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Ch. 5 Solutions

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Basic

Medium

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Basic

B-1. Basic Calculations with the Clifford Algebra

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Solution Strategy

Substitute specific indices into the Clifford algebra \(\{\gamma^\mu, \gamma^\nu\} = 2\eta^{\mu\nu}\mathbf{1}\).

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

Substituting \(\mu = \nu = 0\):

\[ \{\gamma^0, \gamma^0\} = 2\gamma^0\gamma^0 = 2\eta^{00}\mathbf{1} = 2 \cdot (+1) \cdot \mathbf{1} \]

Therefore

\[ \boxed{\gamma^0\gamma^0 = \mathbf{1}} \]

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

Substituting \(\mu = \nu = 2\):

\[ \{\gamma^2, \gamma^2\} = 2\gamma^2\gamma^2 = 2\eta^{22}\mathbf{1} = 2 \cdot (-1) \cdot \mathbf{1} \]

Therefore

\[ \boxed{\gamma^2\gamma^2 = -\mathbf{1}} \]

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

Substituting \(\mu = 1, \nu = 3\):

\[ \{\gamma^1, \gamma^3\} = \gamma^1\gamma^3 + \gamma^3\gamma^1 = 2\eta^{13}\mathbf{1} = 2 \cdot 0 \cdot \mathbf{1} \]

Therefore

\[ \boxed{\gamma^1\gamma^3 + \gamma^3\gamma^1 = 0} \]

(d) \(\gamma^0 \gamma^2 \gamma^0\)

From the result of (c), when \(\mu \neq \nu\) we have \(\gamma^\mu\gamma^\nu = -\gamma^\nu\gamma^\mu\). Since \(\mu = 0, \nu = 2\) are distinct:

\[ \gamma^0\gamma^2 = -\gamma^2\gamma^0 \]

Substituting this:

\[ \gamma^0\gamma^2\gamma^0 = (-\gamma^2\gamma^0)\gamma^0 = -\gamma^2(\gamma^0\gamma^0) = -\gamma^2 \cdot \mathbf{1} \]

Therefore

\[ \boxed{\gamma^0\gamma^2\gamma^0 = -\gamma^2} \]

Verification

The result of (d) is a special case of the general relation \(\gamma^0\gamma^i\gamma^0 = -\gamma^i\) (for spatial components \(i = 1,2,3\)). This follows immediately from the anticommutativity of \(\gamma^0\) and \(\gamma^i\), i.e., \(\gamma^0\gamma^i = -\gamma^i\gamma^0\), together with \((\gamma^0)^2 = \mathbf{1}\), confirming consistency.

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

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Solution Strategy

Using the definition \(\gamma_\mu = \eta_{\mu\nu}\gamma^\nu\), we evaluate \(\gamma^\mu\gamma_\mu\) with the Clifford algebra.

Detailed Calculation

\[ \gamma^\mu\gamma_\mu = \eta_{\mu\nu}\gamma^\mu\gamma^\nu \]

Here \(\eta_{\mu\nu}\) is symmetric in \(\mu, \nu\), and the antisymmetric part of \(\gamma^\mu\gamma^\nu\) vanishes upon contraction with \(\eta_{\mu\nu}\). Therefore:

\[ \eta_{\mu\nu}\gamma^\mu\gamma^\nu = \eta_{\mu\nu} \cdot \frac{1}{2}(\gamma^\mu\gamma^\nu + \gamma^\nu\gamma^\mu) = \eta_{\mu\nu} \cdot \frac{1}{2}\{\gamma^\mu, \gamma^\nu\} \]

Substituting the Clifford algebra \(\{\gamma^\mu, \gamma^\nu\} = 2\eta^{\mu\nu}\mathbf{1}\):

\[ = \eta_{\mu\nu} \cdot \eta^{\mu\nu}\mathbf{1} = \delta^\mu{}_\mu \cdot \mathbf{1} = d \cdot \mathbf{1} \]

In \(d = 4\) dimensions:

\[ \boxed{\gamma^\mu\gamma_\mu = 4\,\mathbf{1}} \]

Verification

We confirm by direct calculation:

\[ \gamma^\mu\gamma_\mu = \gamma^0\gamma_0 + \gamma^1\gamma_1 + \gamma^2\gamma_2 + \gamma^3\gamma_3 \]
\[ = \gamma^0\gamma^0 + \gamma^1(-\gamma^1) + \gamma^2(-\gamma^2) + \gamma^3(-\gamma^3) \]
\[ = \mathbf{1} - (-\mathbf{1}) - (-\mathbf{1}) - (-\mathbf{1}) = \mathbf{1} + \mathbf{1} + \mathbf{1} + \mathbf{1} = 4\,\mathbf{1} \quad \checkmark \]

B-3. Transformation of the Dirac Adjoint

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(a) Proof that \(\overline{(\gamma^\mu \psi)} = \bar{\psi}\gamma^\mu\)

Solution Strategy

First we show that \((\gamma^\mu)^\dagger = \gamma^0\gamma^\mu\gamma^0\), then use this to carry out the calculation.

Detailed Calculation

Lemma: \((\gamma^\mu)^\dagger = \gamma^0\gamma^\mu\gamma^0\)

  • For \(\mu = 0\): \((\gamma^0)^\dagger = \gamma^0\) (Hermitian). Meanwhile \(\gamma^0\gamma^0\gamma^0 = (\gamma^0)^2\gamma^0 = \gamma^0\). Thus it holds.
  • For \(\mu = i\) (spatial components): \((\gamma^i)^\dagger = -\gamma^i\) (anti-Hermitian). Meanwhile \(\gamma^0\gamma^i\gamma^0 = -\gamma^i(\gamma^0)^2 = -\gamma^i\) (using the generalization of D1(d)). Thus it holds.

Using this:

\[ \overline{(\gamma^\mu\psi)} = (\gamma^\mu\psi)^\dagger\gamma^0 = \psi^\dagger(\gamma^\mu)^\dagger\gamma^0 \]
\[ = \psi^\dagger(\gamma^0\gamma^\mu\gamma^0)\gamma^0 = \psi^\dagger\gamma^0\gamma^\mu(\gamma^0)^2 = \psi^\dagger\gamma^0\gamma^\mu \cdot \mathbf{1} \]
\[ = \bar{\psi}\gamma^\mu \]
\[ \boxed{\overline{(\gamma^\mu\psi)} = \bar{\psi}\gamma^\mu} \]

(b) Proof that \(\bar{\psi}\gamma^\mu\psi\) is real

Detailed Calculation

\(\bar{\psi}\gamma^\mu\psi\) is a \(1 \times 1\) scalar quantity (a number) obtained by contracting spinor indices. Taking its Hermitian conjugate:

\[ (\bar{\psi}\gamma^\mu\psi)^\dagger = \psi^\dagger(\gamma^\mu)^\dagger(\bar{\psi})^\dagger \]

Since \(\bar{\psi} = \psi^\dagger\gamma^0\), we have \((\bar{\psi})^\dagger = (\gamma^0)^\dagger\psi = \gamma^0\psi\). Therefore:

\[ (\bar{\psi}\gamma^\mu\psi)^\dagger = \psi^\dagger(\gamma^0\gamma^\mu\gamma^0)\gamma^0\psi = \psi^\dagger\gamma^0\gamma^\mu\psi = \bar{\psi}\gamma^\mu\psi \]
\[ \boxed{(\bar{\psi}\gamma^\mu\psi)^\dagger = \bar{\psi}\gamma^\mu\psi} \]

Therefore \(\bar{\psi}\gamma^\mu\psi\) is Hermitian (classically, real).

Verification

For \(\mu = 0\), we have \(\bar{\psi}\gamma^0\psi = \psi^\dagger(\gamma^0)^2\psi = \psi^\dagger\psi = \sum_\alpha |\psi_\alpha|^2 \geq 0\), which is indeed real. This is also consistent with the positive-definiteness of the probability density.

B-4. Rearrangement of the Lorentz Algebra

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Solution Strategy

Expand \([J^i_+, J^j_-]\) according to its definition and substitute equations (5.3a)–(5.3c).

Detailed Calculation

\[ [J^i_+, J^j_-] = \left[\frac{1}{2}(J^i + iK^i),\, \frac{1}{2}(J^j - iK^j)\right] \]
\[ = \frac{1}{4}\left[(J^i + iK^i),\, (J^j - iK^j)\right] \]

Using the linearity of the commutator, expand into four terms:

\[ = \frac{1}{4}\Bigl([J^i, J^j] + [J^i, -iK^j] + [iK^i, J^j] + [iK^i, -iK^j]\Bigr) \]
\[ = \frac{1}{4}\Bigl([J^i, J^j] - i[J^i, K^j] + i[K^i, J^j] + [K^i, K^j]\Bigr) \]

Substitute equations (5.3a)–(5.3c) into each term:

Term 1: \([J^i, J^j] = i\varepsilon^{ijk}J^k\) (Eq. (5.3a))

Term 2: \(-i[J^i, K^j] = -i \cdot i\varepsilon^{ijk}K^k = -i^2\varepsilon^{ijk}K^k = +\varepsilon^{ijk}K^k\) (Eq. (5.3b))

Term 3: \(i[K^i, J^j] = i \cdot (-[J^j, K^i]) = i \cdot (-i\varepsilon^{jik}K^k) = -i^2\varepsilon^{jik}K^k = +\varepsilon^{jik}K^k\)

Since \(\varepsilon^{jik} = -\varepsilon^{ijk}\):

\[ \text{Term 3} = -\varepsilon^{ijk}K^k \]

Term 4: \([K^i, K^j] = -i\varepsilon^{ijk}J^k\) (Eq. (5.3c))

Adding all terms together:

\[ [J^i_+, J^j_-] = \frac{1}{4}\Bigl(i\varepsilon^{ijk}J^k + \varepsilon^{ijk}K^k - \varepsilon^{ijk}K^k - i\varepsilon^{ijk}J^k\Bigr) \]

\(J^k\) terms: \(i\varepsilon^{ijk}J^k - i\varepsilon^{ijk}J^k = 0\)

\(K^k\) terms: \(\varepsilon^{ijk}K^k - \varepsilon^{ijk}K^k = 0\)

Therefore:

\[ \boxed{[J^i_+, J^j_-] = 0} \]

Verification

This result agrees with equation (5.5c). It is also consistent with the decomposition structure of the Lorentz algebra, in which \(\mathbf{J}_+\) and \(\mathbf{J}_-\) form mutually independent \(\mathfrak{su}(2)\) algebras.

B-5. Boost Generator in Spinor Representation

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Solution Strategy

Taylor expand \(e^{-\frac{\eta}{2}\sigma^1}\) and use \((\sigma^1)^2 = \mathbf{1}\) to separate into even and odd powers.

Detailed Calculation

Taylor expansion:

\[ S_L = e^{-\frac{\eta}{2}\sigma^1} = \sum_{n=0}^{\infty} \frac{1}{n!}\left(-\frac{\eta}{2}\right)^n (\sigma^1)^n \]

From \((\sigma^1)^2 = \mathbf{1}\):

\[ (\sigma^1)^{2k} = \mathbf{1}, \qquad (\sigma^1)^{2k+1} = \sigma^1 \]

Separating into even and odd powers:

\[ S_L = \sum_{k=0}^{\infty} \frac{1}{(2k)!}\left(-\frac{\eta}{2}\right)^{2k}\mathbf{1} + \sum_{k=0}^{\infty} \frac{1}{(2k+1)!}\left(-\frac{\eta}{2}\right)^{2k+1}\sigma^1 \]

Even powers: Since \(\left(-\frac{\eta}{2}\right)^{2k} = \left(\frac{\eta}{2}\right)^{2k}\),

\[ \sum_{k=0}^{\infty} \frac{1}{(2k)!}\left(\frac{\eta}{2}\right)^{2k} = \cosh\frac{\eta}{2} \]

Odd powers: Since \(\left(-\frac{\eta}{2}\right)^{2k+1} = -\left(\frac{\eta}{2}\right)^{2k+1}\),

\[ \sum_{k=0}^{\infty} \frac{1}{(2k+1)!}\left(-\frac{\eta}{2}\right)^{2k+1} = -\sinh\frac{\eta}{2} \]

Therefore:

\[ \boxed{S_L = \cosh\frac{\eta}{2}\,\mathbf{1} - \sinh\frac{\eta}{2}\,\sigma^1} \]

Verification

When \(\eta = 0\) (no boost), \(S_L = \cosh 0 \cdot \mathbf{1} - \sinh 0 \cdot \sigma^1 = \mathbf{1}\). This gives the identity transformation, which is correct.

Furthermore, \(\det S_L = \cosh^2\frac{\eta}{2} - \sinh^2\frac{\eta}{2} = 1\), confirming that it is an element of \(SL(2, \mathbb{C})\).

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

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Solution Strategy

Write \(\mathcal{L} = \bar{\psi}(i\gamma^\mu\partial_\mu - m)\psi\) in component form and apply the Euler-Lagrange equation with respect to \(\bar{\psi}_\alpha\).

Detailed Calculation

In component notation:

\[ \mathcal{L} = i\bar{\psi}_\alpha (\gamma^\mu)_{\alpha\beta}\partial_\mu\psi_\beta - m\bar{\psi}_\alpha\psi_\alpha \]

Partial derivative with respect to \(\bar{\psi}_\alpha\):

\[ \frac{\partial\mathcal{L}}{\partial\bar{\psi}_\alpha} = i(\gamma^\mu)_{\alpha\beta}\partial_\mu\psi_\beta - m\psi_\alpha = \bigl[i\gamma^\mu\partial_\mu\psi - m\psi\bigr]_\alpha \]

Next, we check for terms containing \(\partial_\mu\bar{\psi}_\alpha\). Looking at \(\mathcal{L}\) as written without integration by parts, \(\partial_\mu\bar{\psi}\) does not appear explicitly (the derivative acts on \(\psi\)). Therefore:

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

Thus the Euler-Lagrange equation gives:

\[ i(\gamma^\mu)_{\alpha\beta}\partial_\mu\psi_\beta - m\psi_\alpha = 0 \]

This is precisely the Dirac equation \((i\gamma^\mu\partial_\mu - m)\psi = 0\).

Next, we derive the conjugate equation from the Euler-Lagrange equation with respect to \(\psi\). Integrating \(\mathcal{L}\) by parts to transfer the derivative onto \(\bar{\psi}\):

\[ \mathcal{L} = i\bar{\psi}\gamma^\mu\partial_\mu\psi - m\bar{\psi}\psi \]
\[ = -i(\partial_\mu\bar{\psi})\gamma^\mu\psi - m\bar{\psi}\psi + \partial_\mu(i\bar{\psi}\gamma^\mu\psi) \]

Dropping the total derivative term, the Euler-Lagrange equation with respect to \(\psi_\beta\) gives:

\[ \frac{\partial\mathcal{L}}{\partial\psi_\beta} = -m\bar{\psi}_\beta \]
\[ \frac{\partial\mathcal{L}}{\partial(\partial_\mu\psi_\beta)} = i\bar{\psi}_\alpha(\gamma^\mu)_{\alpha\beta} \]

Therefore:

\[ \frac{\partial\mathcal{L}}{\partial\psi_\beta} - \partial_\mu\frac{\partial\mathcal{L}}{\partial(\partial_\mu\psi_\beta)} = -m\bar{\psi}_\beta - \partial_\mu\bigl(i\bar{\psi}_\alpha(\gamma^\mu)_{\alpha\beta}\bigr) = 0 \]

Returning to matrix notation:

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

This is the conjugate Dirac equation.

Verification

We verify by directly taking the Dirac conjugate of the Dirac equation \((i\gamma^\mu\partial_\mu - m)\psi = 0\). Taking the \(\dagger\) of both sides and multiplying by \(\gamma^0\) from the right:

\[ \psi^\dagger(-i(\gamma^\mu)^\dagger\overleftarrow{\partial}_\mu - m)\gamma^0 = 0 \]
\[ -i\psi^\dagger(\gamma^0\gamma^\mu\gamma^0)\gamma^0\overleftarrow{\partial}_\mu - m\psi^\dagger\gamma^0 = 0 \]
\[ -i\psi^\dagger\gamma^0\gamma^\mu\overleftarrow{\partial}_\mu - m\bar{\psi} = 0 \]
\[ -i(\partial_\mu\bar{\psi})\gamma^\mu - m\bar{\psi} = 0 \]

This agrees with the result obtained above. \(\checkmark\)

B-7. Verification of the Conjugate Momentum

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Strategy

Identify the terms in \(\mathcal{L}\) that contain \(\dot{\psi}_\alpha = \partial_0\psi_\alpha\), then compute \(\Pi_\alpha = \partial\mathcal{L}/\partial\dot{\psi}_\alpha\).

Detailed Calculation

\[ \mathcal{L} = i\psi^\dagger_\alpha(\gamma^0)_{\alpha\beta}(\gamma^\mu)_{\beta\gamma}\partial_\mu\psi_\gamma - m\psi^\dagger_\alpha(\gamma^0)_{\alpha\beta}\psi_\beta \]

The only term containing \(\dot{\psi}_\gamma = \partial_0\psi_\gamma\) is the \(\mu = 0\) term:

\[ \mathcal{L}\big|_{\mu=0} = i\psi^\dagger_\alpha(\gamma^0)_{\alpha\beta}(\gamma^0)_{\beta\gamma}\partial_0\psi_\gamma = i\psi^\dagger_\alpha\bigl[(\gamma^0)^2\bigr]_{\alpha\gamma}\dot{\psi}_\gamma \]

Using \((\gamma^0)^2 = \mathbf{1}\):

\[ \mathcal{L}\big|_{\mu=0} = i\psi^\dagger_\alpha\delta_{\alpha\gamma}\dot{\psi}_\gamma = i\psi^\dagger_\gamma\dot{\psi}_\gamma \]

Therefore:

\[ \Pi_\alpha = \frac{\partial\mathcal{L}}{\partial\dot{\psi}_\alpha} = i\psi^\dagger_\alpha \]
\[ \boxed{\Pi_\alpha = i\psi^\dagger_\alpha} \]

Verification

This agrees with Eq. (5.8) in the text, \(\Pi = i\psi^\dagger\). Unlike the scalar field case where \(\Pi = \dot{\phi}\), the conjugate momentum is proportional to the field itself rather than its time derivative. This reflects the fact that the Dirac Lagrangian is first order in time derivatives.

B-8. Derivation of the Hamiltonian Density

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Solution Strategy

Perform the Legendre transformation \(\mathcal{H} = \Pi\dot{\psi} - \mathcal{L}\), separating \(\mathcal{L}\) into \(\mu = 0\) and \(\mu = j\) components.

Detailed Calculation

First, separate \(\mathcal{L}\) into time and spatial components:

\[ \mathcal{L} = i\bar{\psi}\gamma^0\partial_0\psi + i\bar{\psi}\gamma^j\partial_j\psi - m\bar{\psi}\psi \]

Simplify the first term. Since \(\bar{\psi}\gamma^0 = \psi^\dagger(\gamma^0)^2 = \psi^\dagger\):

\[ i\bar{\psi}\gamma^0\partial_0\psi = i\psi^\dagger\dot{\psi} = \Pi\dot{\psi} \]

(using \(\Pi = i\psi^\dagger\))

Legendre transformation:

\[ \mathcal{H} = \Pi\dot{\psi} - \mathcal{L} = \Pi\dot{\psi} - \Pi\dot{\psi} - i\bar{\psi}\gamma^j\partial_j\psi + m\bar{\psi}\psi \]
\[ = -i\bar{\psi}\gamma^j\partial_j\psi + m\bar{\psi}\psi \]

Substituting \(\bar{\psi} = \psi^\dagger\gamma^0\):

\[ \mathcal{H} = -i\psi^\dagger\gamma^0\gamma^j\partial_j\psi + m\psi^\dagger\gamma^0\psi \]

Writing \(\gamma^0\gamma^j\partial_j = \gamma^0\boldsymbol{\gamma}\cdot\nabla\):

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

This agrees with Eq. (5.9) in the text.

Consistency Check

From the Dirac equation \(i\gamma^0\partial_0\psi + i\gamma^j\partial_j\psi - m\psi = 0\), we have \(i\partial_0\psi = (-i\gamma^0\gamma^j\partial_j + m\gamma^0)\psi\), so \(\mathcal{H}\) can also be written as \(\mathcal{H} = \psi^\dagger(i\partial_0)\psi\). This has the form of an expectation value of the Dirac Hamiltonian \(H_D = -i\gamma^0\boldsymbol{\gamma}\cdot\nabla + m\gamma^0 = \boldsymbol{\alpha}\cdot\boldsymbol{p} + \beta m\) (where \(\boldsymbol{\alpha} = \gamma^0\boldsymbol{\gamma}\), \(\beta = \gamma^0\)), which is consistent with the Dirac theory in quantum mechanics. \(\checkmark\)

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

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Case of commutation relations (causality preserved)

The propagator obtained from commutation relations for a real scalar field is:

\[ [\phi(x), \phi(y)] = i\Delta(x-y) = \int \frac{d^3p}{(2\pi)^3 2\omega_p}\left(e^{-ip\cdot(x-y)} - e^{+ip\cdot(x-y)}\right) \]

The integrand consists of the difference of two terms. The first term corresponds to the amplitude for a particle to propagate from \(y\) to \(x\), and the second to the amplitude for an antiparticle (same particle for a real scalar field) to propagate from \(x\) to \(y\).

For spacelike separated points \((x-y)^2 < 0\), a continuous Lorentz transformation can map \((x-y) \to -(x-y)\). \(\Delta(x-y)\) is an odd function of \((x-y)\):

\[ \Delta(-(x-y)) = -\Delta(x-y) \]

On the other hand, in the spacelike region the Lorentz invariant \((x-y)^2\) is unchanged, so the value of \(\Delta\) is also unchanged. Therefore:

\[ \Delta(x-y) = -\Delta(x-y) \implies \Delta(x-y) = 0 \]

Thus \([\phi(x), \phi(y)] = 0\) in the spacelike region, and causality (microcausality) is preserved.

Case of anticommutation relations (causality violated)

If anticommutation relations are imposed:

\[ \{\phi(x), \phi(y)\} = \int \frac{d^3p}{(2\pi)^3 2\omega_p}\left(e^{-ip\cdot(x-y)} + e^{+ip\cdot(x-y)}\right) \]

Here the integrand becomes the sum of two terms. This is an even function of \((x-y)\):

\[ \Delta_+(x-y) + \Delta_+^*(x-y) = 2\,\text{Re}\,\Delta_+(x-y) \]

Applying the Lorentz transformation \((x-y) \to -(x-y)\) in the spacelike region:

\[ f(-(x-y)) = f(x-y) \]

so the cancellation that occurs for odd functions does not happen. In fact, \(\Delta_+(x-y) = \int \frac{d^3p}{(2\pi)^3 2\omega_p}e^{-ip\cdot(x-y)}\) is a positive-definite Lorentz-invariant function in the spacelike region (specifically expressed in terms of the modified Bessel function \(K_1\)) and does not vanish.

Therefore:

\[ \{\phi(x), \phi(y)\} \neq 0 \quad \text{for} \quad (x-y)^2 < 0 \]
\[ \boxed{\text{Imposing anticommutation relations on a scalar field leads to nontrivial anticommutation at spacelike separation, violating causality (microcausality)}} \]

This is a physical consequence of the spin-statistics theorem. Integer-spin fields must obey Bose statistics (commutation relations) to be consistent with causality, while half-integer-spin fields must obey Fermi statistics (anticommutation relations).

Medium

M-1. Failure of Quantization via Commutation Relations

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(a) Hamiltonian with Commutation Relations Imposed

Strategy

Substitute the mode expansion into the Hamiltonian \(\hat{H} = \int d^3x\,\mathcal{H}\) and pay attention to the sign of the commutation relation in the \(d\) sector.

Detailed Calculation

From the result of D8, the Hamiltonian without normal ordering is:

\[ \hat{H} = \int d^3x\, \hat{\psi}^\dagger(-i\gamma^0\boldsymbol{\gamma}\cdot\nabla + m\gamma^0)\hat{\psi} \]

From the properties of the Dirac equation solutions, substituting the mode expansion and using spatial integration together with the orthogonality and completeness relations of the spinors, the Hamiltonian is organized into the following form:

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

Here the \(d\) sector term is in the order \(\hat{d}\hat{d}^\dagger\) (a result of sign handling originating from the fact that the \(v\) spinor is a negative energy solution).

Using the commutation relation \([\hat{d}^r_{\boldsymbol{p}}, \hat{d}^{s\dagger}_{\boldsymbol{q}}] = -(2\pi)^3\delta^{rs}\delta^{(3)}(\boldsymbol{p}-\boldsymbol{q})\) to reorder:

\[ \hat{d}^s_{\boldsymbol{p}}\hat{d}^{s\dagger}_{\boldsymbol{p}} = \hat{d}^{s\dagger}_{\boldsymbol{p}}\hat{d}^s_{\boldsymbol{p}} + [\hat{d}^s_{\boldsymbol{p}}, \hat{d}^{s\dagger}_{\boldsymbol{p}}] = \hat{d}^{s\dagger}_{\boldsymbol{p}}\hat{d}^s_{\boldsymbol{p}} - (2\pi)^3\delta^{ss}\delta^{(3)}(\boldsymbol{0}) \]

Substituting:

\[ \hat{H} = \sum_s\int\frac{d^3p}{(2\pi)^3}\, E_{\boldsymbol{p}}\left(\hat{b}^{s\dagger}_{\boldsymbol{p}}\hat{b}^s_{\boldsymbol{p}} + \hat{d}^{s\dagger}_{\boldsymbol{p}}\hat{d}^s_{\boldsymbol{p}} - (2\pi)^3\delta^{(3)}(\boldsymbol{0})\right) \]

Removing the constant term:

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

The sign of the \(d\) sector is negative. This directly originates from the sign of \([\hat{d}, \hat{d}^\dagger]\) being \(-1\).

(b) Physical Breakdown

Since the \(d\) sector contribution is \(-E_{\boldsymbol{p}}\hat{d}^{s\dagger}_{\boldsymbol{p}}\hat{d}^s_{\boldsymbol{p}}\), each time \(\hat{d}^{s\dagger}_{\boldsymbol{p}}\) acts on the vacuum, the energy decreases by \(E_{\boldsymbol{p}} > 0\).

Under commutation relations, \((\hat{d}^{s\dagger}_{\boldsymbol{p}})^n \neq 0\) (bosonic statistics places no upper limit on occupation number), so \(\hat{d}^{s\dagger}_{\boldsymbol{p}}\) can be applied arbitrarily many times, allowing the energy to be lowered without bound.

\[ \hat{H}\bigl[(\hat{d}^{s\dagger}_{\boldsymbol{p}})^n|0\rangle\bigr] \sim -nE_{\boldsymbol{p}} \to -\infty \quad (n \to \infty) \]

This means the energy is unbounded from below, and no stable vacuum state exists. This is physically unacceptable.

(c) Resolution via Anticommutation Relations

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

\[ \hat{d}^s_{\boldsymbol{p}}\hat{d}^{s\dagger}_{\boldsymbol{p}} = -\hat{d}^{s\dagger}_{\boldsymbol{p}}\hat{d}^s_{\boldsymbol{p}} + \{\hat{d}^s_{\boldsymbol{p}}, \hat{d}^{s\dagger}_{\boldsymbol{p}}\} = -\hat{d}^{s\dagger}_{\boldsymbol{p}}\hat{d}^s_{\boldsymbol{p}} + (2\pi)^3\delta^{(3)}(\boldsymbol{0}) \]

Substituting into the Hamiltonian:

\[ \hat{H} = \sum_s\int\frac{d^3p}{(2\pi)^3}\, E_{\boldsymbol{p}}\left(\hat{b}^{s\dagger}_{\boldsymbol{p}}\hat{b}^s_{\boldsymbol{p}} - \hat{d}^{s\dagger}_{\boldsymbol{p}}\hat{d}^s_{\boldsymbol{p}} + (2\pi)^3\delta^{(3)}(\boldsymbol{0})\right) \]

The crucial point is that the minus sign from the anticommutation relation cancels the original minus sign in the \(d\) sector when reordering \(\hat{d}\hat{d}^\dagger\). Removing the constant term:

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

Both the \(b\) sector and the \(d\) sector have positive coefficients \(E_{\boldsymbol{p}} > 0\), and the eigenvalues of the number operators \(\hat{b}^{s\dagger}\hat{b}^s\) and \(\hat{d}^{s\dagger}\hat{d}^s\) are \(0\) or \(1\) (due to the anticommutation relations), so the Hamiltonian is positive definite (up to the zero-point energy). A stable vacuum exists.

Consistency Check

With anticommutation relations, \((\hat{d}^{s\dagger}_{\boldsymbol{p}})^2 = 0\) (shown in S2), so the occupation number of \(d\) particles is restricted to 0 or 1, making it impossible to lower the energy indefinitely. The breakdown that occurs with commutation relations is completely resolved. \(\checkmark\)

M-2. Derivation of the Pauli Exclusion Principle

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

Calculation Details

The anticommutation relation between creation operators is:

\[ \{\hat{b}^{r\dagger}_{\boldsymbol{p}}, \hat{b}^{s\dagger}_{\boldsymbol{q}}\} = 0 \]

(From the condition that all anticommutators other than the fundamental anticommutation relation \(\{\hat{b}^r_{\boldsymbol{p}}, \hat{b}^{s\dagger}_{\boldsymbol{q}}\} = (2\pi)^3\delta^{rs}\delta^{(3)}(\boldsymbol{p}-\boldsymbol{q})\) vanish)

Setting \(r = s\), \(\boldsymbol{p} = \boldsymbol{q}\):

\[ \{\hat{b}^{s\dagger}_{\boldsymbol{p}}, \hat{b}^{s\dagger}_{\boldsymbol{p}}\} = 2(\hat{b}^{s\dagger}_{\boldsymbol{p}})^2 = 0 \]
\[ \boxed{(\hat{b}^{s\dagger}_{\boldsymbol{p}})^2 = 0} \]

(b) Pauli Exclusion Principle

For the one-particle state \(|\boldsymbol{p}, s\rangle = \hat{b}^{s\dagger}_{\boldsymbol{p}}|0\rangle\), if we attempt to create another particle with the same quantum numbers:

\[ \hat{b}^{s\dagger}_{\boldsymbol{p}}|\boldsymbol{p}, s\rangle = (\hat{b}^{s\dagger}_{\boldsymbol{p}})^2|0\rangle = 0 \]

That is, it is impossible to place two or more fermions with the same quantum numbers \((\boldsymbol{p}, s)\) in the same state. This is the Pauli exclusion principle.

The occupation number for fermions can only be \(n^s_{\boldsymbol{p}} = 0\) or \(1\), and each mode in Fock space is restricted to two dimensions (empty or occupied).

(c) Comparison with Bosons

For the scalar field commutation relation \([\hat{a}_{\boldsymbol{p}}, \hat{a}^\dagger_{\boldsymbol{q}}] = (2\pi)^3\delta^{(3)}(\boldsymbol{p}-\boldsymbol{q})\):

\[ (\hat{a}^\dagger_{\boldsymbol{p}})^2 \neq 0 \]

In fact, \([\hat{a}_{\boldsymbol{p}}, (\hat{a}^\dagger_{\boldsymbol{p}})^2] = 2(2\pi)^3\delta^{(3)}(\boldsymbol{0})\hat{a}^\dagger_{\boldsymbol{p}} \neq 0\), and in general:

\[ (\hat{a}^\dagger_{\boldsymbol{p}})^n|0\rangle \propto |n_{\boldsymbol{p}}\rangle \neq 0 \quad (\text{for any positive integer } n) \]

In the discretized case, \((\hat{a}^\dagger)^n|0\rangle = \sqrt{n!}\,|n\rangle\), and arbitrarily many bosons can be packed into the same quantum state. This is Bose-Einstein statistics, which is fundamentally different from the Fermi-Dirac statistics of fermions.

Fermions (anticommutation relations) Bosons (commutation relations)
\((\hat{a}^\dagger)^2\) \(= 0\) \(\neq 0\)
Occupation number \(0\) or \(1\) \(0, 1, 2, \ldots\)
Statistics Fermi-Dirac Bose-Einstein

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

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Strategy

Substitute the mode expansion at equal time \(x^0 = y^0 = t\) and compute the anticommutator.

Detailed Calculation

Mode expansion at equal time:

\[ \hat{\psi}_\alpha(\boldsymbol{x}, t) = \sum_s\int\frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2E_{\boldsymbol{p}}}}\left[\hat{b}^s_{\boldsymbol{p}}\, u^s_\alpha(\boldsymbol{p})e^{-iE_{\boldsymbol{p}}t + i\boldsymbol{p}\cdot\boldsymbol{x}} + \hat{d}^{s\dagger}_{\boldsymbol{p}}\, v^s_\alpha(\boldsymbol{p})e^{+iE_{\boldsymbol{p}}t - i\boldsymbol{p}\cdot\boldsymbol{x}}\right] \]
\[ \hat{\psi}^\dagger_\beta(\boldsymbol{y}, t) = \sum_{s'}\int\frac{d^3q}{(2\pi)^3}\frac{1}{\sqrt{2E_{\boldsymbol{q}}}}\left[\hat{b}^{s'\dagger}_{\boldsymbol{q}}\, u^{s'\dagger}_\beta(\boldsymbol{q})e^{+iE_{\boldsymbol{q}}t - i\boldsymbol{q}\cdot\boldsymbol{y}} + \hat{d}^{s'}_{\boldsymbol{q}}\, v^{s'\dagger}_\beta(\boldsymbol{q})e^{-iE_{\boldsymbol{q}}t + i\boldsymbol{q}\cdot\boldsymbol{y}}\right] \]

We compute the anticommutator \(\{\hat{\psi}_\alpha(\boldsymbol{x}, t), \hat{\psi}^\dagger_\beta(\boldsymbol{y}, t)\}\). From the fundamental anticommutation relations, the only non-vanishing contributions come from the \(\{\hat{b}, \hat{b}^\dagger\}\) and \(\{\hat{d}^\dagger, \hat{d}\}\) terms (cross terms such as \(\{\hat{b}, \hat{d}\}\) all vanish).

Contribution from the \(b\) sector:

\[ \sum_{s,s'}\int\frac{d^3p}{(2\pi)^3}\frac{d^3q}{(2\pi)^3}\frac{1}{\sqrt{2E_{\boldsymbol{p}}}}\frac{1}{\sqrt{2E_{\boldsymbol{q}}}}\, u^s_\alpha(\boldsymbol{p})\, u^{s'\dagger}_\beta(\boldsymbol{q})\,\underbrace{\{\hat{b}^s_{\boldsymbol{p}}, \hat{b}^{s'\dagger}_{\boldsymbol{q}}\}}_{(2\pi)^3\delta^{ss'}\delta^{(3)}(\boldsymbol{p}-\boldsymbol{q})}\, e^{i\boldsymbol{p}\cdot\boldsymbol{x} - i\boldsymbol{q}\cdot\boldsymbol{y}} \]

(The time-dependent phases \(e^{-iE_{\boldsymbol{p}}t}\) and \(e^{+iE_{\boldsymbol{q}}t}\) cancel when \(\boldsymbol{p} = \boldsymbol{q}\))

Performing the \(\boldsymbol{q}\) integration and the sum over \(s'\):

\[ = \sum_s\int\frac{d^3p}{(2\pi)^3}\frac{1}{2E_{\boldsymbol{p}}}\, u^s_\alpha(\boldsymbol{p})\, u^{s\dagger}_\beta(\boldsymbol{p})\, e^{i\boldsymbol{p}\cdot(\boldsymbol{x}-\boldsymbol{y})} \]

Contribution from the \(d\) sector:

\[ \sum_{s,s'}\int\frac{d^3p}{(2\pi)^3}\frac{d^3q}{(2\pi)^3}\frac{1}{\sqrt{2E_{\boldsymbol{p}}}}\frac{1}{\sqrt{2E_{\boldsymbol{q}}}}\, v^s_\alpha(\boldsymbol{p})\, v^{s'\dagger}_\beta(\boldsymbol{q})\,\underbrace{\{\hat{d}^{s\dagger}_{\boldsymbol{p}}, \hat{d}^{s'}_{\boldsymbol{q}}\}}_{(2\pi)^3\delta^{ss'}\delta^{(3)}(\boldsymbol{p}-\boldsymbol{q})}\, e^{-i\boldsymbol{p}\cdot\boldsymbol{x} + i\boldsymbol{q}\cdot\boldsymbol{y}} \]

Performing the \(\boldsymbol{q}\) integration and the sum over \(s'\):

\[ = \sum_s\int\frac{d^3p}{(2\pi)^3}\frac{1}{2E_{\boldsymbol{p}}}\, v^s_\alpha(\boldsymbol{p})\, v^{s\dagger}_\beta(\boldsymbol{p})\, e^{-i\boldsymbol{p}\cdot(\boldsymbol{x}-\boldsymbol{y})} \]

In the \(d\) sector, we change the integration variable \(\boldsymbol{p} \to -\boldsymbol{p}\) (noting that \(E_{\boldsymbol{p}} = E_{-\boldsymbol{p}}\) and \(d^3p\) is invariant):

\[ = \sum_s\int\frac{d^3p}{(2\pi)^3}\frac{1}{2E_{\boldsymbol{p}}}\, v^s_\alpha(-\boldsymbol{p})\, v^{s\dagger}_\beta(-\boldsymbol{p})\, e^{i\boldsymbol{p}\cdot(\boldsymbol{x}-\boldsymbol{y})} \]

Adding both sectors:

\[ \{\hat{\psi}_\alpha(\boldsymbol{x}, t), \hat{\psi}^\dagger_\beta(\boldsymbol{y}, t)\} = \int\frac{d^3p}{(2\pi)^3}\frac{1}{2E_{\boldsymbol{p}}}\left[\sum_s u^s_\alpha(\boldsymbol{p})\, u^{s\dagger}_\beta(\boldsymbol{p}) + \sum_s v^s_\alpha(-\boldsymbol{p})\, v^{s\dagger}_\beta(-\boldsymbol{p})\right]e^{i\boldsymbol{p}\cdot(\boldsymbol{x}-\boldsymbol{y})} \]

Applying the spinor completeness relation:

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

Substituting:

\[ \{\hat{\psi}_\alpha(\boldsymbol{x}, t), \hat{\psi}^\dagger_\beta(\boldsymbol{y}, t)\} = \int\frac{d^3p}{(2\pi)^3}\frac{1}{2E_{\boldsymbol{p}}}\cdot 2E_{\boldsymbol{p}}\,\delta_{\alpha\beta}\, e^{i\boldsymbol{p}\cdot(\boldsymbol{x}-\boldsymbol{y})} \]
\[ = \delta_{\alpha\beta}\int\frac{d^3p}{(2\pi)^3}\, e^{i\boldsymbol{p}\cdot(\boldsymbol{x}-\boldsymbol{y})} = \delta_{\alpha\beta}\,\delta^{(3)}(\boldsymbol{x}-\boldsymbol{y}) \]
\[ \boxed{\{\hat{\psi}_\alpha(\boldsymbol{x}, t), \hat{\psi}^\dagger_\beta(\boldsymbol{y}, t)\} = \delta_{\alpha\beta}\,\delta^{(3)}(\boldsymbol{x}-\boldsymbol{y})} \]

Consistency Check

Setting \(\alpha = \beta\) and summing over all spinor components:

\[ \sum_\alpha \{\hat{\psi}_\alpha(\boldsymbol{x}, t), \hat{\psi}^\dagger_\alpha(\boldsymbol{y}, t)\} = 4\,\delta^{(3)}(\boldsymbol{x}-\boldsymbol{y}) \]

This is consistent with the number of degrees of freedom of a 4-component Dirac spinor. \(\checkmark\)

M-4. Noether Current and Conservation of Fermion Number

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(a) Derivation of the Conserved Current

Solution Strategy

We compute the Noether current for the global \(U(1)\) transformation \(\psi \to e^{i\alpha}\psi\), \(\bar{\psi} \to \bar{\psi}e^{-i\alpha}\).

Detailed Calculation

Infinitesimal transformation (\(\alpha\) is an infinitesimal constant):

\[ \delta\psi = i\alpha\psi, \qquad \delta\bar{\psi} = -i\alpha\bar{\psi} \]

The Noether current formula:

\[ j^\mu = \frac{\partial\mathcal{L}}{\partial(\partial_\mu\psi_\alpha)}\delta\psi_\alpha + \delta\bar{\psi}_\alpha\frac{\partial\mathcal{L}}{\partial(\partial_\mu\bar{\psi}_\alpha)} \]

Computing each partial derivative:

\[ \frac{\partial\mathcal{L}}{\partial(\partial_\mu\psi_\alpha)} = i\bar{\psi}_\beta(\gamma^\mu)_{\beta\alpha} \]
\[ \frac{\partial\mathcal{L}}{\partial(\partial_\mu\bar{\psi}_\alpha)} = 0 \]

(since \(\mathcal{L}\) does not contain \(\partial_\mu\bar{\psi}\))

Therefore:

\[ j^\mu = i\bar{\psi}_\beta(\gamma^\mu)_{\beta\alpha}\cdot i\alpha\psi_\alpha + 0 = -\alpha\bar{\psi}\gamma^\mu\psi \]

Removing \(\alpha\), the conserved current is (with the sign convention chosen so that \(j^\mu\) is a positive particle number current):

\[ \boxed{j^\mu = \bar{\psi}\gamma^\mu\psi} \]

The conservation law \(\partial_\mu j^\mu = 0\) can be verified directly from the Dirac equation and its conjugate:

\[ \partial_\mu(\bar{\psi}\gamma^\mu\psi) = (\partial_\mu\bar{\psi})\gamma^\mu\psi + \bar{\psi}\gamma^\mu(\partial_\mu\psi) = (im\bar{\psi})\psi + \bar{\psi}(-im\psi) = 0 \]

(b) Mode Expansion of the Conserved Charge

Detailed Calculation

The conserved charge is the spatial integral of \(j^0 = \bar{\psi}\gamma^0\psi = \psi^\dagger\psi\):

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

Substituting the mode expansion, the expansion of \(\hat{\psi}^\dagger\hat{\psi}\) yields four types of terms: \(b^\dagger b\), \(d\, d^\dagger\), \(b^\dagger d^\dagger\) (cross terms), and \(d\, b\) (cross terms).

Performing the spatial integral \(\int d^3x\, e^{i(\boldsymbol{p}-\boldsymbol{q})\cdot\boldsymbol{x}} = (2\pi)^3\delta^{(3)}(\boldsymbol{p}-\boldsymbol{q})\), the \(b^\dagger b\) and \(d\, d^\dagger\) terms survive at \(\boldsymbol{p} = \boldsymbol{q}\).

The cross terms carry time dependence \(e^{\pm 2iE_{\boldsymbol{p}}t}\) and vanish due to the spinor orthogonality relations

\[ u^{s\dagger}(\boldsymbol{p})v^{s'}(-\boldsymbol{p}) = 0, \qquad v^{s\dagger}(-\boldsymbol{p})u^{s'}(\boldsymbol{p}) = 0 \]

(after the substitution \(\boldsymbol{p} \to -\boldsymbol{p}\)).

The remaining terms are:

\[ \hat{Q} = \sum_s\int\frac{d^3p}{(2\pi)^3}\frac{1}{2E_{\boldsymbol{p}}}\left[\hat{b}^{s\dagger}_{\boldsymbol{p}}\hat{b}^s_{\boldsymbol{p}}\sum_{s'}u^{s\dagger}(\boldsymbol{p})u^s(\boldsymbol{p}) + \hat{d}^s_{\boldsymbol{p}}\hat{d}^{s\dagger}_{\boldsymbol{p}}\sum_{s'}v^{s\dagger}(\boldsymbol{p})v^s(\boldsymbol{p})\right] \]

Using the spinor normalization \(u^{s\dagger}(\boldsymbol{p})u^s(\boldsymbol{p}) = 2E_{\boldsymbol{p}}\), \(v^{s\dagger}(\boldsymbol{p})v^s(\boldsymbol{p}) = 2E_{\boldsymbol{p}}\) (no summation):

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

Using the anticommutation relation \(\hat{d}^s_{\boldsymbol{p}}\hat{d}^{s\dagger}_{\boldsymbol{p}} = -\hat{d}^{s\dagger}_{\boldsymbol{p}}\hat{d}^s_{\boldsymbol{p}} + (2\pi)^3\delta^{(3)}(\boldsymbol{0})\):

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

(c) Interpretation of Antiparticles

From the structure of the conserved charge \(\hat{Q}\):

  • For \(\hat{b}^{s\dagger}_{\boldsymbol{p}}|0\rangle\) (one-particle state of a \(b\) particle): \(\hat{Q}\hat{b}^{s\dagger}_{\boldsymbol{p}}|0\rangle = (+1)\hat{b}^{s\dagger}_{\boldsymbol{p}}|0\rangle + \cdots\) → charge \(+1\)

  • For \(\hat{d}^{s\dagger}_{\boldsymbol{p}}|0\rangle\) (one-particle state of a \(d\) particle): \(\hat{Q}\hat{d}^{s\dagger}_{\boldsymbol{p}}|0\rangle = (-1)\hat{d}^{s\dagger}_{\boldsymbol{p}}|0\rangle + \cdots\) → charge \(-1\)

The \(b\) particles and \(d\) particles have the same mass and the same spin, but opposite charges. This is precisely the particle–antiparticle relationship.

\[ \boxed{\hat{b}^{s\dagger}_{\boldsymbol{p}} \text{ is the particle creation operator,} \quad \hat{d}^{s\dagger}_{\boldsymbol{p}} \text{ is the antiparticle creation operator}} \]

For example, if the \(b\) particle is identified as the electron, then the \(d\) particle corresponds to the positron. The negative-energy solutions of the Dirac equation are naturally reinterpreted, through quantization with anticommutation relations, as antiparticles with positive energy.

Consistency Check

We confirm that \(\hat{Q}\) is time-independent (i.e., it is a conserved quantity). The cross terms vanish due to spinor orthogonality, and the remaining terms have no \(t\) dependence. This is consistent with \(\partial_\mu j^\mu = 0\). \(\checkmark\)

Advanced

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

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(a) Causality of the Scalar Field

From the results of Ch. 4, when commutation relations are imposed on the free scalar field, the Pauli-Jordan function (commutator function)

\[ [\hat{\phi}(x), \hat{\phi}(y)] = i\Delta(x-y) \]

is Lorentz invariant, and for spacelike separations \((x-y)^2 < 0\),

\[ \Delta(x-y) = 0 \]

holds. This is because the propagation amplitude of particles and the propagation amplitude of antiparticles exactly cancel each other at spacelike separations. Physically, this means that measurements at two spatially separated points do not influence each other (causality is preserved).

(b) Violation of Causality When Commutation Relations Are Imposed on the Dirac Field

Consider the case where commutation relations are imposed on the Dirac field. Computing the commutator \([\hat{\psi}_\alpha(x), \bar{\hat{\psi}}_\beta(y)]\) formally gives:

\[ [\hat{\psi}_\alpha(x), \bar{\hat{\psi}}_\beta(y)] = \int\frac{d^3p}{(2\pi)^3}\frac{1}{2E_{\boldsymbol{p}}}\left[(\not\!p + m)_{\alpha\beta}\, e^{-ip\cdot(x-y)} - (\not\!p - m)_{\alpha\beta}\, e^{+ip\cdot(x-y)}\right] \]

Here the first term is the contribution from the \(b\) sector (particle propagation), and the second term is from the \(d\) sector (antiparticle propagation).

In the bosonic case, the particle and antiparticle propagation amplitudes take the same value at spacelike separations, and they cancel due to the sign structure of the commutation relations. However, when commutation relations are imposed on fermions, the sign of the \(d\) sector commutation relation is \(-1\) (see S1(a)), so the relative sign between the two terms changes, and the cancellation does not occur.

Specifically, in the argument using the Lorentz transformation that connects \(e^{-ip\cdot(x-y)}\) and \(e^{+ip\cdot(x-y)}\) at spacelike separations (if \((x-y)\) is spacelike, there exists a transformation that takes \(x-y \to -(x-y)\)), for bosons the two terms have the same sign and cancel, but for fermions with commutation relations they have opposite signs and add up instead.

Therefore, for \((x-y)^2 < 0\), \([\hat{\psi}_\alpha(x), \bar{\hat{\psi}}_\beta(y)] \neq 0\), and causality is violated.

(c) Recovery of Causality When Anticommutation Relations Are Imposed

When anticommutation relations are imposed, computing the anticommutator gives:

\[ \{\hat{\psi}_\alpha(x), \bar{\hat{\psi}}_\beta(y)\} = \int\frac{d^3p}{(2\pi)^3}\frac{1}{2E_{\boldsymbol{p}}}\left[(\not\!p + m)_{\alpha\beta}\, e^{-ip\cdot(x-y)} + (\not\!p - m)_{\alpha\beta}\, e^{+ip\cdot(x-y)}\right] \]

With anticommutation relations, the sign of the \(d\) sector is \(+1\), so the relative sign between the two terms changes. For spacelike separations \((x-y)^2 < 0\), by the Lorentz invariance argument, these two terms exactly cancel:

\[ \{\hat{\psi}_\alpha(x), \bar{\hat{\psi}}_\beta(y)\} = (i\not\!\partial_x + m)_{\alpha\beta}\, i\Delta(x-y) = 0 \quad \text{for } (x-y)^2 < 0 \]

Here \(\Delta(x-y)\) is the Pauli-Jordan function of the scalar field, which vanishes at spacelike separations.

Consistency with causality: At first glance, one might wonder whether \(\{\hat{\psi}(x), \bar{\hat{\psi}}(y)\} = 0\) is sufficient as a condition for causality, since it is only the anticommutator, not the commutator, that vanishes. However, physical observables are written as even products (bilinear forms) of fermion fields. For example:

  • Current density: \(j^\mu = \bar{\psi}\gamma^\mu\psi\)
  • Energy density: \(\mathcal{H} = \psi^\dagger(-i\gamma^0\boldsymbol{\gamma}\cdot\nabla + m\gamma^0)\psi\)
  • Operators appearing in scattering amplitudes

The commutator of such observables \(\mathcal{O}_1(x) = \bar{\psi}(x)\Gamma_1\psi(x)\), \(\mathcal{O}_2(y) = \bar{\psi}(y)\Gamma_2\psi(y)\) is:

\[ [\mathcal{O}_1(x), \mathcal{O}_2(y)] = 0 \quad \text{for } (x-y)^2 < 0 \]

This follows from \(\{\hat{\psi}(x), \bar{\hat{\psi}}(y)\} = 0\). When fermion fields are exchanged twice, the sign changes twice, resulting in the commutator being zero. Therefore, causality is fully preserved at the level of observables.

(d) Summary of the Spin-Statistics Theorem

The above discussion is organized from two perspectives.

(i) Positive-definiteness of energy (results from S1):

Field type With commutation relations With anticommutation relations
Integer spin (bosons) Hamiltonian positive-definite ✓ Hamiltonian identically zero (trivial theory) ✗
Half-integer spin (fermions) Energy unbounded below ✗ Hamiltonian positive-definite ✓

If anticommutation relations are imposed on integer-spin fields, \((\hat{a}^{s\dagger})^2 = 0\) and the degrees of freedom of the bosonic field vanish (the theory becomes trivial). If commutation relations are imposed on half-integer-spin fields, the energy becomes unbounded below (as shown in S1).

(ii) Causality (results of the present problem):

Field type With commutation relations With anticommutation relations
Integer spin (bosons) Causality ✓ Causality ✗
Half-integer spin (fermions) Causality ✗ Causality ✓

The cancellation of propagation amplitudes at spacelike separations depends on the relative sign between particle and antiparticle contributions. The correct cancellation occurs only when commutation relations are imposed for integer spin and anticommutation relations are imposed for half-integer spin.

Conclusion (Spin-Statistics Theorem):

In a Lorentz-invariant local field theory, in order to simultaneously satisfy both positive-definiteness of energy and causality, integer-spin fields must be quantized as bosons (commutation relations) and half-integer-spin fields must be quantized as fermions (anticommutation relations).

This is a theorem rigorously proven by Pauli and is one of the deepest results of quantum field theory.

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

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(a) Invariance of the Dirac Equation under Parity Transformation

Solution Strategy

We show that the transformed field \(\psi'(t, \boldsymbol{x}) = \eta_P\gamma^0\psi(t, -\boldsymbol{x})\) satisfies the Dirac equation.

Detailed Calculation

Assume the original field \(\psi(t, \boldsymbol{x})\) satisfies the Dirac equation:

\[ (i\gamma^0\partial_0 + i\gamma^j\partial_j - m)\psi(t, \boldsymbol{x}) = 0 \]

We write the transformed field as \(\psi'(x') = \eta_P\gamma^0\psi(x)\), where \(x' = (t, -\boldsymbol{x})\) and \(x = (t, \boldsymbol{x})\).

Writing the Dirac equation for \(\psi'\) in the \(x'\) coordinates:

\[ \left(i\gamma^0\frac{\partial}{\partial t} + i\gamma^j\frac{\partial}{\partial x'^j} - m\right)\psi'(t, \boldsymbol{x}') = 0 \]

Since \(x'^j = -x^j\), we have \(\frac{\partial}{\partial x'^j} = -\frac{\partial}{\partial x^j}\). Substituting \(\psi'(t, \boldsymbol{x}') = \eta_P\gamma^0\psi(t, \boldsymbol{x})\):

\[ \eta_P\left(i\gamma^0\partial_0 - i\gamma^j\partial_j - m\right)\gamma^0\psi(t, \boldsymbol{x}) = 0 \]

To move \(\gamma^0\) to the left of the operator, we use the commutation relations of the \(\gamma\) matrices with \(\gamma^0\):

  • \(\gamma^0\gamma^0 = \mathbf{1}\), and \(\gamma^j\gamma^0 = -\gamma^0\gamma^j\)... but here we need to pass \(\gamma^0\) through from the right side of the operator.

More precisely, we move \(\gamma^0\) to the left side of the operator. Looking at each term:

\[ \gamma^0\gamma^0 = \mathbf{1}, \qquad \gamma^j\gamma^0 = -\gamma^0\gamma^j \]

Therefore:

\[ (i\gamma^0\partial_0 - i\gamma^j\partial_j - m)\gamma^0 = i\gamma^0\gamma^0\partial_0 - i\gamma^j\gamma^0\partial_j - m\gamma^0 \]
\[ = i\mathbf{1}\cdot\partial_0 + i\gamma^0\gamma^j\partial_j - m\gamma^0 = \gamma^0(i\gamma^0\partial_0 + i\gamma^j\partial_j - m) \]

The last equality is obtained by factoring out \(\gamma^0\) on the left: \(\gamma^0 \cdot i\gamma^0\partial_0 = i(\gamma^0)^2\partial_0 = i\partial_0\) ✓, \(\gamma^0 \cdot i\gamma^j\partial_j = i\gamma^0\gamma^j\partial_j\) ✓, \(\gamma^0 \cdot (-m) = -m\gamma^0\) ✓.

Therefore:

\[ \eta_P\gamma^0(i\gamma^0\partial_0 + i\gamma^j\partial_j - m)\psi(t, \boldsymbol{x}) = 0 \]

The expression in parentheses is precisely the left-hand side of the original Dirac equation, which vanishes since \(\psi\) satisfies the Dirac equation.

\[ \boxed{\text{The Dirac equation is invariant under parity transformation.}} \]

(b) Charge Conjugation and Particle-Antiparticle Exchange

Solution Strategy

We substitute the mode expansion into the definition of the \(C\) transformation \(\hat{C}\hat{\psi}(x)\hat{C}^{-1} = \eta_C C\bar{\hat{\psi}}^T(x)\) and demonstrate the exchange of \(b\) and \(d\).

Detailed Calculation

We find the mode expansion of \(\bar{\psi}^T\). Since \(\bar{\psi} = \psi^\dagger\gamma^0\):

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

Taking the transpose:

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

Multiplying by the \(C\) matrix:

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

From the property of the charge conjugation matrix \(C\gamma^{\mu T}C^{-1} = -\gamma^\mu\), the following relations hold for solutions of the Dirac equation:

\[ C\bar{u}^{sT}(\boldsymbol{p}) = \eta^s_v\, v^s(\boldsymbol{p}), \qquad C\bar{v}^{sT}(\boldsymbol{p}) = \eta^s_u\, u^s(\boldsymbol{p}) \]

where \(\eta^s_v, \eta^s_u\) are phase factors (\(|\eta| = 1\)). This means that the \(C\) matrix connects positive-energy spinors \(u\) with negative-energy spinors \(v\).

Substituting:

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

Comparing this with the original mode expansion

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

for \(\hat{C}\hat{\psi}\hat{C}^{-1} = \eta_C C\bar{\hat{\psi}}^T\) to hold, we need:

\[ \hat{C}\hat{b}^s_{\boldsymbol{p}}\hat{C}^{-1} = \eta_C\eta^s_u\,\hat{d}^s_{\boldsymbol{p}}, \qquad \hat{C}\hat{d}^{s\dagger}_{\boldsymbol{p}}\hat{C}^{-1} = \eta_C\eta^s_v\,\hat{b}^{s\dagger}_{\boldsymbol{p}} \]

That is (up to phase factors):

\[ \boxed{\hat{C}: \quad \hat{b}^s_{\boldsymbol{p}} \leftrightarrow \hat{d}^s_{\boldsymbol{p}}} \]

Charge conjugation \(C\) exchanges particles and antiparticles.

(c) Invariance of the Lagrangian under \(CPT\) Transformation

Solution Strategy

We apply the \(C\), \(P\), and \(T\) transformations successively and verify that the Lagrangian is invariant.

Detailed Calculation

We summarize the action of each transformation on the Dirac field:

Parity \(P\):

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

Time reversal \(T\): (\(T\) is an anti-unitary operator)

\[ \hat{T}\hat{\psi}(t, \boldsymbol{x})\hat{T}^{-1} = \eta_T\gamma^1\gamma^3\hat{\psi}(-t, \boldsymbol{x}) \]

Charge conjugation \(C\):

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

Consider the composition of the \(CPT\) transformation. Letting \(\Theta = CPT\), with an appropriate choice of phase factors:

\[ \hat{\Theta}\hat{\psi}(x)\hat{\Theta}^{-1} = \eta_\Theta\gamma^5\hat{\psi}^c(-x) \]

where \(\gamma^5 \equiv i\gamma^0\gamma^1\gamma^2\gamma^3\) and \(\hat{\psi}^c\) is the charge-conjugated field. More specifically:

\[ \hat{\Theta}\hat{\psi}(x)\hat{\Theta}^{-1} \propto \gamma^5 C\bar{\hat{\psi}}^T(-x) \]

We verify the \(CPT\) transformation of the Lagrangian \(\mathcal{L} = \bar{\psi}(i\gamma^\mu\partial_\mu - m)\psi\).

Under the \(CPT\) transformation \(x^\mu \to -x^\mu\), so \(\partial_\mu \to -\partial_\mu\).

Substituting the transformations of \(\bar{\psi}\) and \(\psi\), and using the properties of \(\gamma^5\):

\[ \{\gamma^5, \gamma^\mu\} = 0 \quad (\text{i.e., } \gamma^5\gamma^\mu = -\gamma^\mu\gamma^5) \]
\[ (\gamma^5)^2 = \mathbf{1} \]

each term of the Lagrangian transforms as follows:

Kinetic term: \(CPT\) transformation of \(\bar{\psi}(x)i\gamma^\mu\partial_\mu\psi(x)\)

The two sign changes from \(\partial_\mu \to -\partial'_\mu\) (due to \(x \to -x\)) and \(\gamma^5\gamma^\mu = -\gamma^\mu\gamma^5\) cancel each other, leaving the kinetic term invariant.

Mass term: \(CPT\) transformation of \(-m\bar{\psi}(x)\psi(x)\)

Since \(\gamma^5\) appears twice and \((\gamma^5)^2 = \mathbf{1}\), the mass term is also invariant.

Therefore:

\[ \boxed{\mathcal{L}' = \bar{\psi}'(i\gamma^\mu\partial'_\mu - m)\psi' = \bar{\psi}(i\gamma^\mu\partial_\mu - m)\psi = \mathcal{L}} \]

The Dirac Lagrangian is invariant under the \(CPT\) transformation.

On the \(CPT\) Theorem

The \(CPT\) theorem is a general theorem that holds for any field theory satisfying the following conditions:

  1. Lorentz invariance (Poincaré symmetry)
  2. Locality (the Lagrangian is a local function of fields and their finite-order derivatives)
  3. Positive-definiteness of energy (the Hamiltonian is bounded from below)

Under these conditions, it is proven that the \(CPT\) transformation is always a symmetry of the theory (Lüders-Pauli theorem). The Dirac field case demonstrated above is a specific example of this general theorem.

Important consequences of the \(CPT\) theorem include:

  • Particles and antiparticles have the same mass
  • Particles and antiparticles have the same lifetime
  • If \(CP\) symmetry is violated, then \(T\) symmetry is also violated (and vice versa)

Verification

That each of the \(C\), \(P\), \(T\) transformations individually preserves the Dirac equation was confirmed for \(P\) in part (a). The same can be verified for \(C\) and \(T\). That the composition of the three transformations leaves the Lagrangian invariant follows automatically from the invariance under each transformation, but we also confirmed it directly above.

Furthermore, the anticommutativity of \(\gamma^5\), \(\{\gamma^5, \gamma^\mu\} = 0\), is derived from the Clifford algebra:

\[ \gamma^5\gamma^\mu = i\gamma^0\gamma^1\gamma^2\gamma^3\gamma^\mu = -\gamma^\mu\gamma^5 \]

(When anticommuting \(\gamma^\mu\) through all four \(\gamma\) matrices in \(\gamma^5\), the sign depends on whether \(\mu\) is \(0,1,2,3\), but in all cases \(\gamma^5\gamma^\mu = -\gamma^\mu\gamma^5\) holds. For any value of \(\mu\), \(\gamma^\mu\) commutes with itself within \(\gamma^5\) (producing \((\gamma^\mu)^2\)) and anticommutes with the remaining three, giving an overall sign of \((-1)^3 = -1\).) \(\checkmark\)