Skip to content

Ch. 5 Problems

← Back to chapter　|　View solutions

Table of Contents

Basic

Medium

Advanced

Basic

B-1. Basic Calculations with the Clifford Algebra

Using the Clifford algebra \(\{\gamma^\mu, \gamma^\nu\} = 2\eta^{\mu\nu}\mathbf{1}\), compute the following. Here \(\eta^{\mu\nu} = \mathrm{diag}(+1,-1,-1,-1)\).

(a) \(\gamma^0 \gamma^0\)

(b) \(\gamma^2 \gamma^2\)

(c) \(\gamma^1 \gamma^3 + \gamma^3 \gamma^1\)

(d) \(\gamma^0 \gamma^2 \gamma^0\) (compute step by step, referring to the hint)

Hint

For (a)–(c), simply substitute specific values of \(\mu, \nu\) into \(\{\gamma^\mu, \gamma^\nu\} = 2\eta^{\mu\nu}\mathbf{1}\). For (d), first use \(\gamma^0 \gamma^2 = -\gamma^2 \gamma^0\) (anticommutativity for different indices) to move \(\gamma^0\) to the right.

→ View solution

B-2. Computation of \(\gamma^\mu \gamma_\mu\)

Using the anticommutation relations of the Clifford algebra, compute \(\gamma^\mu \gamma_\mu = \eta_{\mu\nu}\gamma^\mu \gamma^\nu\) and show that the result is \(d\,\mathbf{1}\) (where \(d\) is the spacetime dimension). Here, take \(d = 4\) and find the specific value.

Hint

Treat \(\gamma^\mu \gamma_\mu = \eta_{\mu\nu}\gamma^\mu \gamma^\nu\) as a symmetric part. Transform it as \(\eta_{\mu\nu}\gamma^\mu \gamma^\nu = \eta_{\mu\nu} \cdot \frac{1}{2}\{\gamma^\mu, \gamma^\nu\} = \eta_{\mu\nu}\eta^{\mu\nu}\mathbf{1}\), and use \(\eta_{\mu\nu}\eta^{\mu\nu} = \delta^\mu{}_\mu = d\).

→ View solution

B-3. Transformation of the Dirac Adjoint

Using the definition of the Dirac adjoint \(\bar{\psi} \equiv \psi^\dagger \gamma^0\), show the following.

(a) Show that \(\overline{(\gamma^\mu \psi)} = \bar{\psi}\gamma^\mu\). You may use the Hermiticity of \(\gamma^0\), namely \((\gamma^0)^\dagger = \gamma^0\), and the anti-Hermiticity of the spatial components \((\gamma^i)^\dagger = -\gamma^i\).

(b) Using the result above, show that \(\bar{\psi}\gamma^\mu\psi\) is real (i.e., \((\bar{\psi}\gamma^\mu\psi)^\dagger = \bar{\psi}\gamma^\mu\psi\)).

Hint

(a) First verify that \((\gamma^\mu)^\dagger = \gamma^0 \gamma^\mu \gamma^0\) holds by considering the cases \(\mu = 0\) and \(\mu = i\) separately. Using this, you can simplify as \((\gamma^\mu \psi)^\dagger \gamma^0 = \psi^\dagger (\gamma^\mu)^\dagger \gamma^0 = \psi^\dagger \gamma^0 \gamma^\mu \gamma^0 \gamma^0 = \bar{\psi}\gamma^\mu\). (b) Either write out the spinor components explicitly or directly apply the result of (a).

→ View solution

B-4. Rearrangement of the Lorentz Algebra

Using the generators defined in equation (5.4) of the text,

\[ J^i_+ = \frac{1}{2}(J^i + iK^i), \qquad J^i_- = \frac{1}{2}(J^i - iK^i) \]

derive \([J^i_+, J^j_-] = 0\) (equation (5.5c)). Do not omit any intermediate calculations, and substitute the commutation relations (5.3a)–(5.3c) one by one.

Hint

The procedure is exactly the same as computing \([J^i_+, J^j_+]\) in the text. Expand \([J^i_+, J^j_-] = \frac{1}{4}[(J^i + iK^i),(J^j - iK^j)]\) and compute each of the four commutators \([J^i,J^j]\), \([J^i,-iK^j]\), \([iK^i,J^j]\), and \([iK^i,-iK^j]\) separately, then add them together. The \(K\) terms and the \(J\) terms should each cancel out.

→ View solution

B-5. Boost Generator in Spinor Representation

The boost generator for the left-handed Weyl spinor \((1/2, 0)\) is \(\mathbf{K}_L = -i\boldsymbol{\sigma}/2\). Write the transformation matrix for a boost in the \(x\)-direction with rapidity \(\eta\),

\[ S_L = \exp\left(-i\eta \, K_L^1\right) = \exp\left(-\frac{\eta}{2}\sigma^1\right) \]

explicitly in terms of \(\cosh\) and \(\sinh\), using the property \((\sigma^1)^2 = \mathbf{1}\).

Hint

Taylor expand \(e^{-\frac{\eta}{2}\sigma^1}\) and separate the even and odd powers using \((\sigma^1)^{2n} = \mathbf{1}\) and \((\sigma^1)^{2n+1} = \sigma^1\). Compare with the series definitions of \(\cosh\) and \(\sinh\).

→ View solution

B-6. Euler-Lagrange Equation for the Dirac Field (\(\psi\) Variation)

Given the Lagrangian density \(\mathcal{L} = \bar{\psi}(i\gamma^\mu \partial_\mu - m)\psi\), compute the Euler-Lagrange equation with respect to \(\psi\) (not \(\bar{\psi}\)):

\[ \frac{\partial \mathcal{L}}{\partial \bar{\psi}_\alpha} - \partial_\mu \frac{\partial \mathcal{L}}{\partial(\partial_\mu \bar{\psi}_\alpha)} = 0 \]

and derive the conjugate Dirac equation:

\[ i\partial_\mu \bar{\psi}\gamma^\mu + m\bar{\psi} = 0 \]

Here, \(\alpha\) denotes the spinor component index.

Hint

Either integrate by parts to transfer the derivative onto \(\bar{\psi}\) in \(\mathcal{L}\), or directly take the partial derivative with respect to \(\bar{\psi}_\alpha\). Writing out the components explicitly as \(\mathcal{L} = i\bar{\psi}_\alpha (\gamma^\mu)_{\alpha\beta}\partial_\mu \psi_\beta - m\bar{\psi}_\alpha \psi_\alpha\) makes the calculation easier. Be careful not to overlook terms containing \(\partial_\mu \bar{\psi}\) (integration by parts is required).

→ View solution

B-7. Verification of the Conjugate Momentum

In the text, equation (5.8) gives \(\Pi = i\psi^\dagger\). Verify this result in component form. That is, writing \(\mathcal{L} = i\psi^\dagger_\alpha (\gamma^0)_{\alpha\beta}(\gamma^\mu)_{\beta\gamma}\partial_\mu \psi_\gamma - m\psi^\dagger_\alpha (\gamma^0)_{\alpha\beta}\psi_\beta\), compute \(\Pi_\alpha = \partial \mathcal{L}/\partial \dot{\psi}_\alpha\) and confirm that \(\Pi_\alpha = i\psi^\dagger_\alpha\).

Hint

The only term containing \(\dot{\psi}_\alpha = \partial_0 \psi_\alpha\) is the \(\mu = 0\) term. Use \(i\psi^\dagger_\alpha (\gamma^0)_{\alpha\beta}(\gamma^0)_{\beta\gamma}\partial_0 \psi_\gamma = i\psi^\dagger_\alpha [(\gamma^0)^2]_{\alpha\gamma}\partial_0 \psi_\gamma\), and substitute \((\gamma^0)^2 = \mathbf{1}\).

→ View solution

B-8. Derivation of the Hamiltonian Density

Complete the derivation of Eq. (5.9) in the text. Substitute \(\Pi = i\psi^\dagger\) and \(\mathcal{L} = \bar{\psi}(i\gamma^\mu\partial_\mu - m)\psi\) into the Legendre transformation

\[ \mathcal{H} = \Pi \dot{\psi} - \mathcal{L} \]

and separate \(\mathcal{L}\) into the \(\mu = 0\) component and \(\mu = j\) (spatial components) to obtain Eq. (5.9):

\[ \mathcal{H} = \psi^\dagger(-i\gamma^0\boldsymbol{\gamma}\cdot\nabla + m\gamma^0)\psi \]
Hint

Decompose \(\mathcal{L} = i\bar{\psi}\gamma^0\partial_0\psi + i\bar{\psi}\gamma^j\partial_j\psi - m\bar{\psi}\psi\). Since \(\bar{\psi}\gamma^0 = \psi^\dagger(\gamma^0)^2 = \psi^\dagger\), the first term is \(i\psi^\dagger\dot{\psi} = \Pi\dot{\psi}\). This cancels with \(\Pi\dot{\psi}\) in the Legendre transformation, and the remaining terms give \(\mathcal{H}\). Note that \(\bar{\psi}\gamma^j = \psi^\dagger\gamma^0\gamma^j\).

→ View solution

B-9. Anticommutation Relations for a Scalar Field and Violation of Causality

Suppose that instead of commutation relations, one imposes anticommutation relations \(\{\phi(x), \phi(y)\} = i\Delta(x-y)\) on a real scalar field \(\phi(x)\). Show that for spacelike separated points \((x - y)^2 < 0\), \(\{\phi(x), \phi(y)\} \neq 0\) (violation of causality). Contrast this with the case of commutation relations, where causality is preserved.

Hint

\(\Delta(x-y) = \int \frac{d^3p}{(2\pi)^3 2\omega_p}(e^{-ip\cdot(x-y)} - e^{ip\cdot(x-y)})\) is an odd function. For anticommutation relations, the \(+\) sign appears, making it an even function (which does not vanish in the spacelike region).

→ View solution

Medium

M-1. Failure of Quantization via Commutation Relations

Consider the mode expansion of the Dirac field:

\[ \hat{\psi}(x) = \sum_{s=1}^{2}\int \frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2E_{\boldsymbol{p}}}}\left[\hat{b}^s_{\boldsymbol{p}}\, u^s(\boldsymbol{p})e^{-ip\cdot x} + \hat{d}^{s\dagger}_{\boldsymbol{p}}\, v^s(\boldsymbol{p})e^{+ip\cdot x}\right] \]

Here \(u^s(\boldsymbol{p})\) and \(v^s(\boldsymbol{p})\) are the positive- and negative-energy solutions of the Dirac equation, and \(E_{\boldsymbol{p}} = \sqrt{|\boldsymbol{p}|^2 + m^2}\).

(a) Suppose we impose commutation relations on \(\hat{b}^s_{\boldsymbol{p}}\) and \(\hat{d}^s_{\boldsymbol{p}}\):

\[ [\hat{b}^r_{\boldsymbol{p}}, \hat{b}^{s\dagger}_{\boldsymbol{q}}] = (2\pi)^3 \delta^{rs}\delta^{(3)}(\boldsymbol{p}-\boldsymbol{q}), \qquad [\hat{d}^r_{\boldsymbol{p}}, \hat{d}^{s\dagger}_{\boldsymbol{q}}] = -(2\pi)^3 \delta^{rs}\delta^{(3)}(\boldsymbol{p}-\boldsymbol{q}) \]

Show that the Hamiltonian takes the form

\[ \hat{H} = \sum_s \int \frac{d^3p}{(2\pi)^3} E_{\boldsymbol{p}}\left(\hat{b}^{s\dagger}_{\boldsymbol{p}}\hat{b}^s_{\boldsymbol{p}} - \hat{d}^{s\dagger}_{\boldsymbol{p}}\hat{d}^s_{\boldsymbol{p}}\right) + (\text{constant}) \]

(A complete derivation is not required. Focus your discussion on the sign of the \(d\) sector.)

(b) From the above result, explain that repeatedly applying \(\hat{d}^{s\dagger}_{\boldsymbol{p}}\) can lower the energy without bound, and state why this is physically unacceptable.

(c) Show that if we instead impose anticommutation relations

\[ \{\hat{b}^r_{\boldsymbol{p}}, \hat{b}^{s\dagger}_{\boldsymbol{q}}\} = (2\pi)^3 \delta^{rs}\delta^{(3)}(\boldsymbol{p}-\boldsymbol{q}), \qquad \{\hat{d}^r_{\boldsymbol{p}}, \hat{d}^{s\dagger}_{\boldsymbol{q}}\} = (2\pi)^3 \delta^{rs}\delta^{(3)}(\boldsymbol{p}-\boldsymbol{q}) \]

the sign in the \(d\) sector is reversed, and the Hamiltonian becomes positive definite (up to the zero-point energy).

Hint

(a) Substitute the mode expansion into the Hamiltonian and simplify using the spinor completeness relations such as \(\sum_s u^s(\boldsymbol{p})\bar{u}^s(\boldsymbol{p}) = \not\!p + m\). The key point is that in the \(d\) sector, the commutation relation \(\hat{d}\hat{d}^\dagger = \hat{d}^\dagger\hat{d} + [\hat{d},\hat{d}^\dagger]\) carries a sign of \(-1\). (b) Discuss how \(\hat{d}^{s\dagger}\) ends up functioning as a creation operator for "negative-energy particles." (c) Verify that with anticommutation relations, \(\hat{d}\hat{d}^\dagger = -\hat{d}^\dagger\hat{d} + \{\hat{d},\hat{d}^\dagger\}\), which flips the sign.

→ View solution

M-2. Derivation of the Pauli Exclusion Principle

Using the anticommutation relation \(\{\hat{b}^r_{\boldsymbol{p}}, \hat{b}^{s\dagger}_{\boldsymbol{q}}\} = (2\pi)^3\delta^{rs}\delta^{(3)}(\boldsymbol{p}-\boldsymbol{q})\), show the following.

(a) Show that \((\hat{b}^{s\dagger}_{\boldsymbol{p}})^2 = 0\).

(b) Explain why this result implies that two or more fermions with identical quantum numbers \((\boldsymbol{p}, s)\) cannot occupy the same state (the Pauli exclusion principle).

(c) Contrast this with the case of the scalar field commutation relation \([\hat{a}_{\boldsymbol{p}}, \hat{a}^\dagger_{\boldsymbol{q}}] = (2\pi)^3\delta^{(3)}(\boldsymbol{p}-\boldsymbol{q})\), where \((\hat{a}^\dagger_{\boldsymbol{p}})^n |0\rangle \neq 0\) for any positive integer \(n\).

Hint

(a) Try computing \(\{\hat{b}^{s\dagger}_{\boldsymbol{p}}, \hat{b}^{s\dagger}_{\boldsymbol{p}}\} = 2(\hat{b}^{s\dagger}_{\boldsymbol{p}})^2\). In addition to the anticommutation relation with \(r=s\), \(\boldsymbol{p}=\boldsymbol{q}\), you need the relation \(\{\hat{b}^{s\dagger}_{\boldsymbol{p}}, \hat{b}^{r\dagger}_{\boldsymbol{q}}\} = 0\). (c) Recall that for bosons, \((\hat{a}^\dagger)^n |0\rangle = \sqrt{n!}|n\rangle\).

→ View solution

M-3. Equal-Time Anticommutation Relations of the Dirac Field

Starting from the mode expansion and the fundamental anticommutation relations

\[ \{\hat{b}^r_{\boldsymbol{p}}, \hat{b}^{s\dagger}_{\boldsymbol{q}}\} = (2\pi)^3\delta^{rs}\delta^{(3)}(\boldsymbol{p}-\boldsymbol{q}), \qquad \{\hat{d}^r_{\boldsymbol{p}}, \hat{d}^{s\dagger}_{\boldsymbol{q}}\} = (2\pi)^3\delta^{rs}\delta^{(3)}(\boldsymbol{p}-\boldsymbol{q}) \]

(all other anticommutators vanish), derive the equal-time anticommutation relation

\[ \{\hat{\psi}_\alpha(\boldsymbol{x}, t), \hat{\psi}^\dagger_\beta(\boldsymbol{y}, t)\} = \delta_{\alpha\beta}\,\delta^{(3)}(\boldsymbol{x}-\boldsymbol{y}) \]

Here \(\alpha, \beta\) are spinor component indices. You may use the spinor completeness relation

\[ \sum_{s=1}^{2} u^s_\alpha(\boldsymbol{p})\, u^{s\dagger}_\beta(\boldsymbol{p}) + \sum_{s=1}^{2} v^s_\alpha(-\boldsymbol{p})\, v^{s\dagger}_\beta(-\boldsymbol{p}) = 2E_{\boldsymbol{p}}\,\delta_{\alpha\beta} \]
Hint

Substituting the mode expansion and computing the anticommutator, the cross terms between \(b\) and \(d\) vanish (due to \(\{b, d^\dagger\} = 0\), etc.). Add the contributions from the \(b\) sector and the \(d\) sector, and verify that the spinor completeness relation appears inside the momentum integral. For the \(v^s(\boldsymbol{p})e^{+ip\cdot x}\) terms, the substitution of the integration variable \(\boldsymbol{p} \to -\boldsymbol{p}\) is useful.

→ View solution

M-4. Noether Current and Conservation of Fermion Number

The Dirac field Lagrangian \(\mathcal{L} = \bar{\psi}(i\not\!\partial - m)\psi\) is invariant under the global \(U(1)\) transformation

\[ \psi \to e^{i\alpha}\psi, \qquad \bar{\psi} \to \bar{\psi}\, e^{-i\alpha} \]

(a) Using Noether's theorem, derive the conserved current \(j^\mu = \bar{\psi}\gamma^\mu\psi\).

(b) Rewrite the corresponding conserved charge

\[ \hat{Q} = \int d^3x\, \hat{\psi}^\dagger \hat{\psi} \]

using the mode expansion, and show that it takes the form

\[ \hat{Q} = \sum_s \int \frac{d^3p}{(2\pi)^3}\left(\hat{b}^{s\dagger}_{\boldsymbol{p}}\hat{b}^s_{\boldsymbol{p}} - \hat{d}^{s\dagger}_{\boldsymbol{p}}\hat{d}^s_{\boldsymbol{p}}\right) + (\text{constant}) \]

(c) From this result, read off that the particles created by \(\hat{b}^{s\dagger}_{\boldsymbol{p}}\) and those created by \(\hat{d}^{s\dagger}_{\boldsymbol{p}}\) carry charges of opposite sign, and explain why \(\hat{d}^{s\dagger}_{\boldsymbol{p}}\) is the creation operator for antiparticles.

Hint

(a) Use the Noether current formula for the infinitesimal transformation \(\delta\psi = i\alpha\psi\): \(j^\mu = \frac{\partial\mathcal{L}}{\partial(\partial_\mu\psi)}\delta\psi + \delta\bar{\psi}\frac{\partial\mathcal{L}}{\partial(\partial_\mu\bar{\psi})}\). (b) Substitute the mode expansion into \(\hat{Q} = \int d^3x\, \hat{\psi}^\dagger\hat{\psi}\), and use the fact that the spatial integral of \(e^{i(\boldsymbol{p}-\boldsymbol{q})\cdot\boldsymbol{x}}\) produces \(\delta^{(3)}(\boldsymbol{p}-\boldsymbol{q})\). The \(b\)-\(d\) cross terms vanish due to orthogonality (terms of the \(u^\dagger v\) type).

→ View solution

Advanced

A-1. Spin-Statistics Theorem — Argument from Causality

We discuss causality (independence of measurements) for the Dirac field at two points separated by a spacelike interval \((x - y)^2 < 0\).

(a) For a scalar field (boson), state—citing the results of Ch. 4—that when commutation relations are imposed, \([\hat{\phi}(x), \hat{\phi}(y)] = 0\) holds at spacelike separations.

(b) For the Dirac field, argue that if commutation relations were imposed, \([\hat{\psi}_\alpha(x), \bar{\hat{\psi}}_\beta(y)] \neq 0\) at spacelike separations, violating causality. Show this by demonstrating that the cancellation between positive-energy and negative-energy contributions fails to occur.

(c) Show that when anticommutation relations are imposed, \(\{\hat{\psi}_\alpha(x), \bar{\hat{\psi}}_\beta(y)\} = 0\) (for \((x-y)^2 < 0\)) holds, and explain why this is consistent with causality. In particular, argue that the commutators of physical observables (bilinear forms of fermion fields such as \(\bar{\psi}\Gamma\psi\)) vanish at spacelike separations.

(d) Generalizing the above discussion, summarize the physical requirements of the spin-statistics theorem—"integer-spin fields must be bosons (commutation relations) and half-integer-spin fields must be fermions (anticommutation relations)"—from the two viewpoints of (i) positive-definiteness of energy and (ii) causality.

Hint

(a) Recall the Lorentz invariance of \([\hat{\phi}(x), \hat{\phi}(y)]\) shown in Ch. 4 and its vanishing at equal times \(x^0 = y^0\). (b) In the bosonic case, the propagation amplitudes of particle and antiparticle cancel at spacelike separations because both obey the same statistics. If commutation relations are imposed on fermions, the relative sign necessary for this cancellation is not obtained. (c) With anticommutation relations, \(\hat{\psi}(x)\bar{\hat{\psi}}(y) = -\bar{\hat{\psi}}(y)\hat{\psi}(x) + \{\hat{\psi}(x), \bar{\hat{\psi}}(y)\}\), so if the anticommutator vanishes, swapping the field ordering only introduces a sign change. Note that observables constructed from fermion fields are always written as products of an even number of fermion fields. (d) Combine the results of S1 with the results of this problem.

→ View solution

A-2. \(C\), \(P\), \(T\) Transformations and the \(CPT\) Theorem

We consider the discrete symmetries \(C\) (charge conjugation), \(P\) (parity), and \(T\) (time reversal) for the Dirac field.

(a) Suppose the Dirac field transforms under the parity transformation \(P\) as

\[ \hat{P}\hat{\psi}(t, \boldsymbol{x})\hat{P}^{-1} = \eta_P \gamma^0 \hat{\psi}(t, -\boldsymbol{x}) \]

where \(\eta_P\) is a phase factor. Show that the Dirac equation \((i\gamma^\mu\partial_\mu - m)\psi = 0\) is invariant under this transformation rule.

(b) Define the charge conjugation transformation \(C\) as

\[ \hat{C}\hat{\psi}(x)\hat{C}^{-1} = \eta_C \, C\,\bar{\hat{\psi}}^T(x) \]

where \(C\) is the charge conjugation matrix satisfying \(C\gamma^{\mu T}C^{-1} = -\gamma^\mu\). Using the mode expansion, show that the \(C\) transformation exchanges particles and antiparticles (\(\hat{b}^s_{\boldsymbol{p}} \leftrightarrow \hat{d}^s_{\boldsymbol{p}}\)).

(c) Verify that the Lagrangian \(\mathcal{L} = \bar{\psi}(i\not\!\partial - m)\psi\) is invariant under the combined \(CPT\) transformation acting on the Dirac field. While noting that the \(CPT\) theorem is a general theorem derived from Lorentz invariance and locality, demonstrate the invariance explicitly in this specific example.

Hint

(a) Substitute \(\psi'(t, \boldsymbol{x}) = \eta_P \gamma^0 \psi(t, -\boldsymbol{x})\) into the Dirac equation. Note that \(\partial_0\) remains unchanged, but \(\partial_i \to -\partial_i'\). Use \(\gamma^0 \gamma^i \gamma^0 = -\gamma^i\). (b) Write out the mode expansion of \(\bar{\psi}^T\) and use the fact that the \(C\) matrix relates \(u\) spinors and \(v\) spinors (\(C\bar{u}^T \propto v\), etc.). (c) Apply each transformation in sequence, or write out the composition of \(CPT\) transformations at once. Verify that \(CPT\) transforms the field as \(\hat{\psi}(x) \to \gamma^5 \hat{\psi}^c(-x)\) or a similar form. The properties of \(\gamma^5 \equiv i\gamma^0\gamma^1\gamma^2\gamma^3\) will be needed.

→ View solution