Photon Propagator with Retarded Green’s Function

Some prerequisites: complex analysis, Fourier transform, Laplace transform (optional if you are a physicist).

1. Introduction

In this post, I will solve Maxwell’s equation with source term. This would provide an idea on what it means by “propagation”. Of course, as in other posts for this blog, the material is standard, but I present it in the way that I prefer. Note that at some points of this post, the presentations are not rigorous as I still do not completely understand this subject. Furthermore, I am stuck at the final step where I have no idea if the final integral can be integrated out.

2. Green’s function for d’Alembertian operator

Before directly working with Maxwell’s equations, let us consider an important toy model. Consider a field {\phi(t,\vec{x}).} Suppose that it is influenced by the source {{\rho}(t,\vec{x})} through the inhomogeneous differential equation

\displaystyle  (-{\partial}_t^2 + \nabla^2){\phi}(t,\vec{x}) = {\rho}(t,\vec{x}), \ \ \ \ \ (1)

where {(\vec{x}) = (x,y,z),} and

\displaystyle  {\partial}_t^2\equiv \frac{{\partial}^2}{{\partial} t^2},\qquad \nabla^2\equiv \frac{{\partial}^2}{{\partial} x^2}+\frac{{\partial}^2}{{\partial} y^2} + \frac{{\partial}^2}{{\partial} z^2}. \ \ \ \ \ (2)

In fact, I have made use of the convention by setting speed of light {c=1.} Roughly interpreting eq.(1), the source {{\rho}(t,\vec{x})} produces the field {\phi(t,\vec{x}).} In more details but still rough, the field {\phi(t,\vec{x})} is modified in response to the presence of the source {{\rho}(t,\vec{x}).}

Let us suppose that initially, the source is absence and the field is identically zero everywhere. One normally waits for quite a relatively long while from the initial time until one turns on the source. During the time when the source is still off, the field remains zero. But once the source is turned on, the field in influenced by the source through eq.(1). We would like to describe the values of the field after the source is turned on.

Having read the above sentence once let me repeat it again, but this time I include symbols and more details:

“ Let us suppose that initially {t\rightarrow-\infty}, the source is absence {{\rho}(-\infty,\vec{x})=0} and the field is identically zero everywhere {{\phi}(-\infty,\vec{x})=0}. One normally waits for quite a relatively long while from the initial time until one turns on the source. This means that one turns on the source at a finite {t = t_0.} During the time when the source is still off, the field remains zero, i.e. {{\phi}(t,\vec{x}) = 0} for {t<t_0.} But once the source is turned on, the field in influenced by the source through eq.(1): {(-{\partial}_t^2 + \nabla^2){\phi}(t,\vec{x}) = {\rho}(t,\vec{x}).} We would like to describe the values of the field after the source is turned on, i.e. {{\phi}(t,\vec{x}) = ?} for {t>t_0.}

The first step in studying this problem is to solve eq.(1) by Fourier expanding the source and the field as

\displaystyle  {\phi}(t,\vec{x}) =\int \frac{d{\omega}}{2{\pi}}\int \frac{d^3\vec{k}}{(2{\pi})^3} \tilde{\phi}({\omega},\vec{k})e^{-i{\omega} t+i\vec{k}\cdot\vec{x}}, \ \ \ \ \ (3)

\displaystyle  {\rho}(t,\vec{x}) =\int \frac{d{\omega}}{2{\pi}}\int \frac{d^3\vec{k}}{(2{\pi})^3} \tilde{\rho}({\omega},\vec{k})e^{-i{\omega} t+i\vec{k}\cdot\vec{x}}, \ \ \ \ \ (4)

where the integral range for {{\omega}} is {(-\infty,\infty)} and the integral range for {\vec{k}} is {(-\infty,\infty)^3.} But in the calculation, we will see that {{\omega}} is extended to a complex number. I believe that in order to do so properly, one should have used inverse Laplace transform like in the case for mathematicians and engineers. [One difference between Laplace transform and Fourier transform is that for Fourier transform, the transformed variable (e.g. angular frequency {{\omega}}) is a real number whereas the one for Laplace transform is a complex number. But physicists remedy this by saying something like `well, in order to describe the real world, let us allow the frequency in Fourier transform to be complex. The imaginary part will instruct how the amplitude of the wave decays’. I think this is not necessary because one should have used Laplace transform in the first place, and the decaying of the amplitude would already be taken into account.] However, being trained as a physicist for so long, I am not so familiar with Laplace transform. So let me instead keep using physicist’s convention by working non-rigorously with Fourier transform and extends the variable to be complex when necessary.

In fact eq.(3)(4) are still not suitable with the subsequent calculation. In order to do so, needs to impose an extra real-world condition into account. Suppose that when turning on the source, its value does not jump up at instantly. Instead, its value gradually increases. This is known as adiabatic approximation. In order to model this, we introduce a factor {e^{{\epsilon} t}} with small {{\epsilon}>0} to the field and source:

\displaystyle  {\phi}(t,\vec{x}) =\int \frac{d{\omega}}{2{\pi}}\int \frac{d^3\vec{k}}{(2{\pi})^3} \tilde{\phi}({\omega},\vec{k})e^{-i{\omega} t+i\vec{k}\cdot\vec{x}}e^{{\epsilon} t}, \ \ \ \ \ (5)

\displaystyle  {\rho}(t,\vec{x}) =\int \frac{d{\omega}}{2{\pi}}\int \frac{d^3\vec{k}}{(2{\pi})^3} \tilde{\rho}({\omega},\vec{k})e^{-i{\omega} t+i\vec{k}\cdot\vec{x}}e^{{\epsilon} t}. \ \ \ \ \ (6)

This extra factor vanishes at initial time {t\rightarrow-\infty,} but is finite at any time of measurement (we always measure the value of the field at finite {t.}). Then substituting eq.(5)(6) into eq.(1) gives

\displaystyle  \int \frac{d{\omega}}{2{\pi}}\int \frac{d^3\vec{k}}{(2{\pi})^3} (({\omega}+i{\epsilon})^2 - \vec{k}^2)\tilde{\phi}({\omega},\vec{k})e^{-i{\omega} t+i\vec{k}\cdot\vec{x}}e^{{\epsilon} t} = \int \frac{d{\omega}}{2{\pi}}\int \frac{d^3\vec{k}}{(2{\pi})^3} \tilde{\rho}({\omega},\vec{k})e^{-i({\omega}+i{\epsilon}) t+i\vec{k}\cdot\vec{x}}e^{{\epsilon} t}. \ \ \ \ \ (7)

This implies

\displaystyle  \tilde{\phi}({\omega},\vec{k}) =\frac{\tilde{\rho}({\omega},\vec{k})}{({\omega}+i{\epsilon})^2 - \vec{k}^2}. \ \ \ \ \ (8)

Substituting into eq.(3) and using Fourier transform [Until I can find a rigorous or satisfying explanation, let me be sloppy with the factor {e^{{\epsilon} t}} by intentionally forgetting it from time to time.]

\displaystyle  \tilde{\rho}({\omega},\vec{k}) =\int dt'\int d\vec{x}'{\rho}(t',\vec{x}') e^{i{\omega} t' - i\vec{k}\cdot\vec{x}} \ \ \ \ \ (9)


\displaystyle  {\phi}(t,\vec{x}) =\int dt'\int d^3\vec{x}' \int \frac{d{\omega}}{2{\pi}}\int \frac{d^3\vec{k}}{(2{\pi})^3} \frac{{\rho}(t',\vec{x}')}{({\omega}+i{\epsilon})^2 - \vec{k}^2} e^{-i{\omega} (t-t')+i\vec{k}\cdot(\vec{x}-\vec{x}')}. \ \ \ \ \ (10)

This equation is in the form

\displaystyle  {\phi}(t,\vec{x}) =\int dt'\int d^3\vec{x}' G(t-t',\vec{x}-\vec{x}'){\rho}(t',\vec{x}'), \ \ \ \ \ (11)

where {G(t-t',\vec{x}-\vec{x}')} is called Green’s function. In our case, it is given by

\displaystyle  G(t-t',\vec{x}-\vec{x}') =\int \frac{d{\omega}}{2{\pi}}\int \frac{d^3\vec{k}}{(2{\pi})^3} \frac{1}{({\omega}+i{\epsilon})^2 - \vec{k}^2} e^{-i{\omega} (t-t')+i\vec{k}\cdot(\vec{x}-\vec{x}')}. \ \ \ \ \ (12)

Furthermore, it satisfies

\displaystyle  (-{\partial}_t^2 + \nabla^2)G(t-t',\vec{x}-\vec{x}') ={\delta}(t-t'){\delta}^{(3)}(\vec{x}-\vec{x}'). \ \ \ \ \ (13)

Our task now is to simplify eq.(12). So

\displaystyle  \begin{array}{rl} G(t-t',\vec{x}-\vec{x}') &=\displaystyle\int \frac{d{\omega}}{2{\pi}}\int \frac{d^3\vec{k}}{(2{\pi})^3} \frac{1}{2|\vec{k}|}\left(\frac{1}{{\omega}+i{\epsilon} - |\vec{k}|}-\frac{1}{{\omega}+i{\epsilon} + |\vec{k}|}\right) e^{-i{\omega} (t-t')+i\vec{k}\cdot(\vec{x}-\vec{x}')}\\ &\\ &=\displaystyle{\cal P}\int \frac{d{\omega}}{2{\pi}}\int \frac{d^3\vec{k}}{(2{\pi})^3} \frac{1}{2|\vec{k}|}\left(\frac{1}{{\omega} - |\vec{k}|}-\frac{1}{{\omega} + |\vec{k}|}\right) e^{-i{\omega} (t-t')+i\vec{k}\cdot(\vec{x}-\vec{x}')}\\ &\qquad\displaystyle+\int \frac{d{\omega}}{2{\pi}}\int \frac{d^3\vec{k}}{(2{\pi})^3} \frac{i{\pi}}{2|\vec{k}|}\left(-{\delta}({\omega}-|\vec{k}|)+{\delta}({\omega}+|\vec{k}|)\right) e^{-i{\omega} (t-t')+i\vec{k}\cdot(\vec{x}-\vec{x}')}, \end{array} \ \ \ \ \ (14)

where we have made use of the Sokhotski-Plemelj formula

\displaystyle  \frac{1}{x+i{\epsilon}} = {\cal P}\frac{1}{x}-i{\pi}{\delta}(x) \ \ \ \ \ (15)

with {{\cal P}} being the principal part. Let us now proceed by considering

\displaystyle  \begin{array}{rl} \displaystyle{\cal P}\int_{-\infty}^{\infty} \frac{d{\omega}}{2{\pi}} \frac{1}{{\omega} - |\vec{k}|} e^{-i{\omega} (t-t')} &=\displaystyle{\cal P}\int_{-\infty}^{\infty} \frac{d{\omega}}{2{\pi}} \frac{1}{{\omega}} e^{-i({\omega} + |\vec{k}|) (t-t')}\\ &\\ &=\displaystyle-\frac{i}{2}sign(t-t')e^{-i|\vec{k}|(t-t')}, \end{array} \ \ \ \ \ (16)

where we used the identity [I might come back and add how to prove this identity]

\displaystyle  \displaystyle{\cal P}\int_{-\infty}^{\infty} dx \frac{e^{ixs}}{x} =i{\pi}\ sign(s). \ \ \ \ \ (17)


\displaystyle  \begin{array}{rl} \displaystyle{\cal P}\int_{-\infty}^{\infty} \frac{d{\omega}}{2{\pi}} \frac{1}{{\omega} + |\vec{k}|} e^{-i{\omega} (t-t')} &=\displaystyle{\cal P}\int_{-\infty}^{\infty} \frac{d{\omega}}{2{\pi}} \frac{1}{{\omega}} e^{-i({\omega} - |\vec{k}|) (t-t')}\\ &\\ &=\displaystyle-\frac{i}{2}sign(t-t')e^{i|\vec{k}|(t-t')}. \end{array} \ \ \ \ \ (18)


\displaystyle  \begin{array}{rl} G(t-t',\vec{x}-\vec{x}') &=\displaystyle\int \frac{d^3\vec{k}}{(2{\pi})^3}\left(-\frac{i}{2}\right)sign(t-t') \frac{1}{2|\vec{k}|}\left(e^{-i|\vec{k}|(t-t')}-e^{i|\vec{k}|(t-t')}\right) e^{i\vec{k}\cdot(\vec{x}-\vec{x}')}\\ &\qquad\displaystyle+\int \frac{d^3\vec{k}}{(2{\pi})^3} \frac{i}{4|\vec{k}|}\left(-e^{-i|\vec{k}| (t-t')}+e^{i|\vec{k}| (t-t')}\right) e^{i\vec{k}\cdot(\vec{x}-\vec{x}')}\\ &\\ &=\displaystyle-\int \frac{d^3\vec{k}}{(2{\pi})^3}(sign(t-t')+1) \frac{\sin(|\vec{k}|(t-t'))}{2|\vec{k}|} e^{i\vec{k}\cdot(\vec{x}-\vec{x}')}\\ &\\ &=\displaystyle-\int \frac{d^3\vec{k}}{(2{\pi})^3}\Theta(t-t') \frac{\sin(|\vec{k}|(t-t'))}{|\vec{k}|} e^{i\vec{k}\cdot(\vec{x}-\vec{x}')}, \end{array} \ \ \ \ \ (19)

where {\Theta(t-t')} is Heaviside step function. Let us use spherical polar coordinates in the space of {\vec{k}} such that the “{z-}axis” is aligned with {\vec{x}-\vec{x}'.} So the integral above simplifies as follows

\displaystyle  \begin{array}{rl} G(t-t',\vec{x}-\vec{x}') &=\displaystyle-\int_{0}^{2{\pi}}\int_0^{\pi}\int_{0}^\infty \frac{dk d{\theta} d{\phi}}{(2{\pi})^3}\ k^2\sin{\theta}\Theta(t-t') \frac{\sin(k(t-t'))}{k} e^{ikR\cos{\theta}}\\ &\\ &=\displaystyle-\int_{-1}^1\int_{0}^\infty \frac{dk \ d\cos{\theta}}{(2{\pi})^2}\ k\Theta(t-t') \sin(k(t-t')) e^{ikR\cos{\theta}}\\ &\\ &=\displaystyle-\int_{0}^\infty \frac{dk}{(2{\pi})^2}\ k\Theta(t-t') \sin(k(t-t')) \frac{1}{ikR}\left(e^{ikR}-e^{-ikR}\right)\\ &\\ &=\displaystyle-\int_{0}^\infty \frac{dk}{(2{\pi})^2}\ \Theta(t-t') \sin(k(t-t')) \frac{2}{R}\sin(kR)\\ &\\ &=\displaystyle-\int_{-\infty}^\infty \frac{dk}{(2{\pi})^2}\ \Theta(t-t') \sin(k(t-t')) \frac{1}{R}\sin(kR)\\ &\\ &=\displaystyle-\int_{-\infty}^\infty \frac{dk}{(2{\pi})^2}\ \Theta(t-t') \frac{1}{2R}(\cos(k(R-t+t'))-\cos(k(R+t-t')))\\ &\\ &=\displaystyle-\frac{1}{(2{\pi})}\ \Theta(t-t') \frac{1}{2R}({\delta}(R-t+t')-{\delta}(R+t-t')) \end{array} \ \ \ \ \ (20)

where {k\equiv|\vec{k}|, R\equiv |\vec{x}-\vec{x}'|.} In the above equation, each step is a simple manipulation and is left as an exercise for interested readers, especially students. [In fact, I do not feel like explaining the calculations in each step.] Next, note that due to the presence of {\Theta(t-t'),} Green’s function vanishes unless {t-t'>0.} This condition makes {R+t-t'>0,} which in turns implies {{\delta}(R+t-t')=0.} Therefore, the second term in the last line of eq.(20) does not contribute. This leaves us with

\displaystyle  G(t-t',\vec{x}-\vec{x}') =-\dfrac{1}{4{\pi} |\vec{x}-\vec{x}'|}\ \Theta(t-t'){\delta}(|\vec{x}-\vec{x}'|-t+t'). \ \ \ \ \ (21)

Substituting into eq.(11) gives

\displaystyle  \begin{array}{rl} \displaystyle{\phi}(t,\vec{x}) &=\displaystyle-\int dt'\int d^3\vec{x}' \dfrac{1}{4{\pi} |\vec{x}-\vec{x}'|}\ \Theta(t-t'){\delta}(|\vec{x}-\vec{x}'|-t+t'){\rho}(t',\vec{x}')\\ &\\ &=\displaystyle-\int d^3\vec{x}' \dfrac{1}{4{\pi} |\vec{x}-\vec{x}'|}{\rho}(t-|\vec{x}-\vec{x}'|,\vec{x}'). \end{array} \ \ \ \ \ (22)

The interpretation is that {{\phi}(t,\vec{x})} is the value of the field measured at {(t,\vec{x}),} and is obtained from the superposition of the response from the source. The sourse at {(t',\vec{x}')} will spend the time {|\vec{x}-\vec{x}'|/c} to reach and influence the field situated at {\vec{x}} [the speed of light {c} is restored to make the argument clearer]. So one could say that the signal from the source travels with the speed of light.

3. Photon Propagator

Let us now consider Maxwell’s equations

\displaystyle  {\partial}_{\mu} F^{{\mu}{\nu}} = j^{\nu}, \ \ \ \ \ (23)

where {F_{{\mu}{\nu}} = {\partial}_{\mu} A_{\nu} - {\partial}_{\nu} A_{\mu},} and indices are raised and lowered by {({\eta}^{{\mu}{\nu}}) = diag(-1,1,1,1)= ({\eta}_{{\mu}{\nu}}).} It is often convenient to impose Coulomb’s gauge {{\partial}_i A^i = 0.} It is usually useful to separate the components {A_i} into transverse and longitudinal parts. To be more precise, let us expand {A_i(t,\vec{x})} in Fourier modes:

\displaystyle  A_i(t,\vec{x}) =\int\frac{d{\omega}}{2{\pi}}\int \frac{d^3\vec{k}}{(2{\pi})^3}\tilde A_i({\omega},\vec{k}) e^{-i{\omega} t + i\vec{k}\cdot\vec{x}}, \ \ \ \ \ (24)

Here, a `mode’ means a plane wave

\displaystyle  \tilde A_i({\omega},\vec{k}) e^{-i{\omega} t + i\vec{k}\cdot\vec{x}}. \ \ \ \ \ (25)

with fixed {{\omega}} and {\vec{k}.} We see that in fact each mode has {3} components labelled by index {i.} The longitudinal part is defined to be the projection along the direction of {\vec{k}.} By a very simple vector analysis, the longitudinal part of the mode (25) is given by

\displaystyle  \frac{k^j \tilde A_j({\omega},\vec{k})}{|\vec{k}|^2}\ k_i e^{-i{\omega} t + i\vec{k}\cdot\vec{x}}. \ \ \ \ \ (26)

So we define the longitudinal part of {A_i(t,\vec{x})} to be the superposition of the longitudinal part of is mode. That is

\displaystyle  A_i^{\parallel}(t,\vec{x}) \equiv\int\frac{d{\omega}}{2{\pi}}\int \frac{d^3\vec{k}}{(2{\pi})^3}\frac{k^j \tilde A_j({\omega},\vec{k})}{|\vec{k}|^2}\ k_i e^{-i{\omega} t + i\vec{k}\cdot\vec{x}}. \ \ \ \ \ (27)

Now, Coulomb gauge demands

\displaystyle  \int\frac{d{\omega}}{2{\pi}}\int \frac{d^3\vec{k}}{(2{\pi})^3}{\partial}^i\tilde A_i({\omega},\vec{k}) e^{-i{\omega} t + i\vec{k}\cdot\vec{x}} =0, \ \ \ \ \ (28)

which implies {{\partial}^i\tilde A_i({\omega},\vec{k})=0.} This in turn gives {A_i^{\parallel}(t,\vec{x}) = 0.} That is, Coulomb gauge condition demands that the longitudinal part of {A_i} automatically vanishes. Furthermore, in this gauge Maxwell’s equations read

\displaystyle  {\partial}_{\mu}{\partial}^{\mu} A^{\nu} - {\partial}^{\nu}{\partial}_0 A^0 = j^{\nu}. \ \ \ \ \ (29)

The component {{\nu} = 0} is

\displaystyle  \nabla^2 A_0 = j_0. \ \ \ \ \ (30)

By following a similar analysis to the one in the previous section, one obtains, in parallel to eq.(11)(12),

\displaystyle  A_0(t,\vec{x}) =\int dt'\int d^3\vec{x}' G_0(t-t',\vec{x}-\vec{x}')j_0(t',\vec{x}'), \ \ \ \ \ (31)


\displaystyle  \begin{array}{rl} \displaystyle G_0(t-t',\vec{x}-\vec{x}') &=\displaystyle\int \frac{d{\omega}}{2{\pi}}\int \frac{d^3\vec{k}}{(2{\pi})^3} \frac{1}{ - \vec{k}^2} e^{-i{\omega} (t-t')+i\vec{k}\cdot(\vec{x}-\vec{x}')}\\ &\\ &=\displaystyle\int \frac{d^3\vec{k}}{(2{\pi})^3} \frac{1}{ - \vec{k}^2}{\delta}(t-t') e^{i\vec{k}\cdot(\vec{x}-\vec{x}')}\\ &\\ &=\displaystyle-\int_{0}^{2{\pi}}\int_0^{\pi}\int_{0}^\infty \frac{dk d{\theta} d{\phi}}{(2{\pi})^3}k^2\sin{\theta} \frac{1}{k^2}{\delta}(t-t') e^{ikR\cos{\theta}}\\ &\\ &=\displaystyle-\int_{0}^\infty \frac{dk}{2{\pi}^2} {\delta}(t-t')\frac{1}{kR}\sin(kR)\\ &\\ &=\displaystyle-\int_{-\infty}^\infty \frac{dk}{4{\pi}^2} {\delta}(t-t')\frac{1}{kR}\sin(kR)\\ &\\ &=\displaystyle-\int_{-\infty}^\infty \frac{dk}{4{\pi}^2} {\delta}(t-t')\frac{1}{kR}\sin(kR)\\ &\\ &=\displaystyle-\frac{1}{4{\pi}^2R}{\delta}(t-t')\ \textrm{Im}\int_{-\infty}^\infty dk \frac{e^{ikR}}{k}\\ &\\ &=\displaystyle-\frac{1}{4{\pi} R}{\delta}(t-t'), \end{array} \ \ \ \ \ (32)

where in the last step, we used eq.(17). Substituting eq.(32) into eq.(31) gives

\displaystyle  \begin{array}{rl} \displaystyle A_0(t,\vec{x}) &=\displaystyle-\int dt'\int d^3\vec{x}' \frac{1}{4{\pi} |\vec{x}-\vec{x}'|}{\delta}(t-t')j_0(t',\vec{x}'),\\ &\\ &=\displaystyle-\int d^3\vec{x}' \frac{j_0(t,\vec{x}')}{4{\pi} |\vec{x}-\vec{x}'|}. \end{array} \ \ \ \ \ (33)

The interpretation for this result is that {A_0} is only influenced by the {0}th component {j_0} of the source. Furthermore, once the source is turned on, {A_0} is influenced directly. The signal does not need to spend time travelling. So {A_0} is non-physical and is not counted towards degrees of freedom of {A_{\mu}.}

Let us now consider the component {{\nu}=i} of eq.(29). This is

\displaystyle  \begin{array}{rl} \displaystyle (-{\partial}_t^2 + \nabla^2)A_i &=\displaystyle j_i + {\partial}_i{\partial}_0 A^0. \end{array} \ \ \ \ \ (34)


\displaystyle  \begin{array}{rl} \displaystyle {\partial}_i{\partial}^0 A_0(t,\vec{x}) &=\displaystyle-{\partial}_i\int d^3\vec{x}' \frac{{\partial}^0 j_0(t,\vec{x}')}{4{\pi} |\vec{x}-\vec{x}'|}\\ &\\ &=\displaystyle{\partial}_i\int d^3\vec{x}' \frac{{\partial}'^j j_j(t,\vec{x}')}{4{\pi} |\vec{x}-\vec{x}'|}\\ &\\ &=\displaystyle-\int d^3\vec{x}' \frac{{\partial}'^j j_j(t,\vec{x}')}{4{\pi} |\vec{x}-\vec{x}'|^3}(x_i-x_i'). \end{array} \ \ \ \ \ (35)


\displaystyle  j_i(t,\vec{x}) + {\partial}_i{\partial}^0 A_0(t,\vec{x}) =\int d^3\vec{x}'\left({\delta}^{(3)}(\vec{x}-\vec{x}'){\delta}_i^j -\frac{x_i-x_i' }{4{\pi} |\vec{x}-\vec{x}'|^3}{\partial}'^j\right)j_j(t,\vec{x}'). \ \ \ \ \ (36)

Then by using the result of the previous section, one obtains

\displaystyle  \begin{array}{rl} \displaystyle A_i(t,\vec{x}) &=\displaystyle\int d^3\vec{x}'\int d^3\vec{x}''\frac{1}{4{\pi}|\vec{x}-\vec{x}'|}\left({\delta}^{(3)}(\vec{x}'-\vec{x}''){\delta}_i^j -\frac{x_i'-x_i'' }{4{\pi} |\vec{x}'-\vec{x}''|^3}{\partial}''^j\right)j_j(t-|\vec{x}-\vec{x}'|,\vec{x}'')\\ &\\ &=\displaystyle\int d^3\vec{x}'\int d^3\vec{x}''\frac{1}{4{\pi}|\vec{x}-\vec{x}'|}\left({\delta}^{(3)}(\vec{x}'-\vec{x}''){\delta}_i^j -\frac{{\delta}_i^j }{4{\pi} |\vec{x}'-\vec{x}''|^3} +\frac{3(x_i'-x_i'')(x'^j-x''^j) }{4{\pi} |\vec{x}'-\vec{x}''|^5}\right)\times\\ &\qquad\displaystyle j_j(t-|\vec{x}-\vec{x}'|,\vec{x}''). \end{array} \ \ \ \ \ (37)

This form tells us that the transverse part (recall that the longitudinal part vanishes because of the Coulomb gauge) of {A_i} is propagating. The source at {\vec{x}''} needs to spend time travelling to {\vec{x}} with the speed of light {c.} Next, in order to read off Green’s function, let us slightly rewrite this equation as

\displaystyle  \begin{array}{rl} \displaystyle A_i(t,\vec{x})&=\displaystyle\int d^3\vec{x}'\int dt''\int d^3\vec{x}''\frac{1}{4{\pi}|\vec{x}-\vec{x}'|}\left({\delta}^{(3)}(\vec{x}'-\vec{x}''){\delta}_i^j -\frac{{\delta}_i^j }{4{\pi} |\vec{x}'-\vec{x}''|^3} +\frac{3(x_i'-x_i'')(x'^j-x''^j) }{4{\pi} |\vec{x}'-\vec{x}''|^5}\right)\times\\ &\qquad\displaystyle {\delta}(t''-t+|\vec{x}-\vec{x}'|)j_j(t'',\vec{x}''). \end{array} \ \ \ \ \ (38)

From this, one can read off Green’s function as

\displaystyle  \begin{array}{rl} \displaystyle G_i^j(t,\vec{x};t'',\vec{x}'') &=\displaystyle \int d^3\vec{x}'\frac{{\delta}(t''-t+|\vec{x}-\vec{x}'|)}{4{\pi}|\vec{x}-\vec{x}'|}\left({\delta}^{(3)}(\vec{x}'-\vec{x}''){\delta}_i^j -\frac{{\delta}_i^j }{4{\pi} |\vec{x}'-\vec{x}''|^3} +\frac{3(x_i'-x_i'')(x'^j-x''^j) }{4{\pi} |\vec{x}'-\vec{x}''|^5}\right)\\ &\\ &=\displaystyle \int d^3\vec{x}'\frac{{\delta}(t''-t+|\vec{x}-\vec{x}'|)}{4{\pi}|\vec{x}-\vec{x}'|}\left({\delta}^{(3)}(\vec{x}'-\vec{x}''){\delta}_i^j +{\partial}'_i{\partial}'^j\frac{1 }{4{\pi} |\vec{x}'-\vec{x}''|}\right)\\ &\\ &=\displaystyle \frac{{\delta}(t''-t+|\vec{x}-\vec{x}''|)}{4{\pi}|\vec{x}-\vec{x}''|}{\delta}_i^j +\int d^3\vec{x}'\frac{{\delta}(t''-t+|\vec{x}-\vec{x}'|)}{4{\pi}|\vec{x}-\vec{x}'|} {\partial}'_i{\partial}'^j\frac{1 }{4{\pi} |\vec{x}'-\vec{x}''|}. \end{array} \ \ \ \ \ (39)

I have no idea whether the integral in the final stage can be integrated out.

In fact, whenever propagator is mentioned in physics, one almost always mean the Green’s function in (energy-)momentum space. The presentation in this post however concerns how to write the Green’s function in coordinate space. So I will stop this post here but might start a new one in the future concerning Green’s function in momentum space in order to catch up with the standard notion.

Basics of Non-degenerate time-independent quantum perturbation theory (Rayleigh-Schrödinger’s)

Some prerequisites: series expansion, quantum mechanics, bra-ket notation.

1. Introduction

I just recently discovered a satisfied explanation of the manipulations in non-degenerate time-independent quantum perturbation theory. I believe my discovery is not original. Even if it is I believe it is not of the level of publication in physics journals. Maybe, after some further polishes and additions, it might be publishable in some physics teaching journals. Well, I have no experience in publishing in this type of journals and do not wish to try it any time soon. More importantly, I believe it would be beneficial to students who start learning this topic. So it is more urgent to put it in my blog (although not many people would read mine anyway). I am sure that if I am satisfied with the analysis, students would be also satisfied.

2. Analysis often found in textbooks

The starting point of non-degenerate time-independent quantum perturbation theory is an eigenvalue equation

\displaystyle  H|\psi_n\rangle = E_n|\psi_n\rangle, \ \ \ \ \ (1)

where the Hamiltonian is expressed as

\displaystyle  H = H^0 + {\lambda} H^1, \ \ \ \ \ (2)

where {H^0} is the Hamiltonian that we know the spectrum, {H^1} is the perturbation Hamiltonian, and {{\lambda}} is a book-keeping parameter. Knowing the spectrum of {H^0} means knowing all the eigenvalues {E^0_n} and eigenstates {|\psi^0_n\rangle} from

\displaystyle  H^0|\psi_n^0\rangle = E_n^0|\psi_n^0\rangle. \ \ \ \ \ (3)

The eigenstates are orthonormal, i.e. {\langle\psi_n^0|\psi_m^0\rangle = {\delta}_{nm}.} Note that we have limited the analysis to the non-degenerate case only.

Then in order to solve eq.(1), one first expands

\displaystyle  E_n =E_n^0 + {\lambda} E_n^1 + {\lambda}^2 E_n^2 + \cdots, \ \ \ \ \ (4)

\displaystyle  |\psi_n\rangle =|\psi_n^0\rangle + {\lambda} |\psi_n^1\rangle + {\lambda}^2 |\psi_n^2\rangle + \cdots. \ \ \ \ \ (5)

Substituting eq.(2), (4), (5) into eq.(1) gives

\displaystyle  (H^0 + {\lambda} H^1)(|\psi_n^0\rangle + {\lambda} |\psi_n^1\rangle + {\lambda}^2 |\psi_n^2\rangle + \cdots) = (E_n^0 + {\lambda} E_n^1 + {\lambda}^2 E_n^2 + \cdots)(|\psi_n^0\rangle + {\lambda} |\psi_n^1\rangle + {\lambda}^2 |\psi_n^2\rangle + \cdots). \ \ \ \ \ (6)

Expanding both sides of the equation and comparing the orders in {{\lambda}} gives

\displaystyle  {\lambda}^0: H^0 |\psi_n^0\rangle = E_n^0|\psi_n^0\rangle, \ \ \ \ \ (7)

\displaystyle  {\lambda}^1: H^1 |\psi_n^0\rangle+H^0 |\psi_n^1\rangle = E_n^1|\psi_n^0\rangle+E_n^0|\psi_n^1\rangle, \ \ \ \ \ (8)

\displaystyle  {\lambda}^2: H^2 |\psi_n^0\rangle+H^1 |\psi_n^1\rangle +H^0 |\psi_n^2\rangle = E_n^2|\psi_n^0\rangle+E_n^1|\psi_n^1\rangle+E_n^0|\psi_n^2\rangle, \ \ \ \ \ (9)

\displaystyle  etc. \ \ \ \ \ (10)

One sees that the order {{\lambda}^0} simply coincides with eq.(3). So let us consider the order {{\lambda}^1.} Acting on the equation (8) by {\langle\psi_n^0|} and rearranging gives

\displaystyle  E_n^1 =\langle\psi_n^0|H^1|\psi_n^0\rangle. \ \ \ \ \ (11)

One can also determine {|\psi_n^1\rangle} from eq.(8) by first expressing it in the basis of {|\psi_m^0\rangle.} That is

\displaystyle  |\psi_n^1\rangle =c_{nn}|\psi_n^0\rangle +\sum_{m\neq n}c_{mn}|\psi_m^0\rangle. \ \ \ \ \ (12)

Substituting this expression into eq.(8) gives

\displaystyle  H^1 |\psi_n^0\rangle+\sum_{m\neq n}(E_m^0 - E_n^0) c_{mn}|\psi_m^0\rangle = E_n^1|\psi_n^0\rangle. \ \ \ \ \ (13)

Acting on this equation by {\langle\psi_n^0|} with {k\neq n} gives

\displaystyle  \langle\psi_k^0| H^1 |\psi_n^0\rangle+(E_m^0 - E_n^0) c_{kn} = 0. \ \ \ \ \ (14)


\displaystyle  c_{kn} = -\frac{\langle\psi_k^0| H^1 |\psi_n^0\rangle}{E_k^0 - E_n^0};\qquad k\neq n. \ \ \ \ \ (15)

Note however that the expression of {c_{nn}} is not fixed by eq.(8). So up to this point, one has

\displaystyle  |\psi_n^1\rangle =c_{nn}|\psi_n^0\rangle -\sum_{m\neq n}\frac{\langle\psi_m^0| H^1 |\psi_n^0\rangle}{E_m^0 - E_n^0}|\psi_m^0\rangle. \ \ \ \ \ (16)

In principle, {c_{nn}} can be fixed by using an additional condition which does not contradict to eq.(7)(10). The usual choice is {c_{nn} = 0.} I have found from textbooks two alternative explanations for this:

  1. The equation (1) we are allowed to choose any normalisation and phase of {|\psi_n\rangle.} So let us demand the normalisation {\langle\psi_n|\psi_n\rangle = 1.} By expanding in {{\lambda},} one obtains {\langle\psi_n^0|\psi_n^1\rangle = -\langle\psi_n^1|\psi_n^0\rangle = -\langle\psi_n^0|\psi_n^1\rangle^*.} Furthermore, we can choose the phase such that {\langle\psi_n^0|\psi_n^1\rangle} is a real number. But this gives {c_{nn} = \langle\psi_n^0|\psi_n^1\rangle = 0.}
  2. Alternatively, we do not yet wish to normalise {|\psi_n\rangle.} But let us choose its scale such that {\langle\psi_n^0|\psi_n\rangle=1.} By expanding in {{\lambda}} this gives {c_{nn} = \langle\psi_n^0|\psi_n^1\rangle = 0.}

I have always been worried with these explanations as it is not clear to me (except after I learn of the analysis to be given in the next section) why we are still allowed to choose normalisation and phase of {|\psi_n\rangle} after we have fixed the normalisation of its zeroth order {|\psi_n^0\rangle.} How can we see this explicitly? Or maybe how can we count the number of independent conditions against the number of coefficients to really see if we are allowed to impose further conditions; after all these numbers seem to be infinite anyway?

So although the analysis is still need to be continue in order to obtain {E_n^2, E_n^3,\cdots, |\psi_n^2\rangle, |\psi_n^3\rangle, \cdots,} let us end this section here and present an alternative analysis in which the above questions do not arise.

3. The analysis that I find it more satisfying

The basic setup of this alternative analysis is the same as that given in the previous section. Let us start from

\displaystyle  (H^0 + {\lambda} H^1)|\psi_n\rangle = E_n|\psi_n\rangle. \ \ \ \ \ (17)

Acting on this equation by {\langle\psi_n^0|} gives

\displaystyle  E_n =E_n^0 + {\lambda}\dfrac{\langle\psi_n^0|H^1|\psi_n\rangle}{\langle\psi_n^0|\psi_n\rangle}. \ \ \ \ \ (18)

Next, let us act on eq.(17) by {\langle\psi_m^0|} for {m\neq n.} This gives

\displaystyle  (E_m^0 - E_n)\langle\psi_m^0|\psi_n\rangle +{\lambda}\langle\psi_m^0|H^1|\psi_n\rangle =0. \ \ \ \ \ (19)


\displaystyle  \langle\psi_m^0|\psi_n\rangle =-{\lambda}\dfrac{\langle\psi_m^0|H^1|\psi_n\rangle}{E_m^0 - E_n}. \ \ \ \ \ (20)

Applying completeness relation on {|\psi_n\rangle,} we obtain

\displaystyle  \begin{array}{rl} |\psi_n\rangle &=\displaystyle |\psi_n^0\rangle\langle\psi_n^0|\psi_n\rangle +\sum_{m\neq n}|\psi_m^0\rangle\langle\psi_m^0|\psi_n\rangle\\ &\\ &=\displaystyle |\psi_n^0\rangle\langle\psi_n^0|\psi_n\rangle -{\lambda}\sum_{m\neq n}\dfrac{\langle\psi_m^0|H^1|\psi_n\rangle}{E_m^0 - E_n}|\psi_m^0\rangle. \end{array} \ \ \ \ \ (21)

The idea to obtain the spectrum of {H} up to a desired accuracy is to make a repeated use of of eq.(18) and (21). However, we see that {\langle\psi_n^0|\psi_n\rangle} is not restricted by any condition. So we are allowed to fix it. But instead of fixing, I am more comfortable with defining a new ket

\displaystyle  |\varphi_n\rangle =\dfrac{|\psi_n\rangle}{\langle\psi_n^0|\psi_n\rangle}. \ \ \ \ \ (22)

This definition automatically implies {|\varphi_n^0\rangle = |\psi_n^0\rangle.} So eq.(18) and (21) now become

\displaystyle  E_n =E_n^0 + {\lambda}\langle\varphi_n^0|H^1|\varphi_n\rangle. \ \ \ \ \ (23)

\displaystyle  |\varphi_n\rangle =\displaystyle |\varphi_n^0\rangle -{\lambda}\sum_{m\neq n}\dfrac{\langle\varphi_m^0|H^1|\varphi_n\rangle}{E_m^0 - E_n}|\varphi_m^0\rangle. \ \ \ \ \ (24)

In fact, it is better to use

\displaystyle  (E_m^0-E_n)\langle\varphi_m^0|\varphi_n\rangle + {\lambda}\langle\varphi_m^0|H^1|\varphi_n\rangle =0;\qquad m\neq n. \ \ \ \ \ (25)

Furthermore, the definition of {|\varphi_n\rangle} gives

\displaystyle  \langle\varphi_n^0|\varphi_n\rangle = 1. \ \ \ \ \ (26)

In order to obtain {E_n} and {\varphi_n,} we expand in power of {{\lambda}} and repeatedly solve eq.(23)(26) order-by-order. From eq.(23), it is easy to obtain

\displaystyle  E_n^k = \langle\varphi_n^0|H^1|\varphi_n^{k-1}\rangle;\qquad k=1,2,3,\cdots. \ \ \ \ \ (27)

Next, after some analysis eq.(25)(26) give [I might come back and show the detailed calculation later]

\displaystyle  \displaystyle|\varphi_n^k\rangle =\sum_{m\neq n}\frac{|\varphi_m^0\rangle}{E_m^0-E_n^0} \left(\sum_{i=1}^{k-1}E_n^i\langle\varphi_m^0|\varphi_n^{k-i}\rangle - \langle\varphi_m^0|H^1|\varphi_n^{k-1}\rangle\right); \qquad k=1,2,3,\cdots \ \ \ \ \ (28)

By recursively applying eq.(27)(28) starting from smaller values of {k,} one can obtain order-by-order the solutions for {E_n} and {|\varphi_n\rangle.}

Simple geometrical methods on scalar field

Some prerequisites: some exposures with classical field theory would help

1. Introduction

For definiteness, let us study scalar field on flat {{\mathbb R}^4.} Physicists often call a quantity {{\phi}(t,\vec{x})} as scalar field. because it is a function of space and time. To mathematicians’ eyes, I guess [I am not a mathematician, so I can only assume this] that {{\phi}(t,\vec{x})} is viewed merely as a number. Instead, the scalar field is given by a scalar function {{\phi}:{\mathbb R}^4\rightarrow{\mathbb R}.} The arguments {(t,\vec{x})} are in the domain whereas {{\phi}(t,\vec{x})} takes the value in the codomain.

As physicists, we should already be familiar with viewing {{\phi}(t,\vec{x})} as a scalar field. But in this post, let us take a peek on the alternative point of view. I believe this point of view is preferred not only by mathematicians, but also by some physicists.

2. Klein-Gordon Lagrangian

To describe

\displaystyle  {\phi}(t,\vec{x}), \ \ \ \ \ (1)

let us define a map

\displaystyle  {\phi}_t:{\mathbb R}^3\rightarrow{\mathbb R} \ \ \ \ \ (2)

such that {{\phi}_t(\vec{x}) = {\phi}(t,\vec{x}).} There are other maps which are also useful. Define

\displaystyle  {\phi}_t^2:{\mathbb R}^3\rightarrow{\mathbb R} \ \ \ \ \ (3)

such that

\displaystyle  \begin{array}{rl} {\phi}_t^2(\vec{x}) &={\phi}_t(\vec{x}){\phi}_t(\vec{x})\\ &=({\phi}(t,\vec{x}))^2. \end{array} \ \ \ \ \ (4)

In general, a multiplication between two functions {f,g:{\mathbb R}^3\rightarrow{\mathbb R}} is given by the function {fg:{\mathbb R}^3\rightarrow{\mathbb R}} such that

\displaystyle  (fg)(\vec{x}) = f(\vec{x})g(\vec{x}). \ \ \ \ \ (5)

This is a typical definition of multiplication of scalar functions in the context of differential geometry. Let us also define some differentiations. The function

\displaystyle  \dot{\phi}_t:{\mathbb R}^3\rightarrow{\mathbb R} \ \ \ \ \ (6)

is defined as [I hope I do not need to make use of {{\epsilon}-{\delta}}]

\displaystyle  \dot{\phi}_t \equiv \lim_{h\rightarrow 0}\frac{{\phi}_{t+h}-{\phi}_t}{h}. \ \ \ \ \ (7)

Sometimes, it is more convenient to write {{\partial}_0{\phi}_t} instead of {\dot{\phi}_t.} But in any case, we will avoid (at least in this post) writing “{{\partial}_t{\phi}_t.}” This is because {t} is a parameter, and can be replaced by any number. So using this choice for example would lead to

\displaystyle  \textrm{``}\left({\partial}_4{\phi}_4 = \lim_{h\rightarrow 0}\frac{{\phi}_{4+h}-{\phi}_4}{h}\right)\textrm{''}, \ \ \ \ \ (8)

whose LHS is mentally irritating. Let us also define

\displaystyle  {\partial}_1{\phi}_t:{\mathbb R}^3\rightarrow{\mathbb R} \ \ \ \ \ (9)

such that

\displaystyle  {\partial}_1{\phi}_t(x^1,x^2,x^3) =\lim_{h\rightarrow 0}\frac{{\phi}_t(x^1+h,x^2,x^3)-{\phi}_t(x^1,x^2,x^3)}{h}. \ \ \ \ \ (10)

The functions {{\partial}_2{\phi}_t, {\partial}_3{\phi}_t} can also be defined in a similar way. The “trace” of a function {f:{\mathbb R}^3\rightarrow{\mathbb R}} is given by

\displaystyle  \lbrack f\rbrack =\int d^3\vec{x} f(\vec{x}). \ \ \ \ \ (11)

With the above ingredients, the Klein-Gordon Lagrangian is given by

\displaystyle  L_t = \left\lbrack{\dfrac{1}{2}}\dot{\phi}_t^2 - {\dfrac{1}{2}}({\partial}_i{\phi}_t)({\partial}_i{\phi}_t) - {\dfrac{1}{2}} m^2{\phi}_t^2\right\rbrack. \ \ \ \ \ (12)

It is a functional of {{\phi}_t} and {\dot{\phi}_t.} So

\displaystyle  \begin{array}{rl} {\delta} L_t =\left\lbrack \dfrac{{\delta} L_t}{{\delta}\dot{\phi}_t}{\delta}\dot{\phi}_t + \dfrac{{\delta} L_t}{{\delta}{\phi}_t}{\delta}{\phi}_t \right\rbrack. \end{array} \ \ \ \ \ (13)

On the other hand, direct computation gives

\displaystyle  \begin{array}{rl} {\delta} L_t &=\left\lbrack \dot{\phi}_t{\delta}\dot{\phi}_t - {\partial}_i{\phi}_t{\delta}({\partial}_i{\phi}_t) - m^2{\phi}_t{\delta}{\phi}_t\right\rbrack\\ &\\ &=\left\lbrack \dot{\phi}_t{\delta}\dot{\phi}_t + {\partial}_i{\partial}_i{\phi}_t{\delta}{\phi}_t - m^2{\phi}_t{\delta}{\phi}_t \right\rbrack - \left\lbrack {\partial}_i({\partial}_i{\phi}_t{\delta}{\phi}_t)\right\rbrack, \end{array} \ \ \ \ \ (14)

where we used the identity {{\delta}({\partial}_i{\phi}_t) = {\partial}_i({\delta}{\phi}_t),} which can be shown directly from the definition. The last term on RHS of eq.(14) gives surface integral at spatial infinity. Let us assume that this vanishes. So we can now read off

\displaystyle  \dfrac{{\delta} L_t}{{\delta}\dot{\phi}_t} =\dot{\phi}_t,\qquad \dfrac{{\delta} L_t}{{\delta}{\phi}_t} ={\partial}_i{\partial}_i{\phi}_t - m^2{\phi}_t. \ \ \ \ \ (15)

Euler-Lagrange’s equation

\displaystyle  {\partial}_0\dfrac{{\delta} L_t}{{\delta}\dot{\phi}_t} -\dfrac{{\delta} L_t}{{\delta}{\phi}_t} = 0 \ \ \ \ \ (16)


\displaystyle  \ddot{\phi}_t - {\partial}_i^2{\phi}_t + m^2{\phi}_t = 0, \ \ \ \ \ (17)

which is Klein-Gordon equation.

Note the version of Euler-Lagrange’s equation (16). This version is often seen in Classical Mechanics but rarely appeared in Classical Field Theory. Physics students are often introduced to Classical Field Theory along the line of “Like in Classical Mechanics, there is also Euler-Lagrange’s equation. But we have to proceed differently to get Euler-Lagrange’s equation

\displaystyle  \textrm{`` }{\partial}_{\mu}\dfrac{{\partial}{\cal L}}{{\partial}{\partial}_{\mu}{\phi}} - \dfrac{{\partial}{\cal L}}{{\partial}{\phi}} = 0\textrm{ ''} \ \ \ \ \ (18)

which is tailored-made for Classical Field Theory”. From my direct experience, this approach makes Classical Field Theory to look far more complicated than Classical Mechanics and that there is an unforeseeable gap between them which cannot easily be filled in.

So hopefully our use of Euler-Lagrange’s equation (16) should make the connection between Classical Mechanics and Classical Field Theory a bit clearer.

3. Klein-Gordon Hamiltonian

To obtain Hamiltonian, we first look for conjugate momenta of {{\phi}_t.} It is given by

\displaystyle  \begin{array}{rl} {\pi}_t \equiv\dfrac{{\delta} L_t}{{\delta}\dot{\phi}_t} =\dot{\phi}_t. \end{array} \ \ \ \ \ (19)

Hamiltonian is then the functional of {{\phi}_t, {\pi}_t.} It is given by

\displaystyle  \begin{array}{rl} H_t &=\lbrack{\pi}_t\dot{\phi}_t\rbrack - L_t\\ &\\ &=\left\lbrack{\dfrac{1}{2}}{\pi}_t^2 + {\dfrac{1}{2}}({\partial}_i{\phi}_t)({\partial}_i{\phi}_t) + {\dfrac{1}{2}} m^2{\phi}_t^2\right\rbrack. \end{array} \ \ \ \ \ (20)

Let us next compute Poisson’s bracket. In the case of Classical Field Theory, we are often taught to start from the identity

\displaystyle  \{{\phi}_t(\vec{x}),{\phi}_t(\vec{y})\} = 0,\qquad \{{\phi}_t(\vec{x}),{\pi}_t(\vec{y})\} = {\delta}^{(3)}(\vec{x}-\vec{y}),\qquad \{{\pi}_t(\vec{x}),{\pi}_t(\vec{y})\} = 0, \ \ \ \ \ (21)

where {{\delta}^{(3)}(\vec{x}-\vec{y})} is 3d Dirac’s delta function. However, there is an alternative way to avoid direct manipulation with Dirac’s delta function.

For this, let us view {{\phi}_t, {\pi}_t} as coordinates of some manifold {{\cal W}} [NB: physics students exposed to General Relativity would tend to think of manifold as associated only to spacetime. However, this viewpoint is too limited. It is better to think of Manifold as some mathematical concepts which has wide applications. The description of spacetime as manifold is only one application in physics.]. So {{\delta}{\phi}_t, {\delta}{\pi}_t} are one-forms on {{\cal W},} whereas

\displaystyle  \dfrac{{\delta}}{{\delta}{\phi}_t},\qquad \dfrac{{\delta}}{{\delta}{\pi}_t}, \ \ \ \ \ (22)

are vectors on {{\cal W}.} The one-forms and vectors are dual in the sense that

\displaystyle  \lbrack f{\delta}{\phi}_t\rbrack\left(\dfrac{{\delta}}{{\delta}{\phi}_t}\right) =f,\qquad \lbrack f{\delta}{\phi}_t\rbrack\left(\dfrac{{\delta}}{{\delta}{\pi}_t}\right) =0, \ \ \ \ \ (23)

\displaystyle  \lbrack f{\delta}{\pi}_t\rbrack\left(\dfrac{{\delta}}{{\delta}{\phi}_t}\right) =0,\qquad \lbrack f{\delta}{\pi}_t\rbrack\left(\dfrac{{\delta}}{{\delta}{\pi}_t}\right) =f, \ \ \ \ \ (24)

where {f:{\mathbb R}^3\rightarrow{\mathbb R}} is a well-behaved arbitrary function. Next, let us define a linear map {\Omega} which changes a one-form to a vector as follows

\displaystyle  \Omega\left({\delta}{\phi}_t\right) =\dfrac{{\delta}}{{\delta}{\pi}_t},\qquad \Omega\left({\delta}{\pi}_t\right) =-\dfrac{{\delta}}{{\delta}{\phi}_t}. \ \ \ \ \ (25)

Let us define a vector field (called Hamiltonian vector field) of a scalar function {F} on {{\cal W}} as

\displaystyle  X_F\equiv\Omega({\delta} F). \ \ \ \ \ (26)

So it can be shown that Poisson’s bracket of scalar functions {F,G} on {{\cal W}} is given by

\displaystyle  \{F,G\} = {\delta} G(X_F). \ \ \ \ \ (27)

We want to compute {\{{\phi}_t,H_t\}, \{{\pi}_t,H_t\}.} For this, we first compute

\displaystyle  {\delta} H_t =\left\lbrack {\pi}_t{\delta}{\pi}_t + (m^2{\phi}_t - {\partial}_i{\partial}_i{\phi}_t){\delta} {\phi}_t\right\rbrack. \ \ \ \ \ (28)


\displaystyle  \{{\phi}_t,H_t\} ={\delta} H_t\left(\dfrac{{\delta}}{{\delta}{\pi}_t}\right) = {\pi}_t, \ \ \ \ \ (29)

\displaystyle  \{{\pi}_t,H_t\} ={\delta} H_t\left(-\dfrac{{\delta}}{{\delta}{\phi}_t}\right) = {\partial}_i{\partial}_i{\phi}_t-m^2{\phi}_t, \ \ \ \ \ (30)

4. Solution to equation of motion

This section is not so related to the title of the post. This means that we do not make use of any geometrical methods in the analysis [It might be possible to use one, but I still do not have enough knowledge to do so or to see if it is really possible]. However, I include this section just to make this post a bit more complete.

From either Lagrangian or Hamiltonian analyses, we obtain equation of motion for {{\phi}_t:}

\displaystyle  \ddot{\phi}_t - {\partial}_i{\partial}_i{\phi}_t + m^2{\phi}_t = 0. \ \ \ \ \ (31)

Recall that LHS of this equation is in fact a function {{\mathbb R}^3\rightarrow{\mathbb R}.} So let us apply it on {\vec{x}\in{\mathbb R}^3.} For this, let us consider the Fourier transform

\displaystyle  {\phi}_t(\vec{x}) =\int\dfrac{d^3\vec{k}}{(2{\pi})^3}e^{i\vec{k}\cdot\vec{x}}\tilde{\phi}_t(\vec{k}). \ \ \ \ \ (32)

Substituting this into the equation of motion (31) (whose LHS is already applied to {\vec{x}}) gives

\displaystyle  \int\dfrac{d^3\vec{k}}{(2{\pi})^3} e^{i\vec{k}\cdot\vec{x}}(\ddot{\tilde{\phi}}_t(\vec{k}) + (\vec{k}^2+m^2)\tilde{\phi}_t(\vec{k})) =0. \ \ \ \ \ (33)

By inverse Fourier transforming, we obtain

\displaystyle  \ddot{\tilde{\phi}}_t(\vec{k}) + (\vec{k}^2+m^2)\tilde{\phi}_t(\vec{k}) =0. \ \ \ \ \ (34)

For each {\vec{k},} this equation is just SHO equation. So it is easy to see (by high school or beginning undergraduate physics students in Thailand) that the solution is given by

\displaystyle  \tilde{{\phi}}_t(\vec{k}) =A_{\vec{k}}e^{-i\sqrt{\vec{k}^2+m^2}\;t} + B_{\vec{k}}e^{i\sqrt{\vec{k}^2+m^2}\;t}. \ \ \ \ \ (35)

But reality condition {{\phi}_t(\vec{x})^* = {\phi}_t(\vec{x})} implies that {\tilde{\phi}_t(\vec{k}) = \tilde{\phi}_t(-\vec{k})^*,} which in turn implies {B_{\vec{k}} = A_{-\vec{k}}^*,} and hence

\displaystyle  \tilde{{\phi}}_t(\vec{k}) =A_{\vec{k}}e^{-i\sqrt{\vec{k}^2+m^2}\;t} + A_{-\vec{k}}^*e^{i\sqrt{\vec{k}^2+m^2}\;t}. \ \ \ \ \ (36)

This gives

\displaystyle  \begin{array}{rl} {\phi}_t(\vec{x}) &=\displaystyle\int\dfrac{d^3\vec{k}}{(2{\pi})^3}e^{i\vec{k}\cdot\vec{x}}\left(A_{\vec{k}}e^{-i\sqrt{\vec{k}^2+m^2}\;t} + A_{-\vec{k}}^*e^{i\sqrt{\vec{k}^2+m^2}\;t}\right)\\ &\\ &=\displaystyle\int\dfrac{d^3\vec{k}}{(2{\pi})^3}\left(A_{\vec{k}}e^{i\vec{k}\cdot\vec{x}}e^{-i\sqrt{\vec{k}^2+m^2}\;t} + A_{\vec{k}}^*e^{-i\vec{k}\cdot\vec{x}}e^{i\sqrt{\vec{k}^2+m^2}\;t}\right)\\ &\\ &=\displaystyle 2\textrm{Re}\int\dfrac{d^3\vec{k}}{(2{\pi})^3}A_{\vec{k}}e^{i\vec{k}\cdot\vec{x}}e^{-i\sqrt{\vec{k}^2+m^2}\;t}. \end{array} \ \ \ \ \ (37)

Sometimes, it is preferable to write RHS in terms of relativistic quantities. Based on the insight from Special Relativity, one could expect to put {t} and {\vec{x}} on equal footing. Similarly, one would expect to put {{\omega}} and {\vec{k}} on equal footing. For this, let us make use of the trick of Dirac’s delta function to get

\displaystyle  \begin{array}{rl} {\phi}_t(\vec{x}) &=2\textrm{Re}\displaystyle\int\dfrac{d^3\vec{k}}{(2{\pi})^3}\int_0^\infty d{\omega} A_{\vec{k}}e^{-i{\omega} t+i\vec{k}\cdot\vec{x}}{\delta}({\omega}-\sqrt{\vec{k}^2+m^2}). \end{array} \ \ \ \ \ (38)

The exponent is already in the relativistic form: {-i{\omega} t+i\vec{k}\cdot\vec{x} = -ik^{\mu} x_{\mu}.} However, the argument of the Dirac’s delta function is still not. So let us make use of the identity

\displaystyle  {\delta}({\omega}^2-\vec{k}^2 - m^2) =\dfrac{{\delta}({\omega} -\sqrt{\vec{k}^2+m^2}) + {\delta}({\omega} +\sqrt{\vec{k}^2+m^2})}{2\sqrt{\vec{k}^2+m^2}}. \ \ \ \ \ (39)

So by using the fact that {{\delta}({\omega} +\sqrt{\vec{k}^2+m^2})} vanishes for {{\omega}\in[0,\infty),} we obtain

\displaystyle  \begin{array}{rl} {\phi}_t(\vec{x}) &=2\textrm{Re}\displaystyle\int\dfrac{d^3\vec{k}}{(2{\pi})^3}\int_0^\infty d{\omega} A_{\vec{k}}e^{-i{\omega} t+i\vec{k}\cdot\vec{x}}2\sqrt{\vec{k}^2+m^2}{\delta}({\omega}^2-\vec{k}^2-m^2). \end{array} \ \ \ \ \ (40)

The coefficient would look nicer if we set

\displaystyle  c_{\vec{k}} \equiv 2\sqrt{\vec{k}^2 + m^2}A_{\vec{k}}, \ \ \ \ \ (41)

[NB: This choice has no physical motivation. It just only makes the presentation a bit nicer. The choice which often appears in Classical Field Theory is {a_{\vec{k}} \sim (\vec{k}^2 + m^2)^{1/4}A_{\vec{k}}.} In order to discuss this choice, I will need to include other discussions. But I do not intend to do it in this post.] giving

\displaystyle  \begin{array}{rl} {\phi}_t(\vec{x}) &=2\textrm{Re}\displaystyle\int\dfrac{d^3\vec{k}}{(2{\pi})^3}\int_0^\infty d{\omega} c_{\vec{k}}e^{-i{\omega} t+i\vec{k}\cdot\vec{x}}{\delta}({\omega}^2-\vec{k}^2-m^2)\\ &\\ &=4{\pi}\textrm{Re}\displaystyle\int\dfrac{d^4k}{(2{\pi})^4} \Theta({\omega})c_{\vec{k}}e^{-i k^{\mu} x_{\mu}}{\delta}(k^{\mu} k_{\mu}-m^2). \end{array} \ \ \ \ \ (42)

Perhaps this form is the most relativistic that one could get. As a cross-check, one could integrate out {{\omega}} to obtain the original form

\displaystyle  {\phi}_t(\vec{x}) =\displaystyle\int\dfrac{d^3\vec{k}}{(2{\pi})^3}\frac{1}{2\sqrt{\vec{k}^2+m^2}}\left(c_{\vec{k}}e^{i\vec{k}\cdot\vec{x}}e^{-i\sqrt{\vec{k}^2+m^2}\;t} + c_{\vec{k}}^*e^{-i\vec{k}\cdot\vec{x}}e^{i\sqrt{\vec{k}^2+m^2}\;t}\right). \ \ \ \ \ (43)

Let us substitute the solution into the Hamiltonian (20). For this, it is convenient if we first rewrite

\displaystyle  {\phi}_t(\vec{x}) =\displaystyle\int\dfrac{d^3\vec{k}}{(2{\pi})^3}\frac{1}{2\sqrt{\vec{k}^2+m^2}}\left(c_{\vec{k}}e^{-i\sqrt{\vec{k}^2+m^2}\;t} + c_{-\vec{k}}^*e^{i\sqrt{\vec{k}^2+m^2}\;t}\right)e^{i\vec{k}\cdot\vec{x}}. \ \ \ \ \ (44)

Then we compute

\displaystyle  \begin{array}{rl} {\pi}_t(\vec{x}) &=\dot{\phi}_t(\vec{x})\\ &\\ &=\displaystyle\int\dfrac{d^3\vec{k}}{(2{\pi})^3}\frac{i}{2}\left(-c_{\vec{k}}e^{-i\sqrt{\vec{k}^2+m^2}\;t} + c_{-\vec{k}}^*e^{i\sqrt{\vec{k}^2+m^2}\;t}\right)e^{i\vec{k}\cdot\vec{x}}. \end{array} \ \ \ \ \ (45)

Any well-behaved functions {f,g:{\mathbb R}^3\rightarrow{\mathbb R}} can be written as a Fourier transform

\displaystyle  f(\vec{x}) =\displaystyle\int \dfrac{d^3\vec{k}}{(2{\pi})^3}e^{i\vec{k}\cdot\vec{x}}\tilde f(\vec{k}), \ \ \ \ \ (46)

\displaystyle  g(\vec{x}) =\displaystyle\int \dfrac{d^3\vec{k}}{(2{\pi})^3}e^{i\vec{k}\cdot\vec{x}}\tilde g(\vec{k}). \ \ \ \ \ (47)

Direct computation gives

\displaystyle  \lbrack fg\rbrack =\displaystyle\int \dfrac{d^3\vec{k}}{(2{\pi})^3}\tilde f(\vec{k})\tilde g(-\vec{k}). \ \ \ \ \ (48)

Using this identity, it is easy to show that

\displaystyle  \lbrack{\pi}_t^2\rbrack =-\dfrac{1}{4}\int\dfrac{d^3\vec{k}}{(2{\pi})^3} (c_{\vec{k}}c_{-\vec{k}}e^{-2i\sqrt{\vec{k}^2+m^2}\;t} -2|c_{\vec{k}}|^2 + c_{\vec{k}}^*c_{-\vec{k}}^*e^{2i\sqrt{\vec{k}^2+m^2}\;t}), \ \ \ \ \ (49)

\displaystyle  \lbrack{\partial}_i{\phi}_t{\partial}_i{\phi}_t\rbrack =\displaystyle\int\dfrac{d^3\vec{k}}{(2{\pi})^3}\dfrac{\vec{k}^2}{4{\omega}_{\vec{k}}^2} (c_{\vec{k}}c_{-\vec{k}}e^{-2i\sqrt{\vec{k}^2+m^2}\;t} +2|c_{\vec{k}}|^2 + c_{\vec{k}}^*c_{-\vec{k}}^*e^{2i\sqrt{\vec{k}^2+m^2}\;t}), \ \ \ \ \ (50)

\displaystyle  \lbrack{\phi}_t^2\rbrack =\displaystyle\int\dfrac{d^3\vec{k}}{(2{\pi})^3}\dfrac{1}{4{\omega}_{\vec{k}}^2} (c_{\vec{k}}c_{-\vec{k}}e^{-2i\sqrt{\vec{k}^2+m^2}\;t} +2|c_{\vec{k}}|^2 + c_{\vec{k}}^*c_{-\vec{k}}^*e^{2i\sqrt{\vec{k}^2+m^2}\;t}). \ \ \ \ \ (51)

Combining these results give

\displaystyle  H_t ={\dfrac{1}{2}}\displaystyle\int\dfrac{d^3\vec{k}}{(2{\pi})^3} |c_{\vec{k}}|^2. \ \ \ \ \ (52)

So Hamiltonian of the system is time-independent.

Let us simply end this post here.