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Ch. 1 Solutions

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Basic

Medium

Advanced

Basic

B-1. Substituting a plane wave solution into the Klein-Gordon equation to derive the dispersion relation

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Solution Strategy

Compute the derivatives of the plane wave solution \(\phi(\mathbf{x}, t) = A\, e^{i(\mathbf{p} \cdot \mathbf{x} - Et)}\) and substitute them into the Klein-Gordon equation.

Detailed Calculation

Time derivatives:

\[ \frac{\partial \phi}{\partial t} = A \cdot (-iE) \, e^{i(\mathbf{p} \cdot \mathbf{x} - Et)} = -iE\, \phi \]
\[ \frac{\partial^2 \phi}{\partial t^2} = (-iE)^2 \phi = -E^2 \phi \]

Spatial derivatives:

\[ \frac{\partial \phi}{\partial x^k} = i p_k \, \phi \]
\[ \nabla^2 \phi = \sum_{k=1}^{3} \frac{\partial^2 \phi}{\partial (x^k)^2} = \sum_{k=1}^{3} (ip_k)^2 \phi = -|\mathbf{p}|^2 \phi \]

Substitution into the Klein-Gordon equation:

\[ \frac{\partial^2 \phi}{\partial t^2} - \nabla^2 \phi + m^2 \phi = -E^2 \phi - (-|\mathbf{p}|^2 \phi) + m^2 \phi = 0 \]
\[ (-E^2 + |\mathbf{p}|^2 + m^2)\phi = 0 \]

Since \(\phi \neq 0\):

\[ -E^2 + |\mathbf{p}|^2 + m^2 = 0 \]

Final Answer

\[ \boxed{E^2 = |\mathbf{p}|^2 + m^2} \]

Verification

Dimensional analysis: In natural units, \([E] = [p] = [m] = 1\) (mass dimension 1), so \(E^2\), \(|\mathbf{p}|^2\), and \(m^2\) all have mass dimension 2, which is consistent. Additionally, setting \(\mathbf{p} = 0\) gives \(E = \pm m\), which is correct for the energy of a particle at rest.

B-2. Calculation of Probability Density for Negative Energy Solutions

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Solution Strategy

Substitute the negative-energy solution \(\phi = A\, e^{i(\mathbf{p} \cdot \mathbf{x} + |E|t)}\) into the expression for \(\rho\) given by equation (1.6). This is the plane wave \(e^{i(\mathbf{p}\cdot\mathbf{x} - Et)} = e^{i(\mathbf{p}\cdot\mathbf{x} + |E|t)}\) corresponding to \(E = -|E|\).

Detailed Calculation

Computing the time derivatives:

\[ \frac{\partial \phi}{\partial t} = A \cdot (i|E|) \, e^{i(\mathbf{p} \cdot \mathbf{x} + |E|t)} = i|E|\, \phi \]
\[ \frac{\partial \phi^*}{\partial t} = A^* \cdot (-i|E|) \, e^{-i(\mathbf{p} \cdot \mathbf{x} + |E|t)} = -i|E|\, \phi^* \]

Substituting into \(\rho\):

\[ \rho = \frac{i}{2m}\left(\phi^* \frac{\partial \phi}{\partial t} - \phi \frac{\partial \phi^*}{\partial t}\right) \]
\[ = \frac{i}{2m}\left(\phi^* \cdot i|E|\, \phi - \phi \cdot (-i|E|)\, \phi^*\right) \]
\[ = \frac{i}{2m}\left(i|E|\, |\phi|^2 + i|E|\, |\phi|^2\right) \]
\[ = \frac{i}{2m} \cdot 2i|E|\, |A|^2 \]
\[ = \frac{i \cdot 2i|E|}{2m} |A|^2 = \frac{2i^2 |E|}{2m} |A|^2 = \frac{-|E|}{m} |A|^2 \]

Final Answer

\[ \boxed{\rho = -\frac{|E|}{m}|A|^2 < 0} \]

Since \(|E| > 0\), \(m > 0\), and \(|A|^2 > 0\), we indeed have \(\rho < 0\).

Verification

In the text, equation (1.10) gives \(\rho = \frac{E}{m}|A|^2\). Substituting \(E = -|E|\) yields \(\rho = -\frac{|E|}{m}|A|^2\), which agrees with the result above.

B-3. Anticommutation Relations from the Dirac Equation Condition (1.12)

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(a) Proof that \((\alpha^1)^2 = 1\)

Setting \(i = j = 1\) in the anticommutation relation \(\{\alpha^i, \alpha^j\} = 2\delta^{ij}\):

\[ \{\alpha^1, \alpha^1\} = 2\delta^{11} = 2 \]

The left-hand side is:

\[ \{\alpha^1, \alpha^1\} = \alpha^1 \alpha^1 + \alpha^1 \alpha^1 = 2(\alpha^1)^2 \]

Therefore:

\[ 2(\alpha^1)^2 = 2 \implies \boxed{(\alpha^1)^2 = 1} \]

(b) Proof that \(\alpha^1 \beta = -\beta \alpha^1\)

Setting \(i = 1\) in the anticommutation relation \(\{\alpha^i, \beta\} = 0\):

\[ \{\alpha^1, \beta\} = \alpha^1 \beta + \beta \alpha^1 = 0 \]

Therefore:

\[ \boxed{\alpha^1 \beta = -\beta \alpha^1} \]

(c) Proof that the eigenvalues of \(\alpha^i\) are only \(\pm 1\)

Let \(|\mathbf{v}\rangle\) be an eigenvector of \(\alpha^i\) with corresponding eigenvalue \(\lambda\):

\[ \alpha^i |\mathbf{v}\rangle = \lambda |\mathbf{v}\rangle \]

Acting with \(\alpha^i\) from the left on both sides:

\[ (\alpha^i)^2 |\mathbf{v}\rangle = \lambda \cdot \alpha^i |\mathbf{v}\rangle = \lambda^2 |\mathbf{v}\rangle \]

Since \((\alpha^i)^2 = 1\) (from the anticommutation relation with \(i = j\)), as shown in (a):

\[ |\mathbf{v}\rangle = \lambda^2 |\mathbf{v}\rangle \]

Since \(|\mathbf{v}\rangle \neq 0\):

\[ \lambda^2 = 1 \implies \boxed{\lambda = \pm 1} \]

Verification

Similarly for \(\beta\), since \(\beta^2 = 1\), its eigenvalues are also only \(\pm 1\). This is consistent with the fact that the Dirac matrices are both Hermitian and unitary (their squares equal the identity matrix).

B-4. Proof of the Continuity Equation Identity

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Solution Strategy

Differentiate the right-hand side with respect to time, expand, and show that it equals the left-hand side.

Detailed Calculation

Expand the right-hand side using the product rule:

\[ \frac{\partial}{\partial t}\left(\phi^* \frac{\partial \phi}{\partial t} - \phi \frac{\partial \phi^*}{\partial t}\right) \]
\[ = \frac{\partial}{\partial t}\left(\phi^* \frac{\partial \phi}{\partial t}\right) - \frac{\partial}{\partial t}\left(\phi \frac{\partial \phi^*}{\partial t}\right) \]

Expanding the first term:

\[ \frac{\partial}{\partial t}\left(\phi^* \frac{\partial \phi}{\partial t}\right) = \frac{\partial \phi^*}{\partial t}\frac{\partial \phi}{\partial t} + \phi^* \frac{\partial^2 \phi}{\partial t^2} \]

Expanding the second term:

\[ \frac{\partial}{\partial t}\left(\phi \frac{\partial \phi^*}{\partial t}\right) = \frac{\partial \phi}{\partial t}\frac{\partial \phi^*}{\partial t} + \phi \frac{\partial^2 \phi^*}{\partial t^2} \]

Taking the difference:

\[ \left(\frac{\partial \phi^*}{\partial t}\frac{\partial \phi}{\partial t} + \phi^* \frac{\partial^2 \phi}{\partial t^2}\right) - \left(\frac{\partial \phi}{\partial t}\frac{\partial \phi^*}{\partial t} + \phi \frac{\partial^2 \phi^*}{\partial t^2}\right) \]
\[ = \cancel{\frac{\partial \phi^*}{\partial t}\frac{\partial \phi}{\partial t}} + \phi^* \frac{\partial^2 \phi}{\partial t^2} - \cancel{\frac{\partial \phi}{\partial t}\frac{\partial \phi^*}{\partial t}} - \phi \frac{\partial^2 \phi^*}{\partial t^2} \]
\[ = \phi^* \frac{\partial^2 \phi}{\partial t^2} - \phi \frac{\partial^2 \phi^*}{\partial t^2} \]

Final Answer

This is exactly the left-hand side, so the identity is proven:

\[ \boxed{\phi^* \frac{\partial^2 \phi}{\partial t^2} - \phi \frac{\partial^2 \phi^*}{\partial t^2} = \frac{\partial}{\partial t}\left(\phi^* \frac{\partial \phi}{\partial t} - \phi \frac{\partial \phi^*}{\partial t}\right)} \]

Verification

An entirely analogous identity holds for spatial derivatives: \(\phi^* \nabla^2 \phi - \phi \nabla^2 \phi^* = \nabla \cdot (\phi^* \nabla \phi - \phi \nabla \phi^*)\). This is used in the derivation of the continuity equation in the main text.

B-5. Calculation of the Compton Wavelength

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Solution Strategy

Substitute numerical values into \(\lambda_C = \hbar/(mc)\).

Detailed Calculation

Compton wavelength of the electron:

\[ \lambda_C^{(e)} = \frac{\hbar}{m_e c} = \frac{1.055 \times 10^{-34}}{9.109 \times 10^{-31} \times 2.998 \times 10^{8}} \]

Computing the denominator:

\[ m_e c = 9.109 \times 10^{-31} \times 2.998 \times 10^{8} = 2.731 \times 10^{-22} \text{ kg·m/s} \]

Therefore:

\[ \lambda_C^{(e)} = \frac{1.055 \times 10^{-34}}{2.731 \times 10^{-22}} = 3.862 \times 10^{-13} \text{ m} \]
\[ \boxed{\lambda_C^{(e)} \approx 3.86 \times 10^{-13} \text{ m} \approx 386 \text{ fm}} \]

Compton wavelength of the proton:

\[ \lambda_C^{(p)} = \frac{\hbar}{m_p c} = \frac{1.055 \times 10^{-34}}{1.673 \times 10^{-27} \times 2.998 \times 10^{8}} \]

Computing the denominator:

\[ m_p c = 1.673 \times 10^{-27} \times 2.998 \times 10^{8} = 5.015 \times 10^{-19} \text{ kg·m/s} \]

Therefore:

\[ \lambda_C^{(p)} = \frac{1.055 \times 10^{-34}}{5.015 \times 10^{-19}} = 2.103 \times 10^{-16} \text{ m} \]
\[ \boxed{\lambda_C^{(p)} \approx 2.10 \times 10^{-16} \text{ m} \approx 0.210 \text{ fm}} \]

Comparison:

\[ \frac{\lambda_C^{(p)}}{\lambda_C^{(e)}} = \frac{m_e}{m_p} = \frac{9.109 \times 10^{-31}}{1.673 \times 10^{-27}} \approx \frac{1}{1836} \]

The Compton wavelength of the proton is approximately \(1/1836\) times that of the electron; the larger the mass, the shorter the Compton wavelength.

Verification

\(\lambda_C^{(p)} / \lambda_C^{(e)} = m_e / m_p \approx 1/1836\) is simply the mass ratio itself, which is consistent with the known value. Furthermore, the electron Compton wavelength \(\approx 386\) fm agrees with the literature value.

B-6. Mass Dimension in Natural Units

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Solution Strategy

In \(\hbar = c = 1\), fundamental dimensions are expressed solely as powers of energy (mass). We take \([E] = 1\) as the reference.

Detailed Calculation

In \(\hbar = c = 1\), from \([\hbar] = [E][t] = 0\) we get \([t] = -[E] = -1\). From \([c] = [x]/[t] = 0\) we get \([x] = [t] = -1\).

(a) Time \(t\):

\[ \boxed{[t] = -1} \]

(b) Length \(x\):

\[ \boxed{[x] = -1} \]

(c) Mass \(m\):

\[ \boxed{[m] = +1} \]

(d) Klein-Gordon field \(\phi\):

The action \(S = \int d^4x\, \mathcal{L}\) is dimensionless since \([\hbar] = 0\): \([S] = 0\).

\[ [d^4x] = 4 \times [x] = 4 \times (-1) = -4 \]

Therefore:

\[ [\mathcal{L}] = -[d^4x] = +4 \]

Examining each term of the Lagrangian density \(\mathcal{L} = \frac{1}{2}\partial_\mu\phi\,\partial^\mu\phi - \frac{1}{2}m^2\phi^2\). Since \([\partial_\mu] = -[x] = +1\):

\[ [\partial_\mu\phi\,\partial^\mu\phi] = 2[\partial_\mu] + 2[\phi] = 2 + 2[\phi] \]

Setting this equal to \([\mathcal{L}] = 4\):

\[ 2 + 2[\phi] = 4 \implies [\phi] = 1 \]
\[ \boxed{[\phi] = +1} \]

Verification

Dimension of the \(m^2\phi^2\) term: \([m^2\phi^2] = 2[m] + 2[\phi] = 2 + 2 = 4 = [\mathcal{L}]\). ✓

B-7. Time Scale of Pair Creation from the Uncertainty Relation

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(a) Creation time of virtual electron-positron pairs

From the energy-time uncertainty relation \(\Delta E \cdot \Delta t \gtrsim \hbar\), the maximum time during which an energy fluctuation of \(\Delta E = 2m_e c^2\) is permitted is:

\[ \Delta t \sim \frac{\hbar}{\Delta E} = \frac{\hbar}{2m_e c^2} \]

Substituting numerical values:

\[ 2m_e c^2 = 2 \times 9.109 \times 10^{-31} \times (2.998 \times 10^8)^2 = 2 \times 8.187 \times 10^{-14} \text{ J} = 1.637 \times 10^{-13} \text{ J} \]
\[ \Delta t \sim \frac{1.055 \times 10^{-34}}{1.637 \times 10^{-13}} \approx 6.44 \times 10^{-22} \text{ s} \]
\[ \boxed{\Delta t \sim \frac{\hbar}{2m_e c^2} \approx 6.4 \times 10^{-22} \text{ s}} \]

(b) Comparison of the distance traveled by light with the Compton wavelength

\[ c \cdot \Delta t = c \cdot \frac{\hbar}{2m_e c^2} = \frac{\hbar}{2m_e c} = \frac{\lambda_C}{2} \]

Numerical value:

\[ c \cdot \Delta t = 2.998 \times 10^8 \times 6.44 \times 10^{-22} \approx 1.93 \times 10^{-13} \text{ m} \]

This is approximately half the electron Compton wavelength \(\lambda_C \approx 3.86 \times 10^{-13}\) m:

\[ \boxed{c \cdot \Delta t = \frac{\lambda_C}{2} \approx 1.93 \times 10^{-13} \text{ m}} \]

Verification

\(\lambda_C / 2 = 3.86 \times 10^{-13} / 2 = 1.93 \times 10^{-13}\) m agrees with the directly calculated value. This result is consistent with the discussion in the text that "pair creation becomes important at distance scales on the order of the Compton wavelength."

B-8. d'Alembert Operator and the Covariant Form of the Klein-Gordon Equation

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Solution Strategy

Expand \(\partial_\mu \partial^\mu\) in components using the metric tensor \(\eta^{\mu\nu} = \mathrm{diag}(+1, -1, -1, -1)\).

Detailed Calculation

Definition of the four-gradient operator:

\[ \partial_\mu = \frac{\partial}{\partial x^\mu} = \left(\frac{\partial}{\partial t},\, \frac{\partial}{\partial x^1},\, \frac{\partial}{\partial x^2},\, \frac{\partial}{\partial x^3}\right) \]

Raising the index:

\[ \partial^\mu = \eta^{\mu\nu}\partial_\nu \]

Computing each component:

\[ \partial^0 = \eta^{00}\partial_0 = (+1)\frac{\partial}{\partial t} = \frac{\partial}{\partial t} \]
\[ \partial^k = \eta^{kk}\partial_k = (-1)\frac{\partial}{\partial x^k} = -\frac{\partial}{\partial x^k} \quad (k = 1, 2, 3) \]

Therefore, the d'Alembert operator is:

\[ \Box \equiv \partial_\mu \partial^\mu = \partial_0 \partial^0 + \partial_1 \partial^1 + \partial_2 \partial^2 + \partial_3 \partial^3 \]
\[ = \frac{\partial}{\partial t}\cdot\frac{\partial}{\partial t} + \frac{\partial}{\partial x^1}\cdot\left(-\frac{\partial}{\partial x^1}\right) + \frac{\partial}{\partial x^2}\cdot\left(-\frac{\partial}{\partial x^2}\right) + \frac{\partial}{\partial x^3}\cdot\left(-\frac{\partial}{\partial x^3}\right) \]
\[ = \frac{\partial^2}{\partial t^2} - \left(\frac{\partial^2}{\partial (x^1)^2} + \frac{\partial^2}{\partial (x^2)^2} + \frac{\partial^2}{\partial (x^3)^2}\right) \]
\[ = \frac{\partial^2}{\partial t^2} - \nabla^2 \]

Final Answer

The Klein-Gordon equation (1.1) is:

\[ \frac{\partial^2 \phi}{\partial t^2} - \nabla^2 \phi + m^2 \phi = 0 \]
\[ \implies (\Box + m^2)\phi = 0 \]
\[ \boxed{(\partial_\mu \partial^\mu + m^2)\phi = 0} \]

Verification

With the metric signature \((+,-,-,-)\), we have \(p_\mu p^\mu = E^2 - |\mathbf{p}|^2 = m^2\). Applying the operator substitution \(p^\mu \to i\partial^\mu\) gives \(-\partial_\mu\partial^\mu = m^2\), i.e., \((\Box + m^2)\phi = 0\). This is consistent.

Medium

M-1. Lorentz Covariance of the Probability Current Density for the Klein-Gordon Equation

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(a) Verification that \(j^0 = \rho\)

Definition of the 4-current density:

\[ j^\mu = \frac{i}{2m}\left(\phi^* \partial^\mu \phi - \phi\, \partial^\mu \phi^*\right) \]

Taking the \(\mu = 0\) component:

\[ j^0 = \frac{i}{2m}\left(\phi^* \partial^0 \phi - \phi\, \partial^0 \phi^*\right) \]

From the metric \(\eta^{\mu\nu} = \mathrm{diag}(+1,-1,-1,-1)\), we have \(\partial^0 = \partial/\partial t\), so:

\[ j^0 = \frac{i}{2m}\left(\phi^* \frac{\partial \phi}{\partial t} - \phi \frac{\partial \phi^*}{\partial t}\right) \]

This is exactly \(\rho\) as given in equation (1.6).

\[ \boxed{j^0 = \rho} \]

(b) Verification that \(j^k\) agrees with \(\mathbf{j}\) in equation (1.7)

Taking the \(\mu = k\) (\(k = 1, 2, 3\)) components:

\[ j^k = \frac{i}{2m}\left(\phi^* \partial^k \phi - \phi\, \partial^k \phi^*\right) \]

Here \(\partial^k = \eta^{kk}\partial_k = -\partial/\partial x^k = -\nabla_k\), so:

\[ j^k = \frac{i}{2m}\left(\phi^* (-\nabla_k \phi) - \phi\, (-\nabla_k \phi^*)\right) \]
\[ = \frac{i}{2m}\left(-\phi^* \nabla_k \phi + \phi\, \nabla_k \phi^*\right) \]
\[ = \frac{-i}{2m}\left(\phi^* \nabla_k \phi - \phi\, \nabla_k \phi^*\right) \]
\[ = \frac{1}{2mi}\left(\phi^* \nabla_k \phi - \phi\, \nabla_k \phi^*\right) \]

At first glance this appears to have a sign flip relative to equation (1.7), but care is needed.

When defining the 4-current as a contravariant vector \(j^\mu\), the relationship to the 3-dimensional current \(\mathbf{j}\) depends on the metric sign convention. In the \((+,-,-,-)\) convention, \(j_\mu = (\rho, -\mathbf{j})\), and when writing \(j^\mu = (\rho, \mathbf{j})\), the spatial components are:

Checking equation (1.7) in the text:

\[ \mathbf{j} = \frac{1}{2mi}\left(\phi^* \nabla \phi - \phi \nabla \phi^*\right) \]

The \(j^k\) obtained above is exactly this:

\[ \boxed{j^k = \frac{1}{2mi}\left(\phi^* \nabla_k \phi - \phi\, \nabla_k \phi^*\right) = j_k \text{ (the $k$-th component of equation (1.7))}} \]

(Note: We used \(\frac{-i}{2m} = \frac{1}{2mi}\).)

(c) Covariant form of the continuity equation

The continuity equation (1.5) is:

\[ \frac{\partial \rho}{\partial t} + \nabla \cdot \mathbf{j} = 0 \]

Writing this in 4-vector notation:

\[ \frac{\partial j^0}{\partial t} + \sum_{k=1}^{3} \frac{\partial j^k}{\partial x^k} = \partial_0 j^0 + \partial_k j^k = \partial_\mu j^\mu = 0 \]
\[ \boxed{\partial_\mu j^\mu = 0} \]

Discussion of Lorentz covariance:

\(\partial_\mu j^\mu\) is a Lorentz scalar. This follows because: - \(j^\mu\) transforms as a contravariant 4-vector under Lorentz transformations (since \(\phi\) is a scalar field and \(\partial^\mu\) is a contravariant vector) - \(\partial_\mu\) transforms as a covariant vector - The contraction of a contravariant vector with a covariant vector \(\partial_\mu j^\mu\) is a Lorentz scalar

Therefore, the equation \(\partial_\mu j^\mu = 0\) takes the same form in all inertial frames. If it holds in one inertial frame, it holds in any inertial frame. This is the Lorentz covariance of the continuity equation.

Verification

We directly verify that \(\partial_\mu j^\mu = 0\). When \(\phi\) satisfies the Klein-Gordon equation \((\Box + m^2)\phi = 0\):

\[ \partial_\mu j^\mu = \frac{i}{2m}\partial_\mu\left(\phi^* \partial^\mu \phi - \phi\, \partial^\mu \phi^*\right) \]
\[ = \frac{i}{2m}\left((\partial_\mu\phi^*)(\partial^\mu\phi) + \phi^* \Box\phi - (\partial_\mu\phi)(\partial^\mu\phi^*) - \phi\, \Box\phi^*\right) \]

The first and third terms cancel:

\[ = \frac{i}{2m}\left(\phi^* \Box\phi - \phi\, \Box\phi^*\right) = \frac{i}{2m}\left(\phi^*(-m^2\phi) - \phi(-m^2\phi^*)\right) = 0 \]

Indeed \(\partial_\mu j^\mu = 0\) holds. ✓

M-2. Non-negativity of Probability Density in the Dirac Equation

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Strategy

Calculate \(\partial_t(\psi^\dagger\psi)\) from the Dirac equation and its Hermitian conjugate, then arrange it into the form of a continuity equation.

Detailed Calculation

Dirac equation:

\[ i\frac{\partial \psi}{\partial t} = H_D \psi = \left(-i\boldsymbol{\alpha}\cdot\nabla + \beta m\right)\psi \tag{*} \]

Taking the Hermitian conjugate:

Taking the Hermitian conjugate (\(\dagger\)) of both sides of \((*)\):

\[ -i\frac{\partial \psi^\dagger}{\partial t} = \psi^\dagger H_D^\dagger \]

Here \(H_D^\dagger = (-i\boldsymbol{\alpha}\cdot\nabla + \beta m)^\dagger\). Since \(\alpha^i\) and \(\beta\) are Hermitian matrices (\((\alpha^i)^\dagger = \alpha^i\), \(\beta^\dagger = \beta\)) and \(\nabla\) is a real operator:

\[ H_D^\dagger = (i\boldsymbol{\alpha}\cdot\overleftarrow{\nabla} + \beta m) \]

where \(\overleftarrow{\nabla}\) denotes the differential operator acting to the left. Writing more explicitly:

\[ -i\frac{\partial \psi^\dagger}{\partial t} = i(\nabla\psi^\dagger)\cdot\boldsymbol{\alpha} + \psi^\dagger \beta m \tag{**} \]

Or equivalently:

\[ \frac{\partial \psi^\dagger}{\partial t} = (\nabla\psi^\dagger)\cdot\boldsymbol{\alpha} - i\psi^\dagger \beta m \tag{**'} \]

Calculating \(\partial_t \rho\):

\[ \frac{\partial \rho}{\partial t} = \frac{\partial}{\partial t}(\psi^\dagger \psi) = \frac{\partial \psi^\dagger}{\partial t}\psi + \psi^\dagger \frac{\partial \psi}{\partial t} \]

From \((*)\):

\[ \frac{\partial \psi}{\partial t} = -i(-i\boldsymbol{\alpha}\cdot\nabla + \beta m)\psi = (-\boldsymbol{\alpha}\cdot\nabla - i\beta m)\psi \]

From \((**)\):

\[ \frac{\partial \psi^\dagger}{\partial t} = (\nabla\psi^\dagger)\cdot\boldsymbol{\alpha} - i\psi^\dagger \beta m \]

Substituting:

\[ \frac{\partial \rho}{\partial t} = \left[(\nabla\psi^\dagger)\cdot\boldsymbol{\alpha} - i\psi^\dagger \beta m\right]\psi + \psi^\dagger\left[-\boldsymbol{\alpha}\cdot\nabla - i\beta m\right]\psi \]
\[ = (\nabla\psi^\dagger)\cdot\boldsymbol{\alpha}\,\psi - i\psi^\dagger \beta m\,\psi - \psi^\dagger\boldsymbol{\alpha}\cdot\nabla\psi - i\psi^\dagger\beta m\,\psi \]

Looking at the \(\beta m\) terms:

\[ -i\psi^\dagger \beta m\,\psi - i\psi^\dagger\beta m\,\psi = -2i\psi^\dagger\beta m\,\psi \]

This is incorrect. Let us redo this carefully.

From \((*)\):

\[ \frac{\partial \psi}{\partial t} = \frac{1}{i}(-i\boldsymbol{\alpha}\cdot\nabla + \beta m)\psi = (-\boldsymbol{\alpha}\cdot\nabla - i\beta m)\psi \]

Re-deriving \((**)\) correctly. The Hermitian conjugate of \((*)\):

\[ \left(i\frac{\partial \psi}{\partial t}\right)^\dagger = \left[(-i\boldsymbol{\alpha}\cdot\nabla + \beta m)\psi\right]^\dagger \]

Left-hand side:

\[ -i\frac{\partial \psi^\dagger}{\partial t} \]

Right-hand side: takes the form of operators acting from the right on \(\psi^\dagger\). In position representation:

\[ -i\frac{\partial \psi^\dagger}{\partial t} = \psi^\dagger(i\boldsymbol{\alpha}\cdot\overleftarrow{\nabla} + \beta m) \]

where \(\overleftarrow{\nabla}\) acts on \(\psi^\dagger\). That is:

\[ -i\frac{\partial \psi^\dagger}{\partial t} = i(\nabla\psi^\dagger)\cdot\boldsymbol{\alpha} + \psi^\dagger \beta m \]

Therefore:

\[ \frac{\partial \psi^\dagger}{\partial t} = (\nabla\psi^\dagger)\cdot\boldsymbol{\alpha} + i\psi^\dagger \beta m \]

Recalculation:

\[ \frac{\partial \rho}{\partial t} = \frac{\partial \psi^\dagger}{\partial t}\psi + \psi^\dagger \frac{\partial \psi}{\partial t} \]
\[ = \left[(\nabla\psi^\dagger)\cdot\boldsymbol{\alpha} + i\psi^\dagger \beta m\right]\psi + \psi^\dagger\left[-\boldsymbol{\alpha}\cdot\nabla - i\beta m\right]\psi \]
\[ = (\nabla\psi^\dagger)\cdot\boldsymbol{\alpha}\,\psi + i\psi^\dagger \beta m\,\psi - \psi^\dagger\boldsymbol{\alpha}\cdot(\nabla\psi) - i\psi^\dagger\beta m\,\psi \]

The mass terms cancel:

\[ + i\psi^\dagger \beta m\,\psi - i\psi^\dagger\beta m\,\psi = 0 \]

What remains is:

\[ \frac{\partial \rho}{\partial t} = (\nabla\psi^\dagger)\cdot\boldsymbol{\alpha}\,\psi - \psi^\dagger\boldsymbol{\alpha}\cdot(\nabla\psi) \]

Writing in components (for the \(k\)-th component):

\[ = \sum_k \left[(\partial_k \psi^\dagger)\alpha^k\psi - \psi^\dagger\alpha^k(\partial_k\psi)\right] \]
\[ = -\sum_k \partial_k\left(\psi^\dagger \alpha^k \psi\right) + \sum_k\left[(\partial_k\psi^\dagger)\alpha^k\psi + \psi^\dagger\alpha^k(\partial_k\psi)\right] - \sum_k \partial_k(\psi^\dagger\alpha^k\psi) \]

No, more directly:

\[ (\nabla\psi^\dagger)\cdot\boldsymbol{\alpha}\,\psi - \psi^\dagger\boldsymbol{\alpha}\cdot(\nabla\psi) \]

Let us check whether this can be written as \(-\nabla\cdot(\psi^\dagger\boldsymbol{\alpha}\psi)\):

\[ \nabla\cdot(\psi^\dagger\boldsymbol{\alpha}\psi) = \sum_k \partial_k(\psi^\dagger\alpha^k\psi) = \sum_k\left[(\partial_k\psi^\dagger)\alpha^k\psi + \psi^\dagger\alpha^k(\partial_k\psi)\right] \]

On the other hand, what we obtained is:

\[ \frac{\partial\rho}{\partial t} = \sum_k\left[(\partial_k\psi^\dagger)\alpha^k\psi - \psi^\dagger\alpha^k(\partial_k\psi)\right] \]

This differs from \(\nabla\cdot(\psi^\dagger\boldsymbol{\alpha}\psi)\) (the sign is different).

Let us redo this carefully from \((*)\).

Correct derivation:

Dirac equation \((*)\):

\[ i\frac{\partial\psi}{\partial t} = -i\alpha^k\partial_k\psi + \beta m\psi \]
\[ \therefore \frac{\partial\psi}{\partial t} = -\alpha^k\partial_k\psi - i\beta m\psi \tag{A} \]

Hermitian conjugate (\((\alpha^k)^\dagger = \alpha^k\), \(\beta^\dagger = \beta\)):

\[ -i\frac{\partial\psi^\dagger}{\partial t} = i(\partial_k\psi^\dagger)\alpha^k + \psi^\dagger\beta m \]
\[ \therefore \frac{\partial\psi^\dagger}{\partial t} = (\partial_k\psi^\dagger)\alpha^k + i\psi^\dagger\beta m \tag{B} \]

Computing \(\partial_t\rho\):

\[ \frac{\partial\rho}{\partial t} = \left(\frac{\partial\psi^\dagger}{\partial t}\right)\psi + \psi^\dagger\left(\frac{\partial\psi}{\partial t}\right) \]

Substituting (B) and (A):

\[ = \left[(\partial_k\psi^\dagger)\alpha^k + i\psi^\dagger\beta m\right]\psi + \psi^\dagger\left[-\alpha^k\partial_k\psi - i\beta m\psi\right] \]
\[ = (\partial_k\psi^\dagger)\alpha^k\psi + i\psi^\dagger\beta m\psi - \psi^\dagger\alpha^k\partial_k\psi - i\psi^\dagger\beta m\psi \]

The mass terms cancel:

\[ \frac{\partial\rho}{\partial t} = (\partial_k\psi^\dagger)\alpha^k\psi - \psi^\dagger\alpha^k(\partial_k\psi) \]

Let us check the signs. This is:

\[ = -\left[\psi^\dagger\alpha^k(\partial_k\psi) - (\partial_k\psi^\dagger)\alpha^k\psi\right] \]

Now examine the relationship with \(\nabla\cdot(\psi^\dagger\boldsymbol{\alpha}\psi)\):

\[ \nabla\cdot(\psi^\dagger\boldsymbol{\alpha}\psi) = \partial_k(\psi^\dagger\alpha^k\psi) = (\partial_k\psi^\dagger)\alpha^k\psi + \psi^\dagger\alpha^k(\partial_k\psi) \]

The result we obtained is:

\[ \frac{\partial\rho}{\partial t} = (\partial_k\psi^\dagger)\alpha^k\psi - \psi^\dagger\alpha^k(\partial_k\psi) \]

This does not match \(\nabla\cdot(\psi^\dagger\boldsymbol{\alpha}\psi)\).

Let us re-verify (A). Looking again at \((*)\):

\((*)\): \(i\partial_t\psi = (-i\alpha^k\partial_k + \beta m)\psi\)

Dividing both sides by \(i\):

\[ \partial_t\psi = \frac{1}{i}(-i\alpha^k\partial_k + \beta m)\psi = (-\alpha^k\partial_k + \frac{\beta m}{i})\psi = -\alpha^k\partial_k\psi - i\beta m\psi \]

Since \(\frac{1}{i} = -i\), we have \(\frac{\beta m}{i} = -i\beta m\). OK.

Let us re-derive the Hermitian conjugate. Taking \(\dagger\) of both sides of \((*)\):

\[ (i\partial_t\psi)^\dagger = [(-i\alpha^k\partial_k + \beta m)\psi]^\dagger \]

Left-hand side: \((i\partial_t\psi)^\dagger = (\partial_t\psi)^\dagger(-i) = (\partial_t\psi^\dagger)(-i) = -i\partial_t\psi^\dagger\)

Right-hand side: \(\psi^\dagger(-i\alpha^k\partial_k + \beta m)^\dagger\)

Note here: in position representation, \(\partial_k\) is a real operator, but when taking the Hermitian conjugate it is not the case that \(\partial_k \to -\partial_k\) (sign change from integration by parts); rather, we write it in the form acting from the right on \(\psi^\dagger\).

Correctly, the Hermitian conjugate of the Dirac equation is:

\[ -i\frac{\partial\psi^\dagger}{\partial t} = (i\partial_k\psi^\dagger)\alpha^k + \psi^\dagger\beta m \]

This is obtained by taking the Hermitian conjugate of each term in \((*)\): - \((-i\alpha^k\partial_k\psi)^\dagger = (\partial_k\psi)^\dagger(i)(\alpha^k)^\dagger = i(\partial_k\psi^\dagger)\alpha^k\)

Therefore:

\[ -i\partial_t\psi^\dagger = i(\partial_k\psi^\dagger)\alpha^k + \psi^\dagger\beta m \]
\[ \partial_t\psi^\dagger = (\partial_k\psi^\dagger)\alpha^k + i\psi^\dagger\beta m \tag{B} \]

This is the same as before.

Now, the result for \(\partial_t\rho\):

\[ \frac{\partial\rho}{\partial t} = (\partial_k\psi^\dagger)\alpha^k\psi - \psi^\dagger\alpha^k(\partial_k\psi) \]

This should be in the form \(-\nabla\cdot\mathbf{j}\). If we define \(\mathbf{j} = \psi^\dagger\boldsymbol{\alpha}\psi\):

\[ \nabla\cdot\mathbf{j} = \partial_k(\psi^\dagger\alpha^k\psi) = (\partial_k\psi^\dagger)\alpha^k\psi + \psi^\dagger\alpha^k(\partial_k\psi) \]

The \(\partial_t\rho\) we obtained:

\[ \frac{\partial\rho}{\partial t} = (\partial_k\psi^\dagger)\alpha^k\psi - \psi^\dagger\alpha^k(\partial_k\psi) \]

is different from \(\nabla\cdot\mathbf{j}\). What is wrong?

Identifying the issue: Let us verify the sign in (A).

\[ i\partial_t\psi = -i\alpha^k\partial_k\psi + \beta m\psi \]

Multiplying both sides by \(-i\) (\(\frac{1}{i} = -i\)):

\[ \partial_t\psi = -i \cdot \frac{1}{i}(-i\alpha^k\partial_k\psi + \beta m\psi) \]

No, simply:

\[ i\partial_t\psi = -i\alpha^k\partial_k\psi + \beta m\psi \]
\[ \partial_t\psi = \frac{-i\alpha^k\partial_k\psi + \beta m\psi}{i} = \frac{-i\alpha^k\partial_k\psi}{i} + \frac{\beta m\psi}{i} \]
\[ = -\alpha^k\partial_k\psi + (-i)\beta m\psi = -\alpha^k\partial_k\psi - i\beta m\psi \]

OK, (A) is correct.

Let us also verify (B). Taking the Hermitian conjugate of \((*)\):

\(i\partial_t\psi = -i\alpha^k\partial_k\psi + \beta m\psi\), taking \(\dagger\):

LHS \(\dagger\): \((i\partial_t\psi)^\dagger = -i\partial_t\psi^\dagger\)

RHS \(\dagger\): \((-i\alpha^k\partial_k\psi + \beta m\psi)^\dagger\)

Here, computing \((-i\alpha^k\partial_k\psi)^\dagger\). Since this involves the product of a matrix and a differential operator, care is needed.

Rather than considering the Hermitian conjugate in the sense of the Hilbert space inner product, we simply take the matrix \(\dagger\) and complex conjugation (operations in position representation):

\[ (-i\alpha^k\partial_k\psi)^\dagger = (\partial_k\psi)^\dagger (i)(\alpha^k)^\dagger = i(\partial_k\psi^\dagger)\alpha^k \]

\((\beta m\psi)^\dagger = m\psi^\dagger\beta\)

Therefore:

\[ -i\partial_t\psi^\dagger = i(\partial_k\psi^\dagger)\alpha^k + m\psi^\dagger\beta \]
\[ \partial_t\psi^\dagger = -(\partial_k\psi^\dagger)\alpha^k + im\psi^\dagger\beta \]

Wait, the sign changed! Let me redo this.

\[ -i\partial_t\psi^\dagger = i(\partial_k\psi^\dagger)\alpha^k + m\psi^\dagger\beta \]

Multiplying both sides by \(\frac{1}{-i} = i\):

\[ \partial_t\psi^\dagger = i \cdot [i(\partial_k\psi^\dagger)\alpha^k + m\psi^\dagger\beta] \]
\[ = i^2(\partial_k\psi^\dagger)\alpha^k + im\psi^\dagger\beta \]
\[ = -(\partial_k\psi^\dagger)\alpha^k + im\psi^\dagger\beta \]

This has different sign from the previous (B)! The earlier (B) was incorrect. The correct result is:

\[ \frac{\partial\psi^\dagger}{\partial t} = -(\partial_k\psi^\dagger)\alpha^k + im\psi^\dagger\beta \tag{B'} \]

Recalculation:

\[ \frac{\partial\rho}{\partial t} = \left[-(\partial_k\psi^\dagger)\alpha^k + im\psi^\dagger\beta\right]\psi + \psi^\dagger\left[-\alpha^k\partial_k\psi - i\beta m\psi\right] \]
\[ = -(\partial_k\psi^\dagger)\alpha^k\psi + im\psi^\dagger\beta\psi - \psi^\dagger\alpha^k(\partial_k\psi) - im\psi^\dagger\beta\psi \]

The mass terms cancel:

\[ \frac{\partial\rho}{\partial t} = -(\partial_k\psi^\dagger)\alpha^k\psi - \psi^\dagger\alpha^k(\partial_k\psi) \]
\[ = -\left[(\partial_k\psi^\dagger)\alpha^k\psi + \psi^\dagger\alpha^k(\partial_k\psi)\right] \]
\[ = -\partial_k(\psi^\dagger\alpha^k\psi) \]
\[ = -\nabla\cdot(\psi^\dagger\boldsymbol{\alpha}\psi) \]

Final Answer

\[ \boxed{\frac{\partial\rho}{\partial t} + \nabla\cdot\mathbf{j} = 0, \quad \rho = \psi^\dagger\psi \geq 0, \quad \mathbf{j} = \psi^\dagger\boldsymbol{\alpha}\psi} \]

Summary of the derivation:

  1. Dirac equation: \(i\partial_t\psi = (-i\boldsymbol{\alpha}\cdot\nabla + \beta m)\psi\)
  2. Hermitian conjugate: \(-i\partial_t\psi^\dagger = i(\partial_k\psi^\dagger)\alpha^k + m\psi^\dagger\beta\)
  3. Computing \(\partial_t(\psi^\dagger\psi)\), the mass terms cancel and \(-\nabla\cdot(\psi^\dagger\boldsymbol{\alpha}\psi)\) remains

Since \(\rho = \psi^\dagger\psi = \sum_{a=1}^{4}|\psi_a|^2 \geq 0\), the probability density is always non-negative. This is the crucial difference from the Klein-Gordon equation.

Verification

Each component of \(\mathbf{j} = \psi^\dagger\boldsymbol{\alpha}\psi\) is real since it is the expectation value of the Hermitian matrix \(\alpha^k\). Furthermore, integrating \(\rho\) and \(\mathbf{j}\) over all space:

\[ \frac{d}{dt}\int d^3x\, \rho = -\int d^3x\, \nabla\cdot\mathbf{j} = 0 \]

(when the surface terms vanish). This means that the total probability \(\int d^3x\, \psi^\dagger\psi = 1\) is conserved, which is consistent with the probability interpretation.

M-3. Compton Wavelength and Position Localization

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(a) Momentum Uncertainty and Energy Uncertainty

From the uncertainty principle, the momentum uncertainty of a particle confined in a box of width \(L\) is:

\[ \Delta p \sim \frac{\hbar}{L} \]

Since the relativistic energy is \(E = \sqrt{(pc)^2 + (mc^2)^2}\), the energy uncertainty is:

\[ \Delta E \sim \sqrt{(\Delta p \cdot c)^2 + (mc^2)^2} - mc^2 \]

(the uncertainty in kinetic energy after subtracting the rest energy)

However, since our goal here is to compare with the energy required for pair creation, it is more appropriate to estimate \(\Delta E \sim c\,\Delta p\) (when \(\Delta p\) is large).

In the limit \(\Delta p \gg mc\) (ultra-relativistic):

\[ \Delta E \approx c\,\Delta p = \frac{\hbar c}{L} \]

In the limit \(\Delta p \ll mc\) (non-relativistic):

\[ \Delta E \approx \frac{(\Delta p)^2}{2m} = \frac{\hbar^2}{2mL^2} \]

(b) Critical Condition for Pair Creation

For virtual particle-antiparticle pair creation, \(\Delta E \geq 2mc^2\) is required. Using the ultra-relativistic estimate (corresponding to the case of small \(L\)):

\[ \frac{\hbar c}{L} \geq 2mc^2 \]
\[ L \leq \frac{\hbar}{2mc} = \frac{\lambda_C}{2} \]
\[ \boxed{L_{\text{critical}} \sim \frac{\lambda_C}{2} = \frac{\hbar}{2mc}} \]

(c) Consistency with the Breakdown of the Single-Particle Picture

This result means that when one attempts to localize a particle to a distance scale on the order of or below the Compton wavelength \(\lambda_C = \hbar/(mc)\), the energy uncertainty exceeds the pair creation threshold \(2mc^2\).

In this regime: - Even when attempting to observe "a single particle," virtual particle-antiparticle pairs may be created - The number of particles becomes indefinite, and the concept of "the wave function of a single particle" loses its meaning - Measuring position with precision better than \(\lambda_C\) itself requires a many-particle description

This is precisely consistent with the discussion in the main text — "at distance scales shorter than the Compton wavelength, the single-particle picture breaks down and quantum field theory becomes necessary."

Verification

For the electron: \(\lambda_C^{(e)} \approx 3.86 \times 10^{-13}\) m. The Bohr radius of the atom \(a_0 \approx 5.3 \times 10^{-11}\) m is approximately 137 times \(\lambda_C\) (\(= 1/\alpha\), the inverse of the fine-structure constant), which is consistent with the fact that pair creation can be neglected in atomic physics.

M-4. General Solution of the Klein-Gordon Field and Positive/Negative Frequency Modes

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(a) Verifying that the general solution satisfies the Klein-Gordon equation

General solution:

\[ \phi(\mathbf{x}, t) = \int \frac{d^3p}{(2\pi)^3} \frac{1}{\sqrt{2\omega_{\mathbf{p}}}} \left[ a(\mathbf{p})\, e^{i(\mathbf{p}\cdot\mathbf{x} - \omega_{\mathbf{p}} t)} + b^*(\mathbf{p})\, e^{-i(\mathbf{p}\cdot\mathbf{x} - \omega_{\mathbf{p}} t)} \right] \]

We apply \((\Box + m^2)\) to this expression. For each Fourier mode:

Positive-frequency mode \(e^{i(\mathbf{p}\cdot\mathbf{x} - \omega_{\mathbf{p}} t)}\):

\[ \Box\, e^{i(\mathbf{p}\cdot\mathbf{x} - \omega_{\mathbf{p}} t)} = \left(\frac{\partial^2}{\partial t^2} - \nabla^2\right) e^{i(\mathbf{p}\cdot\mathbf{x} - \omega_{\mathbf{p}} t)} \]
\[ = \left[(-i\omega_{\mathbf{p}})^2 - (i\mathbf{p})^2\right] e^{i(\mathbf{p}\cdot\mathbf{x} - \omega_{\mathbf{p}} t)} \]
\[ = (-\omega_{\mathbf{p}}^2 + |\mathbf{p}|^2)\, e^{i(\mathbf{p}\cdot\mathbf{x} - \omega_{\mathbf{p}} t)} \]

Therefore:

\[ (\Box + m^2)\, e^{i(\mathbf{p}\cdot\mathbf{x} - \omega_{\mathbf{p}} t)} = (-\omega_{\mathbf{p}}^2 + |\mathbf{p}|^2 + m^2)\, e^{i(\mathbf{p}\cdot\mathbf{x} - \omega_{\mathbf{p}} t)} \]

Since \(\omega_{\mathbf{p}} = \sqrt{|\mathbf{p}|^2 + m^2}\), we have \(\omega_{\mathbf{p}}^2 = |\mathbf{p}|^2 + m^2\), so:

\[ -\omega_{\mathbf{p}}^2 + |\mathbf{p}|^2 + m^2 = 0 \]

Negative-frequency mode \(e^{-i(\mathbf{p}\cdot\mathbf{x} - \omega_{\mathbf{p}} t)}\):

\[ \Box\, e^{-i(\mathbf{p}\cdot\mathbf{x} - \omega_{\mathbf{p}} t)} = \left[(i\omega_{\mathbf{p}})^2 - (-i\mathbf{p})^2\right] e^{-i(\mathbf{p}\cdot\mathbf{x} - \omega_{\mathbf{p}} t)} \]
\[ = (-\omega_{\mathbf{p}}^2 + |\mathbf{p}|^2)\, e^{-i(\mathbf{p}\cdot\mathbf{x} - \omega_{\mathbf{p}} t)} \]

Similarly, applying \((\Box + m^2)\) yields \(0\).

Therefore, since each Fourier mode satisfies the Klein-Gordon equation, their linear combination \(\phi\) also satisfies it.

\[ \boxed{(\Box + m^2)\phi = 0 \quad \text{is verified}} \]

(b) Condition for a real scalar field

We impose the condition \(\phi = \phi^*\). Computing \(\phi^*\):

\[ \phi^*(\mathbf{x}, t) = \int \frac{d^3p}{(2\pi)^3} \frac{1}{\sqrt{2\omega_{\mathbf{p}}}} \left[ a^*(\mathbf{p})\, e^{-i(\mathbf{p}\cdot\mathbf{x} - \omega_{\mathbf{p}} t)} + b(\mathbf{p})\, e^{i(\mathbf{p}\cdot\mathbf{x} - \omega_{\mathbf{p}} t)} \right] \]

To impose \(\phi = \phi^*\), we substitute the integration variable \(\mathbf{p} \to -\mathbf{p}\) in \(\phi^*\) (noting that \(\omega_{\mathbf{p}} = \omega_{-\mathbf{p}}\) and \(d^3p\) is invariant):

\[ \phi^* = \int \frac{d^3p}{(2\pi)^3} \frac{1}{\sqrt{2\omega_{\mathbf{p}}}} \left[ a^*(-\mathbf{p})\, e^{i(\mathbf{p}\cdot\mathbf{x} + \omega_{\mathbf{p}} t)} + b(-\mathbf{p})\, e^{-i(\mathbf{p}\cdot\mathbf{x} + \omega_{\mathbf{p}} t)} \right] \]

Let us instead compare more directly. Writing \(\phi\) and \(\phi^*\) side by side:

\[ \phi = \int \frac{d^3p}{(2\pi)^3} \frac{1}{\sqrt{2\omega_{\mathbf{p}}}} \left[ a(\mathbf{p})\, e^{i\mathbf{p}\cdot\mathbf{x} - i\omega_{\mathbf{p}} t} + b^*(\mathbf{p})\, e^{-i\mathbf{p}\cdot\mathbf{x} + i\omega_{\mathbf{p}} t} \right] \]
\[ \phi^* = \int \frac{d^3p}{(2\pi)^3} \frac{1}{\sqrt{2\omega_{\mathbf{p}}}} \left[ a^*(\mathbf{p})\, e^{-i\mathbf{p}\cdot\mathbf{x} + i\omega_{\mathbf{p}} t} + b(\mathbf{p})\, e^{i\mathbf{p}\cdot\mathbf{x} - i\omega_{\mathbf{p}} t} \right] \]

The second term of \(\phi^*\), \(b(\mathbf{p})\, e^{i\mathbf{p}\cdot\mathbf{x} - i\omega_{\mathbf{p}} t}\), has the same exponential as the first term of \(\phi\), \(a(\mathbf{p})\, e^{i\mathbf{p}\cdot\mathbf{x} - i\omega_{\mathbf{p}} t}\).

The first term of \(\phi^*\), \(a^*(\mathbf{p})\, e^{-i\mathbf{p}\cdot\mathbf{x} + i\omega_{\mathbf{p}} t}\), has the same exponential as the second term of \(\phi\), \(b^*(\mathbf{p})\, e^{-i\mathbf{p}\cdot\mathbf{x} + i\omega_{\mathbf{p}} t}\).

From the condition \(\phi = \phi^*\), comparing Fourier components:

\[ \boxed{b(\mathbf{p}) = a(\mathbf{p})} \]

That is, for a real scalar field, \(a(\mathbf{p})\) and \(b(\mathbf{p})\) are not independent, but satisfy the relation \(b(\mathbf{p}) = a(\mathbf{p})\). The general solution becomes:

\[ \phi(\mathbf{x}, t) = \int \frac{d^3p}{(2\pi)^3} \frac{1}{\sqrt{2\omega_{\mathbf{p}}}} \left[ a(\mathbf{p})\, e^{i(\mathbf{p}\cdot\mathbf{x} - \omega_{\mathbf{p}} t)} + a^*(\mathbf{p})\, e^{-i(\mathbf{p}\cdot\mathbf{x} - \omega_{\mathbf{p}} t)} \right] \]

(c) Complex scalar field and antiparticles

For a complex scalar field, since \(\phi \neq \phi^*\), \(a(\mathbf{p})\) and \(b(\mathbf{p})\) are independent complex functions.

Physical interpretation: - \(a(\mathbf{p})\): amplitude of the positive-frequency mode (\(e^{-i\omega_{\mathbf{p}} t}\)) → after quantization, becomes the particle annihilation operator - \(b^*(\mathbf{p})\): amplitude of the negative-frequency mode (\(e^{+i\omega_{\mathbf{p}} t}\)) → after quantization, becomes the antiparticle creation operator

Connection to the discussion in the main text: - The negative-energy solutions (\(E < 0\)) of the Klein-Gordon equation are reinterpreted as "positive-energy states of antiparticles" within the framework of quantum field theory - A complex field carries the degrees of freedom to distinguish particles from antiparticles, and there exists a conserved charge (particle number \(-\) antiparticle number) corresponding to the \(U(1)\) symmetry \(\phi \to e^{i\theta}\phi\) - For a real scalar field, particles and antiparticles are identical (\(b = a\)), and the field carries no charge

\[ \boxed{\text{Complex field: } a(\mathbf{p}) \text{ and } b(\mathbf{p}) \text{ are independent. } b^*(\mathbf{p}) \text{ corresponds to antiparticle creation.}} \]

Consistency check

Degrees of freedom for a real scalar field: for each \(\mathbf{p}\), there is \(a(\mathbf{p})\) (one complex number = two real numbers). For a complex scalar field: for each \(\mathbf{p}\), there are \(a(\mathbf{p})\) and \(b(\mathbf{p})\) (two complex numbers = four real numbers). The complex field has twice the degrees of freedom of a real field, which is consistent with the decomposition \(\phi = (\phi_1 + i\phi_2)/\sqrt{2}\).

Advanced

A-1. Causality and the Klein-Gordon Propagator

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(a) Fourier Representation of the Retarded Green's Function

Definition:

\[ (\Box_x + m^2)\, G_R(x - y) = -\delta^4(x - y) \]

Taking the 4-dimensional Fourier transform of \(G_R\):

\[ G_R(x - y) = \int \frac{d^4p}{(2\pi)^4}\, \tilde{G}_R(p)\, e^{-ip\cdot(x-y)} \]

where \(p\cdot x = p^0 x^0 - \mathbf{p}\cdot\mathbf{x}\) (metric \((+,-,-,-)\)).

Fourier representation of \(\delta^4(x-y)\):

\[ \delta^4(x-y) = \int \frac{d^4p}{(2\pi)^4}\, e^{-ip\cdot(x-y)} \]

Computing \(\Box_x\, e^{-ip\cdot(x-y)}\):

\[ \Box_x\, e^{-ip\cdot(x-y)} = \partial_\mu\partial^\mu\, e^{-ip\cdot(x-y)} = (-ip_\mu)(-ip^\mu)\, e^{-ip\cdot(x-y)} = -p_\mu p^\mu\, e^{-ip\cdot(x-y)} \]
\[ = -(p^0)^2 + |\mathbf{p}|^2)\, e^{-ip\cdot(x-y)} = -p^2\, e^{-ip\cdot(x-y)} \]

where \(p^2 \equiv p_\mu p^\mu = (p^0)^2 - |\mathbf{p}|^2\).

Substituting into the defining equation:

\[ \int \frac{d^4p}{(2\pi)^4}\, (-p^2 + m^2)\, \tilde{G}_R(p)\, e^{-ip\cdot(x-y)} = -\int \frac{d^4p}{(2\pi)^4}\, e^{-ip\cdot(x-y)} \]

Comparing Fourier components:

\[ (-p^2 + m^2)\, \tilde{G}_R(p) = -1 \]
\[ \tilde{G}_R(p) = \frac{1}{p^2 - m^2} = \frac{1}{(p^0)^2 - |\mathbf{p}|^2 - m^2} = \frac{1}{(p^0)^2 - \omega_{\mathbf{p}}^2} \]

where \(\omega_{\mathbf{p}} = \sqrt{|\mathbf{p}|^2 + m^2}\).

Pole positions and the \(i\varepsilon\) prescription:

The solutions of \((p^0)^2 - \omega_{\mathbf{p}}^2 = (p^0 - \omega_{\mathbf{p}})(p^0 + \omega_{\mathbf{p}}) = 0\) are \(p^0 = \pm\omega_{\mathbf{p}}\).

To implement the retarded boundary condition (\(G_R = 0\) for \(x^0 < y^0\)), we shift both poles below the real axis in the \(p^0\) integration:

\[ p^0 \to p^0 + i\varepsilon \quad (\varepsilon > 0) \]

That is:

\[ \boxed{\tilde{G}_R(p) = \frac{1}{(p^0 + i\varepsilon)^2 - \omega_{\mathbf{p}}^2}} \]

Realization of the retarded condition:

Consider the \(p^0\) integration in \(G_R(x-y)\):

\[ G_R(x-y) = \int \frac{d^3p}{(2\pi)^3}\, e^{i\mathbf{p}\cdot(\mathbf{x}-\mathbf{y})} \int \frac{dp^0}{2\pi}\, \frac{e^{-ip^0(x^0-y^0)}}{(p^0 + i\varepsilon)^2 - \omega_{\mathbf{p}}^2} \]

The poles are at \(p^0 = \omega_{\mathbf{p}} - i\varepsilon\) and \(p^0 = -\omega_{\mathbf{p}} - i\varepsilon\) (both below the real axis).

  • For \(x^0 - y^0 > 0\): \(e^{-ip^0(x^0-y^0)}\) decays as \(\mathrm{Im}(p^0) \to -\infty\), so we close the contour in the lower half-plane. Both poles are enclosed, so the residue theorem gives a non-zero contribution.

  • For \(x^0 - y^0 < 0\): \(e^{-ip^0(x^0-y^0)} = e^{-ip^0 \cdot (\text{negative})}\) decays as \(\mathrm{Im}(p^0) \to +\infty\), so we close the contour in the upper half-plane. Since there are no poles in the upper half-plane:

\[ G_R(x-y) = 0 \quad \text{for } x^0 < y^0 \]

This is the retarded boundary condition. In summary:

\[ G_R(x-y) = \theta(x^0 - y^0) \cdot [\text{contribution from residues}] \]

Computing the residues:

\[ G_R(x-y) = -\theta(x^0-y^0) \int \frac{d^3p}{(2\pi)^3}\, \frac{\sin[\omega_{\mathbf{p}}(x^0-y^0)]}{\omega_{\mathbf{p}}}\, e^{i\mathbf{p}\cdot(\mathbf{x}-\mathbf{y})} \]

(b) Causality at Spacelike Separations

We show that \(G_R(x-y) = 0\) for spacelike separations \((x-y)^2 = (x^0-y^0)^2 - |\mathbf{x}-\mathbf{y}|^2 < 0\).

Method 1: Direct argument using the \(\theta\) function

\(G_R(x-y)\) contains \(\theta(x^0-y^0)\) as a factor. For spacelike separations, the sign of \(x^0 - y^0\) can be changed by a Lorentz transformation (the time ordering of two spacelike-separated events depends on the observer).

However, since \(G_R\) is a solution of a Lorentz-invariant equation, its value should depend only on the Lorentz invariant \((x-y)^2\) (for the part excluding the \(\theta\) function).

More precisely:

  • For \(x^0 > y^0\): \(G_R\) may be non-zero
  • But for spacelike separations, there exists a frame where \(x^0 < y^0\) via a Lorentz transformation
  • In that frame, \(G_R = 0\) (retarded condition)
  • Since the part of \(G_R\) excluding the \(\theta\) function is Lorentz invariant, \(G_R = 0\) in all frames

Method 2: Direct calculation

Using the expression obtained in (a):

\[ G_R(x-y) = -\theta(x^0-y^0) \int \frac{d^3p}{(2\pi)^3}\, \frac{\sin[\omega_{\mathbf{p}}(x^0-y^0)]}{\omega_{\mathbf{p}}}\, e^{i\mathbf{p}\cdot(\mathbf{x}-\mathbf{y})} \]

The content inside the \(\theta\) function (\(\Delta(x-y) \equiv -\int \frac{d^3p}{(2\pi)^3}\frac{\sin(\omega_{\mathbf{p}} \Delta t)}{\omega_{\mathbf{p}}} e^{i\mathbf{p}\cdot\Delta\mathbf{x}}\)) is a Lorentz-invariant function, and it is known to vanish for spacelike separations \((x-y)^2 < 0\) (a property of the Pauli-Jordan function).

This can be understood as follows: for spacelike separations, there exists a frame where \(\Delta t = 0\) via a Lorentz transformation. In that frame, \(\sin(\omega_{\mathbf{p}} \cdot 0) = 0\), so \(\Delta = 0\). Since \(\Delta\) is Lorentz invariant, \(\Delta = 0\) in all frames.

\[ \boxed{G_R(x-y) = 0 \quad \text{for } (x-y)^2 < 0} \]

Expression of causality: This means that "the influence of a source (a disturbance at point \(y\)) does not propagate outside the light cone (to a point \(x\) that is spacelike separated)." Signals cannot propagate faster than the speed of light.

(c) Virtual Particle Exchange and Causality

The text states that "forces are mediated by the exchange of virtual particles." The consistency of this picture with causality is justified as follows:

  1. Causal property of the retarded propagator: \(G_R(x-y) = 0\) for \((x-y)^2 < 0\) guarantees that classical signals propagate only at or below the speed of light.

  2. Virtual particles and the Feynman propagator: What appears in perturbative calculations in quantum field theory is not the retarded propagator but the Feynman propagator \(G_F\), which does not vanish at spacelike separations. However, \(G_F\) represents "correlations of quantum fluctuations of the field" rather than "signal propagation."

  3. Causality of observables: When physically observable quantities (scattering amplitudes, changes in expectation values, etc.) are calculated, the causal influence between spacelike-separated events vanishes. This is expressed as the field commutation relation \([\hat{\phi}(x), \hat{\phi}(y)] = 0\) for \((x-y)^2 < 0\) (for bosonic fields).

  4. "Propagation" of virtual particles: Virtual particles do not satisfy the mass-shell condition \(p^2 = m^2\) (they are off-shell). The internal lines of virtual particles in Feynman diagrams are not actual particle propagation but rather a mathematical representation of quantum fluctuations of the field. Causality is guaranteed not at the level of individual virtual particle "propagation," but at the level of physical observables obtained by summing all contributions.

Consistency Check

Dimensional analysis: \([G_R] = [1/(\Box + m^2)][\delta^4] = [m^{-2}][m^4] = [m^2]\) (in natural units). The Fourier representation \(\tilde{G}_R = 1/(p^2 - m^2)\) has dimension \([m^{-2}]\), and combined with the dimension of \(\int d^4p/(2\pi)^4\), which is \([m^4]\), we get \([G_R] = [m^2]\). ✓

A-2. Equivalence of the Dirac Sea and Quantum Field Theory

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(a) Divergence of the Vacuum Energy in the Dirac Sea

In the Dirac sea picture, the vacuum state is the state in which all negative energy states are occupied by electrons.

The energy of negative energy solutions of the free Dirac equation is \(E = -\omega_{\mathbf{p}} = -\sqrt{|\mathbf{p}|^2 + m^2}\) (for each momentum \(\mathbf{p}\), each spin \(s = 1, 2\)).

The vacuum energy is the sum of the energies of all occupied negative energy states:

\[ E_{\text{vac}} = \sum_{\mathbf{p}} \sum_{s=1}^{2} (-\omega_{\mathbf{p}}) = -2\sum_{\mathbf{p}} \omega_{\mathbf{p}} \]

In the continuum limit:

\[ E_{\text{vac}} = -2 \int \frac{d^3p}{(2\pi)^3}\, \omega_{\mathbf{p}} = -2 \int \frac{d^3p}{(2\pi)^3}\, \sqrt{|\mathbf{p}|^2 + m^2} \]

This integral diverges as \(|\mathbf{p}| \to \infty\):

\[ \int_0^\Lambda \frac{4\pi p^2\, dp}{(2\pi)^3}\, \sqrt{p^2 + m^2} \sim \int_0^\Lambda \frac{p^3\, dp}{2\pi^2} \sim \frac{\Lambda^4}{8\pi^2} \to \infty \]
\[ \boxed{E_{\text{vac}} = -2\sum_{\mathbf{p}} \omega_{\mathbf{p}} \to -\infty \quad \text{(formally divergent)}} \]

(The factor of 2 comes from the spin degrees of freedom.)

(b) Equivalence of Physical Processes — Electron-Positron Pair Creation

Dirac sea picture:

When energy \(E \geq 2mc^2\) is injected into the vacuum (the state where the sea is completely filled), one electron in the sea is excited to a positive energy state.

  • Initial state: vacuum with a completely filled sea
  • Final state: one positive energy electron + one hole in the sea

Energy balance: - Energy of the positive energy electron: \(+\omega_{\mathbf{p}_1}\) - Energy of the hole: since an electron with energy \(-\omega_{\mathbf{p}_2}\) was removed from the sea, the energy change relative to the vacuum is \(+\omega_{\mathbf{p}_2}\) - Total required energy: \(\omega_{\mathbf{p}_1} + \omega_{\mathbf{p}_2} \geq 2m\)

Charge balance: - Charge of the vacuum: \(Q_{\text{vac}}\) (taken as 0 by convention) - Charge of the positive energy electron: \(-e\) - Charge of the hole: \(-(-e) = +e\) (since one electron carrying charge \(-e\) is missing) - Total charge change: \(-e + e = 0\) (charge conservation)

Quantum field theory picture:

Acting with creation operators on the vacuum state \(|0\rangle\) (Fock vacuum):

\[ \hat{a}^\dagger_{\mathbf{p}_1, s_1} \hat{b}^\dagger_{\mathbf{p}_2, s_2} |0\rangle \]

where \(\hat{a}^\dagger\) is the electron creation operator and \(\hat{b}^\dagger\) is the positron creation operator.

Energy balance: - Energy of the electron: \(+\omega_{\mathbf{p}_1}\) - Energy of the positron: \(+\omega_{\mathbf{p}_2}\) - Total required energy: \(\omega_{\mathbf{p}_1} + \omega_{\mathbf{p}_2} \geq 2m\)

Charge balance: - Charge of the electron: \(-e\) - Charge of the positron: \(+e\) - Total charge change: \(0\)

Equivalence:

\[ \boxed{\text{In both pictures, the energy cost } \omega_{\mathbf{p}_1} + \omega_{\mathbf{p}_2} \text{ and charge conservation } \Delta Q = 0 \text{ agree}} \]

All observable physical quantities (energy differences, charge differences, momenta, etc.) give identical results. The Dirac sea picture is merely a "rephrasing" of quantum field theory.

(c) Impossibility of the Dirac Sea for Bosons

From the perspective of the Pauli exclusion principle:

The construction of the Dirac sea essentially requires the following:

  1. At most one particle can occupy each quantum state (Pauli exclusion principle)
  2. All negative energy states must be completely filled with particles
  3. "Holes" behave as antiparticles

Since scalar fields (spin 0) obey Bose statistics:

  • Any number of particles can occupy each state
  • It is impossible to "fill all negative energy states" — even after placing one particle, another can be added, and the states can never be completely filled
  • Therefore, a "stable vacuum" (a state where the sea is completely filled) cannot be defined
  • The concept of a "hole" is also meaningless
\[ \boxed{\text{Bosons do not obey the Pauli exclusion principle, so the Dirac sea cannot be constructed}} \]

Resolution by quantum field theory:

In quantum field theory, the difference in statistics is automatically incorporated through the choice of commutation relations:

  • Bosonic fields: Creation and annihilation operators satisfy commutation relations
\[ [\hat{a}_{\mathbf{p}}, \hat{a}^\dagger_{\mathbf{q}}] = (2\pi)^3 \delta^3(\mathbf{p} - \mathbf{q}) \]
  • Fermionic fields: Creation and annihilation operators satisfy anticommutation relations
\[ \{\hat{b}_{\mathbf{p}}, \hat{b}^\dagger_{\mathbf{q}}\} = (2\pi)^3 \delta^3(\mathbf{p} - \mathbf{q}) \]

In both cases: - The vacuum \(|0\rangle\) is defined as the state annihilated by all annihilation operators: \(\hat{a}_{\mathbf{p}}|0\rangle = 0\) - Both particles and antiparticles are described as excitations with positive energy - Negative energy solutions are reinterpreted as "positive energy states of antiparticles" - There is no need to assume infinitely many particles as in the Dirac sea

Quantum field theory can treat both fermions and bosons in a unified manner, completely overcoming the limitations of the Dirac sea picture.

Verification

Consistency with the spin-statistics theorem: In quantum field theory, it can be proven from Lorentz invariance and causality (commutation/anticommutation relations of field operators at spacelike separations) that integer-spin fields must obey Bose statistics and half-integer-spin fields must obey Fermi statistics (the spin-statistics theorem). This provides the deep reason why the Dirac sea is applicable only to half-integer spin particles.