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Ch. 7 Problems

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Basic

Medium

Advanced

Basic

B-1. Mass Dimension of Interaction Terms

In \(d\)-dimensional spacetime (for the case \(d = 4\)), in natural units \(\hbar = c = 1\), we have \([\mathcal{L}] = d = 4\) and \([\phi] = 1\). For each of the following interaction terms, determine the mass dimension of the coupling constant.

(a) \(\mathcal{L}_{\text{int}} = -g\,\phi^3\)

(b) \(\mathcal{L}_{\text{int}} = -\frac{\lambda}{4!}\,\phi^4\)

(c) \(\mathcal{L}_{\text{int}} = -\frac{\kappa}{5!}\,\phi^5\)

(d) \(\mathcal{L}_{\text{int}} = -\frac{\eta}{6!}\,\phi^6\)

Hint

From \([\mathcal{L}_{\text{int}}] = 4\), use \([\text{coupling constant}] + n[\phi] = 4\). Since \([\phi] = 1\), we get \([\text{coupling constant}] = 4 - n\). Theories with coupling constants of negative mass dimension are called non-renormalizable.

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B-2. Time Derivative of Operators in the Interaction Picture

Given that the field operator in the interaction picture is defined as \(\hat{\phi}_I(t, \mathbf{x}) = e^{i\hat{H}_0 t}\,\hat{\phi}_S(\mathbf{x})\,e^{-i\hat{H}_0 t}\), show by direct calculation of Eq. (7.3) that

\[ i\frac{\partial}{\partial t}\hat{\phi}_I(t, \mathbf{x}) = [\hat{\phi}_I(t, \mathbf{x}),\, \hat{H}_0] \]

holds. Write out the intermediate differentiation steps without omission.

Hint

Use \(\frac{d}{dt}(e^{i\hat{H}_0 t}) = i\hat{H}_0\,e^{i\hat{H}_0 t}\) and \(\frac{d}{dt}(e^{-i\hat{H}_0 t}) = -i\hat{H}_0\,e^{-i\hat{H}_0 t}\), and apply the product rule for differentiation. Finally, use the definition \(\hat{O}_I = e^{i\hat{H}_0 t}\hat{O}_S e^{-i\hat{H}_0 t}\) to simplify.

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B-3. Explicit Calculation of the Time-Ordered Product

For the scalar field \(\hat{\phi}_I(x)\), use the definition of the time-ordered product (7.14) to calculate the following (assume \(x^0 > y^0\)).

(a) Write \(T[\hat{\phi}_I(x)\,\hat{\phi}_I(y)]\) in terms of the ordinary product.

(b) What happens in the case \(x^0 < y^0\)?

(c) To verify that the vacuum expectation value of the two-point time-ordered product \(\langle 0|T[\hat{\phi}_I(x)\,\hat{\phi}_I(y)]|0\rangle\) equals the Feynman propagator

\[ D_F(x-y) = \int \frac{d^4p}{(2\pi)^4}\,\frac{i}{p^2 - m^2 + i\epsilon} \]

substitute the mode expansion (7.4) of \(\hat{\phi}_I\) and use \(\langle 0|\hat{a}_{\mathbf{p}}\hat{a}_{\mathbf{q}}^\dagger|0\rangle = (2\pi)^3\delta^3(\mathbf{p} - \mathbf{q})\) to write down the expression for the case \(x^0 > y^0\).

Hint

For (a) and (b), simply rearrange according to the definition. For (c), decompose \(\hat{\phi}_I = \hat{\phi}^{(+)} + \hat{\phi}^{(-)}\) (annihilation and creation parts), and use \(\hat{a}|0\rangle = 0\) for the vacuum \(|0\rangle\). When \(x^0 > y^0\), the only nonzero contribution is \(\langle 0|\hat{\phi}^{(+)}(x)\hat{\phi}^{(-)}(y)|0\rangle\).

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B-4. Operator Structure of \(\hat{\phi}^4\)

When we schematically write \(\hat{\phi}_I \sim \hat{a} + \hat{a}^\dagger\), expanding \(\hat{\phi}_I^4\) yields combinations of creation and annihilation operators. For each of the following terms, state "by how much does it change the particle number?"

(a) \(\hat{a}^\dagger\hat{a}^\dagger\hat{a}^\dagger\hat{a}^\dagger\)

(b) \(\hat{a}^\dagger\hat{a}^\dagger\hat{a}^\dagger\hat{a}\)

(c) \(\hat{a}^\dagger\hat{a}^\dagger\hat{a}\,\hat{a}\)

(d) \(\hat{a}^\dagger\hat{a}\,\hat{a}\,\hat{a}\)

(e) \(\hat{a}\,\hat{a}\,\hat{a}\,\hat{a}\)

Hint

\(\hat{a}^\dagger\) increases the particle number by \(+1\), and \(\hat{a}\) decreases the particle number by \(-1\). The change in particle number is the difference between the number of \(\hat{a}^\dagger\)'s and the number of \(\hat{a}\)'s in each term.

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B-5. First-Order Term of the Dyson Series

In \(\phi^4\) theory, let \(\hat{H}_I(t) = \frac{\lambda}{4!}\int d^3x\,\hat{\phi}_I^4(t, \mathbf{x})\). Write down explicitly the first-order term of the Dyson series (7.16). That is, rewrite

\[ \hat{S}^{(1)} = (-i)\int_{-\infty}^{+\infty} dt_1\,\hat{H}_I(t_1) \]

in the form of a four-dimensional integral \(\int d^4x\), expressing it as a Lorentz-invariant expression.

Hint

Combine as \(\int dt_1\,\hat{H}_I(t_1) = \frac{\lambda}{4!}\int dt_1\int d^3x\,\hat{\phi}_I^4(x) = \frac{\lambda}{4!}\int d^4x\,\hat{\phi}_I^4(x)\). Pay attention to the sign: since \(\hat{H}_I = -\int d^3x\,\mathcal{L}_{\text{int}}\) with \(\mathcal{L}_{\text{int}} = -\frac{\lambda}{4!}\phi^4\), we have \(\hat{H}_I = +\frac{\lambda}{4!}\int d^3x\,\phi^4\).

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B-6. Symmetry of the Time-Ordered Product

Consider the time-ordered product \(T[\hat{H}_I(t_1)\hat{H}_I(t_2)\hat{H}_I(t_3)]\) for three times \(t_1, t_2, t_3\).

(a) When \(t_1 > t_2 > t_3\), write \(T[\hat{H}_I(t_1)\hat{H}_I(t_2)\hat{H}_I(t_3)]\) as an ordinary product.

(b) When \(t_3 > t_1 > t_2\), write it similarly.

(c) How many permutations of the three time variables are there? Confirm that this is the origin of the \(1/3!\) factor in the third-order term of the Dyson series.

Hint

The time-ordered product follows the rule of "placing the operator at the latest time furthest to the left." For bosons, no sign is introduced upon interchange. There are \(3! = 6\) permutations of three variables.

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B-7. Decomposition of \(\hat{S} = \mathbb{1} + i\hat{T}\)

When the S-operator is written as \(\hat{S} = \mathbb{1} + i\hat{T}\), show the following.

(a) From the unitarity of the S-matrix \(\hat{S}^\dagger\hat{S} = \mathbb{1}\), derive the condition that \(\hat{T}\) must satisfy (the starting point of the optical theorem).

(b) When the initial and final states are the same \(|i\rangle = |f\rangle\), give the lowest-order (\(\lambda^0\)) value of \(\langle i|\hat{S}|i\rangle\).

Hint

(a) Expand \(\hat{S}^\dagger\hat{S} = (\mathbb{1} - i\hat{T}^\dagger)(\mathbb{1} + i\hat{T}) = \mathbb{1}\). (b) Since \(\hat{S} = \mathbb{1} + O(\lambda)\), the lowest order is the contribution from \(\mathbb{1}\).

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B-8. Picture Transformation of the Interaction Hamiltonian

Given that the interaction Hamiltonian in the Schrödinger picture is \(\hat{H}' = \frac{\lambda}{4!}\int d^3x\,\hat{\phi}_S^4(\mathbf{x})\), show that \(\hat{H}_I(t)\) in the interaction picture (Eq. (7.7)) becomes

\[ \hat{H}_I(t) = \frac{\lambda}{4!}\int d^3x\,\hat{\phi}_I^4(t, \mathbf{x}) \]

using the definition \(\hat{\phi}_I(t, \mathbf{x}) = e^{i\hat{H}_0 t}\hat{\phi}_S(\mathbf{x})e^{-i\hat{H}_0 t}\).

Hint

Insert the identity operator \(e^{-i\hat{H}_0 t}e^{i\hat{H}_0 t} = \mathbb{1}\) three times between each pair of \(\hat{\phi}_S\) factors in \(e^{i\hat{H}_0 t}\hat{\phi}_S^4(\mathbf{x})e^{-i\hat{H}_0 t}\).

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Medium

M-1. Derivation of the Equation of Motion for States in the Interaction Picture

Starting from the definition in Eq. (7.5), \(|\psi_I(t)\rangle = e^{i\hat{H}_0 t}|\psi(t)\rangle_S\), carefully derive Eq. (7.6) following the steps below.

(a) Differentiate \(|\psi_I(t)\rangle\) with respect to \(t\) and substitute the Schrödinger equation \(i\frac{d}{dt}|\psi(t)\rangle_S = (\hat{H}_0 + \hat{H}')|\psi(t)\rangle_S\).

(b) Verify that the \(\hat{H}_0\) terms cancel, and show that \(\hat{H}_I(t) = e^{i\hat{H}_0 t}\hat{H}'e^{-i\hat{H}_0 t}\) naturally emerges from the remaining terms.

(c) If \([\hat{H}_0, \hat{H}'] = 0\) holds, what does \(\hat{H}_I(t)\) become? Explain physically why perturbation theory becomes unnecessary in this case.

Hint

(c) If \([\hat{H}_0, \hat{H}'] = 0\), then \(\hat{H}_I(t) = \hat{H}'\) (time-independent). Furthermore, since \(\hat{H}_0\) and \(\hat{H}'\) can be simultaneously diagonalized, an exact solution can be obtained. However, in quantum field theory, \([\hat{H}_0, \hat{H}'] \neq 0\) is usually the case.

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M-2. Second-Order Term of the Dyson Series and Time-Ordered Product

For the second-order term of the Dyson series

\[ \hat{U}_I^{(2)} = (-i)^2\int_{t_0}^{t}dt_1\int_{t_0}^{t_1}dt_2\,\hat{H}_I(t_1)\hat{H}_I(t_2) \]

show the following.

(a) Exchange the integration variables \(t_1 \leftrightarrow t_2\) and confirm that \(\hat{H}_I(t_2)\hat{H}_I(t_1)\) appears in the region \(t_0 \le t_1 \le t_2 \le t\).

(b) Show that adding the two contributions above gives

\[ \hat{U}_I^{(2)} = \frac{(-i)^2}{2!}\int_{t_0}^{t}dt_1\int_{t_0}^{t}dt_2\,T[\hat{H}_I(t_1)\hat{H}_I(t_2)] \]

(c) Generalize this result to \(n\)-th order to obtain the form of the Dyson series (7.16). Describe the logic of the generalization (a rigorous proof is not required).

Hint

(a) The original integration region corresponds to the triangle \(t_0 \le t_2 \le t_1 \le t\) in the \((t_1, t_2)\) plane. Under \(t_1 \leftrightarrow t_2\), it maps to the other triangle \(t_0 \le t_1 \le t_2 \le t\). (b) Combining the two triangles gives the entire square \([t_0, t]^2\). The time-ordered product ensures the correct ordering in both triangles.

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M-3. 2→2 Scattering Amplitude in \(\phi^4\) Theory (Leading Order)

In \(\phi^4\) theory, with initial state \(|i\rangle = |\mathbf{p}_1, \mathbf{p}_2\rangle = \hat{a}_{\mathbf{p}_1}^\dagger\hat{a}_{\mathbf{p}_2}^\dagger|0\rangle\) and final state \(|f\rangle = |\mathbf{p}_3, \mathbf{p}_4\rangle = \hat{a}_{\mathbf{p}_3}^\dagger\hat{a}_{\mathbf{p}_4}^\dagger|0\rangle\), compute the lowest-order (first order in \(\lambda\)) contribution to the S-matrix

\[ \langle f|\hat{S}^{(1)}|i\rangle = \langle \mathbf{p}_3, \mathbf{p}_4|\left(\frac{-i\lambda}{4!}\int d^4x\,\hat{\phi}_I^4(x)\right)|\mathbf{p}_1, \mathbf{p}_2\rangle \]

following the steps below.

(a) Substitute the mode expansion (7.4) of \(\hat{\phi}_I(x)\) and extract from \(\hat{\phi}_I^4(x)\) the terms that annihilate 2 particles and create 2 particles.

(b) Using the commutation relations of the creation and annihilation operators, show that the result is proportional to

\[ \langle f|\hat{S}^{(1)}|i\rangle = -i\lambda\,(2\pi)^4\delta^4(p_1 + p_2 - p_3 - p_4)\,\frac{1}{\sqrt{2\omega_{\mathbf{p}_1}}}\frac{1}{\sqrt{2\omega_{\mathbf{p}_2}}}\frac{1}{\sqrt{2\omega_{\mathbf{p}_3}}}\frac{1}{\sqrt{2\omega_{\mathbf{p}_4}}} \]

(in a form including external leg factors). Explicitly show how the \(4!\) factor cancels.

(c) From this result, read off that the invariant scattering amplitude at leading order is \(\mathcal{M} = -\lambda\).

Hint

(a) When expanding \(\hat{\phi}_I^4\), out of the four fields, two act as annihilation operators to destroy the particles in the initial state, and the remaining two act as creation operators to produce the particles in the final state. (b) The number of ways to choose which of the four fields annihilates \(\mathbf{p}_1\), which annihilates \(\mathbf{p}_2\), and so on, is \(4!/(2!\cdot 2!) \times 2! \times 2! = 4!\), which cancels the \(1/4!\). (c) In the convention of the LSZ reduction formula, the part with the external leg factors \(1/\sqrt{2\omega}\) removed gives \(i\mathcal{M}\).

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M-4. Normal Ordering and Wick's Theorem (Two-Field Case)

Decompose the free scalar field \(\hat{\phi}_I(x)\) into its positive-frequency part \(\hat{\phi}^{(+)}(x)\) (containing annihilation operators) and negative-frequency part \(\hat{\phi}^{(-)}(x)\) (containing creation operators), so that \(\hat{\phi}_I = \hat{\phi}^{(+)} + \hat{\phi}^{(-)}\).

(a) State the definition of normal ordering \(:\hat{\phi}_I(x)\hat{\phi}_I(y):\) and write it out explicitly using \(\hat{\phi}^{(\pm)}\).

(b) Define the contraction as

\[ \underbrace{\hat{\phi}_I(x)\hat{\phi}_I(y)} \equiv T[\hat{\phi}_I(x)\hat{\phi}_I(y)] - :\hat{\phi}_I(x)\hat{\phi}_I(y): \]

Show that this is a c-number (not an operator), and verify that it equals the Feynman propagator \(D_F(x-y)\).

(c) From the above, derive Wick's theorem for two fields:

\[ T[\hat{\phi}_I(x)\hat{\phi}_I(y)] = :\hat{\phi}_I(x)\hat{\phi}_I(y): + D_F(x-y) \]
Hint

(a) Normal ordering is the operation of placing creation operators to the left and annihilation operators to the right. (b) Split \(T[\hat{\phi}(x)\hat{\phi}(y)]\) into the cases \(x^0 > y^0\) and \(y^0 > x^0\), and use the fact that the commutator \([\hat{\phi}^{(+)}(x), \hat{\phi}^{(-)}(y)]\) is a c-number.

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M-5. Unitarity of the S-Matrix and Probability Conservation

Verify that the S operator is unitary, \(\hat{S}^\dagger\hat{S} = \hat{S}\hat{S}^\dagger = \mathbb{1}\), by following the steps below.

(a) From the definition (7.9) of \(\hat{U}_I(t, t_0)\), show that \(\hat{U}_I^\dagger(t, t_0) = \hat{U}_I(t_0, t)\) (time evolution in the reverse direction) holds.

(b) Show that \(\hat{U}_I(t, t_0)\hat{U}_I(t_0, t) = \mathbb{1}\), and obtain \(\hat{S}^\dagger\hat{S} = \mathbb{1}\) in the limit \(t \to +\infty\), \(t_0 \to -\infty\).

(c) What does unitarity physically mean? Insert the completeness relation \(\sum_f |f\rangle\langle f| = \mathbb{1}\) and show that the sum of probabilities equals 1.

Hint

(a) Taking the Hermitian conjugate of the differential equation (7.9) satisfied by \(\hat{U}_I(t,t_0)\) gives \(-i\frac{\partial}{\partial t}\hat{U}_I^\dagger = \hat{U}_I^\dagger\hat{H}_I(t)\). This coincides with the equation satisfied by \(\hat{U}_I(t_0, t)\) (using \(\hat{H}_I = \hat{H}_I^\dagger\)).

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Advanced

A-1. Extension to Yukawa Theory and Application of Wick's Theorem

Consider the Yukawa interaction between a scalar field \(\phi\) and a Dirac field \(\psi\):

\[ \mathcal{L}_{\text{int}} = -g\,\bar{\psi}\psi\phi \]

where \(g\) is the Yukawa coupling constant.

(a) Determine the mass dimension of the coupling constant \(g\) (in 4-dimensional spacetime, with \([\psi] = 3/2\), \([\phi] = 1\)).

(b) Write down the first-order term \(\hat{S}^{(1)}\) of the S-matrix in this theory.

(c) At what order in \(\hat{S}\) does the lowest-order contribution to fermion-fermion scattering \(\psi + \psi \to \psi + \psi\) first appear? Explain the reason based on the operator structure of \(\hat{S}^{(1)}\).

(d) When Wick's theorem is applied to the second-order contribution \(\hat{S}^{(2)}\), the scalar field contraction \(D_F(x-y)\) appears. Provide a physical explanation of why this represents the force due to "\(\phi\) exchange"—that is, the field-theoretic origin of the Yukawa potential.

Hint

(a) Determine it from \([g] + [\bar{\psi}] + [\psi] + [\phi] = 4\). (c) \(\hat{S}^{(1)} \propto \int d^4x\,\bar{\psi}\psi\phi\) annihilates 1 \(\psi\) and creates 1 \(\psi\), and creates or annihilates 1 \(\phi\). For the process of 2 fermions → 2 fermions, \(\phi\) must propagate as an internal line, so \(\hat{S}^{(2)}\) is required. (d) Recall the relationship between the potential in the Born approximation and the propagator from perturbation theory in quantum mechanics Ch. 13.

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A-2. Adiabatic Hypothesis and the Gell-Mann–Low Theorem

In this chapter, we implicitly assumed that the interaction "vanishes" as \(t \to \pm\infty\). To make this rigorous, we introduce adiabatic switching:

\[ \hat{H}_I(t) \to \hat{H}_I(t)\,e^{-\epsilon|t|} \]

(where \(\epsilon > 0\) is a small parameter, with \(\epsilon \to 0^+\) taken at the end).

(a) Confirm that under this prescription, \(\hat{H}_I(t) \to 0\) as \(t \to \pm\infty\).

(b) The Gell-Mann–Low theorem asserts that "the vacuum state \(|\Omega\rangle\) of the interacting theory is adiabatically generated from the free-theory vacuum \(|0\rangle\)." Formally, this can be written as

\[ |\Omega\rangle = \lim_{\epsilon \to 0^+} \frac{\hat{U}_I(0, -\infty)|0\rangle}{\langle 0|\hat{U}_I(0, -\infty)|0\rangle} \]

Discuss why this holds under the adiabatic hypothesis. Explain the role of the denominator (phase removal and normalization).

(c) Explain why this theorem provides the justification for "ignoring the contribution of vacuum bubbles" in scattering amplitude calculations. Show conceptually that the vacuum bubble diagram contributions appearing at each order of the Dyson series are cancelled by the denominator \(\langle 0|\hat{S}|0\rangle\).

Hint

(b) At \(t = -\infty\), the state is \(|0\rangle\) (the free vacuum). Evolving to time 0 with \(\hat{U}_I(0, -\infty)\), in the limit \(\epsilon \to 0\), one arrives at the interacting vacuum \(|\Omega\rangle\). The denominator corrects for the phase and magnitude of the overlap between \(|0\rangle\) and \(|\Omega\rangle\). (c) Use the linked-cluster theorem. The vacuum bubbles appearing in the numerator of S-matrix elements factorize exponentially and cancel with the denominator.

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