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Ch. 7 Solutions

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Basic

B-1. Mass Dimension of Interaction Terms

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Problem:

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

Solution Strategy:

In natural units in 4-dimensional spacetime, \([\mathcal{L}] = 4\) and \([\phi] = 1\). We determine the mass dimension of the coupling constants from the condition that the interaction term \(\mathcal{L}_{\text{int}}\) has mass dimension 4.

Calculation:

From \([\text{coupling constant}] + n[\phi] = 4\), we get \([\text{coupling constant}] = 4 - n\).

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

\[ [g] + 3[\phi] = 4 \implies [g] = 4 - 3 = 1 \]

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

\[ [\lambda] + 4[\phi] = 4 \implies [\lambda] = 4 - 4 = 0 \]

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

\[ [\kappa] + 5[\phi] = 4 \implies [\kappa] = 4 - 5 = -1 \]

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

\[ [\eta] + 6[\phi] = 4 \implies [\eta] = 4 - 6 = -2 \]

Final Answer:

Interaction term Mass dimension of coupling constant Renormalizability
\(g\,\phi^3\) \([g] = 1\) super-renormalizable
\(\frac{\lambda}{4!}\,\phi^4\) \([\lambda] = 0\) marginal (renormalizable)
\(\frac{\kappa}{5!}\,\phi^5\) \([\kappa] = -1\) non-renormalizable
\(\frac{\eta}{6!}\,\phi^6\) \([\eta] = -2\) non-renormalizable

Verification:

Theories with coupling constants of negative mass dimension develop new divergent structures at each increasing order of loop corrections and cannot be renormalized with a finite number of counterterms (non-renormalizable). The fact that the \(\phi^4\) theory with \([\lambda] = 0\) is renormalizable is consistent with the standard result in 4-dimensional quantum field theory.

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

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Problem:

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.

Solution Strategy:

Differentiate \(\hat{\phi}_I(t, \mathbf{x}) = e^{i\hat{H}_0 t}\,\hat{\phi}_S(\mathbf{x})\,e^{-i\hat{H}_0 t}\) with respect to \(t\), applying the product rule.

Detailed Calculation:

Differentiate \(\hat{\phi}_I(t, \mathbf{x})\) with respect to \(t\). Since it is a product of three factors, we apply the product rule:

\[ \frac{\partial}{\partial t}\hat{\phi}_I(t, \mathbf{x}) = \frac{\partial}{\partial t}\left(e^{i\hat{H}_0 t}\right)\,\hat{\phi}_S(\mathbf{x})\,e^{-i\hat{H}_0 t} + e^{i\hat{H}_0 t}\,\hat{\phi}_S(\mathbf{x})\,\frac{\partial}{\partial t}\left(e^{-i\hat{H}_0 t}\right) \]

(Since \(\hat{\phi}_S(\mathbf{x})\) has no time dependence, the middle term does not contribute.)

The derivatives of each exponential are:

\[ \frac{\partial}{\partial t}\left(e^{i\hat{H}_0 t}\right) = i\hat{H}_0\,e^{i\hat{H}_0 t} \]
\[ \frac{\partial}{\partial t}\left(e^{-i\hat{H}_0 t}\right) = -i\hat{H}_0\,e^{-i\hat{H}_0 t} \]

Substituting:

\[ \frac{\partial}{\partial t}\hat{\phi}_I = i\hat{H}_0\,e^{i\hat{H}_0 t}\,\hat{\phi}_S\,e^{-i\hat{H}_0 t} + e^{i\hat{H}_0 t}\,\hat{\phi}_S\,(-i\hat{H}_0)\,e^{-i\hat{H}_0 t} \]

The first term can be recognized as \(i\hat{H}_0\,\hat{\phi}_I(t, \mathbf{x})\), and the second term as \(-i\,\hat{\phi}_I(t, \mathbf{x})\,\hat{H}_0\) (since \(\hat{H}_0 = e^{i\hat{H}_0 t}\hat{H}_0 e^{-i\hat{H}_0 t}\) because \(\hat{H}_0\) commutes with itself).

Therefore:

\[ \frac{\partial}{\partial t}\hat{\phi}_I = i\hat{H}_0\,\hat{\phi}_I - i\,\hat{\phi}_I\,\hat{H}_0 = i[\hat{H}_0,\, \hat{\phi}_I] \]

Multiplying both sides by \(i\):

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

Verification:

Since the commutator on the right-hand side satisfies \([\hat{\phi}_I, \hat{H}_0] = \hat{\phi}_I \hat{H}_0 - \hat{H}_0 \hat{\phi}_I = -[\hat{H}_0, \hat{\phi}_I]\), we have \(i\frac{\partial}{\partial t}\hat{\phi}_I = -i[\hat{H}_0, \hat{\phi}_I] \cdot (-1) = [\hat{\phi}_I, \hat{H}_0]\). We have confirmed that the signs are consistent. Furthermore, this result corresponds to the Heisenberg picture equation of motion \(i\frac{d}{dt}\hat{O}_H = [\hat{O}_H, \hat{H}]\) with the replacement \(\hat{H} \to \hat{H}_0\), which is consistent with the property that operators in the interaction picture evolve under the free Hamiltonian.

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

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Problem:

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 \(x^0 > y^0\).

(a) Case \(x^0 > y^0\):

From the definition of the time-ordered product (7.14), the operator at later time is placed to the left:

\[ T[\hat{\phi}_I(x)\,\hat{\phi}_I(y)] = \hat{\phi}_I(x)\,\hat{\phi}_I(y) \quad (x^0 > y^0) \]

(b) Case \(x^0 < y^0\):

\[ T[\hat{\phi}_I(x)\,\hat{\phi}_I(y)] = \hat{\phi}_I(y)\,\hat{\phi}_I(x) \quad (x^0 < y^0) \]

(c) Vacuum expectation value and the Feynman propagator:

We decompose \(\hat{\phi}_I(x)\) into its positive-frequency part (containing annihilation operators) and negative-frequency part (containing creation operators):

\[ \hat{\phi}_I(x) = \hat{\phi}^{(+)}(x) + \hat{\phi}^{(-)}(x) \]

where

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

For \(x^0 > y^0\):

\[ \langle 0|T[\hat{\phi}_I(x)\,\hat{\phi}_I(y)]|0\rangle = \langle 0|\hat{\phi}_I(x)\,\hat{\phi}_I(y)|0\rangle \]

Since \(\hat{a}_{\mathbf{p}}|0\rangle = 0\), terms where \(\hat{\phi}^{(+)}\) appears at the rightmost position or \(\hat{\phi}^{(-)}\) appears at the leftmost position vanish. The non-zero contribution is:

\[ \langle 0|\hat{\phi}^{(+)}(x)\,\hat{\phi}^{(-)}(y)|0\rangle \]

Substituting the mode expansions:

\[ = \int \frac{d^3p}{(2\pi)^3}\,\frac{1}{\sqrt{2\omega_{\mathbf{p}}}}\,e^{-ip\cdot x} \int \frac{d^3q}{(2\pi)^3}\,\frac{1}{\sqrt{2\omega_{\mathbf{q}}}}\,e^{+iq\cdot y}\,\langle 0|\hat{a}_{\mathbf{p}}\,\hat{a}_{\mathbf{q}}^\dagger|0\rangle \]

Using \(\langle 0|\hat{a}_{\mathbf{p}}\,\hat{a}_{\mathbf{q}}^\dagger|0\rangle = (2\pi)^3\delta^3(\mathbf{p} - \mathbf{q})\):

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

Here \(p\cdot(x-y) = \omega_{\mathbf{p}}(x^0 - y^0) - \mathbf{p}\cdot(\mathbf{x} - \mathbf{y})\) with \(p^0 = \omega_{\mathbf{p}}\).

\[ \boxed{\langle 0|T[\hat{\phi}_I(x)\,\hat{\phi}_I(y)]|0\rangle \Big|_{x^0 > y^0} = \int \frac{d^3p}{(2\pi)^3}\,\frac{1}{2\omega_{\mathbf{p}}}\,e^{-i\omega_{\mathbf{p}}(x^0-y^0)+i\mathbf{p}\cdot(\mathbf{x}-\mathbf{y})}} \]

Verification:

This expression agrees with the result obtained by performing the \(p^0\) integration of the Feynman propagator using the residue theorem for the case \(x^0 > y^0\). Indeed, when evaluating the \(p^0\) integral of \(D_F(x-y) = \int \frac{d^4p}{(2\pi)^4}\,\frac{i}{p^2 - m^2 + i\epsilon}\) by picking up the pole at \(p^0 = \omega_{\mathbf{p}} - i\epsilon\) in the lower half-plane, one obtains the above expression for \(x^0 > y^0\).

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

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Problem:

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

Solution Strategy:

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

Final Answer:

(a) \(\hat{a}^\dagger\hat{a}^\dagger\hat{a}^\dagger\hat{a}^\dagger\): \(4 - 0 = +4\) (creates 4 particles)

(b) \(\hat{a}^\dagger\hat{a}^\dagger\hat{a}^\dagger\hat{a}\): \(3 - 1 = +2\) (increases particle number by 2)

(c) \(\hat{a}^\dagger\hat{a}^\dagger\hat{a}\,\hat{a}\): \(2 - 2 = 0\) (particle number unchanged, 2→2 scattering)

(d) \(\hat{a}^\dagger\hat{a}\,\hat{a}\,\hat{a}\): \(1 - 3 = -2\) (decreases particle number by 2)

(e) \(\hat{a}\,\hat{a}\,\hat{a}\,\hat{a}\): \(0 - 4 = -4\) (annihilates 4 particles)

Verification:

Combining all cases, the expansion of \(\hat{\phi}^4\) contains terms that change the particle number by \(+4, +2, 0, -2, -4\). This corresponds to the binomial expansion of \((\hat{a} + \hat{a}^\dagger)^4\) with terms containing \(k\) factors of \(\hat{a}^\dagger\) and \(4-k\) factors of \(\hat{a}\) (\(k = 0, 1, 2, 3, 4\)), where the particle number change \(2k - 4\) gives \(-4, -2, 0, +2, +4\), which is consistent.

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

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Problem:

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.

Solution Strategy:

Substitute \(\hat{H}_I(t) = \frac{\lambda}{4!}\int d^3x\,\hat{\phi}_I^4(t, \mathbf{x})\) into the first-order term of the Dyson series and combine into a four-dimensional integral.

Calculation:

\[ \hat{S}^{(1)} = (-i)\int_{-\infty}^{+\infty} dt_1\,\hat{H}_I(t_1) = (-i)\int_{-\infty}^{+\infty} dt_1\,\frac{\lambda}{4!}\int d^3x\,\hat{\phi}_I^4(t_1, \mathbf{x}) \]

Combining \(dt_1\,d^3x = d^4x\):

\[ \boxed{\hat{S}^{(1)} = \frac{-i\lambda}{4!}\int d^4x\,\hat{\phi}_I^4(x)} \]

Sign Verification:

From \(\mathcal{L}_{\text{int}} = -\frac{\lambda}{4!}\phi^4\), we have \(\hat{H}' = -\int d^3x\,\mathcal{L}_{\text{int}} = +\frac{\lambda}{4!}\int d^3x\,\phi^4\). Therefore \(\hat{H}_I(t) = +\frac{\lambda}{4!}\int d^3x\,\hat{\phi}_I^4\). Thus \(\hat{S}^{(1)} = (-i)\int dt\,\hat{H}_I = \frac{-i\lambda}{4!}\int d^4x\,\hat{\phi}_I^4\).

Alternatively, this can be written as \(\hat{S}^{(1)} = i\int d^4x\,\mathcal{L}_{\text{int}}\):

\[ \hat{S}^{(1)} = i\int d^4x\,\left(-\frac{\lambda}{4!}\hat{\phi}_I^4\right) = \frac{-i\lambda}{4!}\int d^4x\,\hat{\phi}_I^4 \]

This is consistent. This expression is Lorentz invariant (both \(d^4x\) and \(\hat{\phi}_I^4(x)\) are Lorentz scalars).

B-6. Symmetry of the Time-Ordered Product

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Problem:

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.

(a) When \(t_1 > t_2 > t_3\):

The time-ordered product places operators at later times to the left:

\[ T[\hat{H}_I(t_1)\hat{H}_I(t_2)\hat{H}_I(t_3)] = \hat{H}_I(t_1)\hat{H}_I(t_2)\hat{H}_I(t_3) \]

(Already in time order)

(b) When \(t_3 > t_1 > t_2\):

The latest \(t_3\) goes to the left, then \(t_1\), and finally \(t_2\):

\[ T[\hat{H}_I(t_1)\hat{H}_I(t_2)\hat{H}_I(t_3)] = \hat{H}_I(t_3)\hat{H}_I(t_1)\hat{H}_I(t_2) \]

(c) Number of permutations:

There are \(3! = 6\) possible orderings of the three time variables \(t_1, t_2, t_3\).

The third-order term of the Dyson series originally has the integral

\[ (-i)^3\int_{t_0}^{t}dt_1\int_{t_0}^{t_1}dt_2\int_{t_0}^{t_2}dt_3\,\hat{H}_I(t_1)\hat{H}_I(t_2)\hat{H}_I(t_3) \]

which is integrated only over the time-ordered region (\(t_3 \le t_2 \le t_1\)). By using the time-ordered product, the integration region can be extended to the entire cube \([t_0, t]^3\), and since all 6 permutations give the same contribution, we divide by \(1/3!\):

\[ \hat{U}_I^{(3)} = \frac{(-i)^3}{3!}\int_{t_0}^{t}dt_1\int_{t_0}^{t}dt_2\int_{t_0}^{t}dt_3\,T[\hat{H}_I(t_1)\hat{H}_I(t_2)\hat{H}_I(t_3)] \]

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

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Problem:

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 \(\langle i|\hat{S}|i\rangle\).

(a) Condition that \(\hat{T}\) must satisfy:

Substituting \(\hat{S} = \mathbb{1} + i\hat{T}\) into \(\hat{S}^\dagger\hat{S} = \mathbb{1}\):

\[ \hat{S}^\dagger = \mathbb{1} - i\hat{T}^\dagger \]
\[ \hat{S}^\dagger\hat{S} = (\mathbb{1} - i\hat{T}^\dagger)(\mathbb{1} + i\hat{T}) = \mathbb{1} + i\hat{T} - i\hat{T}^\dagger + \hat{T}^\dagger\hat{T} = \mathbb{1} \]

Therefore:

\[ \boxed{i(\hat{T} - \hat{T}^\dagger) + \hat{T}^\dagger\hat{T} = 0} \]

Or equivalently:

\[ -i(\hat{T} - \hat{T}^\dagger) = \hat{T}^\dagger\hat{T} \]

This is the operator version of the optical theorem. Taking the matrix element with a specific state \(|i\rangle\) and inserting the completeness relation \(\sum_f |f\rangle\langle f| = \mathbb{1}\) on the right-hand side:

\[ -i(\langle i|\hat{T}|i\rangle - \langle i|\hat{T}^\dagger|i\rangle) = \sum_f |\langle f|\hat{T}|i\rangle|^2 \]

The left-hand side is twice the imaginary part of the forward scattering amplitude, and the right-hand side is proportional to the total cross section.

(b) Lowest-order value:

Since \(\hat{S} = \mathbb{1} + i\hat{T}\) with \(\hat{T} = O(\lambda)\), when \(|i\rangle = |f\rangle\):

\[ \langle i|\hat{S}|i\rangle = \langle i|\mathbb{1}|i\rangle + i\langle i|\hat{T}|i\rangle = 1 + O(\lambda) \]
\[ \boxed{\langle i|\hat{S}|i\rangle\Big|_{\lambda^0} = 1} \]

Consistency check:

This is consistent with the physical requirement that "if there is no interaction, no scattering occurs and the state remains unchanged."

B-8. Picture Transformation of the Interaction Hamiltonian

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Problem:

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

Solution Strategy:

Insert the identity operator into \(e^{i\hat{H}_0 t}\hat{\phi}_S^4(\mathbf{x})e^{-i\hat{H}_0 t}\) to obtain \(\hat{\phi}_I^4\).

Detailed Calculation:

\[ \hat{H}_I(t) = e^{i\hat{H}_0 t}\,\hat{H}'\,e^{-i\hat{H}_0 t} = e^{i\hat{H}_0 t}\left(\frac{\lambda}{4!}\int d^3x\,\hat{\phi}_S^4(\mathbf{x})\right)e^{-i\hat{H}_0 t} \]
\[ = \frac{\lambda}{4!}\int d^3x\,e^{i\hat{H}_0 t}\,\hat{\phi}_S(\mathbf{x})\hat{\phi}_S(\mathbf{x})\hat{\phi}_S(\mathbf{x})\hat{\phi}_S(\mathbf{x})\,e^{-i\hat{H}_0 t} \]

Insert \(e^{-i\hat{H}_0 t}e^{i\hat{H}_0 t} = \mathbb{1}\) three times between the \(\hat{\phi}_S\) factors:

\[ = \frac{\lambda}{4!}\int d^3x\,\underbrace{e^{i\hat{H}_0 t}\hat{\phi}_S(\mathbf{x})e^{-i\hat{H}_0 t}}_{\hat{\phi}_I(t,\mathbf{x})}\underbrace{e^{i\hat{H}_0 t}\hat{\phi}_S(\mathbf{x})e^{-i\hat{H}_0 t}}_{\hat{\phi}_I(t,\mathbf{x})}\underbrace{e^{i\hat{H}_0 t}\hat{\phi}_S(\mathbf{x})e^{-i\hat{H}_0 t}}_{\hat{\phi}_I(t,\mathbf{x})}\underbrace{e^{i\hat{H}_0 t}\hat{\phi}_S(\mathbf{x})e^{-i\hat{H}_0 t}}_{\hat{\phi}_I(t,\mathbf{x})} \]
\[ \boxed{\hat{H}_I(t) = \frac{\lambda}{4!}\int d^3x\,\hat{\phi}_I^4(t, \mathbf{x})} \]

Verification:

Since we simply used the definition \(\hat{\phi}_I(t, \mathbf{x}) = e^{i\hat{H}_0 t}\hat{\phi}_S(\mathbf{x})e^{-i\hat{H}_0 t}\) four times, the result is trivially correct. Moreover, when \(\lambda = 0\), we have \(\hat{H}_I = 0\), which is consistent with the fact that the state does not evolve in the absence of interactions.

Medium

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

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Problem:

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.

(a) Time derivative of \(|\psi_I(t)\rangle\):

Differentiating

\[ |\psi_I(t)\rangle = e^{i\hat{H}_0 t}|\psi(t)\rangle_S \]

with respect to \(t\):

\[ i\frac{d}{dt}|\psi_I(t)\rangle = i\frac{d}{dt}\left(e^{i\hat{H}_0 t}\right)|\psi(t)\rangle_S + e^{i\hat{H}_0 t}\left(i\frac{d}{dt}|\psi(t)\rangle_S\right) \]

First term:

\[ i \cdot i\hat{H}_0\,e^{i\hat{H}_0 t}|\psi(t)\rangle_S = -\hat{H}_0\,e^{i\hat{H}_0 t}|\psi(t)\rangle_S \]

Substituting the Schrödinger equation \(i\frac{d}{dt}|\psi(t)\rangle_S = (\hat{H}_0 + \hat{H}')|\psi(t)\rangle_S\) into the second term:

\[ e^{i\hat{H}_0 t}(\hat{H}_0 + \hat{H}')|\psi(t)\rangle_S \]

Combining:

\[ i\frac{d}{dt}|\psi_I(t)\rangle = -\hat{H}_0\,e^{i\hat{H}_0 t}|\psi(t)\rangle_S + e^{i\hat{H}_0 t}\hat{H}_0|\psi(t)\rangle_S + e^{i\hat{H}_0 t}\hat{H}'|\psi(t)\rangle_S \]

(b) Cancellation of the \(\hat{H}_0\) terms:

Looking at the first and second terms:

\[ -\hat{H}_0\,e^{i\hat{H}_0 t}|\psi(t)\rangle_S + e^{i\hat{H}_0 t}\hat{H}_0|\psi(t)\rangle_S \]

Since \(\hat{H}_0\) commutes with \(e^{i\hat{H}_0 t}\) (\([\hat{H}_0, e^{i\hat{H}_0 t}] = 0\)):

\[ -\hat{H}_0\,e^{i\hat{H}_0 t}|\psi(t)\rangle_S + \hat{H}_0\,e^{i\hat{H}_0 t}|\psi(t)\rangle_S = 0 \]

The remaining term is:

\[ i\frac{d}{dt}|\psi_I(t)\rangle = e^{i\hat{H}_0 t}\hat{H}'|\psi(t)\rangle_S \]

Substituting \(|\psi(t)\rangle_S = e^{-i\hat{H}_0 t}|\psi_I(t)\rangle\):

\[ i\frac{d}{dt}|\psi_I(t)\rangle = e^{i\hat{H}_0 t}\hat{H}'\,e^{-i\hat{H}_0 t}|\psi_I(t)\rangle = \hat{H}_I(t)|\psi_I(t)\rangle \]
\[ \boxed{i\frac{d}{dt}|\psi_I(t)\rangle = \hat{H}_I(t)|\psi_I(t)\rangle} \]

Here \(\hat{H}_I(t) = e^{i\hat{H}_0 t}\hat{H}'e^{-i\hat{H}_0 t}\) arises naturally.

(c) The case \([\hat{H}_0, \hat{H}'] = 0\):

If \([\hat{H}_0, \hat{H}'] = 0\), then:

\[ \hat{H}_I(t) = e^{i\hat{H}_0 t}\hat{H}'e^{-i\hat{H}_0 t} = \hat{H}' \]

(Since \(\hat{H}'\) commutes with \(\hat{H}_0\), it is invariant under the unitary transformation.)

In this case, \(\hat{H}_I(t)\) becomes a time-independent constant operator \(\hat{H}'\).

Physical reason: When \([\hat{H}_0, \hat{H}'] = 0\) holds, \(\hat{H}_0\) and \(\hat{H}'\) are simultaneously diagonalizable. Therefore, the eigenstates of the full Hamiltonian \(\hat{H} = \hat{H}_0 + \hat{H}'\) can be written in the same basis as the eigenstates of \(\hat{H}_0\), and the energy eigenvalues are simply additive: \(E_n = E_n^{(0)} + E_n'\). Since an exact solution is obtainable, a perturbative expansion is unnecessary.

However, in quantum field theory, \(\hat{H}_{\text{int}}\) contains cubic and higher-order terms in the fields and changes particle number, so in general \([\hat{H}_0, \hat{H}'] \neq 0\), and perturbation theory becomes necessary.

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

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Problem:

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).

(a) Swapping Variables:

The original second-order term is:

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

The integration region is the triangle \(\mathcal{R}_1: t_0 \le t_2 \le t_1 \le t\) in the \((t_1, t_2)\) plane.

Now we swap the integration variables \(t_1 \leftrightarrow t_2\) (simply exchanging the names of the dummy variables):

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

The integration region is \(\mathcal{R}_2: t_0 \le t_1 \le t_2 \le t\), and in this region \(t_2 > t_1\), so \(\hat{H}_I(t_2)\) is later in time—meaning \(\hat{H}_I(t_2)\hat{H}_I(t_1)\) is in time-ordered form.

(b) Combining the Two Contributions:

In the original integral (region \(\mathcal{R}_1\)), we have \(t_1 > t_2\), so:

\[ \hat{H}_I(t_1)\hat{H}_I(t_2) = T[\hat{H}_I(t_1)\hat{H}_I(t_2)] \quad \text{on } \mathcal{R}_1 \]

In the swapped integral (region \(\mathcal{R}_2\)), we have \(t_2 > t_1\), so:

\[ \hat{H}_I(t_2)\hat{H}_I(t_1) = T[\hat{H}_I(t_1)\hat{H}_I(t_2)] \quad \text{on } \mathcal{R}_2 \]

The two triangles \(\mathcal{R}_1 \cup \mathcal{R}_2\) cover the entire square \([t_0, t]^2\) (the diagonal \(t_1 = t_2\) has measure zero and can be ignored).

The original integral is over \(\mathcal{R}_1\) only and is equal to the swapped integral (renaming the dummy variables back gives the same expression). Therefore:

\[ \hat{U}_I^{(2)} = (-i)^2\int_{\mathcal{R}_1}dt_1\,dt_2\,T[\hat{H}_I(t_1)\hat{H}_I(t_2)] \]

Since the contributions from \(\mathcal{R}_1\) and \(\mathcal{R}_2\) are equal:

\[ 2\,\hat{U}_I^{(2)} = (-i)^2\int_{\mathcal{R}_1 \cup \mathcal{R}_2}dt_1\,dt_2\,T[\hat{H}_I(t_1)\hat{H}_I(t_2)] = (-i)^2\int_{t_0}^{t}dt_1\int_{t_0}^{t}dt_2\,T[\hat{H}_I(t_1)\hat{H}_I(t_2)] \]

Therefore:

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

(c) Generalization to \(n\)-th Order:

The \(n\)-th order term is originally:

\[ (-i)^n\int_{t_0}^{t}dt_1\int_{t_0}^{t_1}dt_2\cdots\int_{t_0}^{t_{n-1}}dt_n\,\hat{H}_I(t_1)\hat{H}_I(t_2)\cdots\hat{H}_I(t_n) \]

which is an integral over the time-ordered region \(t_n \le \cdots \le t_2 \le t_1\).

There are \(n!\) permutations of the \(n\) time variables, and using the time-ordered product \(T\) automatically guarantees the correct operator ordering for any permutation. Combining the \(n!\) triangular regions gives the entire hypercube \([t_0, t]^n\), and each region contributes equally, so:

\[ \boxed{\hat{U}_I(t, t_0) = \sum_{n=0}^{\infty}\frac{(-i)^n}{n!}\int_{t_0}^{t}dt_1\cdots\int_{t_0}^{t}dt_n\,T[\hat{H}_I(t_1)\cdots\hat{H}_I(t_n)]} \]

This is the Dyson series. Taking \(t_0 \to -\infty\), \(t \to +\infty\) yields the S-operator \(\hat{S} = T\exp\left(-i\int_{-\infty}^{+\infty}dt\,\hat{H}_I(t)\right)\).

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

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Problem:

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 \(\mathcal{M} = -\lambda\).

(a) Extracting the "2 annihilation, 2 creation" terms from \(\hat{\phi}_I^4(x)\):

Write \(\hat{\phi}_I(x) = \hat{\phi}^{(+)}(x) + \hat{\phi}^{(-)}(x)\). Expanding \(\hat{\phi}_I^4(x)\), terms appear with \(k\) factors of \(\hat{\phi}^{(+)}\) and \(4-k\) factors of \(\hat{\phi}^{(-)}\) (for \(k = 0, 1, 2, 3, 4\)).

The initial state \(|i\rangle = |\mathbf{p}_1, \mathbf{p}_2\rangle\) is a 2-particle state, and the final state \(|f\rangle = |\mathbf{p}_3, \mathbf{p}_4\rangle\) is also a 2-particle state. For the matrix element \(\langle f|\hat{\phi}_I^4|i\rangle\) to be non-zero, \(\hat{\phi}_I^4\) must annihilate the 2 particles of the initial state (requiring 2 factors of \(\hat{\phi}^{(+)}\)) and create the 2 particles of the final state (requiring 2 factors of \(\hat{\phi}^{(-)}\)).

Therefore the required terms have \(k = 2\) (2 annihilation parts, 2 creation parts):

\[ \hat{\phi}^{(-)}(x)\hat{\phi}^{(-)}(x)\hat{\phi}^{(+)}(x)\hat{\phi}^{(+)}(x) \quad \text{and its permutations} \]

The number of ways to choose which 2 of the 4 fields are \(\hat{\phi}^{(+)}\) and which 2 are \(\hat{\phi}^{(-)}\) is \(\binom{4}{2} = 6\).

(b) Calculation of the combinatorial factor and the result:

Substituting the mode expansion: \(\hat{a}_{\mathbf{k}}\) in \(\hat{\phi}^{(+)}(x)\) annihilates particles using commutation relations with \(\hat{a}_{\mathbf{p}_1}^\dagger\) or \(\hat{a}_{\mathbf{p}_2}^\dagger\) in the initial state. Similarly, \(\hat{a}_{\mathbf{k}}^\dagger\) in \(\hat{\phi}^{(-)}(x)\) corresponds to \(\hat{a}_{\mathbf{p}_3}^\dagger\) or \(\hat{a}_{\mathbf{p}_4}^\dagger\) in the final state.

Counting the combinatorial factor:

  1. Which 2 of the 4 fields provide the annihilation operators: \(\binom{4}{2} = 6\) ways
  2. Of the 2 chosen annihilation operators, which one annihilates \(\mathbf{p}_1\) and which annihilates \(\mathbf{p}_2\): \(2! = 2\) ways
  3. Of the remaining 2 creation operators, which one creates \(\mathbf{p}_3\) and which creates \(\mathbf{p}_4\): \(2! = 2\) ways

Total: \(6 \times 2 \times 2 = 24 = 4!\) ways

This exactly cancels the \(\frac{1}{4!}\) prefactor.

Computing explicitly, when \(\hat{\phi}^{(+)}(x)\) annihilates \(\mathbf{p}_1\):

\[ \hat{a}_{\mathbf{k}}|\mathbf{p}_1, \mathbf{p}_2\rangle \to (2\pi)^3\delta^3(\mathbf{k} - \mathbf{p}_1)|\mathbf{p}_2\rangle + (2\pi)^3\delta^3(\mathbf{k} - \mathbf{p}_2)|\mathbf{p}_1\rangle \]

Collecting all contributions, the \(x\) integration yields the momentum-conserving delta function:

\[ \int d^4x\,e^{i(p_1 + p_2 - p_3 - p_4)\cdot x} = (2\pi)^4\delta^4(p_1 + p_2 - p_3 - p_4) \]

The final result:

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

(c) Invariant scattering amplitude \(\mathcal{M}\):

The standard decomposition of the S-matrix element is:

\[ \langle f|\hat{S}|i\rangle = \langle f|i\rangle + i(2\pi)^4\delta^4(p_f - p_i)\,\prod_{\text{external}}\frac{1}{\sqrt{2\omega_{\mathbf{p}}}}\,\mathcal{M} \]

(The conventions for external line factors vary between textbooks; here we adopt the form that includes \(1/\sqrt{2\omega}\) as external line factors.)

Assuming \(\langle f|i\rangle = 0\) (i.e., \(|i\rangle \neq |f\rangle\)) and comparing:

\[ i\mathcal{M} \times i(2\pi)^4\delta^4(\cdots)\prod\frac{1}{\sqrt{2\omega}} = -i\lambda\,(2\pi)^4\delta^4(\cdots)\prod\frac{1}{\sqrt{2\omega}} \]

Organizing the conventions, if we define \(\langle f|i\hat{T}|i\rangle = i\mathcal{M}\,(2\pi)^4\delta^4(p_i - p_f)\prod\frac{1}{\sqrt{2\omega}}\), then:

\[ i\mathcal{M} = -i\lambda \]
\[ \boxed{\mathcal{M} = -\lambda} \]

Verification:

  • Dimensional analysis: Since \(\lambda\) is dimensionless, \(\mathcal{M}\) is also dimensionless. For 2→2 scattering in 4 dimensions, \([\mathcal{M}] = 0\) is correct.
  • Momentum conservation: \((2\pi)^4\delta^4(p_1 + p_2 - p_3 - p_4)\) appears, ensuring conservation of 4-momentum.
  • Cancellation of \(4!\): The \(1/4!\) from \(\phi^4\) and the 24 combinatorial arrangements cancel exactly, yielding simply \(-\lambda\). This is consistent with the Feynman rule for the \(\phi^4\) vertex (vertex factor \(-i\lambda\)).

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

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Problem:

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:

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

(a) Definition of Normal Ordering:

Normal ordering \(:\hat{O}:\) is the operation of rearranging all creation operators \(\hat{a}^\dagger\) (i.e., \(\hat{\phi}^{(-)}\)) to the left of all annihilation operators \(\hat{a}\) (i.e., \(\hat{\phi}^{(+)}\)) in a product of operators. For bosons, there is no sign change associated with the rearrangement.

Specifically:

\[ :\hat{\phi}_I(x)\hat{\phi}_I(y): \;= \;:\!(\hat{\phi}^{(+)}(x) + \hat{\phi}^{(-)}(x))(\hat{\phi}^{(+)}(y) + \hat{\phi}^{(-)}(y))\!: \]

Expanding and applying normal ordering:

\[ :\hat{\phi}_I(x)\hat{\phi}_I(y): \;= \;\hat{\phi}^{(-)}(x)\hat{\phi}^{(-)}(y) + \hat{\phi}^{(-)}(x)\hat{\phi}^{(+)}(y) + \hat{\phi}^{(-)}(y)\hat{\phi}^{(+)}(x) + \hat{\phi}^{(+)}(x)\hat{\phi}^{(+)}(y) \]

Note: In the terms \(\hat{\phi}^{(-)}(x)\hat{\phi}^{(+)}(y)\) and \(\hat{\phi}^{(-)}(y)\hat{\phi}^{(+)}(x)\), the creation parts are already on the left and the annihilation parts on the right.

(b) Proof that the Contraction is a c-number:

Definition of contraction:

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

Case \(x^0 > y^0\):

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

Expanding:

\[ = \hat{\phi}^{(+)}(x)\hat{\phi}^{(+)}(y) + \hat{\phi}^{(+)}(x)\hat{\phi}^{(-)}(y) + \hat{\phi}^{(-)}(x)\hat{\phi}^{(+)}(y) + \hat{\phi}^{(-)}(x)\hat{\phi}^{(-)}(y) \]

Taking the difference with the normal-ordered expression:

\[ T[\hat{\phi}(x)\hat{\phi}(y)] - :\hat{\phi}(x)\hat{\phi}(y): = \hat{\phi}^{(+)}(x)\hat{\phi}^{(-)}(y) - \hat{\phi}^{(-)}(y)\hat{\phi}^{(+)}(x) \]
\[ = [\hat{\phi}^{(+)}(x),\, \hat{\phi}^{(-)}(y)] \]

Computing this commutator:

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

This is a c-number (a function containing no operators).

Similarly, for the case \(x^0 < y^0\), the commutator \([\hat{\phi}^{(+)}(y),\, \hat{\phi}^{(-)}(x)]\) appears, which is also a c-number.

As shown in D3(c), when \(x^0 > y^0\):

\[ [\hat{\phi}^{(+)}(x),\, \hat{\phi}^{(-)}(y)] = \int\frac{d^3p}{(2\pi)^3}\frac{1}{2\omega_{\mathbf{p}}}e^{-ip\cdot(x-y)} = \langle 0|T[\hat{\phi}(x)\hat{\phi}(y)]|0\rangle = D_F(x-y) \]

(since the vacuum expectation value of the normal-ordered product is \(\langle 0|:\hat{\phi}(x)\hat{\phi}(y):|0\rangle = 0\))

Therefore:

\[ \boxed{\underbrace{\hat{\phi}_I(x)\hat{\phi}_I(y)} = D_F(x-y)} \]

(c) Wick's Theorem for Two Fields:

Rearranging the result of (b) immediately gives:

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

This is Wick's theorem for two fields.

Verification:

Taking the vacuum expectation value of both sides:

  • Left-hand side: \(\langle 0|T[\hat{\phi}(x)\hat{\phi}(y)]|0\rangle = D_F(x-y)\)
  • Right-hand side: \(\langle 0|:\hat{\phi}(x)\hat{\phi}(y):|0\rangle + D_F(x-y) = 0 + D_F(x-y)\)

They agree. ✓

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

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Problem:

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.

(a) Proof that \(\hat{U}_I^\dagger(t, t_0) = \hat{U}_I(t_0, t)\):

\(\hat{U}_I(t, t_0)\) satisfies the differential equation (7.9):

\[ i\frac{\partial}{\partial t}\hat{U}_I(t, t_0) = \hat{H}_I(t)\,\hat{U}_I(t, t_0), \quad \hat{U}_I(t_0, t_0) = \mathbb{1} \]

Taking the adjoint (using \(\hat{H}_I^\dagger = \hat{H}_I\)):

\[ -i\frac{\partial}{\partial t}\hat{U}_I^\dagger(t, t_0) = \hat{U}_I^\dagger(t, t_0)\,\hat{H}_I(t) \]

On the other hand, consider the equation satisfied by \(\hat{U}_I(t_0, t)\). Since \(\hat{U}_I(t_0, t)\) represents time evolution "from \(t\) to \(t_0\)," the differential equation with respect to \(t\) is:

\[ i\frac{\partial}{\partial t}\hat{U}_I(t_0, t) = -\hat{U}_I(t_0, t)\,\hat{H}_I(t) \]

(The sign is reversed because in \(\hat{U}_I(t_0, t)\), the variable \(t\) functions not as the "departure time" but as the "reverse of the arrival time." Formally, this can be derived by differentiating \(\hat{U}_I(t, t_0)\hat{U}_I(t_0, t) = \mathbb{1}\) with respect to \(t\).)

Rearranging:

\[ -i\frac{\partial}{\partial t}\hat{U}_I(t_0, t) = \hat{U}_I(t_0, t)\,\hat{H}_I(t) \]

This is identical to the equation satisfied by \(\hat{U}_I^\dagger(t, t_0)\), and the initial conditions also agree: \(\hat{U}_I^\dagger(t_0, t_0) = \mathbb{1} = \hat{U}_I(t_0, t_0)\). By the uniqueness of solutions to differential equations:

\[ \boxed{\hat{U}_I^\dagger(t, t_0) = \hat{U}_I(t_0, t)} \]

(b) Proof of Unitarity:

In the group property of the time evolution operator \(\hat{U}_I(t, t_1)\hat{U}_I(t_1, t_0) = \hat{U}_I(t, t_0)\), setting \(t = t_0\):

\[ \hat{U}_I(t_0, t_1)\hat{U}_I(t_1, t_0) = \hat{U}_I(t_0, t_0) = \mathbb{1} \]

Substituting the result from (a), \(\hat{U}_I(t_0, t_1) = \hat{U}_I^\dagger(t_1, t_0)\):

\[ \hat{U}_I^\dagger(t_1, t_0)\hat{U}_I(t_1, t_0) = \mathbb{1} \]

Taking the limit \(t_1 \to +\infty\), \(t_0 \to -\infty\):

\[ \boxed{\hat{S}^\dagger\hat{S} = \mathbb{1}} \]

Similarly, from \(\hat{U}_I(t_1, t_0)\hat{U}_I(t_0, t_1) = \mathbb{1}\), we also obtain \(\hat{S}\hat{S}^\dagger = \mathbb{1}\).

(c) Physical Meaning of Probability Conservation:

Unitarity implies conservation of probability. It shows that starting from an initial state \(|i\rangle\), the sum of transition probabilities to all possible final states \(|f\rangle\) equals 1.

Inserting the completeness relation \(\sum_f |f\rangle\langle f| = \mathbb{1}\):

\[ \sum_f |\langle f|\hat{S}|i\rangle|^2 = \sum_f \langle i|\hat{S}^\dagger|f\rangle\langle f|\hat{S}|i\rangle = \langle i|\hat{S}^\dagger\left(\sum_f|f\rangle\langle f|\right)\hat{S}|i\rangle \]
\[ = \langle i|\hat{S}^\dagger\hat{S}|i\rangle = \langle i|\mathbb{1}|i\rangle = 1 \]
\[ \boxed{\sum_f |\langle f|\hat{S}|i\rangle|^2 = 1} \]

This is precisely the requirement of probability conservation — "particles must go somewhere" — meaning that particles cannot simply vanish after scattering.

Advanced

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

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Problem:

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.

(a) Mass dimension of the coupling constant \(g\):

In 4-dimensional spacetime, \([\mathcal{L}] = 4\), \([\psi] = 3/2\), \([\phi] = 1\).

\[ [\mathcal{L}_{\text{int}}] = [g] + [\bar{\psi}] + [\psi] + [\phi] = 4 \]
\[ [g] + \frac{3}{2} + \frac{3}{2} + 1 = 4 \]
\[ \boxed{[g] = 4 - 4 = 0} \]

\(g\) is dimensionless, and Yukawa theory is renormalizable.

(b) First-order term of the S-matrix:

From \(\hat{H}' = -\int d^3x\,\mathcal{L}_{\text{int}} = g\int d^3x\,\hat{\bar{\psi}}\hat{\psi}\hat{\phi}\):

\[ \hat{S}^{(1)} = (-i)\int_{-\infty}^{+\infty}dt\,\hat{H}_I(t) = (-i)\int d^4x\,g\,\hat{\bar{\psi}}_I(x)\hat{\psi}_I(x)\hat{\phi}_I(x) \]
\[ \boxed{\hat{S}^{(1)} = -ig\int d^4x\,\hat{\bar{\psi}}_I(x)\hat{\psi}_I(x)\hat{\phi}_I(x)} \]

(c) Lowest order for fermion-fermion scattering:

We analyze the operator structure of \(\hat{S}^{(1)}\). \(\hat{\bar{\psi}} \sim \hat{b}^\dagger + \hat{d}\) (fermion creation or antifermion annihilation), \(\hat{\psi} \sim \hat{b} + \hat{d}^\dagger\) (fermion annihilation or antifermion creation), \(\hat{\phi} \sim \hat{a} + \hat{a}^\dagger\) (scalar particle creation or annihilation).

Each vertex of \(\hat{S}^{(1)}\): - Annihilates one fermion and creates one fermion - Creates or annihilates one scalar particle

For the process \(\psi + \psi \to \psi + \psi\), the initial state has 2 fermions, the final state has 2 fermions, and there are no scalar particles on external lines. \(\hat{S}^{(1)}\) has only one vertex and can only annihilate one fermion, so it cannot accommodate the 2 fermions in the initial state.

Therefore, the lowest-order contribution appears from \(\hat{S}^{(2)}\) (second order in \(g^2\)). With two vertices, each vertex can annihilate and create one fermion each, and the two vertices can be connected by an internal scalar field line (propagator).

(d) Contraction of the scalar field and the Yukawa potential:

Applying Wick's theorem to \(\hat{S}^{(2)}\):

\[ \hat{S}^{(2)} = \frac{(-ig)^2}{2!}\int d^4x\,d^4y\,T[\hat{\bar{\psi}}(x)\hat{\psi}(x)\hat{\phi}(x)\,\hat{\bar{\psi}}(y)\hat{\psi}(y)\hat{\phi}(y)] \]

In fermion-fermion scattering, since there are no scalar particles on external lines, \(\hat{\phi}(x)\) and \(\hat{\phi}(y)\) are contracted with each other:

\[ \underbrace{\hat{\phi}(x)\hat{\phi}(y)} = D_F(x-y) = \int\frac{d^4k}{(2\pi)^4}\,\frac{i}{k^2 - m_\phi^2 + i\epsilon} \]

Physical interpretation: This contraction represents the process in which a scalar particle is "virtually" created at spacetime point \(x\), propagates to \(y\), and is annihilated there (or vice versa). This is precisely the mediation of force through scalar particle exchange.

Taking the non-relativistic limit (\(|\mathbf{k}|^2 \ll m_\phi^2\), \(k^0 \approx 0\)) in momentum space, the scalar propagator becomes:

\[ \frac{i}{k^2 - m_\phi^2} \approx \frac{-i}{|\mathbf{k}|^2 + m_\phi^2} \]

Fourier transforming this to coordinate space:

\[ V(r) \propto -\frac{g^2}{4\pi}\frac{e^{-m_\phi r}}{r} \]

This is the Yukawa potential. Recalling the relationship between scattering amplitudes and potentials in the Born approximation from quantum mechanics Ch. 13, we see that "particle exchange" in quantum field theory reproduces a Yukawa-type potential in the non-relativistic limit. The mass \(m_\phi\) of the scalar particle determines the range of the force \(\sim 1/m_\phi\).

Consistency checks:

  • In the limit \(m_\phi \to 0\), we recover \(V(r) \propto -g^2/(4\pi r)\) (Coulomb type).
  • In the limit \(m_\phi \to \infty\), \(V(r) \to 0\) (a heavy mediator transmits only short-range forces).
  • Since \([g] = 0\), we have \([g^2/r] = 1\) (dimension of energy). The dimension of the potential is correct.

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

→ Back to problem

Problem:

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 \(\langle 0|\hat{S}|0\rangle\).

(a) Confirming \(\hat{H}_I(t) \to 0\) via Adiabatic Switching:

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

As \(t \to +\infty\): \(e^{-\epsilon|t|} = e^{-\epsilon t} \to 0\) (since \(\epsilon > 0\))

As \(t \to -\infty\): \(e^{-\epsilon|t|} = e^{+\epsilon t} \to 0\) (since \(\epsilon > 0\) and \(t < 0\), so \(\epsilon t \to -\infty\))

Therefore:

\[ \boxed{\lim_{t \to \pm\infty}\hat{H}_I(t)\,e^{-\epsilon|t|} = 0} \]

In the distant past and future, the interaction adiabatically vanishes, and the system behaves as a free theory.

(b) The Gell-Mann–Low Theorem:

Structure of the argument:

Under adiabatic switching, the state at \(t = -\infty\) is the free-theory vacuum \(|0\rangle\) (since the interaction is turned off). Evolving to \(t = 0\) with the time-evolution operator \(\hat{U}_I^\epsilon(0, -\infty)\) gives:

\[ \hat{U}_I^\epsilon(0, -\infty)|0\rangle \]

This state is one that has been continuously deformed from \(|0\rangle\) through the process of adiabatically turning "on" the interaction. By the adiabatic theorem (a generalization of the adiabatic approximation in quantum mechanics), if the system changes sufficiently slowly, the ground state remains the ground state. Therefore, in the limit \(\epsilon \to 0^+\), this state coincides (up to a phase factor) with the vacuum \(|\Omega\rangle\) of the interacting theory.

However, \(\hat{U}_I^\epsilon(0, -\infty)|0\rangle\) is generally not normalized and carries an undetermined phase. The role of the denominator \(\langle 0|\hat{U}_I^\epsilon(0, -\infty)|0\rangle\) is to correct for this:

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

Role of the denominator:

  1. Normalization: It normalizes the norm of \(\hat{U}_I^\epsilon(0, -\infty)|0\rangle\) to 1.
  2. Phase removal: It removes the relative phase between \(|0\rangle\) and \(|\Omega\rangle\) (the \(e^{i\alpha}\) part in \(\langle 0|\Omega\rangle = |\langle 0|\Omega\rangle|e^{i\alpha}\)), adopting the convention that \(\langle 0|\Omega\rangle\) is real and positive.
  3. Absorption of energy shift: The denominator cancels the phase \(e^{-i(E_\Omega - E_0)T}\) (where \(T\) is the time interval) associated with the vacuum energy shift \(E_\Omega - E_0\) due to the interaction.

(c) Cancellation of Vacuum Bubbles:

A vacuum bubble is a Feynman diagram with no external lines—that is, a closed loop diagram not connected to any external particles.

When expanding the Dyson series, at each order in the S-matrix element \(\langle f|\hat{S}|i\rangle\), both "connected diagrams describing the physical scattering process" and "accompanying vacuum bubbles" appear.

Linked-cluster theorem:

The S-matrix element factorizes as follows:

\[ \langle f|\hat{S}|i\rangle = \langle f|\hat{S}|i\rangle_{\text{connected}} \times \langle 0|\hat{S}|0\rangle \]

Here \(\langle 0|\hat{S}|0\rangle\) is the sum of all vacuum bubble contributions, which factorizes exponentially:

\[ \langle 0|\hat{S}|0\rangle = \exp\left(\sum_{\text{connected vacuum bubbles}}\right) \]

Meanwhile, the denominator from the Gell-Mann–Low theorem is:

\[ \langle 0|\hat{U}_I(+\infty, -\infty)|0\rangle = \langle 0|\hat{S}|0\rangle \]

Therefore, when computing physical scattering amplitudes:

\[ \frac{\langle f|\hat{S}|i\rangle}{\langle 0|\hat{S}|0\rangle} = \langle f|\hat{S}|i\rangle_{\text{connected}} \]

In other words, the vacuum bubble contributions are exactly cancelled by the denominator, and only connected diagrams contribute to physical scattering amplitudes.

More precisely, for correlation functions:

\[ \langle \Omega|T[\hat{\phi}(x_1)\cdots\hat{\phi}(x_n)]|\Omega\rangle = \frac{\langle 0|T[\hat{\phi}_I(x_1)\cdots\hat{\phi}_I(x_n)\hat{S}]|0\rangle}{\langle 0|\hat{S}|0\rangle} \]

The disconnected diagrams (connected parts × vacuum bubbles) that appear in the Wick expansion of the numerator cancel against \(\langle 0|\hat{S}|0\rangle\) in the denominator, and ultimately only connected diagrams remain.

Consistency Checks:

  • When \(\lambda = 0\) (free theory), \(|\Omega\rangle = |0\rangle\) and \(\langle 0|\hat{S}|0\rangle = 1\), so everything holds trivially.
  • The cancellation of vacuum bubbles is consistent with the physical requirement that observables (cross sections and decay rates) do not depend on the vacuum energy.
  • This theorem holds to all orders in perturbation theory and provides the theoretical foundation for the practical prescription in the Feynman rules that "one only needs to compute connected diagrams."