4.1.1 Gauge Principle#

Prompts

Why does the wavefunction \(\psi\) and the phase-rotated version \(\mathrm{e}^{\mathrm{i}\alpha}\psi\) describe the same physics? What is the global U(1) symmetry?
What goes wrong when you try to make the phase local, \(\alpha \to \alpha(x)\)? Why does the spatial derivative cause the Schrödinger equation to break?
How does introducing a gauge field \(\boldsymbol{A}\) fix the gauge problem? What is the covariant derivative, and why must it transform covariantly?
Explain why demanding local phase invariance forces the Hamiltonian to take the form \(\hat{H} = \frac{1}{2m}(\hat{\boldsymbol{p}} - q\boldsymbol{A})^2 + q\Phi\). Is this an assumption or a consequence?
Does gauge invariance derive electromagnetism from pure symmetry, or is there hidden physical input in the argument?

Lecture Notes#

Overview#

The gauge principle is a profound statement: the fact that global phase is unobservable suggests that local phase should also be unobservable. Promoting this symmetry from global to local requires introducing a new field—the gauge field—that couples to the wavefunction and restores the equations of motion. This principle lies at the heart of all modern physics, showing how fundamental forces emerge from the requirement that quantum mechanics be invariant under local phase rotations.

Global Phase Symmetry#

Why Global Phase is Unobservable#

Consider two wavefunctions differing by a constant (global) phase:

\[ \psi'(\boldsymbol{r}) = \mathrm{e}^{\mathrm{i}\alpha}\psi(\boldsymbol{r}) \]

where \(\alpha\) is independent of position and time. The Born rule gives:

\[ |\psi'(\boldsymbol{r})|^2 = |\mathrm{e}^{\mathrm{i}\alpha}|^2|\psi(\boldsymbol{r})|^2 = |\psi(\boldsymbol{r})|^2 \]

Global Phase Redundancy

The global phase \(\alpha\) is completely unobservable:

Probability densities are identical
All expectation values \(\langle\psi'|\hat{O}|\psi'\rangle = \langle\psi|\hat{O}|\psi\rangle\) are identical
No measurement can distinguish between \(\psi\) and \(\mathrm{e}^{\mathrm{i}\alpha}\psi\)

This freedom is the U(1) global gauge symmetry of quantum mechanics.

Local Phase Invariance: The Problem#

Now suppose the phase depends on position: \(\alpha \to \alpha(\boldsymbol{r},t)\). We want to ask: is the physics still unchanged under

\[ \psi(\boldsymbol{r},t) \to \psi'(\boldsymbol{r},t) = \mathrm{e}^{\mathrm{i}\alpha(\boldsymbol{r},t)}\psi(\boldsymbol{r},t)? \]

The probability density is still invariant. But the equations of motion are not.

The Kinetic Energy Breaks Covariance#

When we compute spatial derivatives:

\[ \partial_i\psi' = \partial_i\left(\mathrm{e}^{\mathrm{i}\alpha}\psi\right) = \mathrm{e}^{\mathrm{i}\alpha}\left(\mathrm{i}\partial_i\alpha + \partial_i\right)\psi \]

The kinetic energy term in the Schrödinger equation involves \(\nabla^2\psi\), which becomes:

\[ \nabla^2\psi' = \mathrm{e}^{\mathrm{i}\alpha}\left[(\nabla\alpha) \cdot (\nabla\alpha) + 2\mathrm{i}\nabla\alpha \cdot \nabla + \nabla^2\right]\psi \]

The Gauge Problem

Under a local phase transformation \(\psi \to \mathrm{e}^{\mathrm{i}\alpha(\boldsymbol{r},t)}\psi\), the Schrödinger equation is not covariant—it does not maintain the same form:

\[ \mathrm{i}\hbar\frac{\partial\psi'}{{\partial t}} \neq -\frac{\hbar^2}{2m}\nabla^2\psi' + V(\boldsymbol{r})\psi' \]

The spatial derivatives of the varying phase generate extra terms that cannot be absorbed into \(V(\boldsymbol{r})\).

This is the central puzzle: if global phase is unobservable, why should local phase break the fundamental equation of motion?

The Fix: Introducing a Gauge Field#

The resolution is to introduce a new field—the gauge field \(\boldsymbol{A}(\boldsymbol{r},t)\)—that transforms to cancel the unwanted terms.

The Covariant Derivative#

Define:

\[ D_i = \partial_i - \mathrm{i}\frac{q}{\hbar}A_i \]

where \(q\) is the particle’s charge and \(\hbar\) is Planck’s constant divided by \(2\pi\).

Covariant Derivative and Gauge Field

Under a local phase transformation \(\psi \to \mathrm{e}^{\mathrm{i}q\alpha(\boldsymbol{r},t)/\hbar}\psi\), the gauge field must transform as:

\[ \boldsymbol{A} \to \boldsymbol{A} + \nabla\alpha \]

With this choice, the covariant derivative transforms covariantly:

\[ D_i\psi \to \mathrm{e}^{\mathrm{i}q\alpha/\hbar}(D_i\psi) \]

This means \((\nabla\alpha) \cdot \psi\) cancels the unwanted gradient terms from the phase transformation, leaving the kinetic energy with the same form.

Derivation: Why Covariance Works

Under \(\psi' = \mathrm{e}^{\mathrm{i}q\alpha/\hbar}\psi\) and \(\boldsymbol{A}' = \boldsymbol{A} + \nabla\alpha\):

\[ D_i'\psi' = \left(\partial_i - \mathrm{i}\frac{q}{\hbar}A_i'\right)\psi' = \left(\partial_i - \mathrm{i}\frac{q}{\hbar}(A_i + \partial_i\alpha)\right)\mathrm{e}^{\mathrm{i}q\alpha/\hbar}\psi \]

\[ = \mathrm{i}\frac{q}{\hbar}(\partial_i\alpha)\mathrm{e}^{\mathrm{i}q\alpha/\hbar}\psi + \mathrm{e}^{\mathrm{i}q\alpha/\hbar}(\partial_i\psi) - \mathrm{i}\frac{q}{\hbar}A_i\mathrm{e}^{\mathrm{i}q\alpha/\hbar}\psi - \mathrm{i}\frac{q}{\hbar}(\partial_i\alpha)\mathrm{e}^{\mathrm{i}q\alpha/\hbar}\psi \]

The first and last terms cancel:

\[ D_i'\psi' = \mathrm{e}^{\mathrm{i}q\alpha/\hbar}\left(\partial_i - \mathrm{i}\frac{q}{\hbar}A_i\right)\psi = \mathrm{e}^{\mathrm{i}q\alpha/\hbar}(D_i\psi) \]

This is the key: the covariant derivative transforms exactly like the wavefunction itself.

Gauge-Invariant Hamiltonian#

The requirement of local phase covariance forces us to replace the ordinary momentum operator with the covariant derivative everywhere:

\[ \hat{\boldsymbol{p}} \to \hat{\boldsymbol{p}} - q\boldsymbol{A} \]

This is the minimal coupling prescription. The Hamiltonian becomes:

\[ \hat{H} = \frac{1}{2m}(\hat{\boldsymbol{p}} - q\boldsymbol{A})^2 + V(\boldsymbol{r}) + q\Phi(\boldsymbol{r},t) \]

where \(\Phi(\boldsymbol{r},t)\) is the scalar potential (the time component of the four-vector gauge field).

Gauge-Invariant Hamiltonian (Minimal Coupling)

The form of the electromagnetic Hamiltonian is not assumed—it is forced by the requirement that the Schrödinger equation remain covariant under local \(U(1)\) gauge transformations.

This is the deepest insight of gauge theory: forces emerge from symmetry principles, not from external assumptions about how fields should couple to matter.

Expansion of the Kinetic Energy

Problem. Expand \((\hat{\boldsymbol{p}} - q\boldsymbol{A})^2\) and identify the different couplings to the gauge field.

Solution.

\[ (\hat{\boldsymbol{p}} - q\boldsymbol{A})^2 = \hat{\boldsymbol{p}}^2 - q(\hat{\boldsymbol{p}} \cdot \boldsymbol{A} + \boldsymbol{A} \cdot \hat{\boldsymbol{p}}) + q^2\boldsymbol{A}^2 \]

Using \(\hat{\boldsymbol{p}} = -\mathrm{i}\hbar\nabla\), the middle term becomes:

\[ \hat{\boldsymbol{p}} \cdot \boldsymbol{A} + \boldsymbol{A} \cdot \hat{\boldsymbol{p}} = -\mathrm{i}\hbar(\nabla \cdot \boldsymbol{A} + 2\boldsymbol{A} \cdot \nabla) \]

For the Coulomb gauge where \(\nabla \cdot \boldsymbol{A} = 0\):

\[ \frac{1}{2m}(\hat{\boldsymbol{p}} - q\boldsymbol{A})^2 = \frac{\hat{\boldsymbol{p}}^2}{2m} - \frac{q}{m}\boldsymbol{A} \cdot \hat{\boldsymbol{p}} + \frac{q^2}{2m}\boldsymbol{A}^2 \]

The second term is linear in \(\boldsymbol{A}\) (dipole coupling); the third is quadratic (paramagnetic term).

From Gauge Symmetry to Electromagnetism#

The gauge-invariant Hamiltonian \(\hat{H} = \frac{1}{2m}(\hat{\boldsymbol{p}} - q\boldsymbol{A})^2 + q\Phi\) contains more physics than its origin suggests. Computing equations of motion in the Heisenberg picture extracts the Lorentz force \(\boldsymbol{F} = q(\boldsymbol{E} + \boldsymbol{v} \times \boldsymbol{B})\) directly from the gauge structure — with no additional assumptions about electromagnetic forces. This derivation is carried out in §4.1.2.

Discussion: Is Electromagnetism Derived?

A remarkable fact: if we demand that quantum mechanics respect local phase invariance (that local phase should be as unobservable as global phase), we are forced to introduce electromagnetic fields coupled in exactly the way nature couples them.

This suggests that electromagnetism is not an independent phenomenon—it is a consequence of quantum mechanics plus the requirement of local gauge invariance. But is there hidden input? The assumption of U(1) gauge structure and the definition of “unobservable” are themselves non-trivial choices.

Question for discussion: Could other gauge structures (non-Abelian groups) arise from the same principle applied to different symmetries?

Summary#

Global phase \(\psi \to \mathrm{e}^{\mathrm{i}\alpha}\psi\) is unobservable because probability and all observables are unchanged. This is the U(1) global gauge symmetry.
Local phase transformations \(\psi(\boldsymbol{r}) \to \mathrm{e}^{\mathrm{i}\alpha(\boldsymbol{r})}\psi(\boldsymbol{r})\) break the Schrödinger equation because spatial derivatives of the varying phase generate new terms.
To restore covariance, we introduce a gauge field \(\boldsymbol{A}\) that transforms as \(\boldsymbol{A} \to \boldsymbol{A} + \nabla\alpha\).
The covariant derivative \(D_i = \partial_i - \mathrm{i}(q/\hbar)A_i\) transforms covariantly: \(D_i\psi \to \mathrm{e}^{\mathrm{i}q\alpha/\hbar}(D_i\psi)\).
The requirement of local gauge invariance forces the Hamiltonian to have the form \(\hat{H} = \frac{1}{2m}(\hat{\boldsymbol{p}} - q\boldsymbol{A})^2 + V + q\Phi\). This is not an assumption but a consequence of the symmetry principle.
The gauge principle shows that fundamental forces emerge from symmetry; electromagnetism is derived from demanding local phase invariance in quantum mechanics.

Homework#

1. Show that the global phase transformation \(\psi \to \mathrm{e}^{\mathrm{i}\alpha}\psi\) leaves all measurable quantities unchanged. Explicitly compute (a) the probability density \(|\psi'|^2\), and (b) the expectation value \(\langle\psi'|\hat{O}|\psi'\rangle\) for an arbitrary observable \(\hat{O}\). Why is global phase called “unobservable”?

2. Consider a local phase transformation \(\psi(\boldsymbol{r}) \to \mathrm{e}^{\mathrm{i}\alpha(\boldsymbol{r})}\psi(\boldsymbol{r})\). Show that: (a) The probability density \(|\psi|^2\) remains invariant. (b) The spatial derivatives transform as \(\partial_i\psi' = \mathrm{e}^{\mathrm{i}\alpha}(\mathrm{i}\partial_i\alpha + \partial_i)\psi\). (c) The kinetic energy term \(\hat{T}\psi\) does NOT transform as \(\mathrm{e}^{\mathrm{i}\alpha}\hat{T}\psi\). What extra terms appear?

3. The covariant derivative is \(D_i = \partial_i - \mathrm{i}(q/\hbar)A_i\). Show that if the gauge field transforms as \(\boldsymbol{A} \to \boldsymbol{A} + \nabla\alpha\) under \(\psi \to \mathrm{e}^{\mathrm{i}q\alpha/\hbar}\psi\), then the covariant derivative transforms covariantly:

\[D_i\psi \to \mathrm{e}^{\mathrm{i}q\alpha/\hbar}(D_i\psi)\]

4. The minimal coupling prescription says: replace \(\hat{\boldsymbol{p}} \to \hat{\boldsymbol{p}} - q\boldsymbol{A}\) in the free-particle Hamiltonian. Show that the resulting Hamiltonian is invariant under the combined gauge transformation: \(\psi \to \mathrm{e}^{\mathrm{i}q\alpha/\hbar}\psi\) and \(\boldsymbol{A} \to \boldsymbol{A} + \nabla\alpha\).

5. Expand \((\hat{\boldsymbol{p}} - q\boldsymbol{A})^2\) explicitly. In the Coulomb gauge (\(\nabla \cdot \boldsymbol{A} = 0\)), identify: (a) The linear coupling term (proportional to \(\boldsymbol{A}\)). (b) The quadratic term (proportional to \(\boldsymbol{A}^2\)). (c) In what regime might one term dominate over the other?

6. Consider two gauge choices for a uniform magnetic field \(\boldsymbol{B} = B\hat{z}\):

Symmetric gauge: \(\boldsymbol{A} = \frac{B}{2}(-y, x, 0)\)
Landau gauge: \(\boldsymbol{A} = (0, Bx, 0)\)

Show that both yield the same magnetic field \(\boldsymbol{B} = \nabla \times \boldsymbol{A}\). Are the resulting Schrödinger equations the same, or different? Discuss why gauge freedom is useful despite this.

7. Conceptual: A student argues: “Since local phase is supposedly unobservable, the gauge field \(\boldsymbol{A}\) must be unobservable too. But then how can it have physical effects?” Address this concern. Describe one quantum phenomenon where \(\boldsymbol{A}\) has observable consequences even when the magnetic field \(\boldsymbol{B} = \nabla \times \boldsymbol{A}\) vanishes in a region.

8. Show that the canonical momentum \(\boldsymbol{\pi} = \hat{\boldsymbol{p}} - q\boldsymbol{A}\) is invariant (unchanged) under a gauge transformation. Explain why this quantity, rather than the kinetic momentum \(\hat{\boldsymbol{p}} = -\mathrm{i}\hbar\nabla\), is the natural momentum in the presence of a gauge field.