Zee QFT – Part I

Includes only up to I.4. Reorganization pending.



We need to use \hat{H} = \hat{p}^2/2m + V(\hat{q}) as the new Hamiltonian, and we can start by replacing it in \langle q_{j+1}|e^{-i\delta t\hat{H}}|q_j\rangle:

\displaystyle \langle q_{j+1}|e^{-i\delta t\hat{H}}|q_j\rangle = \langle q_{j+1}|e^{-i\delta t(\hat{p}^2/2m + V(\hat{q}))}|q_j\rangle.

In general \hat{p}^2 and V(\hat{q}) don’t commute, but as \delta t is an infinitesimal, we can assume that they do commute:

\displaystyle \langle q_{j+1}|e^{-i\delta t\hat{H}}|q_j\rangle = \langle q_{j+1}|e^{-i\delta t \hat{p}^2/2m}e^{-i\delta t V(\hat{q})}|q_j\rangle

\displaystyle = \langle q_{j+1}|e^{-i\delta t \hat{p}^2/2m}e^{-i\delta t V(q_j)}|q_j\rangle

\displaystyle = e^{-i\delta t V(q_j)}\langle q_{j+1}|e^{-i\delta t \hat{p}^2/2m}|q_j\rangle

\displaystyle = \left(\frac{-i m}{2\pi\delta t}\right)^\frac{1}{2}e^{-i\delta t V(q_j)}e^{i\delta t(m/2)((q_{j+1}-q_j)/\delta t)^2} (using (21))

\displaystyle = \left(\frac{-i m}{2\pi\delta t}\right)^\frac{1}{2}e^{i\delta t(m/2)((q_{j+1}-q_j)/\delta t)^2 - i\delta t V(q_j)}

Applying the same operations as the text we get

\displaystyle \langle q_F|e^{-iHT}|q_I\rangle = \left(\frac{-i m}{2\pi\delta t}\right)^\frac{N}{2}\left(\prod_{k=1}^{N-1}\int dq_k\right)e^{i\delta t\sum_{j=0}^{N-1} (m/2)((q_{j+1}-q_j)/\delta t)^2 - V(q_j)}

and, when going to the continuum limit,

\displaystyle \langle q_F|e^{-iHT}|q_I\rangle = \int Dq(t) e^{i\int_0^T dt \frac{1}{2}m\dot{q}^2 - V(q)}.


Let’s start by checking (18):

\displaystyle \int_{-\infty}^{+\infty} dx\,e^{-\frac{1}{2}a x^2} = \left(\frac{2\pi}{a}\right)^\frac{1}{2} (17)

If we apply n times -2(d/da) to the left side we get

\displaystyle \left(-2\frac{d}{da}\right)^n \int_{-\infty}^{+\infty} dx\,e^{-\frac{1}{2}a x^2} = \left(-2\frac{d}{da}\right)^{n-1} \left(-2\frac{d}{da}\right)\int_{-\infty}^{+\infty} dx\,e^{-\frac{1}{2}a x^2}

\displaystyle = \left(-2\frac{d}{da}\right)^{n-1} (-2)\int_{-\infty}^{+\infty} dx\,\frac{d}{da}e^{-\frac{1}{2}a x^2}

\displaystyle = \left(-2\frac{d}{da}\right)^{n-1} (-2)\int_{-\infty}^{+\infty} dx\,\left(-\frac{1}{2}x^2\right)e^{-\frac{1}{2}a x^2}

\displaystyle = \left(-2\frac{d}{da}\right)^{n-1} \int_{-\infty}^{+\infty} dx\,x^2 e^{-\frac{1}{2}a x^2}

\displaystyle \dots

\displaystyle = \int_{-\infty}^{+\infty} dx\,x^{2n} e^{-\frac{1}{2}a x^2}.

And, if we do the same thing to the right side, we obtain

\displaystyle \left(-2\frac{d}{da}\right)^n \left(\frac{2\pi}{a}\right)^\frac{1}{2} = \left(-2\frac{d}{da}\right)^n (2\pi)^\frac{1}{2} a^{-\frac{1}{2}}

\displaystyle = (2\pi)^\frac{1}{2} (-2)^n \left(\frac{d}{da}\right)^{n-1} \frac{d}{da} a^{-\frac{1}{2}}

\displaystyle = (2\pi)^\frac{1}{2} (-2)^n \left(\frac{d}{da}\right)^{n-1} \left(-\frac{1}{2}\right) a^{-\frac{3}{2}}

\displaystyle = (2\pi)^\frac{1}{2} (-2)^n \left(\frac{d}{da}\right)^{n-2} \left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right) a^{-\frac{5}{2}}

\displaystyle \dots

\displaystyle = (2\pi)^\frac{1}{2} (-2)^n \left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right) \dots\left(-\frac{2n-1}{2}\right) a^{-\frac{2n+1}{2}}

\displaystyle = (2\pi)^\frac{1}{2} (2n-1)!!\,a^{-\frac{2n+1}{2}}.

Dividing this result by (17), we get:

\displaystyle \frac{\int_{-\infty}^{+\infty} dx\,x^{2n} e^{-\frac{1}{2}a x^2}}{\int_{-\infty}^{+\infty} dx\,e^{-\frac{1}{2}a x^2}} = \frac{(2\pi)^\frac{1}{2} (2n-1)!!\,a^{-\frac{2n+1}{2}}}{(2\pi)^\frac{1}{2}\,a^{-\frac{1}{2}}}

\displaystyle = \frac{(2n-1)!!\,a^{-\frac{2n+1}{2}}}{a^{-\frac{1}{2}}}

\displaystyle = (2n-1)!!\,a^{-\frac{2n+1}{2}+\frac{1}{2}}

\displaystyle = (2n-1)!!\,a^{-n},

that matches with (18).

Now we can start with (24). Let’s copy equation (22):

\displaystyle \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} \dots \int_{-\infty}^{+\infty} dx_1 dx_2 \dots dx_N\,e^{-\frac{1}{2}x\cdot A\cdot x + J\cdot x} = \left(\frac{(2\pi)^N}{\det(A)}\right)^\frac{1}{2}e^{\frac{1}{2}J\cdot A^{-1}\cdot J}.

If we derive the left side with respect to J_i and J_j we get:

\displaystyle \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} \dots \int_{-\infty}^{+\infty} dx_1 dx_2 \dots dx_N\,\frac{\partial\,}{\partial J_i}\frac{\partial\,}{\partial J_j} e^{-\frac{1}{2}x\cdot A\cdot x + J\cdot x} =

\displaystyle = \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} \dots \int_{-\infty}^{+\infty} dx_1 dx_2 \dots dx_N\,x_i x_j\,e^{-\frac{1}{2}x\cdot A\cdot x + J\cdot x},

and if we do the same with the right side we are left with

\displaystyle \frac{\partial}{\partial J_i}\frac{\partial}{\partial J_j}\left(\frac{(2\pi)^N}{\det(A)}\right)^\frac{1}{2}e^{\frac{1}{2}J\cdot A^{-1}\cdot J} = \left(\frac{(2\pi)^N}{\det(A)}\right)^\frac{1}{2}\frac{\partial}{\partial J_i}\frac{\partial}{\partial J_j}e^{\frac{1}{2}J\cdot A^{-1}\cdot J}

\displaystyle = \left(\frac{(2\pi)^N}{\det(A)}\right)^\frac{1}{2}\frac{\partial}{\partial J_i}\frac{\partial}{\partial J_j}e^{\frac{1}{2}J_k (A^{-1})_{kl} J_l}

\displaystyle = \left(\frac{(2\pi)^N}{\det(A)}\right)^\frac{1}{2}\frac{\partial}{\partial J_i}\left(e^{\frac{1}{2}J_k (A^{-1})_{kl} J_l}\frac{\partial}{\partial J_j}\left(\frac{1}{2}J_k (A^{-1})_{kl} J_l\right)\right)

\displaystyle = \left(\frac{(2\pi)^N}{\det(A)}\right)^\frac{1}{2}\frac{\partial}{\partial J_i}\left(e^{\frac{1}{2}J_k (A^{-1})_{kl} J_l}\left(\frac{1}{2}\delta_{jk} (A^{-1})_{kl} J_l + \frac{1}{2}J_k (A^{-1})_{kl} \delta{jl}\right)\right)

\displaystyle = \left(\frac{(2\pi)^N}{\det(A)}\right)^\frac{1}{2}\frac{\partial}{\partial J_i}\left(e^{\frac{1}{2}J_k (A^{-1})_{kl} J_l}\left(\frac{1}{2}(A^{-1})_{jl} J_l + \frac{1}{2}J_k (A^{-1})_{kj}\right)\right)

\displaystyle = \left(\frac{(2\pi)^N}{\det(A)}\right)^\frac{1}{2}\frac{\partial}{\partial J_i}\left(e^{\frac{1}{2}J_k (A^{-1})_{kl} J_l}(A^{-1})_{jl} J_l\right)\quad(A^{-1}\,\mathrm{symmetric})

\displaystyle = \left(\frac{(2\pi)^N}{\det(A)}\right)^\frac{1}{2}\left(e^{\frac{1}{2}J_k (A^{-1})_{kl} J_l}(A^{-1})_{ik} J_k (A^{-1})_{jl} J_l + e^{\frac{1}{2}J_k (A^{-1})_{kl} J_l}(A^{-1})_{jl} \delta_{il}\right)

\displaystyle = \left(\frac{(2\pi)^N}{\det(A)}\right)^\frac{1}{2}\left(e^{\frac{1}{2}J_k (A^{-1})_{kl} J_l}(A^{-1})_{ik} J_k (A^{-1})_{jl} J_l + e^{\frac{1}{2}J_k (A^{-1})_{kl} J_l}(A^{-1})_{ji}\right)

If we not set J to zero, we can find the following equality:

\displaystyle \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} \dots \int_{-\infty}^{+\infty} dx_1 dx_2 \dots dx_N\,x_i x_j\,e^{-\frac{1}{2}x\cdot A\cdot x + J\cdot x} =

\displaystyle = \left(\frac{(2\pi)^N}{\det(A)}\right)^\frac{1}{2}(A^{-1})_{ji}.

This can turn quite messy for higher derivatives, but we can use the series expansion of the exponential to save a lot of work: if we do k derivatives, only the terms with k variables can survive after we set J to zero. Taking this into account, let’s try to apply this to a fourth derivative:

\displaystyle \partial_i \partial_j \partial_k \partial_l\,e^{\frac{1}{2}J_m (A^{-1})_{mn} J_n} = \frac{1}{2!\,2^2} \partial_i \partial_j \partial_k \partial_l (J_m (A^{-1})_{mn} J_n)^2

\displaystyle = \frac{1}{8} \partial_i \partial_j \partial_k 2\cdot 2(J_m (A^{-1})_{mn} J_n) (A^{-1})_{lo} J_o

\displaystyle = \frac{1}{2} \partial_i \partial_j \partial_k (J_m (A^{-1})_{mn} J_n) (A^{-1})_{lo} J_o

\displaystyle = \frac{1}{2} \partial_i \partial_j \left[2 (A^{-1})_{kp} J_p (A^{-1})_{lo} J_o + (J_m (A^{-1})_{mn} J_n) (A^{-1})_{lk}\right]

\displaystyle = \frac{1}{2} \partial_i \left[2 (A^{-1})_{kj} (A^{-1})_{lo} J_o + 2 (A^{-1})_{kp} J_p (A^{-1})_{lj} +2 (A^{-1})_{jn} J_n (A^{-1})_{lk}\right]

\displaystyle = \frac{1}{2} \left[2 (A^{-1})_{kj} (A^{-1})_{li} + 2 (A^{-1})_{ki} (A^{-1})_{lj} +2 (A^{-1})_{ji} (A^{-1})_{lk}\right]

\displaystyle = (A^{-1})_{jk} (A^{-1})_{il} + (A^{-1})_{ik} (A^{-1})_{jl} + (A^{-1})_{ij} (A^{-1})_{kl}.

Zee asks to do a sixth derivative:

\displaystyle \partial_i \partial_j \partial_k \partial_l \partial_m \partial_n \,e^{\frac{1}{2}J_o (A^{-1})_{op} J_p} = \frac{1}{3!\,2^3} \partial_i \partial_j \partial_k \partial_l \partial_m \partial_n (J_o (A^{-1})_{op} J_p)^3

\displaystyle = \frac{1}{3!\,2^3} \partial_i \partial_j \partial_k \partial_l \partial_m\,3\cdot (J_o (A^{-1})_{op} J_p)^2\cdot 2 \cdot(A^{-1})_{nq} J_q

\displaystyle = \frac{1}{8} \partial_i \partial_j \partial_k \partial_l \partial_m\,(J_o (A^{-1})_{op} J_p)^2 (A^{-1})_{nq} J_q

\displaystyle = \frac{1}{8} \partial_i \partial_j \partial_k \partial_l \,\left[2\cdot (J_o (A^{-1})_{op} J_p)\cdot 2 \cdot (A^{-1})_{mr} J_r (A^{-1})_{nq} J_q +\right.

\displaystyle \left. (J_o (A^{-1})_{op} J_p)^2 (A^{-1})_{nm} \right]

We can concentrate in the first term, as the second term is the same that was previously done:

\displaystyle \frac{1}{8} \partial_i \partial_j \partial_k \partial_l \,2\cdot (J_o (A^{-1})_{op} J_p)\cdot 2 \cdot (A^{-1})_{mr} J_r (A^{-1})_{nq} J_q =

\displaystyle = \frac{1}{2} \partial_i \partial_j \partial_k \partial_l \,(J_o (A^{-1})_{op} J_p) (A^{-1})_{mr} J_r (A^{-1})_{nq} J_q

\displaystyle = \frac{1}{2} \partial_i \partial_j \partial_k \,\left[2 (A^{-1})_{ls} J_s (A^{-1})_{mr} J_r (A^{-1})_{nq} J_q + \right.

\displaystyle \left. (J_o (A^{-1})_{op} J_p) (A^{-1})_{ml} (A^{-1})_{nq} J_q + (J_o (A^{-1})_{op} J_p) (A^{-1})_{mr} J_r (A^{-1})_{nl}\right]

\displaystyle = \frac{1}{2} \partial_i \partial_j \,\left[2 (A^{-1})_{lk} (A^{-1})_{mr} J_r (A^{-1})_{nq} J_q +\right.

\displaystyle \left. 2 (A^{-1})_{ls} J_s (A^{-1})_{mk} (A^{-1})_{nq} J_q + 2 (A^{-1})_{ls} J_s (A^{-1})_{mr} J_r (A^{-1})_{nk} +\right.

\displaystyle \left. 2 (A^{-1})_{kt} J_t (A^{-1})_{ml} (A^{-1})_{nq} J_q + (J_o (A^{-1})_{op} J_p) (A^{-1})_{ml} (A^{-1})_{nk}\right.

\displaystyle \left. 2 (A^{-1})_{kt} J_t (A^{-1})_{mr} J_r (A^{-1})_{nl} + (J_o (A^{-1})_{op} J_p) (A^{-1})_{mk} (A^{-1})_{nl}\right]

\displaystyle = \frac{1}{2} \partial_i \,\left[2 (A^{-1})_{lk} (A^{-1})_{mj} (A^{-1})_{nq} J_q +\right.

\displaystyle \left. 2 (A^{-1})_{lk} (A^{-1})_{mr} J_r (A^{-1})_{nj} + 2 (A^{-1})_{lj} (A^{-1})_{mk} (A^{-1})_{nq} J_q +\right.

\displaystyle \left. 2 (A^{-1})_{ls} J_s (A^{-1})_{mk} (A^{-1})_{nj} + 2 (A^{-1})_{lj} (A^{-1})_{mr} J_r (A^{-1})_{nk} +\right.

\displaystyle \left. 2 (A^{-1})_{ls} J_s (A^{-1})_{mj} (A^{-1})_{nk} + 2 (A^{-1})_{kj} (A^{-1})_{ml} (A^{-1})_{nq} J_q +\right.

\displaystyle \left. 2 (A^{-1})_{kt} J_t (A^{-1})_{ml} (A^{-1})_{nj} + 2 (A^{-1})_{ju} J_u (A^{-1})_{ml} (A^{-1})_{nk} +\right.

\displaystyle \left. 2 (A^{-1})_{kj} (A^{-1})_{mr} J_r (A^{-1})_{nl} + 2 (A^{-1})_{kt} J_t (A^{-1})_{mj} (A^{-1})_{nl} +\right.

\displaystyle \left. 2 (A^{-1})_{jv} J_v (A^{-1})_{mk} (A^{-1})_{nl} \right]

\displaystyle = \frac{1}{2} \,\left[2 (A^{-1})_{lk} (A^{-1})_{mj} (A^{-1})_{ni} +\right.

\displaystyle \left. 2 (A^{-1})_{lk} (A^{-1})_{mi} (A^{-1})_{nj} + 2 (A^{-1})_{lj} (A^{-1})_{mk} (A^{-1})_{ni} +\right.

\displaystyle \left. 2 (A^{-1})_{li} (A^{-1})_{mk} (A^{-1})_{nj} + 2 (A^{-1})_{lj} (A^{-1})_{mi} (A^{-1})_{nk} +\right.

\displaystyle \left. 2 (A^{-1})_{li} (A^{-1})_{mj} (A^{-1})_{nk} + 2 (A^{-1})_{kj} (A^{-1})_{ml} (A^{-1})_{ni} +\right.

\displaystyle \left. 2 (A^{-1})_{ki} (A^{-1})_{ml} (A^{-1})_{nj} + 2 (A^{-1})_{ji} (A^{-1})_{ml} (A^{-1})_{nk} +\right.

\displaystyle \left. 2 (A^{-1})_{kj} (A^{-1})_{mi} (A^{-1})_{nl} + 2 (A^{-1})_{ki} (A^{-1})_{mj} (A^{-1})_{nl} +\right.

\displaystyle \left. 2 (A^{-1})_{ji} (A^{-1})_{mk} (A^{-1})_{nl} \right]

Combining this with the previous result:

\displaystyle \partial_i \partial_j \partial_k \partial_l \partial_m \partial_n \,e^{\frac{1}{2}J_o (A^{-1})_{op} J_p} = (A^{-1})_{lk} (A^{-1})_{mj} (A^{-1})_{ni} +

\displaystyle (A^{-1})_{lk} (A^{-1})_{mi} (A^{-1})_{nj} + (A^{-1})_{lj} (A^{-1})_{mk} (A^{-1})_{ni} +

\displaystyle (A^{-1})_{li} (A^{-1})_{mk} (A^{-1})_{nj} + (A^{-1})_{lj} (A^{-1})_{mi} (A^{-1})_{nk} +

\displaystyle (A^{-1})_{li} (A^{-1})_{mj} (A^{-1})_{nk} + (A^{-1})_{kj} (A^{-1})_{ml} (A^{-1})_{ni} +

\displaystyle (A^{-1})_{ki} (A^{-1})_{ml} (A^{-1})_{nj} + (A^{-1})_{ji} (A^{-1})_{ml} (A^{-1})_{nk} +

\displaystyle (A^{-1})_{kj} (A^{-1})_{mi} (A^{-1})_{nl} + (A^{-1})_{ki} (A^{-1})_{mj} (A^{-1})_{nl} +

\displaystyle (A^{-1})_{ji} (A^{-1})_{mk} (A^{-1})_{nl} + (A^{-1})_{jk} (A^{-1})_{il} (A^{-1})_{mn} +

\displaystyle (A^{-1})_{ik} (A^{-1})_{jl} (A^{-1})_{mn} + (A^{-1})_{ij} (A^{-1})_{kl} (A^{-1})_{mn}

Ordering the terms:

\displaystyle \partial_i \partial_j \partial_k \partial_l \partial_m \partial_n \,e^{\frac{1}{2}J_o (A^{-1})_{op} J_p} = (A^{-1})_{ij} (A^{-1})_{kl} (A^{-1})_{mn} +

\displaystyle (A^{-1})_{ij} (A^{-1})_{km} (A^{-1})_{ln} + (A^{-1})_{ij} (A^{-1})_{kn} (A^{-1})_{lm} +

\displaystyle (A^{-1})_{ik} (A^{-1})_{jl} (A^{-1})_{mn} + (A^{-1})_{ik} (A^{-1})_{jm} (A^{-1})_{ln} +

\displaystyle (A^{-1})_{ik} (A^{-1})_{jn} (A^{-1})_{lm} + (A^{-1})_{il} (A^{-1})_{jk} (A^{-1})_{mn} +

\displaystyle (A^{-1})_{il} (A^{-1})_{jm} (A^{-1})_{kn} + (A^{-1})_{il} (A^{-1})_{jn} (A^{-1})_{km} +

\displaystyle (A^{-1})_{im} (A^{-1})_{jk} (A^{-1})_{ln} + (A^{-1})_{im} (A^{-1})_{jl} (A^{-1})_{kn} +

\displaystyle (A^{-1})_{im} (A^{-1})_{jn} (A^{-1})_{kl} + (A^{-1})_{in} (A^{-1})_{jk} (A^{-1})_{lm} +

\displaystyle (A^{-1})_{in} (A^{-1})_{jl} (A^{-1})_{km} + (A^{-1})_{in} (A^{-1})_{jm} (A^{-1})_{kl}



The propagator is

\displaystyle D(x) = -i\int\frac{d^3k}{(2\pi)^3 2\omega_k}\left[e^{-i(\omega_k t - \mathbf{k}\cdot\mathbf{x})}\theta(x^0) + e^{i(\omega_k t - \mathbf{k}\cdot\mathbf{x})}\theta(-x^0)\right].

As the integral is Lorentz invariant, we can choose x^0 = +\epsilon, \mathbf{x} = (0, 0, |\mathbf{x}|):

\displaystyle D(x) = -i\int\frac{d^3k}{(2\pi)^3 2\omega_k}e^{i(\mathbf{k}_z|\mathbf{x}|)}.

If we express the 3-momentum volume element as

\displaystyle d^3k = \mathbf{k}^2 \sin \theta d|\mathbf{k}| d\theta d\phi,

we get

\displaystyle D(x) = -i\frac{1}{2(2\pi)^3}\int d|\mathbf{k}| d\theta d\phi \sin \theta\frac{\mathbf{k}^2}{\sqrt{\mathbf{k}^2 + m^2}}  e^{i(|\mathbf{k}||\mathbf{x}|\cos \theta)}.

Integrating out \phi and reordering factors:

\displaystyle D(x) = i\frac{1}{2(2\pi)^2}\int d|\mathbf{k}| \frac{\mathbf{k}^2}{\sqrt{\mathbf{k}^2 + m^2}} \int (-d\theta \sin \theta) e^{i(|\mathbf{k}||\mathbf{x}|\cos \theta)}

\displaystyle D(x) = -i\frac{1}{2(2\pi)^2}\int d|\mathbf{k}| \frac{\mathbf{k}^2}{\sqrt{\mathbf{k}^2 + m^2}} \int_{-1}^1 d(\cos \theta) e^{i(|\mathbf{k}||\mathbf{x}|\cos \theta)}.

Integrating over \cos \theta:

\displaystyle D(x) = i\frac{1}{2(2\pi)^2}\int d|\mathbf{k}| \frac{\mathbf{k}^2}{\sqrt{\mathbf{k}^2 + m^2}} \left[\frac{e^{i(|\mathbf{k}||\mathbf{x}| u)}}{i|\mathbf{k}||\mathbf{x}|}\right]_{-1}^1

\displaystyle D(x) = i\frac{1}{(2\pi)^2}\int d|\mathbf{k}| \frac{\mathbf{k}^2}{|\mathbf{k}||\mathbf{x}|\sqrt{\mathbf{k}^2 + m^2}} \left[\frac{e^{i(|\mathbf{k}||\mathbf{x}|)}-e^{-i(|\mathbf{k}||\mathbf{x}|)}}{2i}\right]

\displaystyle D(x) = i\frac{1}{(2\pi)^2}\int d|\mathbf{k}| \frac{\mathbf{k}^2}{|\mathbf{k}||\mathbf{x}|\sqrt{\mathbf{k}^2 + m^2}} \sin(|\mathbf{k}||\mathbf{x}|)

\displaystyle D(x) = \frac{1}{(2\pi)^2|\mathbf{x}|}\int_0^{+\infty} d|\mathbf{k}| \frac{|\mathbf{k}|}{\sqrt{\mathbf{k}^2 + m^2}} (i\sin(|\mathbf{k}||\mathbf{x}|)).

Completing the integral, following Michael Kachelrieß and using r = |\mathbf{x}|:

\displaystyle D(x) = \frac{1}{2(2\pi)^2r}\int_{-\infty}^{+\infty} dk \frac{k}{\sqrt{k^2 + m^2}} (\cos(k r) + i\sin(k r))

\displaystyle D(x) = \frac{1}{2(2\pi)^2r}\int_{-\infty}^{+\infty} dk \frac{k}{\sqrt{k^2 + m^2}} e^{i k r}

\displaystyle D(x) = -\frac{i}{2(2\pi)^2r}\frac{\partial}{\partial r}\int_{-\infty}^{+\infty} dk \frac{1}{\sqrt{k^2 + m^2}} e^{i k r}

From Mathematica:

\displaystyle D(x) = \frac{2im}{2(2\pi)^2r}K_1(m r)

\displaystyle D(x) = \frac{im}{(2\pi)^2r}K_1(m r)

From Wikipedia: D(x) is exponentially decaying.


The propagator must be given by

\displaystyle D(x) = \int \frac{d^2k}{(2\pi)^2}\frac{e^{ikx}}{k^2-m^2+i\epsilon}.

Setting x^0 to zero, we get:

\displaystyle D(x^1) = \int \frac{d^2k}{(2\pi)^2}\frac{e^{-ik_1x^1}}{k_0^2-k_1^2-m^2+i\epsilon}

\displaystyle = \int \frac{dk_1}{(2\pi)^2}e^{-ik_1x^1} \int dk_0\frac{1}{k_0^2-k_1^2-m^2+i\epsilon},

and, using \omega_k^2 = k_1^2 + m^2,

\displaystyle = \int \frac{dk_1}{(2\pi)^2}e^{-ik_1x^1} \int dk_0\frac{1}{k_0^2-\omega_k^2+i\epsilon}.

As in the case in the book, we have two poles: one at \omega_k - i\epsilon and another at -\omega_k + i\epsilon. But in this case we don’t have the exponential divergence problem, so we can choose any contour. Let’s try both to check that they give the same result.

Counterclockwise, upper half plane, simple pole at -\omega_k + i\epsilon:

\displaystyle \int_{-\infty}^{+\infty} dk_0\frac{1}{k_0^2-\omega_k^2+i\epsilon} = 2\pi i \mathrm{Res}\left(\frac{1}{k_0^2-\omega_k^2+i\epsilon}, -\omega_k + i\epsilon\right)

\displaystyle \approx 2\pi i \frac{1}{2(-\omega_k)}

\displaystyle = -\frac{i\pi}{\omega_k}.

Clockwise, lower half plane, simple pole at \omega_k - i\epsilon:

\displaystyle \int_{-\infty}^{+\infty} dk_0\frac{1}{k_0^2-\omega_k^2+i\epsilon} = -2\pi i \mathrm{Res}\left(\frac{1}{k_0^2-\omega_k^2+i\epsilon}, \omega_k - i\epsilon\right)

\displaystyle \approx -2\pi i \frac{1}{2\omega_k}

\displaystyle = -\frac{i\pi}{\omega_k}.

Yes, they match. Adding this result to the bigger integral:

\displaystyle D(x^1) = -\int \frac{dk_1}{(2\pi)^2}e^{-ik_1x^1} \frac{i\pi}{\omega_k}

\displaystyle = -i\int \frac{dk_1}{(2\pi)2\omega_k}e^{-ik_1x^1}

\displaystyle = -\frac{i}{4\pi}\int dk_1 \frac{e^{-ix^1k_1}}{\sqrt{k_1^2 + m^2}}.

Using Wolfram Alpha we get

\displaystyle D(x^1) = -\frac{i}{2\pi}K_0(m|x^1|),

where K_0 is a modified Bessel function that decays exponentially.


We need to show that the advanced propagator,

\displaystyle D_{adv}(x) = \int \frac{d^4k}{(2\pi)^4}\frac{e^{ikx}}{k^2-m^2+i \,\mathrm{sgn}(k_0)\epsilon},

only propagates to the future. Let’s start as Zee does, integrating over k_0 by the method of contours and, first of all , getting the position of the poles:

\displaystyle k^2 - m^2 - i \,\mathrm{sgn}(k_0)\epsilon = 0

\displaystyle k_0^2 - (\mathbf{k}^2 + m^2) - i \,\mathrm{sgn}(k_0)\epsilon = 0

\displaystyle k_0^2  = (\mathbf{k}^2 + m^2) + i \,\mathrm{sgn}(k_0)\epsilon

Using \pm and \mathrm{sgn} in an open sense and defining \omega_k^2 = \mathbf{k}^2 + m^2, we get:

\displaystyle k_0  = \pm\sqrt{\omega_k^2 + i \,\mathrm{sgn}(k_0)\epsilon}

\displaystyle = \pm\sqrt{\omega_k^2 \pm i \,\epsilon}.

If \epsilon = 0, we have k_0 = \pm \omega_k and, as \epsilon is small, we can linearize the function around the old pole:

\displaystyle \sqrt{z} \approx \sqrt{z_0} + \frac{1}{2\sqrt{z_0}}(z-z_0)

\displaystyle \sqrt{\omega_k^2 \pm i \,\epsilon} \approx \sqrt{\omega_k^2} + \frac{1}{2\sqrt{\omega_k^2}}(\pm i \,\epsilon)

\displaystyle \sqrt{\omega_k^2 \pm i \,\epsilon} \approx \omega_k \pm \frac{ i \,\epsilon}{2\omega_k}.

Replacing into the original expression and taking into account that \epsilon is a generical term for “small positive value”:

\displaystyle k_0  = \pm (\omega_k \pm  i \,\epsilon)

\displaystyle = \pm \omega_k + i \,\epsilon,

so the two poles are in the upper half of the complex plane.

When closing the contour we need to be sure that the integral over the contour converges, so the contour must be closed over the side that makes the exponential small (because an exponential divergence would overwhelm the 1/k_0^2 decay). As the integral is

\displaystyle D_{adv}(x) = \int \frac{d^3\mathbf{k}}{(2\pi)^4}e^{-i\mathbf{k}\cdot\mathbf{x}}\int dk_0 \frac{e^{ik_0 x^0}}{k^2-m^2-i \,\mathrm{sgn}(k_0)\epsilon},

positive values of x^0 give us the requirement of values of k_0 with positive imaginary component to ensure an exponent with a negative real component. So in this case we would catch the poles inside the contour and get a nonzero result. In the other case, x^0 < 0m the contour would have to be closed in the lower half plane, missing the poles and getting a zero result.

The retarded propagator is symmetrical, only switching the position of the poles.



Let’s start with a derivation inside the chapter:

\displaystyle W(J) = -\frac{1}{2}\int\int d^4x d^4y\,J(x)D(x-y)J(y)

\displaystyle = -\frac{1}{2}\int\int d^4x d^4y\,\left(\int \frac{d^4k}{(2\pi)^4} e^{-ikx} J^*(k)\right)\left(\int \frac{d^4k'}{(2\pi)^4} e^{ik'(x-y)} D(k')\right)

\displaystyle \left(\int \frac{d^4k''}{(2\pi)^4} e^{ik''y} J(k'')\right)

Moving all the differentials together, removing the multiple integral signs and reordering the exponentials:

\displaystyle W(J) = -\frac{1}{2}\int d^4x d^4y\frac{d^4k}{(2\pi)^4}\frac{d^4k'}{(2\pi)^4}\frac{d^4k''}{(2\pi)^4} \, e^{-ikx} J^*(k) e^{ik'(x-y)} D(k') e^{ik''y} J(k'')

\displaystyle = -\frac{1}{2}\int d^4x \frac{d^4k}{(2\pi)^4}\frac{d^4k'}{(2\pi)^4}\frac{d^4k''}{(2\pi)^4} \, e^{-ikx} J^*(k) e^{ik'x} D(k') J(k'')\int d^4y\, e^{i(k''-k')y}.

Doing the last integration (observe that a (2\pi)^4 factor is given by the integration of the exponential):

\displaystyle W(J) = -\frac{1}{2}\int d^4x \frac{d^4k}{(2\pi)^4}\frac{d^4k'}{(2\pi)^4}d^4k'' \, e^{-ikx} J^*(k) e^{ik'x} D(k') J(k'')\delta(k''-k')

\displaystyle = -\frac{1}{2}\int d^4x \frac{d^4k}{(2\pi)^4}\frac{d^4k'}{(2\pi)^4} \, e^{-ikx} J^*(k) e^{ik'x} D(k') J(k').

Combining the exponentials and integrating:

\displaystyle W(J) = -\frac{1}{2}\int \frac{d^4k}{(2\pi)^4}\frac{d^4k'}{(2\pi)^4}\, J^*(k) D(k') J(k') \int d^4x\,e^{i(k'-k)x}

\displaystyle = -\frac{1}{2}\int \frac{d^4k}{(2\pi)^4}\frac{d^4k'}{(2\pi)^4}\, J^*(k) D(k') J(k') (2\pi)^4\delta(k'-k)

\displaystyle = -\frac{1}{2}\int \frac{d^4k}{(2\pi)^4}d^4k'\, J^*(k) D(k') J(k') \delta(k'-k)

\displaystyle = -\frac{1}{2}\int \frac{d^4k}{(2\pi)^4}\, J^*(k) D(k) J(k)

\displaystyle = -\frac{1}{2}\int \frac{d^4k}{(2\pi)^4}\, J^*(k) \frac{1}{k^2 - m^2 + i\epsilon} J(k)


Now let’s do the energy integral without consulting the appendix 😀

\displaystyle E = -\int \frac{d^3k}{(2\pi)^3} \frac{e^{i\mathbf{k}\cdot(\mathbf{x}_1 - \mathbf{x}_2})}{\mathbf{k}^2 + m^2}.

Let’s define \mathbf{r} = \mathbf{x}_1 - \mathbf{x}_2 and do the substitution d^3k = k^2 \sin\theta dk d\theta d\phi. By spherical symmetry, we can use \mathbf{r} as the “north pole”:

\displaystyle E = -\frac{1}{(2\pi)^3}\int \mathbf{k}^2 \sin\theta d|\mathbf{k}| d\theta d\phi \frac{e^{i |\mathbf{k}|r\,\cos\theta}}{\mathbf{k}^2 + m^2}

\displaystyle = \frac{1}{(2\pi)^2}\int \mathbf{k}^2  d|\mathbf{k}| \frac{1}{\mathbf{k}^2 + m^2}\int_0^\pi d(\cos\theta) \,e^{i |\mathbf{k}|r\,\cos\theta}

\displaystyle = -\frac{1}{(2\pi)^2}\int \mathbf{k}^2  d|\mathbf{k}| \frac{1}{\mathbf{k}^2 + m^2} \left[ \frac{\,e^{i |\mathbf{k}|r\,u}}{i |\mathbf{k}|r}\right]_{-1}^1

\displaystyle = -\frac{1}{(2\pi)^2r}\int |\mathbf{k}|  d|\mathbf{k}| \frac{1}{\mathbf{k}^2 + m^2} \left[ \frac{\,e^{i |\mathbf{k}|r}-e^{-i |\mathbf{k}|r}}{i}\right]

\displaystyle = -\frac{2}{(2\pi)^2r}\int |\mathbf{k}|  d|\mathbf{k}| \frac{1}{\mathbf{k}^2 + m^2} \frac{e^{i |\mathbf{k}|r}-e^{-i |\mathbf{k}|r}}{2i}

\displaystyle = -\frac{2}{(2\pi)^2r}\int_0^{+\infty} p \,dp \frac{\sin pr}{p^2 + m^2}

\displaystyle = -\frac{1}{i(2\pi)^2r}\int_{-\infty}^{+\infty} \,dp \frac{ip \sin pr}{p^2 + m^2}

\displaystyle = -\frac{1}{i(2\pi)^2r}\int_{-\infty}^{+\infty} \,dp \frac{p (\cos pr + i\sin pr)}{p^2 + m^2} (adds an odd function)

\displaystyle = -\frac{1}{i(2\pi)^2r}\int_{-\infty}^{+\infty} \,dp \frac{p e^{ipr}}{p^2 + m^2}

Closing the contour (it’s not so easy to prove that the complex section doesn’t contribute but it’s true) through the upper half plane (a positive imaginary part of p gives a negative exponent) we pick up the pole at p = +im:

\displaystyle \int_{-\infty}^{+\infty} \,dp \frac{p e^{ipr}}{p^2 + m^2} = 2\pi i \mathrm{Res}\left(\frac{p e^{ipr}}{p^2 + m^2}, im\right)

\displaystyle = 2\pi i \frac{i m\, e^{-mr}}{2 i m}

\displaystyle = i \pi e^{-mr}.

Replacing this value for the integral:

\displaystyle E = -\frac{1}{i(2\pi)^2r} i \pi e^{-mr}

\displaystyle = -\frac{e^{-mr}}{4\pi r}


We can start with the 2-dimensional form of (6):

\displaystyle E = -\int \frac{d^2k}{(2\pi)^2} \frac{e^{i\mathbf{k}\cdot\mathbf{r}}}{\mathbf{k}^2+m^2}

\displaystyle = -\int \frac{d^2k}{(2\pi)^2} \frac{e^{i p r\cos\theta}}{p^2+m^2}

\displaystyle = -\frac{1}{(2\pi)^2}\int_0^{+\infty} dp \frac{p}{p^2+m^2} \int_0^{2\pi}d\theta\,e^{i p r\cos\theta}

Expressing the angular part of the integral as a Bessel Function:

\displaystyle \int_0^{2\pi}d\theta\,e^{i p r\cos\theta} = 2\pi\frac{1}{\pi}\int_0^{\pi}d\theta\,e^{i p r\cos\theta} = 2\pi J_0(p r)

Putting this result in the previous expression:

\displaystyle E = -\frac{1}{2\pi}\int_0^{+\infty} dp \frac{p}{p^2+m^2} J_0(p r)


\displaystyle E = -\frac{1}{2\pi} K_0(m r),

where K_0(z) is a modified Bessel function of the second kind and m \neq 0. If m = 0 the energy would seem constant, but agreement with electrostatic theory would seem to require a logarithmic potential and the integral certainly is divergent for m = 0. Let’s use a reference at r = 1 to avoid this problems:

\displaystyle E' = -\frac{1}{2\pi}\int_0^{+\infty} dp \frac{p}{p^2} J_0(p r) + \frac{1}{2\pi}\int_0^{+\infty} dp \frac{p}{p^2} J_0(p)

\displaystyle = \frac{1}{2\pi}\int_0^{+\infty} dp \frac{J_0(p) - J_0(p r)}{p}

Using Mathematica:

\displaystyle E' = -\frac{1}{2\pi}\ln r

Let’s analyze now the D-dimensional form of (6):

\displaystyle E = -\int \frac{d^Dk}{(2\pi)^D} \frac{e^{i\mathbf{k}\cdot\mathbf{r}}}{\mathbf{k}^2+m^2}

We can express the denominator as an integral over an auxiliary variable:

\displaystyle \frac{1}{\mathbf{k}^2+m^2} = \int_0^{+\infty} dx e^{-x(\mathbf{k}^2 + m^2)}

\displaystyle E = -\int_0^{+\infty} dx \int \frac{d^Dk}{(2\pi)^D} e^{-x(\mathbf{k}^2 + m^2)} e^{i\mathbf{k}\cdot\mathbf{r}}

\displaystyle = -\int_0^{+\infty} dx \int \frac{d^Dk}{(2\pi)^D} e^{-x \mathbf{k}^2 + i\mathbf{k}\cdot\mathbf{r} - x m^2}

\displaystyle = -\int_0^{+\infty} dx \int \frac{d^Dk}{(2\pi)^D} e^{-x \left(\mathbf{k}^2 + \frac{i}{x}\mathbf{k}\cdot\mathbf{r} + \frac{1}{4}\frac{i^2}{x^2}\mathbf{r}^2\right) - \frac{1}{4}x\,\frac{i^2}{x^2}\mathbf{r}^2 - x m^2}

\displaystyle = -\int_0^{+\infty} dx \int \frac{d^Dk}{(2\pi)^D} e^{-x \left(\mathbf{k}^2 + \frac{i}{2x}\mathbf{r}\right)^2 + \frac{1}{4x}\,\mathbf{r}^2 - x m^2}

Setting aside the \mathbf{k} dependent terms:

\displaystyle E = -\int_0^{+\infty} dx e^{-\frac{1}{4x}\,\mathbf{r}^2 - x m^2} \int \frac{d^Dk}{(2\pi)^D} e^{-x \left(\mathbf{k} + \frac{i}{2x}\mathbf{r}\right)^2}

Substituting \mathbf{l} = \mathbf{k} + (i / 2x)\mathbf{r}:

\displaystyle E = -\int_0^{+\infty} dx e^{-\frac{1}{4x}\,\mathbf{r}^2 - x m^2} \int \frac{d^Dl}{(2\pi)^D} e^{-x \mathbf{l}^2}

Doing the n-dimensional gaussian integral:

\displaystyle E = -\int_0^{+\infty} dx e^{-\frac{1}{4x}\,\mathbf{r}^2 - x m^2} \frac{1}{(2\pi)^D}\left(\frac{(2\pi)^D}{(2x)^D}\right)^\frac{1}{2}

\displaystyle = -\int_0^{+\infty} dx e^{-\frac{1}{4x}\,\mathbf{r}^2 - x m^2} \frac{1}{(2\pi)^\frac{D}{2}(2x)^\frac{D}{2}}

\displaystyle = -2^{-D}\pi^{-\frac{D}{2}}\int_0^{+\infty} dx\,x^{-\frac{D}{2}} e^{-\frac{1}{4x}\,\mathbf{r}^2 - x m^2}

Using Wolfram Alpha:

\displaystyle \int_0^{+\infty} dx\,x^A e^{-\frac{B}{4x} - Cx} = 2^{-A} B^\frac{A+1}{2} C^\frac{-A-1}{2} K_{-A-1}\left(\sqrt{B} \sqrt{C}\right)

\displaystyle \int_0^{+\infty} dx\,x^{-\frac{D}{2}} e^{-\frac{1}{4x}\,\mathbf{r}^2 - x m^2} = 2^\frac{D}{2} \left(\mathbf{r}^2\right)^\frac{1-D/2}{2} \left(m^2\right)^\frac{D/2-1}{2} K_{D/2-1}\left(|\mathbf{r}| m\right)

\displaystyle = 2^\frac{D}{2} r^{1-\frac{D}{2}} m^{\frac{D}{2}-1} K_{\frac{D}{2}-1}(mr)

Putting the result of the integral in the expression for the energy:

\displaystyle E = -2^{-D}\pi^{-\frac{D}{2}} 2^\frac{D}{2} r^{1-\frac{D}{2}} m^{\frac{D}{2}-1} K_{\frac{D}{2}-1}(mr)

\displaystyle = -2^{-\frac{D}{2}}\pi^{-\frac{D}{2}} r^{1-\frac{D}{2}} m^{\frac{D}{2}-1} K_{\frac{D}{2}-1}(mr)

\displaystyle = -(2\pi)^{-\frac{D}{2}} r^{1-\frac{D}{2}} m^{\frac{D}{2}-1} K_{\frac{D}{2}-1}(mr)

Checking for values of D with known results:

\displaystyle E_{D=2} = -(2\pi)^{-\frac{2}{2}} r^{1-\frac{2}{2}} m^{\frac{2}{2}-1} K_{\frac{2}{2}-1}(mr)

\displaystyle = -(2\pi)^{-1} r^{1-1} m^{1-1} K_{1-1}(mr)

\displaystyle = -\frac{1}{2\pi}K_0(mr)

\displaystyle E_{D=3} = -(2\pi)^{-\frac{3}{2}} r^{1-\frac{3}{2}} m^{\frac{3}{2}-1} K_{\frac{3}{2}-1}(mr)

\displaystyle = -(2\pi)^{-\frac{3}{2}} r^-\frac{1}{2} m^\frac{1}{2} K_\frac{1}{2}(mr)

Using Wolfram Alpha:

\displaystyle E_{D=3} = - 2^{-\frac{3}{2}} \pi^{-\frac{3}{2}} r^{-\frac{1}{2}} m^\frac{1}{2} \pi^\frac{1}{2} 2^{-\frac{1}{2}} m^{-\frac{1}{2}} r^{-\frac{1}{2}} e^{-mr}

\displaystyle = - 2^{-\frac{4}{2}} \pi^{-\frac{2}{2}} r^{-\frac{2}{2}} e^{-mr}

\displaystyle = - \frac{1}{4\pi r} e^{-mr}

2 thoughts on “Zee QFT – Part I

  1. Eden says:

    Hi, I want to know how to use Mathematica to calculate the last integration in 1.3.1. Can you teach me how? Many thanks!

    • mchouza says:

      You can just evaluate the integral, though you will get a cleaner result if you restrict the domain of r and m:


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