The Klein-Gordon Propagator

1. Klein-Gordon Equation

The Klein-Gordon equation is a relativistic wave equation, related to the Schrödinger equation. It is second order in space and time and manifestly Lorentz covariant. It is a quantized version of the relativistic energy–momentum relation. Its solutions include a quantum scalar or pseudoscalar field, a field whose quanta are spinless particles.

The free Klein-Gordon equation is

(1)   \begin{equation*} (\partial^{2}+m^2)\phi = 0 \end{equation*}

where \partial ^2 = \partial^{\mu}\partial_{\mu} = \frac{\partial^2}{\partial t^2}-\nabla^2. In the Heisenberg picture, the field operator can be written as (Each Fourier mode of the field is treated as an independent oscillator)

(2)   \begin{equation*} \phi(x) = \int \frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2E_{\mathbf{p}}}} \left.\left(a_{\mathbf{p}}e^{-ip\cdot x}- a^{\dagger}_{\mathbf{p}}e^{ip\cdot x}\right) \right|_{p^0=E_{\mathbf{p}}} \end{equation*}

where p = (E, \mathbf{p}), x = (t, \mathbf{x}), p\cdot x = g_{\mu\nu}p^{\mu}x^{\nu} = p^0x^0-\mathbf{p\cdot x}, E_{\mathbf{p}} = \sqrt{|\mathbf{p}|^2+m^2}. \phi(x) acting on the vacuum creates a particle at x. Then, the amplitude for a particle to propagate from y to x is \langle 0 | \phi(x)\phi(y) | 0 \rangle, that is

(3)   \begin{equation*} D(x-y) = \langle 0 | \phi(x)\phi(y) | 0 \rangle = \int \frac{d^3p}{(2\pi)^3}\frac{1}{2E_{\mathbf{p}}} e^{-ip\cdot (x-y)} |_{p^0=E_{\mathbf{p}}} \end{equation*}

Acting (\partial^{2}+m^2) on D(x-y) will produce 0, thus D(x-y) is a solution of Klein-Gordon equation.

 

2. The Green’s Function

We may introduce a source to Klein-Gordon equation if the Klein-Gordon field coupled to an external field

(4)   \begin{equation*}  (\partial^{2}+m^2)\phi(x) = j(x) \end{equation*}

And then, we need the Green’s function which satisfy

(5)   \begin{equation*} (\partial^{2}+m^2)G(x) = -i\delta^{(4)}(x) \end{equation*}

With the help of Fourier’s Transformation

(6)   \begin{equation*} \tilde{f}(p) = \int d^4x e^{ip\cdot x} f(x) \end{equation*}

(7)   \begin{equation*} f(x) = \int \frac{d^4p}{(2\pi)^4} e^{-ip\cdot x} \tilde{f}(x) \end{equation*}

Eq.4 can be transformed into

(8)   \begin{equation*} (-p^2+m^2)\tilde{G}(p) = -i \end{equation*}

G(x) is just the inverse transformation of \tilde{G}(p)

(9)   \begin{equation*} G(x) = \int \frac{d^4p}{(2\pi)^4} \frac{i}{p^2-m^2} e^{-ip\cdot x} \end{equation*}

or

(10)   \begin{equation*}  G(x-y) = \int \frac{d^4p}{(2\pi)^4} \frac{i}{p^2-m^2} e^{-ip\cdot (x-y)} = \int \frac{d\mathbf{p}}{(2\pi)^3} \int \frac{dp^0}{2\pi} \frac{i}{{p^{0}}^2-E_{\mathbf{p}}} e^{-ip\cdot (x-y)} \end{equation*}

We can use contour integral to find out the above integration over p^0

 

3. The Contour Integration

Contour integration is an application of residue theorem. An integral over a closed path on the complex plane can be evaluated by its residues within this path

(11)   \begin{equation*} \ointctrclockwise f(z)dz = 2\pi i \sum \text{Res}(f,a_k) \end{equation*}

Let’s denote the integral over p^0 as I

(12)   \begin{align*} I &= \int dp^0 f(p^0) \\ &= \int \frac{dp^0}{2\pi} \frac{i}{{p^{0}}^2-E_{\mathbf{p}}} e^{-ip\cdot (x-y)} \end{align*}

Depends on whether x^0 is larger or smaller than y^0, there are two sets of contour as be chosen.

Contours for x^0>y^0

a. If x^0>y^0, we chosen the contour with C_R beneath the real axis. Since e^{-ip\cdot (x-y)} will contribute a e^{-r} factor on C_R, which make the integral on this curve becomes 0. Besides the C_R, we also have the freedom to choose C_{\epsilon_1} and C_{\epsilon_2}. But it turns out that it will make no difference. Take the contour (a) for example, the residue theorem is

(13)   \begin{equation*}  \varointctrclockwise f(z)dz = -2\pi i \left[ \text{Res}(f,-E_{\mathbf{p}}) + \text{Res}(f,E_{\mathbf{p}})\right] \end{equation*}

The contour integral can be decomposed into

(14)   \begin{equation*} \varointctrclockwise f(z)dz = \left(\int\limits_{\text{on real axis}} + \int\limits_{C_{\epsilon_1}} + \int\limits_{C_{\epsilon_2}} + \int\limits_{C_R} \right)f(z)dz \end{equation*}

In the limit R \to \infty,

(15)   \begin{equation*} \int\limits_{C_R} = 0 \end{equation*}

In the limit \epsilon \to 0,

(16)   \begin{equation*} \int\limits_{\text{on real axis}} = I \end{equation*}

(17)   \begin{equation*} \int\limits_{C_{\epsilon_1}} = -\pi i \text{Res}(f,-E_{\mathbf{p}}) \end{equation*}

(18)   \begin{equation*} \int\limits_{C_{\epsilon_2}} = -\pi i \text{Res}(f,E_{\mathbf{p}}) \end{equation*}

Take these results into Eq.13, we can get

(19)   \begin{align*} I &= -\pi i \left[ \text{Res}(f,-E_{\mathbf{p}}) + \text{Res}(f,E_{\mathbf{p}})\right] \\ &= \frac{1}{2}\left(\frac{1}{2E_{\mathbf{p}}} e^{-ip\cdot(x-y)} |_{p^0=E_{\mathbf{p}}} -\frac{1}{2E_{\mathbf{p}}} e^{-ip\cdot(x-y)} |_{p^0=-E_{\mathbf{p}}} \right) \end{align*}

We can get the same result if we take contour (b), although these is no singular points within this contour

(20)   \begin{equation*} \varointctrclockwise f(z)dz = 0 \end{equation*}

The integral on C_{\epsilon_1} and C_{\epsilon_2} will be different since they are on the other half circles.

(21)   \begin{equation*} \int\limits_{C_{\epsilon_1}} = \pi i \text{Res}(f,-E_{\mathbf{p}}) \end{equation*}

(22)   \begin{equation*} \int\limits_{C_{\epsilon_2}} = \pi i \text{Res}(f,E_{\mathbf{p}}) \end{equation*}

considering the above adjustment, I is still -\pi i \left[ \text{Res}(f,-E_{\mathbf{p}}) + \text{Res}(f,E_{\mathbf{p}})\right]. The contour (c) and (d) also give us the same result.

Contours for x^0<y^0

b. If x^0<y^0, we chosen the contour with C_R above the real axis to insure \int\limits_{C_R} = 0. The contour (a) will give us

(23)   \begin{equation*} \ointctrclockwise f(z)dz = 0 \end{equation*}

(24)   \begin{equation*} \left(\int\limits_{\text{on real axis}} + \int\limits_{C_{\epsilon_1}} + \int\limits_{C_{\epsilon_2}} + \int\limits_{C_R} \right)f(z)dz = 0 \end{equation*}

(25)   \begin{equation*} I -\pi i \text{Res}(f,-E_{\mathbf{p}}) - \pi i \text{Res}(f,E_{\mathbf{p}}) + 0 = 0 \end{equation*}

thus

(26)   \begin{align*} I &= \pi i \left[ \text{Res}(f,-E_{\mathbf{p}}) + \text{Res}(f,E_{\mathbf{p}})\right] \\ &= -\frac{1}{2}\left(\frac{1}{2E_{\mathbf{p}}} e^{-ip\cdot(x-y)} |_{p^0=E_{\mathbf{p}}} -\frac{1}{2E_{\mathbf{p}}} e^{-ip\cdot(x-y)} |_{p^0=-E_{\mathbf{p}}} \right) \end{align*}

Contour (b),(c),(d) will give us the same result.

Take I into Eq.10, we can get the Green’s function

(27)   \begin{equation*} G(x-y)= \left\{\begin{array}{rl} \frac{1}{2}\left[D(x-y)-D(y-x)\right] & \text{if } x^0 > y^0 \\ -\frac{1}{2}\left[D(x-y)-D(y-x)\right] & \text{if } x^0 < y^0 \end{array}\right. \end{equation*}

 

4. Feynman Propagator

Green’s function plus any solutions of the homogeneous equation is still Green’s function. Add a \frac{1}{2}\left[D(x-y)-D(y-x)\right] to G(x-y) will give us the retarded Green’s function

(28)   \begin{equation*} D_R(x-y) \equiv \theta(x^0-y^0)\left[D(x-y)-D(y-x)\right] = \langle 0 |[\phi(x), \phi(y)] | 0 \rangle \end{equation*}

We can get the Feynman propagator if we add \frac{1}{2}\left[D(x-y)+D(y-x)\right] to G(x-y),

(29)   \begin{align*} D_F(x-y) &=\left\{\begin{array}{rl} D(x-y) & \text{if } x^0 > y^0 \\ D(y-x) & \text{if } x^0 < y^0 \end{array}\right. \\ & \equiv \langle 0 |T\phi(x)\phi(y) | 0 \rangle \end{align*}

Feynman propagator can also be written as

(30)   \begin{equation*} D_F(x-y) = \int \frac{d^4p}{(2\pi)^4} \frac{i}{p^2-m^2+2i\epsilon} e^{-ip\cdot (x-y)} \end{equation*}

which has poles at p^0=\pm(E_{\mathbf{p}}-i\epsilon) and no poles on the real axis. This integral can be find out directly with the help of contour integral.
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Refferences

[1] https://en.wikipedia.org/wiki/Klein–Gordon_equation

[2] Peskin, Michael E., and Daniel V. Schroeder. “An Introduction to quantum field theory.” (1995).

[3] 梁昆淼. 数学物理方法(第四版). 高等教育出版社. 2010.

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