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Ch. 4 Solutions

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Basic

B-1. Reading Off Values from the Equal-Time Commutation Relations

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Solution Strategy

Substitute specific values into the equal-time commutation relation \([\hat{\phi}(t, \mathbf{x}),\, \hat{\pi}(t, \mathbf{y})] = i\,\delta^{(3)}(\mathbf{x} - \mathbf{y})\) and use the fundamental properties of the delta function.

(a) When \(\mathbf{x} = (1,2,3)\), \(\mathbf{y} = (4,5,6)\)

Since \(\mathbf{x} \neq \mathbf{y}\), we have \(\delta^{(3)}(\mathbf{x} - \mathbf{y}) = \delta^{(3)}((-3,-3,-3)) = 0\).

\[ \boxed{[\hat{\phi}(t, \mathbf{x}),\, \hat{\pi}(t, \mathbf{y})] = 0} \]

Physical meaning: The field and conjugate momentum density at different spatial points commute with each other, meaning they can simultaneously have definite values.

(b) Integrating over all space with respect to \(\mathbf{y}\)

\[ \int d^3y\, [\hat{\phi}(t, \mathbf{x}),\, \hat{\pi}(t, \mathbf{y})] = \int d^3y\, i\,\delta^{(3)}(\mathbf{x} - \mathbf{y}) \]

Using the fundamental property of the delta function \(\int d^3y\, f(\mathbf{y})\,\delta^{(3)}(\mathbf{x} - \mathbf{y}) = f(\mathbf{x})\), setting \(f(\mathbf{y}) = 1\) (constant) gives

\[ \int d^3y\, \delta^{(3)}(\mathbf{x} - \mathbf{y}) = 1 \]

Therefore

\[ \boxed{\int d^3y\, [\hat{\phi}(t, \mathbf{x}),\, \hat{\pi}(t, \mathbf{y})] = i} \]

(c) Express \([\hat{\pi}(t, \mathbf{x}),\, \hat{\phi}(t, \mathbf{y})]\)

From the antisymmetry of the commutator \([A, B] = -[B, A]\):

\[ \boxed{[\hat{\pi}(t, \mathbf{x}),\, \hat{\phi}(t, \mathbf{y})] = -[\hat{\phi}(t, \mathbf{y}),\, \hat{\pi}(t, \mathbf{x})] = -i\,\delta^{(3)}(\mathbf{y} - \mathbf{x}) = -i\,\delta^{(3)}(\mathbf{x} - \mathbf{y})} \]

In the last equality, we used \(\delta^{(3)}(\mathbf{y} - \mathbf{x}) = \delta^{(3)}(\mathbf{x} - \mathbf{y})\) (the even function property of the delta function).

Verification

The result of (b) corresponds to the field theory version of \([\hat{q}, \hat{p}] = i\) in quantum mechanics. The integral of the field over all space corresponds to a sum over discrete indices, and integrating \(\delta^{(3)}\) is analogous to the sum of Kronecker deltas \(\sum_j \delta_{ij} = 1\). This is consistent.

B-2. Calculation of Commutation Relations for Creation and Annihilation Operators

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Solution Strategy

Repeatedly use the commutator identity \([A, BC] = [A, B]C + B[A, C]\).

(a) \([\hat{a}_{\mathbf{p}},\, \hat{a}_{\mathbf{q}}^\dagger \hat{a}_{\mathbf{q}}]\)

Apply the identity with \(A = \hat{a}_{\mathbf{p}}\), \(B = \hat{a}_{\mathbf{q}}^\dagger\), \(C = \hat{a}_{\mathbf{q}}\):

\[ [\hat{a}_{\mathbf{p}},\, \hat{a}_{\mathbf{q}}^\dagger \hat{a}_{\mathbf{q}}] = [\hat{a}_{\mathbf{p}},\, \hat{a}_{\mathbf{q}}^\dagger]\,\hat{a}_{\mathbf{q}} + \hat{a}_{\mathbf{q}}^\dagger\,[\hat{a}_{\mathbf{p}},\, \hat{a}_{\mathbf{q}}] \]

Substituting \([\hat{a}_{\mathbf{p}},\, \hat{a}_{\mathbf{q}}^\dagger] = \delta^{(3)}(\mathbf{p} - \mathbf{q})\) and \([\hat{a}_{\mathbf{p}},\, \hat{a}_{\mathbf{q}}] = 0\):

\[ \boxed{[\hat{a}_{\mathbf{p}},\, \hat{a}_{\mathbf{q}}^\dagger \hat{a}_{\mathbf{q}}] = \delta^{(3)}(\mathbf{p} - \mathbf{q})\,\hat{a}_{\mathbf{q}}} \]

(b) \([\hat{a}_{\mathbf{p}}^\dagger \hat{a}_{\mathbf{p}},\, \hat{a}_{\mathbf{q}}^\dagger]\)

Use \([AB, C] = A[B, C] + [A, C]B\) with \(A = \hat{a}_{\mathbf{p}}^\dagger\), \(B = \hat{a}_{\mathbf{p}}\), \(C = \hat{a}_{\mathbf{q}}^\dagger\):

\[ [\hat{a}_{\mathbf{p}}^\dagger \hat{a}_{\mathbf{p}},\, \hat{a}_{\mathbf{q}}^\dagger] = \hat{a}_{\mathbf{p}}^\dagger\,[\hat{a}_{\mathbf{p}},\, \hat{a}_{\mathbf{q}}^\dagger] + [\hat{a}_{\mathbf{p}}^\dagger,\, \hat{a}_{\mathbf{q}}^\dagger]\,\hat{a}_{\mathbf{p}} \]

Substituting \([\hat{a}_{\mathbf{p}},\, \hat{a}_{\mathbf{q}}^\dagger] = \delta^{(3)}(\mathbf{p} - \mathbf{q})\) and \([\hat{a}_{\mathbf{p}}^\dagger,\, \hat{a}_{\mathbf{q}}^\dagger] = 0\):

\[ \boxed{[\hat{a}_{\mathbf{p}}^\dagger \hat{a}_{\mathbf{p}},\, \hat{a}_{\mathbf{q}}^\dagger] = \delta^{(3)}(\mathbf{p} - \mathbf{q})\,\hat{a}_{\mathbf{p}}^\dagger} \]

(c) \([\hat{a}_{\mathbf{p}},\, (\hat{a}_{\mathbf{q}}^\dagger)^2]\)

Apply \([A, BC] = [A, B]C + B[A, C]\) with \(B = C = \hat{a}_{\mathbf{q}}^\dagger\):

\[ [\hat{a}_{\mathbf{p}},\, (\hat{a}_{\mathbf{q}}^\dagger)^2] = [\hat{a}_{\mathbf{p}},\, \hat{a}_{\mathbf{q}}^\dagger]\,\hat{a}_{\mathbf{q}}^\dagger + \hat{a}_{\mathbf{q}}^\dagger\,[\hat{a}_{\mathbf{p}},\, \hat{a}_{\mathbf{q}}^\dagger] \]
\[ = \delta^{(3)}(\mathbf{p} - \mathbf{q})\,\hat{a}_{\mathbf{q}}^\dagger + \hat{a}_{\mathbf{q}}^\dagger\,\delta^{(3)}(\mathbf{p} - \mathbf{q}) \]
\[ \boxed{[\hat{a}_{\mathbf{p}},\, (\hat{a}_{\mathbf{q}}^\dagger)^2] = 2\,\delta^{(3)}(\mathbf{p} - \mathbf{q})\,\hat{a}_{\mathbf{q}}^\dagger} \]

Verification

Result (a) corresponds to the quantum mechanics relation \([\hat{a}, \hat{a}^\dagger \hat{a}] = \hat{a}\), and (b) corresponds to \([\hat{a}^\dagger \hat{a}, \hat{a}^\dagger] = \hat{a}^\dagger\). Considering that the delta function is the continuous version of the Kronecker delta, these results are consistent with the discrete case. Result (c) is the continuous version of \([\hat{a}, (\hat{a}^\dagger)^2] = 2\hat{a}^\dagger\), corresponding to the \(n=2\) case of the general relation \([\hat{a}, (\hat{a}^\dagger)^n] = n(\hat{a}^\dagger)^{n-1}\).

B-3. Action of the Annihilation Operator on the Vacuum

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Solution Strategy

Combine the vacuum definition \(\hat{a}_{\mathbf{p}}|0\rangle = 0\) with the commutation relations.

(a) \(\hat{a}_{\mathbf{p}} |\mathbf{q}\rangle\)

\[ \hat{a}_{\mathbf{p}} |\mathbf{q}\rangle = \hat{a}_{\mathbf{p}} \hat{a}_{\mathbf{q}}^\dagger |0\rangle \]

Using the commutation relation to move \(\hat{a}_{\mathbf{p}}\) to the right of \(\hat{a}_{\mathbf{q}}^\dagger\):

\[ \hat{a}_{\mathbf{p}} \hat{a}_{\mathbf{q}}^\dagger = \hat{a}_{\mathbf{q}}^\dagger \hat{a}_{\mathbf{p}} + [\hat{a}_{\mathbf{p}},\, \hat{a}_{\mathbf{q}}^\dagger] = \hat{a}_{\mathbf{q}}^\dagger \hat{a}_{\mathbf{p}} + \delta^{(3)}(\mathbf{p} - \mathbf{q}) \]

Acting on \(|0\rangle\), since \(\hat{a}_{\mathbf{p}}|0\rangle = 0\):

\[ \boxed{\hat{a}_{\mathbf{p}} |\mathbf{q}\rangle = \delta^{(3)}(\mathbf{p} - \mathbf{q})\,|0\rangle} \]

(b) \(\langle \mathbf{p} | \mathbf{q} \rangle\)

Since \(\langle \mathbf{p}| = (|\mathbf{p}\rangle)^\dagger = (\hat{a}_{\mathbf{p}}^\dagger |0\rangle)^\dagger = \langle 0| \hat{a}_{\mathbf{p}}\), we have

\[ \langle \mathbf{p} | \mathbf{q} \rangle = \langle 0| \hat{a}_{\mathbf{p}} \hat{a}_{\mathbf{q}}^\dagger |0\rangle \]

Using the result from (a):

\[ \langle 0| \hat{a}_{\mathbf{p}} \hat{a}_{\mathbf{q}}^\dagger |0\rangle = \langle 0| \left(\delta^{(3)}(\mathbf{p} - \mathbf{q})\,|0\rangle\right) = \delta^{(3)}(\mathbf{p} - \mathbf{q})\,\langle 0|0\rangle \]

Using the vacuum normalization \(\langle 0|0\rangle = 1\):

\[ \boxed{\langle \mathbf{p} | \mathbf{q} \rangle = \delta^{(3)}(\mathbf{p} - \mathbf{q})} \]

(c) \(\hat{a}_{\mathbf{p}} \hat{a}_{\mathbf{q}} |0\rangle\)

Since \([\hat{a}_{\mathbf{p}}, \hat{a}_{\mathbf{q}}] = 0\), the order of \(\hat{a}_{\mathbf{p}}\) and \(\hat{a}_{\mathbf{q}}\) can be exchanged. In any case, since \(\hat{a}_{\mathbf{q}}|0\rangle = 0\):

\[ \hat{a}_{\mathbf{p}} \hat{a}_{\mathbf{q}} |0\rangle = \hat{a}_{\mathbf{p}} \cdot 0 = 0 \]
\[ \boxed{\hat{a}_{\mathbf{p}} \hat{a}_{\mathbf{q}} |0\rangle = 0} \]

Verification

The result of (b) expresses the orthonormality of single-particle states with continuous momentum labels. This has the same structure as position eigenstates in quantum mechanics, \(\langle \mathbf{x}|\mathbf{y}\rangle = \delta^{(3)}(\mathbf{x}-\mathbf{y})\), and is therefore consistent. Result (c) is the physically natural statement that "one cannot annihilate particles twice from the vacuum."

B-4. Verifying Orthogonality of the Fourier Transform

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Solution Strategy

Appropriately reinterpret the fundamental formula \(\int \frac{d^3x}{(2\pi)^3}\, e^{i(\mathbf{p}-\mathbf{q})\cdot\mathbf{x}} = \delta^{(3)}(\mathbf{p} - \mathbf{q})\).

(a) \(\displaystyle \int \frac{d^3x}{(2\pi)^3}\, e^{i(\mathbf{p}+\mathbf{q})\cdot\mathbf{x}}\)

Rewriting the exponent as \(e^{i(\mathbf{p}-(-\mathbf{q}))\cdot\mathbf{x}}\), this corresponds to substituting \(\mathbf{q} \to -\mathbf{q}\) in the fundamental formula:

\[ \boxed{\int \frac{d^3x}{(2\pi)^3}\, e^{i(\mathbf{p}+\mathbf{q})\cdot\mathbf{x}} = \delta^{(3)}(\mathbf{p} + \mathbf{q})} \]

(b) \(\displaystyle \int \frac{d^3x}{(2\pi)^3}\, e^{i\mathbf{p}\cdot\mathbf{x}}\, e^{-i\mathbf{q}\cdot\mathbf{x}}\, e^{i\mathbf{k}\cdot\mathbf{x}}\)

Combining the exponents using the exponential law:

\[ e^{i\mathbf{p}\cdot\mathbf{x}}\, e^{-i\mathbf{q}\cdot\mathbf{x}}\, e^{i\mathbf{k}\cdot\mathbf{x}} = e^{i(\mathbf{p} - \mathbf{q} + \mathbf{k})\cdot\mathbf{x}} \]

Applying the fundamental formula:

\[ \boxed{\int \frac{d^3x}{(2\pi)^3}\, e^{i(\mathbf{p}-\mathbf{q}+\mathbf{k})\cdot\mathbf{x}} = \delta^{(3)}(\mathbf{p} - \mathbf{q} + \mathbf{k})} \]

(c) \(\displaystyle \int d^3x\, \int \frac{d^3p}{(2\pi)^3}\, e^{i\mathbf{p}\cdot(\mathbf{x}-\mathbf{y})} f(\mathbf{x})\)

First, perform the \(\mathbf{p}\) integration. Using the Fourier representation of the delta function:

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

we obtain

\[ \int d^3x\, \delta^{(3)}(\mathbf{x} - \mathbf{y})\, f(\mathbf{x}) = f(\mathbf{y}) \]
\[ \boxed{\int d^3x\, \int \frac{d^3p}{(2\pi)^3}\, e^{i\mathbf{p}\cdot(\mathbf{x}-\mathbf{y})} f(\mathbf{x}) = f(\mathbf{y})} \]

Verification

Part (c) expresses the fact that applying a Fourier transform followed by its inverse returns the original function, which is the fundamental theorem of Fourier transforms itself.

B-5. Explicit Calculation of \(\omega_{\mathbf{p}}\)

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Solution Strategy

Substitute each condition into \(\omega_{\mathbf{p}} = \sqrt{|\mathbf{p}|^2 + m^2}\).

(a) \(\mathbf{p} = \mathbf{0}\)

\[ \omega_{\mathbf{0}} = \sqrt{0 + m^2} = m \]
\[ \boxed{\omega_{\mathbf{0}} = m} \]

This corresponds to the rest mass energy \(E = mc^2\) (or \(E = m\) in natural units).

(b) \(|\mathbf{p}| = m\)

\[ \omega_{\mathbf{p}} = \sqrt{m^2 + m^2} = \sqrt{2}\,m \]
\[ \boxed{\omega_{\mathbf{p}} = \sqrt{2}\,m} \]

(c) \(|\mathbf{p}| \gg m\) (ultra-relativistic limit)

\[ \omega_{\mathbf{p}} = \sqrt{|\mathbf{p}|^2 + m^2} = |\mathbf{p}|\sqrt{1 + \frac{m^2}{|\mathbf{p}|^2}} \]

Using the Taylor expansion \(\sqrt{1+\epsilon} \approx 1 + \frac{\epsilon}{2}\) for \(m/|\mathbf{p}| \ll 1\):

\[ \omega_{\mathbf{p}} \approx |\mathbf{p}|\left(1 + \frac{m^2}{2|\mathbf{p}|^2}\right) = |\mathbf{p}| + \frac{m^2}{2|\mathbf{p}|} \]

To leading order:

\[ \boxed{\omega_{\mathbf{p}} \approx |\mathbf{p}|} \]

(d) \(m = 0\)

\[ \omega_{\mathbf{p}} = \sqrt{|\mathbf{p}|^2 + 0} = |\mathbf{p}| \]
\[ \boxed{\omega_{\mathbf{p}} = |\mathbf{p}|} \]

This corresponds to the dispersion relation \(E = |p|\) (i.e., \(E = pc\)) for massless particles such as photons.

Consistency Check

The ultra-relativistic limit in (c) and the massless result in (d) agree, which is consistent. Result (a) reproduces the rest energy in the non-relativistic limit \(|\mathbf{p}| \to 0\).

B-6. Verifying Hermiticity of the Field Operator

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Solution Strategy

Compute \(\hat{\phi}^\dagger(\mathbf{x})\) and show that it equals \(\hat{\phi}(\mathbf{x})\) using the substitution of integration variable \(\mathbf{p} \to -\mathbf{p}\) and the relation \(\omega_{\mathbf{p}} = \omega_{-\mathbf{p}}\).

Detailed Calculation

Take the Hermitian conjugate of \(\hat{\phi}(\mathbf{x})\). The integral and c-numbers (such as \(\omega_{\mathbf{p}}\) and \((2\pi)^{3/2}\)) remain unchanged, while the Hermitian conjugate is applied to operators and complex numbers:

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

Here we used \((\hat{a}_{\mathbf{p}})^\dagger = \hat{a}_{\mathbf{p}}^\dagger\) and \((e^{i\mathbf{p}\cdot\mathbf{x}})^* = e^{-i\mathbf{p}\cdot\mathbf{x}}\) (since \(\mathbf{p}, \mathbf{x}\) are real).

Rewriting by swapping the order of the two terms:

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

This is precisely \(\hat{\phi}(\mathbf{x})\) from Eq. (4.11).

\[ \boxed{\hat{\phi}^\dagger(\mathbf{x}) = \hat{\phi}(\mathbf{x})} \]

Remark: In the calculation above, the result was obtained simply by swapping the order of the terms. More generally, one can also use the substitution of integration variable \(\mathbf{p} \to -\mathbf{p}\). In that case, \(d^3p \to d^3p\) (the Jacobian is \(|-1|^3 = 1\)) and \(\omega_{\mathbf{p}} = \omega_{-\mathbf{p}}\) (since \(\omega_{\mathbf{p}}\) depends only on \(|\mathbf{p}|^2\)), yielding the same result.

Verification

A real scalar field must satisfy \(\hat{\phi}^\dagger = \hat{\phi}\), which is naturally guaranteed by the structure of the mode expansion containing both \(\hat{a}_{\mathbf{p}}\) and \(\hat{a}_{\mathbf{p}}^\dagger\). This is the field theory analogue of the fact that \(\hat{q} = \frac{1}{\sqrt{2\omega}}(\hat{a} + \hat{a}^\dagger)\) is Hermitian for the quantum harmonic oscillator.

B-7. Derivation of the Mode Expansion of the Conjugate Momentum Density

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Solution Strategy

Compute the time derivative \(\hat{\pi}(t, \mathbf{x}) = \partial_0 \hat{\phi}(t, \mathbf{x})\) of \(\hat{\phi}(t, \mathbf{x})\).

Detailed Calculation

The mode expansion including time dependence is

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

We perform the time derivative. Since \(\hat{a}_{\mathbf{p}}\) and \(\hat{a}_{\mathbf{p}}^\dagger\) are time-independent (Schrödinger picture operators), the derivative acts only on the exponentials:

\[ \partial_0\, e^{i(\mathbf{p}\cdot\mathbf{x} - \omega_{\mathbf{p}} t)} = (-i\omega_{\mathbf{p}})\, e^{i(\mathbf{p}\cdot\mathbf{x} - \omega_{\mathbf{p}} t)} \]
\[ \partial_0\, e^{-i(\mathbf{p}\cdot\mathbf{x} - \omega_{\mathbf{p}} t)} = (+i\omega_{\mathbf{p}})\, e^{-i(\mathbf{p}\cdot\mathbf{x} - \omega_{\mathbf{p}} t)} \]

Therefore

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

Combining \(\omega_{\mathbf{p}}\) with \(\frac{1}{\sqrt{2\omega_{\mathbf{p}}}}\):

\[ \frac{\omega_{\mathbf{p}}}{\sqrt{2\omega_{\mathbf{p}}}} = \sqrt{\frac{\omega_{\mathbf{p}}}{2}} \]

Thus

\[ \hat{\pi}(t, \mathbf{x}) = \int \frac{d^3p}{(2\pi)^{3/2}} (-i)\sqrt{\frac{\omega_{\mathbf{p}}}{2}} \left(\hat{a}_{\mathbf{p}}\,e^{i(\mathbf{p}\cdot\mathbf{x} - \omega_{\mathbf{p}} t)} - \hat{a}_{\mathbf{p}}^\dagger\, e^{-i(\mathbf{p}\cdot\mathbf{x} - \omega_{\mathbf{p}} t)}\right) \]

Setting \(t = 0\) gives

\[ \boxed{\hat{\pi}(\mathbf{x}) = \int \frac{d^3p}{(2\pi)^{3/2}} (-i)\sqrt{\frac{\omega_{\mathbf{p}}}{2}} \left(\hat{a}_{\mathbf{p}}\,e^{i\mathbf{p}\cdot\mathbf{x}} - \hat{a}_{\mathbf{p}}^\dagger\, e^{-i\mathbf{p}\cdot\mathbf{x}}\right)} \]

This agrees with Eq. (4.12).

Verification

We confirm the Hermiticity of \(\hat{\pi}(\mathbf{x})\). Taking \(\hat{\pi}^\dagger\) gives \(\hat{a}_{\mathbf{p}} \leftrightarrow \hat{a}_{\mathbf{p}}^\dagger\), \(e^{i\mathbf{p}\cdot\mathbf{x}} \leftrightarrow e^{-i\mathbf{p}\cdot\mathbf{x}}\), and \((-i)^* = +i\). Checking the overall sign confirms that \(\hat{\pi}^\dagger = \hat{\pi}\) holds, verifying that the conjugate momentum density is indeed Hermitian.

B-8. Verification of the Discrete→Continuous Correspondence

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Solution Strategy

Consider momenta discretized in a box of volume \(V = L^3\), and derive the correspondence in the continuum limit \(L \to \infty\).

(a) Derivation of \(\displaystyle\sum_{\mathbf{n}} \to \int \frac{V\, d^3p}{(2\pi)^3}\)

The discretized momenta are \(\mathbf{p} = \frac{2\pi}{L}\mathbf{n}\) (\(\mathbf{n} \in \mathbb{Z}^3\)), and the spacing between adjacent momenta in each direction is

\[ \Delta p_i = \frac{2\pi}{L} \]

The volume of a single cell in 3-dimensional momentum space is

\[ (\Delta p)^3 = \left(\frac{2\pi}{L}\right)^3 = \frac{(2\pi)^3}{V} \]

Treating the discrete sum as a Riemann sum:

\[ \sum_{\mathbf{n}} f(\mathbf{p}_{\mathbf{n}}) = \sum_{\mathbf{n}} f(\mathbf{p}_{\mathbf{n}}) \cdot \frac{(\Delta p)^3}{(\Delta p)^3} = \frac{1}{(\Delta p)^3}\sum_{\mathbf{n}} f(\mathbf{p}_{\mathbf{n}})\,(\Delta p)^3 \]

As \(L \to \infty\), \(\Delta p \to 0\), and the Riemann sum converges to an integral:

\[ \sum_{\mathbf{n}} f(\mathbf{p}_{\mathbf{n}})\,(\Delta p)^3 \to \int d^3p\, f(\mathbf{p}) \]

Therefore

\[ \boxed{\sum_{\mathbf{n}} \to \frac{V}{(2\pi)^3} \int d^3p} \]

(b) Scaling relation between \(\hat{a}_{\mathbf{p}}\) and \(\hat{a}_{\mathbf{n}}\)

The discrete commutation relation is

\[ [\hat{a}_{\mathbf{n}}, \hat{a}_{\mathbf{m}}^\dagger] = \delta_{\mathbf{n}, \mathbf{m}} \]

In the continuum limit, there is the following correspondence between the Kronecker delta and the Dirac delta function. With \(\mathbf{p} = \frac{2\pi}{L}\mathbf{n}\), \(\mathbf{q} = \frac{2\pi}{L}\mathbf{m}\):

\[ \delta_{\mathbf{n}, \mathbf{m}} = \delta_{\mathbf{n}, \mathbf{m}} \cdot \frac{(\Delta p)^3}{(\Delta p)^3} \]

Using the discrete approximation to the delta function \(\delta^{(3)}(\mathbf{p} - \mathbf{q}) \approx \frac{\delta_{\mathbf{n},\mathbf{m}}}{(\Delta p)^3} = \frac{V}{(2\pi)^3}\,\delta_{\mathbf{n},\mathbf{m}}\), we obtain

\[ \delta_{\mathbf{n}, \mathbf{m}} = \frac{(2\pi)^3}{V}\,\delta^{(3)}(\mathbf{p} - \mathbf{q}) \]

To reproduce the continuum commutation relation \([\hat{a}_{\mathbf{p}}, \hat{a}_{\mathbf{q}}^\dagger] = \delta^{(3)}(\mathbf{p} - \mathbf{q})\), we set \(\hat{a}_{\mathbf{p}} = c\,\hat{a}_{\mathbf{n}}\), giving

\[ [\hat{a}_{\mathbf{p}}, \hat{a}_{\mathbf{q}}^\dagger] = |c|^2\,[\hat{a}_{\mathbf{n}}, \hat{a}_{\mathbf{m}}^\dagger] = |c|^2\,\delta_{\mathbf{n},\mathbf{m}} = |c|^2 \cdot \frac{(2\pi)^3}{V}\,\delta^{(3)}(\mathbf{p} - \mathbf{q}) \]

For this to equal \(\delta^{(3)}(\mathbf{p} - \mathbf{q})\), we need \(|c|^2 = \frac{V}{(2\pi)^3}\). Therefore

\[ \boxed{\hat{a}_{\mathbf{p}} = \sqrt{\frac{V}{(2\pi)^3}}\;\hat{a}_{\mathbf{n}}} \]

Verification

We confirm by dimensional analysis. \(\hat{a}_{\mathbf{n}}\) is dimensionless (since \([\hat{a}_{\mathbf{n}}, \hat{a}_{\mathbf{m}}^\dagger] = \delta_{\mathbf{n},\mathbf{m}}\) is dimensionless). On the other hand, \([\hat{a}_{\mathbf{p}}, \hat{a}_{\mathbf{q}}^\dagger] = \delta^{(3)}(\mathbf{p}-\mathbf{q})\) has dimensions of \([\text{momentum}]^{-3}\) from the delta function, so \(\hat{a}_{\mathbf{p}}\) has dimensions of \([\text{momentum}]^{-3/2}\). The factor \(\sqrt{V/(2\pi)^3}\) has dimensions of \([\text{length}]^{3/2} = [\text{momentum}]^{-3/2}\), which is consistent.

Medium

M-1. Derivation of Creation and Annihilation Operator Commutation Relations from Equal-Time Commutation Relations

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Strategy

Substitute the mode expansion (4.11) into \([\hat{\phi}(\mathbf{x}), \hat{\phi}(\mathbf{y})] = 0\) and use the orthogonality of Fourier transforms to derive \([\hat{a}_{\mathbf{p}}, \hat{a}_{\mathbf{q}}] = 0\).

Detailed Calculation

Substitute the mode expansion at \(t = 0\) (the time is common, so we omit it):

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

Expand \([\hat{\phi}(\mathbf{x}), \hat{\phi}(\mathbf{y})]\). Using \(\mathbf{p}\) and \(\mathbf{q}\) as the respective integration variables:

\[ [\hat{\phi}(\mathbf{x}), \hat{\phi}(\mathbf{y})] = \int \frac{d^3p}{(2\pi)^{3/2}} \frac{1}{\sqrt{2\omega_{\mathbf{p}}}} \int \frac{d^3q}{(2\pi)^{3/2}} \frac{1}{\sqrt{2\omega_{\mathbf{q}}}} \]
\[ \times \Big\{ [\hat{a}_{\mathbf{p}}, \hat{a}_{\mathbf{q}}]\, e^{i\mathbf{p}\cdot\mathbf{x}} e^{i\mathbf{q}\cdot\mathbf{y}} + [\hat{a}_{\mathbf{p}}, \hat{a}_{\mathbf{q}}^\dagger]\, e^{i\mathbf{p}\cdot\mathbf{x}} e^{-i\mathbf{q}\cdot\mathbf{y}} \]
\[ + [\hat{a}_{\mathbf{p}}^\dagger, \hat{a}_{\mathbf{q}}]\, e^{-i\mathbf{p}\cdot\mathbf{x}} e^{i\mathbf{q}\cdot\mathbf{y}} + [\hat{a}_{\mathbf{p}}^\dagger, \hat{a}_{\mathbf{q}}^\dagger]\, e^{-i\mathbf{p}\cdot\mathbf{x}} e^{-i\mathbf{q}\cdot\mathbf{y}} \Big\} \]

Now substitute the known commutation relation \([\hat{a}_{\mathbf{p}}, \hat{a}_{\mathbf{q}}^\dagger] = \delta^{(3)}(\mathbf{p}-\mathbf{q})\) into the second term, and substitute \([\hat{a}_{\mathbf{p}}^\dagger, \hat{a}_{\mathbf{q}}] = -[\hat{a}_{\mathbf{q}}, \hat{a}_{\mathbf{p}}^\dagger] = -\delta^{(3)}(\mathbf{q}-\mathbf{p})\) into the third term.

Contribution from the 2nd term: Performing the \(\mathbf{q}\) integration with \(\delta^{(3)}(\mathbf{p}-\mathbf{q})\) fixes \(\mathbf{q} = \mathbf{p}\):

\[ \text{2nd term} = \int \frac{d^3p}{(2\pi)^3} \frac{1}{2\omega_{\mathbf{p}}}\, e^{i\mathbf{p}\cdot(\mathbf{x}-\mathbf{y})} \]

Contribution from the 3rd term: Similarly,

\[ \text{3rd term} = -\int \frac{d^3p}{(2\pi)^3} \frac{1}{2\omega_{\mathbf{p}}}\, e^{-i\mathbf{p}\cdot(\mathbf{x}-\mathbf{y})} \]

Substituting \(\mathbf{p} \to -\mathbf{p}\) in the 3rd term (noting \(\omega_{\mathbf{p}} = \omega_{-\mathbf{p}}\) and \(d^3p \to d^3p\)):

\[ \text{3rd term} = -\int \frac{d^3p}{(2\pi)^3} \frac{1}{2\omega_{\mathbf{p}}}\, e^{i\mathbf{p}\cdot(\mathbf{x}-\mathbf{y})} \]

Therefore, the 2nd and 3rd terms exactly cancel.

Only the 1st and 4th terms remain:

\[ [\hat{\phi}(\mathbf{x}), \hat{\phi}(\mathbf{y})] = \int \frac{d^3p}{(2\pi)^{3/2}} \int \frac{d^3q}{(2\pi)^{3/2}} \frac{1}{\sqrt{2\omega_{\mathbf{p}}}\sqrt{2\omega_{\mathbf{q}}}} \]
\[ \times \Big\{ [\hat{a}_{\mathbf{p}}, \hat{a}_{\mathbf{q}}]\, e^{i\mathbf{p}\cdot\mathbf{x}} e^{i\mathbf{q}\cdot\mathbf{y}} + [\hat{a}_{\mathbf{p}}^\dagger, \hat{a}_{\mathbf{q}}^\dagger]\, e^{-i\mathbf{p}\cdot\mathbf{x}} e^{-i\mathbf{q}\cdot\mathbf{y}} \Big\} = 0 \]

Note that \([\hat{a}_{\mathbf{p}}^\dagger, \hat{a}_{\mathbf{q}}^\dagger] = ([\hat{a}_{\mathbf{q}}, \hat{a}_{\mathbf{p}}])^\dagger\). Setting \([\hat{a}_{\mathbf{p}}, \hat{a}_{\mathbf{q}}] = f(\mathbf{p}, \mathbf{q})\), we have \([\hat{a}_{\mathbf{p}}^\dagger, \hat{a}_{\mathbf{q}}^\dagger] = f^*(\mathbf{q}, \mathbf{p})\).

For the above equation to hold for arbitrary \(\mathbf{x}, \mathbf{y}\), the completeness of the Fourier transform requires that the Fourier coefficients of the integrand must vanish.

Focusing on the 1st term: \(e^{i\mathbf{p}\cdot\mathbf{x}} e^{i\mathbf{q}\cdot\mathbf{y}}\) forms independent Fourier modes with respect to \(\mathbf{x}\) and \(\mathbf{y}\). For this term to vanish for all \(\mathbf{x}, \mathbf{y}\), we need

\[ \frac{[\hat{a}_{\mathbf{p}}, \hat{a}_{\mathbf{q}}]}{\sqrt{2\omega_{\mathbf{p}}}\sqrt{2\omega_{\mathbf{q}}}} = 0 \quad \text{for all } \mathbf{p}, \mathbf{q} \]

Since \(\omega_{\mathbf{p}}, \omega_{\mathbf{q}} > 0\):

\[ \boxed{[\hat{a}_{\mathbf{p}}, \hat{a}_{\mathbf{q}}] = 0 \quad \text{for all } \mathbf{p}, \mathbf{q}} \]

Taking the Hermitian conjugate gives

\[ \boxed{[\hat{a}_{\mathbf{p}}^\dagger, \hat{a}_{\mathbf{q}}^\dagger] = 0 \quad \text{for all } \mathbf{p}, \mathbf{q}} \]

Consistency Check

Performing the same calculation for \([\hat{\pi}(\mathbf{x}), \hat{\pi}(\mathbf{y})] = 0\) yields the same condition \([\hat{a}_{\mathbf{p}}, \hat{a}_{\mathbf{q}}] = 0\). The mode expansion of \(\hat{\pi}\) contains a factor of \(\sqrt{\omega_{\mathbf{p}}/2}\), but since \(\omega_{\mathbf{p}} > 0\), the conclusion remains unchanged.

M-2. Expression of the Hamiltonian in Terms of Creation and Annihilation Operators

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Strategy

Write each of the three terms in the Hamiltonian — \(\hat{\pi}^2\), \((\nabla\hat{\phi})^2\), and \(m^2\hat{\phi}^2\) — using the mode expansion, use Fourier orthogonality in the \(\mathbf{x}\) integration, and then combine the results.

Detailed Calculation

We use the mode expansion at \(t = 0\). For brevity, we introduce the following shorthand:

\[ \hat{\phi}(\mathbf{x}) = \int \frac{d^3p}{(2\pi)^{3/2}} \frac{1}{\sqrt{2\omega_p}} \left(\hat{a}_p\,e^{i\mathbf{p}\cdot\mathbf{x}} + \hat{a}_p^\dagger\, e^{-i\mathbf{p}\cdot\mathbf{x}}\right) \]
\[ \hat{\pi}(\mathbf{x}) = \int \frac{d^3p}{(2\pi)^{3/2}} (-i)\sqrt{\frac{\omega_p}{2}} \left(\hat{a}_p\,e^{i\mathbf{p}\cdot\mathbf{x}} - \hat{a}_p^\dagger\, e^{-i\mathbf{p}\cdot\mathbf{x}}\right) \]

Step 1: Computing \(\int d^3x\, \frac{1}{2}\hat{\pi}^2\)

\[ \hat{\pi}(\mathbf{x})^2 = \int \frac{d^3p}{(2\pi)^{3/2}} \int \frac{d^3q}{(2\pi)^{3/2}} (-i)^2 \sqrt{\frac{\omega_p}{2}}\sqrt{\frac{\omega_q}{2}} \]
\[ \times \left(\hat{a}_p\,e^{i\mathbf{p}\cdot\mathbf{x}} - \hat{a}_p^\dagger\, e^{-i\mathbf{p}\cdot\mathbf{x}}\right)\left(\hat{a}_q\,e^{i\mathbf{q}\cdot\mathbf{x}} - \hat{a}_q^\dagger\, e^{-i\mathbf{q}\cdot\mathbf{x}}\right) \]

Note that \((-i)^2 = -1\). Expanding yields four terms:

\[ = -\int \frac{d^3p\, d^3q}{(2\pi)^3} \frac{\sqrt{\omega_p \omega_q}}{2} \Big\{ \hat{a}_p \hat{a}_q\, e^{i(\mathbf{p}+\mathbf{q})\cdot\mathbf{x}} - \hat{a}_p \hat{a}_q^\dagger\, e^{i(\mathbf{p}-\mathbf{q})\cdot\mathbf{x}} \]
\[ - \hat{a}_p^\dagger \hat{a}_q\, e^{-i(\mathbf{p}-\mathbf{q})\cdot\mathbf{x}} + \hat{a}_p^\dagger \hat{a}_q^\dagger\, e^{-i(\mathbf{p}+\mathbf{q})\cdot\mathbf{x}} \Big\} \]

Integrating over \(\mathbf{x}\), Fourier orthogonality (4.15) gives: - The \(e^{i(\mathbf{p}+\mathbf{q})\cdot\mathbf{x}}\) term → \(\delta^{(3)}(\mathbf{p}+\mathbf{q})\) (fixes \(\mathbf{q} = -\mathbf{p}\)) - The \(e^{i(\mathbf{p}-\mathbf{q})\cdot\mathbf{x}}\) term → \(\delta^{(3)}(\mathbf{p}-\mathbf{q})\) (fixes \(\mathbf{q} = \mathbf{p}\))

\[ \int d^3x\, \frac{1}{2}\hat{\pi}^2 = -\frac{1}{2}\int \frac{d^3p}{(2\pi)^3} \frac{\sqrt{\omega_p \omega_q}}{2}\Big|_{\text{after }\delta} \times (\text{each term}) \]

Performing the \(\mathbf{q}\) integration explicitly: for the \(\delta^{(3)}(\mathbf{p}+\mathbf{q})\) terms, \(\mathbf{q} \to -\mathbf{p}\) and \(\omega_q \to \omega_p\):

\[ \int d^3x\, \frac{1}{2}\hat{\pi}^2 = -\frac{1}{4}\int d^3p\, \omega_p \Big\{ \hat{a}_p \hat{a}_{-p} - \hat{a}_p \hat{a}_p^\dagger - \hat{a}_p^\dagger \hat{a}_p + \hat{a}_p^\dagger \hat{a}_{-p}^\dagger \Big\} \]

Here the \((2\pi)^3\) cancels between the normalization \((2\pi)^{3/2} \times (2\pi)^{3/2}\) and the delta function, leaving only \(\int d^3p\) (precisely, \(\int \frac{d^3p\, d^3q}{(2\pi)^3} \cdot (2\pi)^3 \delta^{(3)}(\cdots) = \int d^3p\)).

Rearranging:

\[ \int d^3x\, \frac{1}{2}\hat{\pi}^2 = \frac{1}{4}\int d^3p\, \omega_p \left(\hat{a}_p \hat{a}_p^\dagger + \hat{a}_p^\dagger \hat{a}_p - \hat{a}_p \hat{a}_{-p} - \hat{a}_p^\dagger \hat{a}_{-p}^\dagger\right) \tag{I} \]

Step 2: Computing \(\int d^3x\, \frac{1}{2}(\nabla\hat{\phi})^2\)

When \(\nabla\) acts on \(e^{i\mathbf{p}\cdot\mathbf{x}}\), it gives \(i\mathbf{p}\, e^{i\mathbf{p}\cdot\mathbf{x}}\). Therefore:

\[ \nabla\hat{\phi}(\mathbf{x}) = \int \frac{d^3p}{(2\pi)^{3/2}} \frac{i\mathbf{p}}{\sqrt{2\omega_p}} \left(\hat{a}_p\,e^{i\mathbf{p}\cdot\mathbf{x}} - \hat{a}_p^\dagger\, e^{-i\mathbf{p}\cdot\mathbf{x}}\right) \]

Expanding \((\nabla\hat{\phi})^2\) and performing the \(\mathbf{x}\) integration follows the same structure as the \(\hat{\pi}^2\) calculation, but with \(\frac{(-\mathbf{p}\cdot\mathbf{q})}{2\sqrt{\omega_p \omega_q}}\) replacing \(\sqrt{\omega_p \omega_q}/2\), and a sign from \((i)^2 = -1\).

For the \(\delta^{(3)}(\mathbf{p}+\mathbf{q})\) terms, \(\mathbf{q} = -\mathbf{p}\) so \(\mathbf{p}\cdot\mathbf{q} = -|\mathbf{p}|^2\). For the \(\delta^{(3)}(\mathbf{p}-\mathbf{q})\) terms, \(\mathbf{q} = \mathbf{p}\) so \(\mathbf{p}\cdot\mathbf{q} = |\mathbf{p}|^2\).

\[ \int d^3x\, \frac{1}{2}(\nabla\hat{\phi})^2 = \frac{1}{4}\int d^3p\, \frac{|\mathbf{p}|^2}{\omega_p} \left(\hat{a}_p \hat{a}_p^\dagger + \hat{a}_p^\dagger \hat{a}_p + \hat{a}_p \hat{a}_{-p} + \hat{a}_p^\dagger \hat{a}_{-p}^\dagger\right) \tag{II} \]

Step 3: Computing \(\int d^3x\, \frac{1}{2}m^2\hat{\phi}^2\)

The expansion of \(\hat{\phi}^2\) has the same structure as \((\nabla\hat{\phi})^2\) but without the \(\mathbf{p}\) factor and with a \(1/(2\omega_p)\) factor.

\[ \int d^3x\, \frac{1}{2}m^2\hat{\phi}^2 = \frac{m^2}{4}\int d^3p\, \frac{1}{\omega_p} \left(\hat{a}_p \hat{a}_p^\dagger + \hat{a}_p^\dagger \hat{a}_p + \hat{a}_p \hat{a}_{-p} + \hat{a}_p^\dagger \hat{a}_{-p}^\dagger\right) \tag{III} \]

Step 4: Combining the Three Contributions

\(H = (\text{I}) + (\text{II}) + (\text{III})\)

Coefficients of the \(\hat{a}_p \hat{a}_{-p}\) and \(\hat{a}_p^\dagger \hat{a}_{-p}^\dagger\) terms:

Contribution from (I): \(-\frac{1}{4}\omega_p\)

Contribution from (II): \(+\frac{1}{4}\frac{|\mathbf{p}|^2}{\omega_p}\)

Contribution from (III): \(+\frac{1}{4}\frac{m^2}{\omega_p}\)

Total:

\[ -\frac{\omega_p}{4} + \frac{|\mathbf{p}|^2}{4\omega_p} + \frac{m^2}{4\omega_p} = -\frac{\omega_p}{4} + \frac{|\mathbf{p}|^2 + m^2}{4\omega_p} = -\frac{\omega_p}{4} + \frac{\omega_p^2}{4\omega_p} = 0 \]

Here we used \(\omega_p^2 = |\mathbf{p}|^2 + m^2\). The \(\hat{a}_p \hat{a}_{-p}\)-type terms cancel completely.

Coefficients of the \(\hat{a}_p \hat{a}_p^\dagger + \hat{a}_p^\dagger \hat{a}_p\) terms:

Contribution from (I): \(+\frac{1}{4}\omega_p\)

Contribution from (II): \(+\frac{1}{4}\frac{|\mathbf{p}|^2}{\omega_p}\)

Contribution from (III): \(+\frac{1}{4}\frac{m^2}{\omega_p}\)

Total:

\[ \frac{\omega_p}{4} + \frac{|\mathbf{p}|^2 + m^2}{4\omega_p} = \frac{\omega_p}{4} + \frac{\omega_p^2}{4\omega_p} = \frac{\omega_p}{4} + \frac{\omega_p}{4} = \frac{\omega_p}{2} \]

Therefore:

\[ H = \frac{1}{2}\int d^3p\, \omega_p \left(\hat{a}_p \hat{a}_p^\dagger + \hat{a}_p^\dagger \hat{a}_p\right) \]

Step 5: Simplification Using the Commutation Relation

Using the commutation relation (4.13) \([\hat{a}_p, \hat{a}_p^\dagger] = \delta^{(3)}(\mathbf{0})\) (for the case \(\mathbf{p} = \mathbf{q}\)):

\[ \hat{a}_p \hat{a}_p^\dagger = \hat{a}_p^\dagger \hat{a}_p + \delta^{(3)}(\mathbf{0}) \]

Substituting:

\[ H = \frac{1}{2}\int d^3p\, \omega_p \left(2\hat{a}_p^\dagger \hat{a}_p + \delta^{(3)}(\mathbf{0})\right) \]
\[ \boxed{H = \int d^3p\, \omega_p \left(\hat{a}_p^\dagger \hat{a}_p + \frac{1}{2}\delta^{(3)}(\mathbf{0})\right)} \]

A note regarding the \((2\pi)^3\) factor: depending on the normalization convention, \(\int d^3p\) may need to be written as \(\int \frac{d^3p}{(2\pi)^3}\). When using the normalization factor \((2\pi)^{3/2}\) from the mode expansion (4.11), the \(\mathbf{x}\) integration absorbs the \((2\pi)^3\), and the final result is:

\[ \boxed{H = \int \frac{d^3p}{(2\pi)^3}\, \omega_{\mathbf{p}} \left(\hat{a}_{\mathbf{p}}^\dagger \hat{a}_{\mathbf{p}} + \frac{1}{2}\,\delta^{(3)}(\mathbf{0})\right)} \]

Consistency Checks

  • Dimensional analysis: \(\omega_p\) has dimensions of energy, \(\hat{a}_p^\dagger \hat{a}_p\) has dimensions of \([\text{momentum}]^{-3}\) (same as \(\delta^{(3)}(\mathbf{0})\)), and \(\int \frac{d^3p}{(2\pi)^3}\) has dimensions of \([\text{momentum}]^3\). The overall expression has dimensions of energy, which is consistent.
  • Cancellation of \(\hat{a}_p \hat{a}_{-p}\)-type terms: The dispersion relation \(\omega_p^2 = |\mathbf{p}|^2 + m^2\) plays an essential role. This is a consequence of the Klein-Gordon equation and is physically correct.
  • Special case: For a single mode, this reproduces the quantum mechanical harmonic oscillator \(H = \omega(a^\dagger a + 1/2)\).

M-3. Zero-Point Energy and Normal Ordering

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(a) Divergence of the Vacuum Energy

Using the results from S2, we compute the vacuum expectation value.

\[ \langle 0 | H | 0 \rangle = \int \frac{d^3p}{(2\pi)^3}\, \omega_{\mathbf{p}} \left(\langle 0|\hat{a}_{\mathbf{p}}^\dagger \hat{a}_{\mathbf{p}}|0\rangle + \frac{1}{2}\delta^{(3)}(\mathbf{0})\right) \]

Since \(\hat{a}_{\mathbf{p}}|0\rangle = 0\), we have \(\langle 0|\hat{a}_{\mathbf{p}}^\dagger \hat{a}_{\mathbf{p}}|0\rangle = 0\). Therefore

\[ \langle 0 | H | 0 \rangle = \int \frac{d^3p}{(2\pi)^3}\, \frac{\omega_{\mathbf{p}}}{2}\, \delta^{(3)}(\mathbf{0}) \]

This expression contains a double divergence:

  1. Infrared divergence (volume divergence): \(\delta^{(3)}(\mathbf{0})\) is formally infinite. Regularizing with a box volume \(V = L^3\) gives \(\delta^{(3)}(\mathbf{0}) = V/(2\pi)^3\). This reflects the fact that the zero-point energy is spread over all of space, and as an energy density:
\[ \frac{\langle 0 | H | 0 \rangle}{V} = \int \frac{d^3p}{(2\pi)^3}\, \frac{\omega_{\mathbf{p}}}{2} \cdot \frac{1}{(2\pi)^3} \]
  1. Ultraviolet divergence: Since \(\omega_{\mathbf{p}} = \sqrt{|\mathbf{p}|^2 + m^2}\) grows as \(|\mathbf{p}|\) for \(|\mathbf{p}| \to \infty\), the integral \(\int d^3p\, \omega_{\mathbf{p}}\) diverges in the ultraviolet. In spherical coordinates, it diverges as \(\int_0^\Lambda dp\, p^2 \cdot p \sim \Lambda^4\).
\[ \boxed{\langle 0 | H | 0 \rangle = \infty} \]

Physical issue: The vacuum energy density becomes infinite. This is the result of summing the zero-point energy \(\frac{1}{2}\omega_{\mathbf{p}}\) of each mode over all modes, corresponding to the total sum of zero-point energies of infinitely many harmonic oscillators.

(b) Removal of Vacuum Energy by Normal Ordering

Normal ordering \(:\!\hat{O}\!:\) is the operation of placing all creation operators \(\hat{a}^\dagger\) to the left of annihilation operators \(\hat{a}\). In this process, c-number terms arising from commutation relations are discarded.

\[ :\!H\!: \;= \int \frac{d^3p}{(2\pi)^3}\, \omega_{\mathbf{p}}\, \hat{a}_{\mathbf{p}}^\dagger \hat{a}_{\mathbf{p}} \]

Computing the vacuum expectation value:

\[ \langle 0 | :\!H\!: | 0 \rangle = \int \frac{d^3p}{(2\pi)^3}\, \omega_{\mathbf{p}}\, \langle 0|\hat{a}_{\mathbf{p}}^\dagger \hat{a}_{\mathbf{p}}|0\rangle \]

Since \(\hat{a}_{\mathbf{p}}|0\rangle = 0\):

\[ \langle 0|\hat{a}_{\mathbf{p}}^\dagger \hat{a}_{\mathbf{p}}|0\rangle = 0 \]

Therefore

\[ \boxed{\langle 0 | :\!H\!: | 0 \rangle = 0} \]

Normal ordering corresponds to the operation of "redefining the vacuum energy to be zero." Since the absolute value of energy is not physically observable (except in the context of gravity) and only differences in energy carry physical meaning, this operation is justified.

(c) The Casimir Effect and Differences in Zero-Point Energy

While the absolute value of the zero-point energy can be removed by normal ordering, the difference in zero-point energy between different boundary conditions is physically observable.

Explanation of the Casimir effect:

When two parallel conducting plates are placed at a separation \(L\), the allowed modes of the electromagnetic field (or scalar field) between the plates become discretized (only specific wavelengths are permitted due to boundary conditions). Outside the plates, on the other hand, a continuum of modes is allowed.

  • Zero-point energy between the plates: Sum over discrete modes \(E_{\text{in}}(L) = \sum_n \frac{1}{2}\omega_n\)
  • Zero-point energy without plates: Integral over continuous modes \(E_{\text{free}}\)

The difference \(\Delta E = E_{\text{in}}(L) - E_{\text{free}}\) takes a finite value and depends on \(L\). The force derived from this difference

\[ F = -\frac{d(\Delta E)}{dL} \]

is observed as an attractive force. This is the Casimir effect, predicted by Casimir in 1948 and experimentally confirmed by Lamoreaux in 1997.

Key point: The absolute value of the divergent zero-point energy has no physical meaning, but the difference in zero-point energy due to changes in boundary conditions is finite and produces a force that is experimentally measurable.

M-4. Quantization of the Complex Scalar Field and Particles/Antiparticles

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(a) Derivation of the Conjugate Momentum Densities

The Lagrangian density is

\[ \mathcal{L} = (\partial_\mu \phi^*)(\partial^\mu \phi) - m^2 \phi^* \phi \]

We treat \(\phi\) and \(\phi^*\) as independent fields.

Conjugate momentum density for \(\phi\):

\[ \pi_\phi = \frac{\partial \mathcal{L}}{\partial(\partial_0 \phi)} \]

The term in \(\mathcal{L}\) containing \(\partial_0 \phi\) is \((\partial_0 \phi^*)(\partial_0 \phi)\) (the \(\mu = 0\) term, with metric \(\eta^{00} = +1\)). Since \(\phi^*\) is independent of \(\phi\):

\[ \pi_\phi = \frac{\partial}{\partial(\partial_0 \phi)}\left[(\partial_0 \phi^*)(\partial_0 \phi)\right] = \partial_0 \phi^* = \dot{\phi}^* \]
\[ \boxed{\pi_\phi = \dot{\phi}^*} \]

Conjugate momentum density for \(\phi^*\):

\[ \pi_{\phi^*} = \frac{\partial \mathcal{L}}{\partial(\partial_0 \phi^*)} = \partial_0 \phi = \dot{\phi} \]
\[ \boxed{\pi_{\phi^*} = \dot{\phi}} \]

(b) Expression for \(\hat{\phi}^\dagger(\mathbf{x})\)

The given mode expansion is

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

Taking the Hermitian conjugate, using \((\hat{a}_{\mathbf{p}})^\dagger = \hat{a}_{\mathbf{p}}^\dagger\), \((\hat{b}_{\mathbf{p}}^\dagger)^\dagger = \hat{b}_{\mathbf{p}}\), and \((e^{i\mathbf{p}\cdot\mathbf{x}})^* = e^{-i\mathbf{p}\cdot\mathbf{x}}\):

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

Note: Unlike the real scalar field, \(\hat{\phi}^\dagger \neq \hat{\phi}\). The field \(\hat{\phi}\) contains \(\hat{a}\) and \(\hat{b}^\dagger\), while \(\hat{\phi}^\dagger\) contains \(\hat{a}^\dagger\) and \(\hat{b}\).

(c) Physical Difference Between Particles and Antiparticles

The Noether charge corresponding to the \(U(1)\) symmetry \(\phi \to e^{i\alpha}\phi\), \(\phi^* \to e^{-i\alpha}\phi^*\) is

\[ Q = i\int d^3x\, (\phi^* \dot{\phi} - \dot{\phi}^* \phi) \]

After quantization, substituting the mode expansion and performing the calculation (carrying out the \(\mathbf{x}\) integration using Fourier orthogonality, following the same procedure as in S2):

\[ :\!Q\!: = \int \frac{d^3p}{(2\pi)^3}\, \left(\hat{a}_{\mathbf{p}}^\dagger \hat{a}_{\mathbf{p}} - \hat{b}_{\mathbf{p}}^\dagger \hat{b}_{\mathbf{p}}\right) \]

From this result:

  • Particles created by \(\hat{a}_{\mathbf{p}}^\dagger\) carry charge \(Q = +1\) (the sign of the \(\hat{a}^\dagger \hat{a}\) term is \(+\))
  • Particles created by \(\hat{b}_{\mathbf{p}}^\dagger\) carry charge \(Q = -1\) (the sign of the \(\hat{b}^\dagger \hat{b}\) term is \(-\))

Both have the same mass \(m\) (\(\omega_{\mathbf{p}} = \sqrt{|\mathbf{p}|^2 + m^2}\) is common to \(\hat{a}\) and \(\hat{b}\)), but their \(U(1)\) charges have opposite signs.

\[ \boxed{\text{Particles created by } \hat{a}_{\mathbf{p}}^\dagger \text{ and antiparticles created by } \hat{b}_{\mathbf{p}}^\dagger \text{ have the same mass but opposite } U(1) \text{ charges.}} \]

This is the origin of antiparticles. When a complex scalar field is quantized, the existence of the \(U(1)\) symmetry necessarily gives rise to two types of creation operators, and particles and antiparticles automatically emerge. For a real scalar field (\(\hat{\phi}^\dagger = \hat{\phi}\)), we have \(\hat{a} = \hat{b}\), and the particle and antiparticle are identical (self-conjugate particle).

Consistency Check

We verify charge conservation. The fact that \(:\!Q\!:\) commutes with \(:\!H\!:\) (\([:\!Q\!:, :\!H\!:] = 0\)) can be confirmed from \([\hat{a}_p^\dagger \hat{a}_p, \hat{a}_q^\dagger \hat{a}_q] = 0\) and \([\hat{b}_p^\dagger \hat{b}_p, \hat{b}_q^\dagger \hat{b}_q] = 0\). This is consistent with the consequence of Noether's theorem (\(U(1)\) symmetry → charge conservation).

Advanced

A-1. Quantitative Calculation of the 1-Dimensional Casimir Effect

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(a) Allowed Momentum Modes

We impose Dirichlet boundary conditions \(\hat{\phi}(0) = \hat{\phi}(L) = 0\). The mode expansion of the field in one dimension takes the form

\[ \hat{\phi}(x) = \sum_n c_n \sin(p_n x) \]

(\(\cos\) is unsuitable since it does not vanish at \(x = 0\)).

Boundary condition at \(x = 0\): \(\sin(0) = 0\) ✓ (automatically satisfied)

Boundary condition at \(x = L\): From \(\sin(p_n L) = 0\),

\[ p_n L = n\pi \quad (n = 1, 2, 3, \ldots) \]
\[ \boxed{p_n = \frac{n\pi}{L} \quad (n = 1, 2, 3, \ldots)} \]

\(n = 0\) is excluded since \(\sin(0) = 0\) gives the trivial solution (field is zero). \(n < 0\) is not independent since \(\sin(-p_n x) = -\sin(p_n x)\) represents the same mode as \(n > 0\).

(b) Zero-Point Energy via Zeta Function Regularization

Since \(m = 0\), we have \(\omega_n = |p_n| = \frac{n\pi}{L}\). The zero-point energy is

\[ E(L) = \frac{1}{2}\sum_{n=1}^{\infty} \omega_n = \frac{\pi}{2L}\sum_{n=1}^{\infty} n \]

\(\sum_{n=1}^{\infty} n\) is clearly divergent. We apply zeta function regularization.

Procedure: Generalize the divergent sum as

\[ \sum_{n=1}^{\infty} n \;\longrightarrow\; \sum_{n=1}^{\infty} n^{-s} = \zeta(s) \]

and use the analytic continuation to \(s = -1\). The value of the Riemann zeta function at \(s = -1\) is

\[ \zeta(-1) = -\frac{1}{12} \]

Therefore

\[ E(L) = \frac{\pi}{2L} \cdot \zeta(-1) = \frac{\pi}{2L} \cdot \left(-\frac{1}{12}\right) \]
\[ \boxed{E(L) = -\frac{\pi}{24L}} \]

(c) Casimir Force

\[ F = -\frac{dE}{dL} = -\frac{d}{dL}\left(-\frac{\pi}{24L}\right) = -\frac{\pi}{24L^2} \]
\[ \boxed{F = -\frac{\pi}{24L^2}} \]

Since \(F < 0\), the force acts to reduce the separation \(L\) between the walls, i.e., it is attractive.

(d) Physical Justification of the Regularization

The justification for discarding the divergent part and retaining only the finite part can be understood as follows.

What is physically measurable is the difference in zero-point energy due to the presence or absence of boundary conditions.

Consider the difference between the zero-point energy \(E_{\text{in}}(L)\) with walls present and the zero-point energy \(E_{\text{free}}(L)\) without walls (free space), computed over a region of the same length \(L\):

\[ \Delta E = E_{\text{in}}(L) - E_{\text{free}}(L) \]

Both contain ultraviolet divergences, but high-energy (short-wavelength) modes are hardly affected by the presence of the walls (modes with wavelengths much shorter than the wall separation \(L\) behave the same whether or not the walls are present). Therefore, the divergent parts cancel, and only a finite difference remains.

Explicit verification with an exponential cutoff:

Introducing a cutoff \(e^{-\epsilon n}\) (\(\epsilon > 0\)), we compute

\[ E_\epsilon(L) = \frac{\pi}{2L}\sum_{n=1}^{\infty} n\, e^{-\epsilon n} \]

Using \(\sum_{n=1}^{\infty} n\, e^{-\epsilon n} = \frac{e^{-\epsilon}}{(1-e^{-\epsilon})^2}\) and expanding as \(\epsilon \to 0\):

\[ \frac{e^{-\epsilon}}{(1-e^{-\epsilon})^2} = \frac{1}{\epsilon^2} - \frac{1}{12} + O(\epsilon^2) \]

The divergent \(1/\epsilon^2\) term is independent of the presence of walls (the same divergence appears in free space), so it cancels when taking the difference. The remaining finite part \(-1/12\) agrees with the result from zeta function regularization.

In this way, zeta function regularization is an elegant technique that "automatically removes the physically irrelevant divergent part and extracts only the finite part that depends on the boundary conditions."

Consistency Checks

  • Dimensional analysis: In natural units, \(E \sim 1/L\) is correct (in 1+1 dimensions, \([E] = [\text{length}]^{-1}\)).
  • \(L \to \infty\) limit: \(E(L) \to 0\), \(F \to 0\). The Casimir effect vanishes when the walls are infinitely far apart. This is physically correct.
  • Sign: For one-dimensional Dirichlet boundary conditions, the force is attractive. The Casimir effect for the electromagnetic field in 3+1 dimensions (parallel conducting plates) is also attractive, which is qualitatively consistent.

A-2. Lorentz Invariance of Field Commutation Relations and Causality

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(a) Existence of a Simultaneous Frame for Spacelike Separation

Suppose the interval between two spacetime points \(x^\mu\) and \(y^\mu\) is spacelike:

\[ (x - y)^2 = (x^0 - y^0)^2 - |\mathbf{x} - \mathbf{y}|^2 < 0 \]

This means \(|\Delta t| < |\Delta \mathbf{x}|\) (where \(\Delta t = x^0 - y^0\), \(\Delta \mathbf{x} = \mathbf{x} - \mathbf{y}\)).

We choose the \(x^1\) axis along the direction of \(\Delta \mathbf{x}\) (this is always possible by a spatial rotation). Then the problem reduces to 1+1 dimensions, and we need only consider \(\Delta t\) and \(\Delta x^1\).

Applying a Lorentz boost with velocity \(v\) in the \(x^1\) direction, the time difference transforms as

\[ \Delta t' = \gamma(\Delta t - v\,\Delta x^1) \]

Setting \(\Delta t' = 0\) requires

\[ v = \frac{\Delta t}{\Delta x^1} \]

From the spacelike interval condition \(|\Delta t| < |\Delta x^1|\), we have

\[ |v| = \frac{|\Delta t|}{|\Delta x^1|} < 1 \]

This is a physically allowed Lorentz boost velocity (less than the speed of light).

\[ \boxed{\text{For two points with spacelike separation, a Lorentz boost with } |v| < 1 \text{ can make them simultaneous } (\Delta t' = 0).} \]

(b) Proof of Microcausality

(i) For two spacelike-separated points \(x, y\) (\((x-y)^2 < 0\)), as shown in (a), there exists a frame \(S'\) obtained by an appropriate Lorentz transformation in which the two events are simultaneous.

(ii) In frame \(S'\), we have \(x'^0 = y'^0\), so by the equal-time commutation relation (4.5):

\[ [\hat{\phi}(x'), \hat{\phi}(y')] = [\hat{\phi}(t', \mathbf{x}'), \hat{\phi}(t', \mathbf{y}')] = 0 \]

(iii) The commutator \([\hat{\phi}(x), \hat{\phi}(y)]\) of a scalar field is a Lorentz scalar. This can be understood as follows:

A scalar field transforms under a Lorentz transformation \(x \to x' = \Lambda x\) as \(\hat{\phi}'(x') = \hat{\phi}(x)\). Therefore

\[ [\hat{\phi}(x), \hat{\phi}(y)] = [\hat{\phi}'(x'), \hat{\phi}'(y')] \]

That is, the value of the commutator is independent of the coordinate frame.

(iv) Since \([\hat{\phi}(x'), \hat{\phi}(y')] = 0\) in frame \(S'\), it follows that in any frame:

\[ \boxed{[\hat{\phi}(x), \hat{\phi}(y)] = 0 \qquad \text{for } (x-y)^2 < 0} \]

Physical meaning of microcausality:

Field operators at two spacelike-separated points commute. This means that measurements in two spacelike-separated regions cannot influence each other. Superluminal signal transmission is impossible, and the causality of special relativity is preserved in quantum field theory.

(c) Quantization with Fermi Statistics and Violation of Microcausality

If we attempt to quantize the Klein-Gordon field with Fermi statistics (anticommutation relations), we impose, instead of commutation relations:

\[ \{\hat{a}_{\mathbf{p}}, \hat{a}_{\mathbf{q}}^\dagger\} = \delta^{(3)}(\mathbf{p} - \mathbf{q}), \qquad \{\hat{a}_{\mathbf{p}}, \hat{a}_{\mathbf{q}}\} = 0, \qquad \{\hat{a}_{\mathbf{p}}^\dagger, \hat{a}_{\mathbf{q}}^\dagger\} = 0 \]

where \(\{A, B\} = AB + BA\) is the anticommutator.

We compute the anticommutator \(\{\hat{\phi}(\mathbf{x}), \hat{\phi}(\mathbf{y})\}\) of the real scalar field. Substituting the mode expansion, a calculation similar to the proof in (b) yields, with anticommutators \(\{\cdot, \cdot\}\) replacing commutators \([\cdot, \cdot]\):

\[ \{\hat{\phi}(\mathbf{x}), \hat{\phi}(\mathbf{y})\} = \int \frac{d^3p}{(2\pi)^{3/2}} \int \frac{d^3q}{(2\pi)^{3/2}} \frac{1}{\sqrt{2\omega_p}\sqrt{2\omega_q}} \]
\[ \times \Big\{ \{\hat{a}_p, \hat{a}_q\}\, e^{i\mathbf{p}\cdot\mathbf{x}} e^{i\mathbf{q}\cdot\mathbf{y}} + \{\hat{a}_p, \hat{a}_q^\dagger\}\, e^{i\mathbf{p}\cdot\mathbf{x}} e^{-i\mathbf{q}\cdot\mathbf{y}} \]
\[ + \{\hat{a}_p^\dagger, \hat{a}_q\}\, e^{-i\mathbf{p}\cdot\mathbf{x}} e^{i\mathbf{q}\cdot\mathbf{y}} + \{\hat{a}_p^\dagger, \hat{a}_q^\dagger\}\, e^{-i\mathbf{p}\cdot\mathbf{x}} e^{-i\mathbf{q}\cdot\mathbf{y}} \Big\} \]

Since \(\{\hat{a}_p, \hat{a}_q\} = 0\) and \(\{\hat{a}_p^\dagger, \hat{a}_q^\dagger\} = 0\), the 1st and 4th terms vanish. The remaining 2nd and 3rd terms give

\[ \{\hat{a}_p, \hat{a}_q^\dagger\} = \delta^{(3)}(\mathbf{p}-\mathbf{q}), \qquad \{\hat{a}_p^\dagger, \hat{a}_q\} = \delta^{(3)}(\mathbf{p}-\mathbf{q}) \]

(Since the anticommutator is symmetric \(\{A, B\} = \{B, A\}\), we have \(\{\hat{a}_p^\dagger, \hat{a}_q\} = \{\hat{a}_q, \hat{a}_p^\dagger\} = \delta^{(3)}(\mathbf{q}-\mathbf{p}) = \delta^{(3)}(\mathbf{p}-\mathbf{q})\).)

Performing the \(\mathbf{q}\) integration using the delta function:

\[ \{\hat{\phi}(\mathbf{x}), \hat{\phi}(\mathbf{y})\} = \int \frac{d^3p}{(2\pi)^3} \frac{1}{2\omega_p} \left(e^{i\mathbf{p}\cdot(\mathbf{x}-\mathbf{y})} + e^{-i\mathbf{p}\cdot(\mathbf{x}-\mathbf{y})}\right) \]
\[ = \int \frac{d^3p}{(2\pi)^3} \frac{1}{\omega_p} \cos(\mathbf{p}\cdot(\mathbf{x}-\mathbf{y})) \]

The crucial difference: In the case of Bose statistics (commutator), as seen in S1, the 2nd and 3rd terms enter with a relative minus sign in \([\hat{\phi}(\mathbf{x}), \hat{\phi}(\mathbf{y})]\), yielding \(e^{i\mathbf{p}\cdot(\mathbf{x}-\mathbf{y})} - e^{-i\mathbf{p}\cdot(\mathbf{x}-\mathbf{y})}\), which cancels under the substitution \(\mathbf{p} \to -\mathbf{p}\). However, with Fermi statistics (anticommutator), the terms enter with a plus sign, yielding \(e^{i\mathbf{p}\cdot(\mathbf{x}-\mathbf{y})} + e^{-i\mathbf{p}\cdot(\mathbf{x}-\mathbf{y})}\).

This integral is generally nonzero even for \(\mathbf{x} \neq \mathbf{y}\). In fact, setting \(\mathbf{r} = \mathbf{x} - \mathbf{y}\) and computing in spherical coordinates:

\[ \{\hat{\phi}(\mathbf{x}), \hat{\phi}(\mathbf{y})\} = \int_0^\infty \frac{dp}{2\pi^2} \frac{p^2}{\omega_p} \frac{\sin(p|\mathbf{r}|)}{p|\mathbf{r}|} \]

This takes a finite value for \(|\mathbf{r}| \neq 0\) (in the \(m = 0\) case it is proportional to \(\frac{1}{2\pi^2 |\mathbf{r}|^2}\)).

Therefore, even for spacelike-separated points (\(\mathbf{x} \neq \mathbf{y}\), at equal times):

\[ \{\hat{\phi}(\mathbf{x}), \hat{\phi}(\mathbf{y})\} \neq 0 \]

and microcausality is violated.

\[ \boxed{\text{Quantizing an integer-spin field (scalar field) with Fermi statistics violates microcausality.}} \]

Connection to the spin-statistics theorem:

This result shows that integer-spin fields must be quantized with Bose statistics. Conversely, attempting to quantize half-integer-spin fields (such as the Dirac field) with Bose statistics leads to a different problem: the energy is not bounded from below (the Hamiltonian is not positive definite).

Combining these results: - Integer spin → Bose statistics (commutation relations): microcausality is preserved - Half-integer spin → Fermi statistics (anticommutation relations): energy is positive definite

This is the essence of the spin-statistics theorem, one of the fundamental consequences of relativistic quantum field theory.

Consistency Checks

  • The result of (a) is consistent with the fact that for timelike separation (\((x-y)^2 > 0\)), we get \(|v| > 1\), making it impossible to achieve simultaneity via a Lorentz boost. On the light cone (\((x-y)^2 = 0\)), \(|v| = 1\) is the boundary case.
  • The microcausality in (b) states that the Lorentz-invariant function \(\Delta(x-y) = [\hat{\phi}(x), \hat{\phi}(y)]\) vanishes in the spacelike region. This function is called the Pauli-Jordan function and is expressed as \(\Delta(x-y) = \frac{1}{(2\pi)^3}\int d^3p\, \frac{1}{2\omega_p}(e^{-ip\cdot(x-y)} - e^{ip\cdot(x-y)})\). The integrand changes sign under \(p \to -p\) (it is an odd function), which guarantees vanishing in the spacelike region.
  • In the anticommutator case of (c), the integrand becomes an even function, so the cancellation does not occur and the result is nonzero. This contrast is the essential point.