution.

(a) Write the rephased state as \(\vert\psi'\rangle = \hat{U}\vert\psi\rangle\) with the unitary multiplication operator \(\hat{U} = \mathrm{e}^{\mathrm{i}q\alpha(\hat{\boldsymbol{x}})/\hbar}\). Holding the operator \(\hat{O}\) fixed, its expectation in the new state is

\[ \langle\hat{O}\rangle' = \langle\psi'\vert\hat{O}\vert\psi'\rangle = \langle\psi\vert\hat{U}^\dagger\hat{O}\hat{U}\vert\psi\rangle . \]

If \([\hat{O},\hat{U}] = 0\) then \(\hat{U}^\dagger\hat{O}\hat{U} = \hat{U}^\dagger\hat{U}\hat{O} = \hat{O}\), so \(\langle\hat{O}\rangle' = \langle\hat{O}\rangle\) for every state — invariance holds. Conversely, suppose \(\langle\hat{O}\rangle' = \langle\hat{O}\rangle\) for all \(\vert\psi\rangle\). Then the Hermitian operator \(\hat{U}^\dagger\hat{O}\hat{U} - \hat{O}\) has vanishing expectation in every state, and an operator with zero expectation in every state is the zero operator. Hence \(\hat{U}^\dagger\hat{O}\hat{U} = \hat{O}\), i.e. \(\hat{O}\hat{U} = \hat{U}\hat{O}\), which is \([\hat{O},\hat{U}] = 0\). This proves the “if and only if.”

Since \(\hat{U} = \mathrm{e}^{\mathrm{i}q\alpha(\hat{\boldsymbol{x}})/\hbar}\) is a function of position, the operators that commute with it are exactly the functions of position: \(\hat{\boldsymbol{x}}\) itself, any potential \(V(\hat{\boldsymbol{x}})\), the projector \(\vert\boldsymbol{r}\rangle\langle\boldsymbol{r}\vert\) onto a position eigenstate (hence the probability density \(\rho = \vert\psi\vert^2\)). Operators built from \(\hat{\boldsymbol{p}} = -\mathrm{i}\hbar\nabla\) do not commute with \(\hat{U}\): the canonical momentum \(\hat{\boldsymbol{p}}\), the kinetic energy, the canonical Hamiltonian. The lesson is that “gauge invariance of an observable” cannot mean “its expectation in the rephased state is unchanged while the operator stays fixed” — for almost every interesting operator that fails. The operator must itself be re-expressed in the new gauge (see part (c)).

(b) Take the two standard gauges for \(\boldsymbol{B} = B\boldsymbol{e}_z\): the symmetric gauge \(\boldsymbol{A}_{\mathrm{S}} = \tfrac{B}{2}(-y,x,0)\) and the Landau gauge \(\boldsymbol{A}_{\mathrm{L}} = (0,Bx,0)\). They are related by \(\boldsymbol{A}_{\mathrm{L}} = \boldsymbol{A}_{\mathrm{S}} + \nabla\alpha\) with

\[ \alpha = \tfrac{B}{2}xy, \qquad \nabla\alpha = \tfrac{B}{2}(y,x,0), \]

since \(\boldsymbol{A}_{\mathrm{S}} + \nabla\alpha = \tfrac{B}{2}(-y,x,0) + \tfrac{B}{2}(y,x,0) = (0,Bx,0)\). The same physical state is described by \(\psi_{\mathrm{S}}\) in gauge S and by \(\psi_{\mathrm{L}} = \mathrm{e}^{\mathrm{i}q\alpha/\hbar}\psi_{\mathrm{S}}\) in gauge L. Using the conjugation identity from the preamble,

\[\begin{split} \begin{split} \langle\hat{\boldsymbol{p}}\rangle_{\mathrm{L}} &= \langle\psi_{\mathrm{L}}\vert\hat{\boldsymbol{p}}\vert\psi_{\mathrm{L}}\rangle = \langle\psi_{\mathrm{S}}\vert\,\mathrm{e}^{-\mathrm{i}q\alpha/\hbar}\hat{\boldsymbol{p}}\,\mathrm{e}^{\mathrm{i}q\alpha/\hbar}\,\vert\psi_{\mathrm{S}}\rangle \\ &= \langle\psi_{\mathrm{S}}\vert\,(\hat{\boldsymbol{p}} + q\nabla\alpha)\,\vert\psi_{\mathrm{S}}\rangle = \langle\hat{\boldsymbol{p}}\rangle_{\mathrm{S}} + q\langle\nabla\alpha\rangle . \end{split} \end{split}\]

With \(\nabla\alpha = \tfrac{B}{2}(y,x,0)\) this gives \(\langle\hat{\boldsymbol{p}}\rangle_{\mathrm{L}} = \langle\hat{\boldsymbol{p}}\rangle_{\mathrm{S}} + \tfrac{qB}{2}(\langle y\rangle,\langle x\rangle,0)\), which differs from \(\langle\hat{\boldsymbol{p}}\rangle_{\mathrm{S}}\) whenever \(\langle x\rangle\) or \(\langle y\rangle\) is nonzero (the shift vanishes only for states centred at the origin, with \(\langle x\rangle = \langle y\rangle = 0\)). So \(\langle\hat{\boldsymbol{p}}\rangle\) is gauge-dependent. The flaw in the claim is the identification of \(\hat{\boldsymbol{p}} = -\mathrm{i}\hbar\nabla\) with “physical momentum.” The canonical momentum measures the phase gradient of the wavefunction, and the gauge phase \(\mathrm{e}^{\mathrm{i}q\alpha/\hbar}\) changes that gradient. The physical (mechanical) momentum is \(m\hat{\boldsymbol{v}} = \hat{\boldsymbol{\pi}}\).

(c) The gauge-invariant momentum is the kinetic momentum

\[ \hat{\boldsymbol{\pi}} = \hat{\boldsymbol{p}} - q\boldsymbol{A} = m\hat{\boldsymbol{v}} . \]

Under the gauge transformation the operator must be rebuilt with the new potential, \(\hat{\boldsymbol{\pi}}' = \hat{\boldsymbol{p}} - q\boldsymbol{A}' = \hat{\boldsymbol{p}} - q\boldsymbol{A} - q\nabla\alpha\). Conjugating by the phase,

\[ \mathrm{e}^{-\mathrm{i}q\alpha/\hbar}\,\hat{\boldsymbol{\pi}}'\,\mathrm{e}^{\mathrm{i}q\alpha/\hbar} = (\hat{\boldsymbol{p}} + q\nabla\alpha) - q\boldsymbol{A} - q\nabla\alpha = \hat{\boldsymbol{p}} - q\boldsymbol{A} = \hat{\boldsymbol{\pi}} , \]

because \(q\boldsymbol{A}\) and \(q\nabla\alpha\) are functions of position and pass through the phase untouched. Therefore

\[ \langle\hat{\boldsymbol{\pi}}'\rangle' = \langle\psi'\vert\hat{\boldsymbol{\pi}}'\vert\psi'\rangle = \langle\psi\vert\,\mathrm{e}^{-\mathrm{i}q\alpha/\hbar}\hat{\boldsymbol{\pi}}'\mathrm{e}^{\mathrm{i}q\alpha/\hbar}\,\vert\psi\rangle = \langle\hat{\boldsymbol{\pi}}\rangle . \]

This is the precise sense of gauge invariance for an operator: \(\hat{\boldsymbol{\pi}}\) transforms by conjugation, \(\hat{\boldsymbol{\pi}}' = \hat{U}\hat{\boldsymbol{\pi}}\hat{U}^\dagger\), so that its expectation value is the same number in every gauge.

2. Probability current. The probability current is \(\boldsymbol{j} = \frac{1}{m}\mathrm{Re}(\psi^*\hat{\boldsymbol{\pi}}\psi)\) with \(\hat{\boldsymbol{\pi}} = \hat{\boldsymbol{p}} - q\boldsymbol{A}\).

(a) Show that \(\boldsymbol{j}\) is gauge-invariant.

(b) Show that the continuity equation \(\partial_t\vert\psi\vert^2 + \nabla \cdot \boldsymbol{j} = 0\) holds in every gauge.

Solution.

(a) Expanding \(\hat{\boldsymbol{\pi}} = -\mathrm{i}\hbar\nabla - q\boldsymbol{A}\), the current is

\[ \boldsymbol{j} = \frac{1}{m}\mathrm{Re}\bigl(\psi^*(-\mathrm{i}\hbar\nabla - q\boldsymbol{A})\psi\bigr) = \frac{\hbar}{m}\mathrm{Im}(\psi^*\nabla\psi) - \frac{q}{m}\boldsymbol{A}\,\vert\psi\vert^2 , \]

using \(\mathrm{Re}(-\mathrm{i}\hbar\,\psi^*\nabla\psi) = \hbar\,\mathrm{Im}(\psi^*\nabla\psi)\) and that \(-q\boldsymbol{A}\vert\psi\vert^2\) is already real. The fastest invariance proof uses the homogeneous transformation of \(\hat{\boldsymbol{\pi}}\psi\) established for the covariant derivative in §4.1.1: under the gauge transformation

\[ \hat{\boldsymbol{\pi}}'\psi' = \mathrm{e}^{\mathrm{i}q\alpha/\hbar}\,\hat{\boldsymbol{\pi}}\psi . \]

Then \(\psi'^*\,\hat{\boldsymbol{\pi}}'\psi' = \bigl(\mathrm{e}^{\mathrm{i}q\alpha/\hbar}\psi\bigr)^*\,\mathrm{e}^{\mathrm{i}q\alpha/\hbar}\hat{\boldsymbol{\pi}}\psi = \psi^*\,\mathrm{e}^{-\mathrm{i}q\alpha/\hbar}\mathrm{e}^{\mathrm{i}q\alpha/\hbar}\,\hat{\boldsymbol{\pi}}\psi = \psi^*\,\hat{\boldsymbol{\pi}}\psi\), so \(\boldsymbol{j}' = \boldsymbol{j}\) immediately.

It is instructive to see the cancellation term by term as well. Under \(\psi \to \mathrm{e}^{\mathrm{i}q\alpha/\hbar}\psi\),

\[ \nabla\psi' = \mathrm{e}^{\mathrm{i}q\alpha/\hbar}\Bigl(\nabla\psi + \tfrac{\mathrm{i}q}{\hbar}(\nabla\alpha)\psi\Bigr), \qquad \psi'^*\nabla\psi' = \psi^*\nabla\psi + \tfrac{\mathrm{i}q}{\hbar}(\nabla\alpha)\vert\psi\vert^2 , \]

so \(\mathrm{Im}(\psi'^*\nabla\psi') = \mathrm{Im}(\psi^*\nabla\psi) + \tfrac{q}{\hbar}(\nabla\alpha)\vert\psi\vert^2\). The first term of \(\boldsymbol{j}\) therefore gains \(\tfrac{q}{m}(\nabla\alpha)\vert\psi\vert^2\), while the second term, with \(\boldsymbol{A}' = \boldsymbol{A} + \nabla\alpha\) and \(\vert\psi'\vert^2 = \vert\psi\vert^2\), gains \(-\tfrac{q}{m}(\nabla\alpha)\vert\psi\vert^2\). The two changes cancel exactly:

\[ \boldsymbol{j}' = \frac{\hbar}{m}\mathrm{Im}(\psi^*\nabla\psi) + \frac{q}{m}(\nabla\alpha)\vert\psi\vert^2 - \frac{q}{m}(\boldsymbol{A} + \nabla\alpha)\vert\psi\vert^2 = \boldsymbol{j} . \]

The current is gauge-invariant because the \(\nabla\alpha\) injected by the rephased wavefunction is precisely the \(\nabla\alpha\) absorbed into \(\boldsymbol{A}\).

(b) Write the current with the covariant derivative \(\boldsymbol{D} = \nabla - \tfrac{\mathrm{i}q}{\hbar}\boldsymbol{A}\), so that \(\hat{\boldsymbol{\pi}}\psi = -\mathrm{i}\hbar\,\boldsymbol{D}\psi\) and

\[ \boldsymbol{j} = \frac{\hbar}{m}\mathrm{Im}(\psi^*\boldsymbol{D}\psi) = \frac{\hbar}{2\mathrm{i}m}\bigl(\psi^*\boldsymbol{D}\psi - \psi\,\boldsymbol{D}^*\psi^*\bigr), \qquad \boldsymbol{D}^* = \nabla + \tfrac{\mathrm{i}q}{\hbar}\boldsymbol{A} . \]

Differentiate the density using the Schrödinger equation \(\mathrm{i}\hbar\partial_t\psi = \hat{H}\psi\) with \(\hat{H} = \hat{\boldsymbol{\pi}}^2/2m + q\phi\):

\[ \partial_t\vert\psi\vert^2 = \psi^*\partial_t\psi + \psi\,\partial_t\psi^* = \frac{1}{\mathrm{i}\hbar}\bigl[\psi^*(\hat{H}\psi) - \psi(\hat{H}\psi)^*\bigr] = \frac{2}{\hbar}\,\mathrm{Im}\bigl(\psi^*\hat{H}\psi\bigr) . \]

The scalar-potential term contributes nothing: \(\psi^*(q\phi)\psi = q\phi\vert\psi\vert^2\) is real, so \(\mathrm{Im}\) kills it. Only the kinetic term survives, and with \(\hat{\boldsymbol{\pi}}^2 = -\hbar^2\boldsymbol{D}^2\),

\[ \partial_t\vert\psi\vert^2 = \frac{1}{\hbar m}\,\mathrm{Im}\bigl(\psi^*\hat{\boldsymbol{\pi}}^2\psi\bigr) = -\frac{\hbar}{m}\,\mathrm{Im}\bigl(\psi^*\boldsymbol{D}^2\psi\bigr) = -\frac{\hbar}{2\mathrm{i}m}\bigl(\psi^*\boldsymbol{D}^2\psi - \psi\,(\boldsymbol{D}^*)^2\psi^*\bigr). \]

The covariant derivative obeys a Leibniz identity in which \(\boldsymbol{D}\) on a ket pairs with \(\boldsymbol{D}^*\) on a bra, the \(\boldsymbol{A}\) terms cancelling: \(\nabla\cdot(\psi^*\boldsymbol{V}) = (\boldsymbol{D}^*\psi^*)\cdot\boldsymbol{V} + \psi^*(\boldsymbol{D}\cdot\boldsymbol{V})\). Applying it with \(\boldsymbol{V} = \boldsymbol{D}\psi\) and with \(\boldsymbol{V} = \boldsymbol{D}^*\psi^*\),

\[\begin{split} \begin{split} \nabla\cdot(\psi^*\boldsymbol{D}\psi) &= (\boldsymbol{D}^*\psi^*)\cdot(\boldsymbol{D}\psi) + \psi^*\boldsymbol{D}^2\psi , \\ \nabla\cdot(\psi\,\boldsymbol{D}^*\psi^*) &= (\boldsymbol{D}\psi)\cdot(\boldsymbol{D}^*\psi^*) + \psi\,(\boldsymbol{D}^*)^2\psi^* . \end{split} \end{split}\]

Subtracting, the symmetric cross terms cancel and

\[ \nabla\cdot\bigl(\psi^*\boldsymbol{D}\psi - \psi\,\boldsymbol{D}^*\psi^*\bigr) = \psi^*\boldsymbol{D}^2\psi - \psi\,(\boldsymbol{D}^*)^2\psi^* . \]

Multiplying by \(\hbar/(2\mathrm{i}m)\), the right side is the negative of the expression for \(\partial_t\vert\psi\vert^2\), while the left side is \(\nabla\cdot\boldsymbol{j}\). Hence

\[ \partial_t\vert\psi\vert^2 + \nabla\cdot\boldsymbol{j} = 0 . \]

This holds in every gauge for two independent reasons. First, the derivation above never selected a gauge — it used only the Schrödinger equation with arbitrary \((\boldsymbol{A},\phi)\), so it is valid for any choice. Second, by part (a) both \(\rho = \vert\psi\vert^2\) and \(\boldsymbol{j}\) are gauge-invariant; the continuity equation is therefore literally the same equation, relating the same two gauge-invariant fields, no matter which gauge one computes in.

3. Canonical momentum shifts. Two physicists compute \(\langle\hat{\boldsymbol{p}}\rangle\) for the same electron in a uniform magnetic field \(\boldsymbol{B} = B\boldsymbol{e}_z\). Physicist A uses the symmetric gauge \(\boldsymbol{A} = \frac{B}{2}(-y, x, 0)\); physicist B uses the Landau gauge \(\boldsymbol{A} = (0, Bx, 0)\).

(a) Explain why they get different values of \(\langle\hat{\boldsymbol{p}}\rangle\).

(b) Show that both agree on \(\langle\hat{\boldsymbol{\pi}}\rangle\).

Solution.

(a) The two gauges describe the same magnetic field — both have \(\nabla\times\boldsymbol{A} = B\boldsymbol{e}_z\) — and therefore the same physical electron. They are connected by the gauge transformation found in Problem 1(b),

\[ \boldsymbol{A}_{\mathrm{B}} = \boldsymbol{A}_{\mathrm{A}} + \nabla\alpha, \qquad \alpha = \tfrac{B}{2}xy, \qquad \nabla\alpha = \tfrac{B}{2}(y,x,0), \]

so the electron’s wavefunction is \(\psi_{\mathrm{A}}\) for physicist A and \(\psi_{\mathrm{B}} = \mathrm{e}^{\mathrm{i}q\alpha/\hbar}\psi_{\mathrm{A}}\) for physicist B. Their canonical-momentum expectations differ by

\[ \langle\hat{\boldsymbol{p}}\rangle_{\mathrm{B}} - \langle\hat{\boldsymbol{p}}\rangle_{\mathrm{A}} = q\langle\nabla\alpha\rangle = \frac{qB}{2}\bigl(\langle y\rangle,\,\langle x\rangle,\,0\bigr), \]

which is nonzero whenever \(\langle x\rangle\) or \(\langle y\rangle\) is nonzero (only states with \(\langle x\rangle = \langle y\rangle = 0\) give the same canonical-momentum expectation in both gauges).

The reason is structural: \(\hat{\boldsymbol{p}} = -\mathrm{i}\hbar\nabla\) reads off the phase gradient of the wavefunction, and the two physicists’ wavefunctions differ by the position-dependent phase \(\mathrm{e}^{\mathrm{i}q\alpha/\hbar}\). A different gauge is a different bookkeeping convention for splitting the phase between \(\psi\) and \(\boldsymbol{A}\); the canonical momentum is sensitive to that split. It is a gauge-dependent label, not a physical quantity.

(b) The kinetic momentum is \(\hat{\boldsymbol{\pi}} = \hat{\boldsymbol{p}} - q\boldsymbol{A}\), built with each physicist’s own potential. For physicist B,

\[\begin{split} \begin{split} \langle\hat{\boldsymbol{\pi}}\rangle_{\mathrm{B}} &= \langle\hat{\boldsymbol{p}}\rangle_{\mathrm{B}} - q\langle\boldsymbol{A}_{\mathrm{B}}\rangle = \bigl(\langle\hat{\boldsymbol{p}}\rangle_{\mathrm{A}} + q\langle\nabla\alpha\rangle\bigr) - q\langle\boldsymbol{A}_{\mathrm{A}} + \nabla\alpha\rangle \\ &= \langle\hat{\boldsymbol{p}}\rangle_{\mathrm{A}} - q\langle\boldsymbol{A}_{\mathrm{A}}\rangle = \langle\hat{\boldsymbol{\pi}}\rangle_{\mathrm{A}} . \end{split} \end{split}\]

The \(q\langle\nabla\alpha\rangle\) picked up by the canonical momentum is exactly the \(q\langle\nabla\alpha\rangle\) added to \(q\boldsymbol{A}\); they cancel. Both physicists agree on \(\langle\hat{\boldsymbol{\pi}}\rangle = m\langle\hat{\boldsymbol{v}}\rangle\).

(c) An experiment measures the kinetic momentum \(\langle\hat{\boldsymbol{\pi}}\rangle = m\langle\hat{\boldsymbol{v}}\rangle\) — the quantity tied to the electron’s actual motion (its velocity, its cyclotron radius \(r_c = m v_\perp/qB\), the current it carries, its time-of-flight). This is gauge-invariant, so both physicists predict the same measurable result. The canonical momentum \(\langle\hat{\boldsymbol{p}}\rangle\) is never measured directly: it is not even well-defined until a gauge is chosen, and different legitimate choices give different numbers.

4. Energy under gauge transformation. One might argue: “The eigenvalue equation \(\hat{H}\psi_n = E_n\psi_n\) contains \(\hat{H}\), which depends on \(\boldsymbol{A}\) and \(\phi\). So changing gauge must change the energy eigenvalues.”

(a) Refute this by showing explicitly that if \(\hat{H}\psi_n = E_n\psi_n\), then \(\hat{H}'\psi_n' = E_n\psi_n'\) with the transformed \(\hat{H}'\) and \(\psi_n' = \mathrm{e}^{\mathrm{i}q\alpha/\hbar}\psi_n\).

(b) The kinetic energy \(\frac{1}{2m}\langle\hat{\boldsymbol{\pi}}^2\rangle\) is gauge-invariant, but \(q\langle\phi\rangle\) is not. How can \(E_n\) be gauge-invariant if one of its terms is not? Resolve the apparent paradox.

(c) An experimentalist measures the hydrogen atom spectrum. In what sense is the result “gauge-invariant” even though the calculation was done in a specific gauge?

Solution.

(a) Stationary states exist for static fields, so take a time-independent gauge function \(\alpha(\boldsymbol{r})\). Then \(\phi' = \phi - \partial_t\alpha = \phi\) is unchanged and only \(\boldsymbol{A}' = \boldsymbol{A} + \nabla\alpha\) shifts. The claim is that the new Hamiltonian is a unitary similarity transform of the old one,

\[ \hat{H}' = \mathrm{e}^{\mathrm{i}q\alpha/\hbar}\,\hat{H}\,\mathrm{e}^{-\mathrm{i}q\alpha/\hbar} . \]

To see this, conjugate the kinetic momentum. Using the preamble identity \(\mathrm{e}^{\mathrm{i}q\alpha/\hbar}\hat{\boldsymbol{p}}\,\mathrm{e}^{-\mathrm{i}q\alpha/\hbar} = \hat{\boldsymbol{p}} - q\nabla\alpha\), and that \(q\boldsymbol{A}\) commutes with the phase,

\[ \mathrm{e}^{\mathrm{i}q\alpha/\hbar}\,\hat{\boldsymbol{\pi}}\,\mathrm{e}^{-\mathrm{i}q\alpha/\hbar} = (\hat{\boldsymbol{p}} - q\nabla\alpha) - q\boldsymbol{A} = \hat{\boldsymbol{p}} - q\boldsymbol{A}' = \hat{\boldsymbol{\pi}}' . \]

Squaring, \(\hat{\boldsymbol{\pi}}'^2 = \mathrm{e}^{\mathrm{i}q\alpha/\hbar}\hat{\boldsymbol{\pi}}^2\mathrm{e}^{-\mathrm{i}q\alpha/\hbar}\) (the phase factors between the two \(\hat{\boldsymbol{\pi}}\)’s cancel), and \(q\phi' = q\phi\) commutes with the phase, so indeed \(\hat{H}' = \mathrm{e}^{\mathrm{i}q\alpha/\hbar}\hat{H}\,\mathrm{e}^{-\mathrm{i}q\alpha/\hbar}\). Now act on the rephased eigenstate \(\psi_n' = \mathrm{e}^{\mathrm{i}q\alpha/\hbar}\psi_n\):

\[ \hat{H}'\psi_n' = \mathrm{e}^{\mathrm{i}q\alpha/\hbar}\hat{H}\,\mathrm{e}^{-\mathrm{i}q\alpha/\hbar}\,\mathrm{e}^{\mathrm{i}q\alpha/\hbar}\psi_n = \mathrm{e}^{\mathrm{i}q\alpha/\hbar}\hat{H}\psi_n = \mathrm{e}^{\mathrm{i}q\alpha/\hbar}E_n\psi_n = E_n\psi_n' . \]

The transformed state is an eigenstate of the transformed Hamiltonian with the same eigenvalue \(E_n\). This is inevitable: a unitary similarity transformation never changes a spectrum. The argument in quotation marks confuses “\(\hat{H}\) looks different” with “\(\hat{H}\) has different eigenvalues” — but \(\hat{H}' = \hat{U}\hat{H}\hat{U}^\dagger\) is unitarily equivalent to \(\hat{H}\), so the two Hamiltonians have the same spectrum even though their matrix elements in a fixed basis differ.

(b) The decomposition \(E_n = \tfrac{1}{2m}\langle\hat{\boldsymbol{\pi}}^2\rangle + q\langle\phi\rangle\) is a split into expectation values, and it is not a gauge-covariant split. Two points resolve the paradox.

First, restrict to gauge transformations that preserve the static structure of the problem (time-independent \(\alpha\), as in part (a)). For these, \(\phi\) does not change at all, so \(q\langle\phi\rangle\) is in fact invariant — and \(\langle\hat{\boldsymbol{\pi}}^2\rangle\) is invariant, and so is their sum \(E_n\). There is no paradox within the class of transformations that keep “stationary state” meaningful.

Second, the statement “\(q\langle\phi\rangle\) is not gauge-invariant” refers to the full gauge group, which includes time-dependent \(\alpha\). The simplest such transformation, a purely time-dependent \(\alpha(t)\) with \(\partial_t\alpha = -c\), leaves \(\boldsymbol{A}\) untouched (\(\nabla\alpha = 0\)) but shifts \(\phi \to \phi + c\), hence \(q\langle\phi\rangle \to q\langle\phi\rangle + qc\) and \(E_n \to E_n + qc\). This shifts every level by the same constant — it is nothing but the freedom to choose the zero of the potential energy. The kinetic part \(\tfrac{1}{2m}\langle\hat{\boldsymbol{\pi}}^2\rangle\) is untouched. So the resolution is: \(E_n\) is gauge-invariant up to a global, \(n\)-independent additive constant fixed by the energy-zero convention. The genuinely physical, fully gauge-invariant quantities are the energy differences \(E_n - E_m\), in which the constant cancels. The “non-invariant term” \(q\langle\phi\rangle\) encodes exactly the unphysical freedom of where to put the energy origin — the familiar statement that absolute potential energy has no meaning.

(c) Spectroscopy of hydrogen measures spectral line frequencies, which are energy differences: \(\omega_{nm} = (E_n - E_m)/\hbar\). The calculation is typically done in the Coulomb gauge (\(\boldsymbol{A} = 0\), \(\phi = -e/4\pi\epsilon_0 r\)), but any other gauge yields wavefunctions \(\psi_n' = \mathrm{e}^{\mathrm{i}q\alpha/\hbar}\psi_n\) related by a unitary transformation, hence the same spectrum up to the global constant of part (b). Since the measured frequencies are differences, the constant drops out and every gauge predicts identical line positions. More broadly, every genuine observable — transition frequencies, transition rates, cross-sections, the probability density — comes out gauge-invariant. “The result is gauge-invariant” means precisely that the physics extracted from the calculation does not depend on the gauge chosen to perform it, even though intermediate objects like \(\boldsymbol{A}\), \(\hat{\boldsymbol{p}}\), or the absolute \(E_n\) do.

5. Coulomb gauge residual freedom. In the Coulomb gauge \(\nabla \cdot \boldsymbol{A} = 0\), show that the gauge is not fully fixed: \(\boldsymbol{A}' = \boldsymbol{A} + \nabla\alpha\) still satisfies \(\nabla \cdot \boldsymbol{A}' = 0\) provided \(\nabla^2\alpha = 0\). Give an explicit example of such a residual transformation.

Solution.

A gauge condition is meant to single out one representative from each class of physically equivalent potentials. The Coulomb (transverse) condition imposes \(\nabla\cdot\boldsymbol{A} = 0\). Take a potential \(\boldsymbol{A}\) already satisfying it and apply a further gauge transformation \(\boldsymbol{A}' = \boldsymbol{A} + \nabla\alpha\). Its divergence is

\[ \nabla\cdot\boldsymbol{A}' = \nabla\cdot\boldsymbol{A} + \nabla\cdot(\nabla\alpha) = 0 + \nabla^2\alpha . \]

Hence \(\boldsymbol{A}'\) is still in the Coulomb gauge if and only if

\[ \nabla^2\alpha = 0 , \]

i.e. \(\alpha\) is a harmonic function. The Coulomb condition therefore does not fix the gauge completely: the whole space of harmonic functions \(\alpha\) generates residual gauge transformations that move \(\boldsymbol{A}\) around within the Coulomb gauge. The condition \(\nabla\cdot\boldsymbol{A} = 0\) removes the longitudinal part of the gauge freedom but leaves the harmonic part.

Explicit examples of residual transformations (each has \(\nabla^2\alpha = 0\)):

\(\alpha = \boldsymbol{c}\cdot\boldsymbol{r}\) with \(\boldsymbol{c}\) a constant vector. Then \(\nabla\alpha = \boldsymbol{c}\), so \(\boldsymbol{A}' = \boldsymbol{A} + \boldsymbol{c}\) — adding a uniform constant to the vector potential keeps it transverse.
\(\alpha = xy\), giving \(\nabla\alpha = (y,x,0)\); or \(\alpha = x^2 - y^2\), giving \(\nabla\alpha = (2x,-2y,0)\). Both satisfy \(\nabla^2\alpha = 0\), as does any real or imaginary part of an analytic function of \(x+\mathrm{i}y\).
A spatially uniform but time-dependent \(\alpha = f(t)\): trivially harmonic, leaves \(\boldsymbol{A}\) unchanged, and shifts \(\phi \to \phi - \dot f\) — the energy-zero freedom of Problem 4(b).

A remark on boundary conditions: which residual transformations are admissible depends on the domain. For fields that must vanish at infinity, the only harmonic \(\alpha\) bounded everywhere is a constant (\(\nabla\alpha = 0\)), so the Coulomb gauge is then effectively fixed up to that triviality. On bounded domains, or in multiply connected regions, nontrivial harmonic \(\alpha\) survive and the residual freedom is genuine.

6. Gauge-invariant classification. For each quantity below, check the box indicating whether it is gauge-invariant or gauge-dependent under \(\boldsymbol{A}\to\boldsymbol{A}+\nabla\alpha\), \(\phi\to\phi-\partial_t\alpha\), \(\psi\to\mathrm{e}^{\mathrm{i}q\alpha/\hbar}\psi\). For each gauge-dependent entry, identify a related gauge-invariant quantity.

Quantity	Gauge invariant	Gauge-dependent
\(\vert\psi(\boldsymbol{r})\vert^2\)	☐	☐
\(\arg\psi(\boldsymbol{r})\)	☐	☐
\(\langle\hat{\boldsymbol{p}}\rangle\)	☐	☐
\(\boldsymbol{B} = \nabla\times\boldsymbol{A}\)	☐	☐
\(\boldsymbol{A}(\boldsymbol{r})\)	☐	☐
\(\oint\boldsymbol{A}\cdot\mathrm{d}\boldsymbol{l}\) (closed loop)	☐	☐
Energy eigenvalue \(E_n\)	☐	☐
Canonical momentum eigenvalue \(\hbar k\)	☐	☐
Electric field \(\boldsymbol{E} = -\nabla\phi - \partial_t\boldsymbol{A}\)	☐	☐

Solution.

Quantity	Gauge invariant	Gauge-dependent
\(\vert\psi(\boldsymbol{r})\vert^2\)	☑
\(\arg\psi(\boldsymbol{r})\)		☑
\(\langle\hat{\boldsymbol{p}}\rangle\)		☑
\(\boldsymbol{B} = \nabla\times\boldsymbol{A}\)	☑
\(\boldsymbol{A}(\boldsymbol{r})\)		☑
\(\oint\boldsymbol{A}\cdot\mathrm{d}\boldsymbol{l}\) (closed loop)	☑
Energy eigenvalue \(E_n\)	☑
Canonical momentum eigenvalue \(\hbar k\)		☑
Electric field \(\boldsymbol{E} = -\nabla\phi - \partial_t\boldsymbol{A}\)	☑

Reasoning, entry by entry.

\(\vert\psi(\boldsymbol{r})\vert^2\) — invariant: \(\vert\mathrm{e}^{\mathrm{i}q\alpha/\hbar}\psi\vert^2 = \vert\psi\vert^2\) (probability density is physical).
\(\arg\psi(\boldsymbol{r})\) — shifts by \(q\alpha(\boldsymbol{r})/\hbar\). Repair: the covariant phase gradient \(\hbar\nabla\arg\psi - q\boldsymbol{A}\) is invariant (both terms pick up \(q\nabla\alpha\) and cancel). It equals the local kinetic momentum, with \(\boldsymbol{j} = (\rho/m)(\hbar\nabla\arg\psi - q\boldsymbol{A})\).
\(\langle\hat{\boldsymbol{p}}\rangle\) — shifts under the rephasing (Problems 1b, 3). Repair: the kinetic-momentum expectation \(\langle\hat{\boldsymbol{\pi}}\rangle = \langle\hat{\boldsymbol{p}}-q\boldsymbol{A}\rangle = m\langle\hat{\boldsymbol{v}}\rangle\).
\(\boldsymbol{B} = \nabla\times\boldsymbol{A}\) — invariant: \(\nabla\times(\boldsymbol{A}+\nabla\alpha) = \nabla\times\boldsymbol{A}\) since \(\nabla\times\nabla\alpha = 0\).
\(\boldsymbol{A}(\boldsymbol{r})\) — shifts by \(\nabla\alpha\). Repair: its curl \(\boldsymbol{B}\) (local) or its closed-loop integral (nonlocal).
\(\oint\boldsymbol{A}\cdot\mathrm{d}\boldsymbol{l}\) — invariant: under \(\boldsymbol{A}\to\boldsymbol{A}+\nabla\alpha\) the change is \(\oint\nabla\alpha\cdot\mathrm{d}\boldsymbol{l} = \alpha(\boldsymbol{r}_\text{end}) - \alpha(\boldsymbol{r}_\text{start}) = 0\) for single-valued \(\alpha\) on a closed loop. By Stokes it equals the enclosed magnetic flux \(\Phi\).
\(E_n\) — invariant under transformations preserving the static structure (Problem 4a); fully physical content is the set of differences \(E_n - E_m\) (Problem 4b).
Canonical momentum eigenvalue \(\hbar k\) — shifts under rephasing of a plane-wave eigenstate of \(\hat{\boldsymbol{p}}\): \(\hbar k \to \hbar k + q\,\partial_x\alpha\). The same gauge shift afflicts any canonical-momentum matrix element. Repair: the kinetic-momentum eigenvalue \(\hbar k - qA\) (eigenvalue of \(\hat{\boldsymbol{\pi}}\)), equivalently \(mv\).
\(\boldsymbol{E} = -\nabla\phi - \partial_t\boldsymbol{A}\) — invariant: the \(\partial_t\alpha\) from \(\phi\) and the \(\nabla\alpha\) from \(\boldsymbol{A}\) cancel, \(-\nabla(-\partial_t\alpha) - \partial_t(\nabla\alpha) = 0\), using \(\partial_t\nabla\alpha = \nabla\partial_t\alpha\).

Pattern. Moduli, curls, and closed-loop integrals are gauge-invariant; phases, potentials, canonical labels, and open-path integrals are gauge-dependent. Each gauge-dependent quantity is repaired either by subtracting the matching potential (\(\hat{\boldsymbol{p}}\to\hat{\boldsymbol{\pi}}\)) or by closing the loop.