## What happens to the simple harmonic oscillator path integral at half periods?

Previously we had stated but not elucidated the result for the transition amplitude of the simple harmonic oscillator (SHO) at half periods. The SHO Hamiltonian, with $X$ and $p$ denoting the position and momentum operators respectively, reads

(1):    $H_\text{SHO} \equiv \frac{1}{2} p^2 + \frac{1}{2} \omega^2 X^2 .$

The classical solution for $X$ is a linear combination of $\sin[\omega t]$ and $\cos[\omega t]$. The period $T$ is therefore $\omega T = 2\pi \Rightarrow T = 2\pi/\omega$. Over a half period $\pi/\omega$, i.e., upon replacing $t \to t \pm \pi/\omega$, we have $\sin[\omega t] \to - \sin[\omega t]$ and $\cos[\omega t] \to - \cos[\omega t]$. In other words, starting at time $t$ the motion performs a parity flip $X \to -X$ over a half-oscillation.

We will now discuss why the quantum SHO particle does exactly the same thing. If it begins from the location $x'$ at time $t$, then it has to be at $x \equiv (-)^n x'$ at time $t + n \pi/\omega$ for integer $n$. Namely, we have the path integral (aka transition amplitude)

(2):    $\langle x \vert \exp[-i H_{\text{SHO}} \Delta t_n ] \vert x' \rangle = \frac{1}{i^n} \delta[x - (-)^n x'] ;$

with the time interval

(2′):     $\Delta t_n \equiv \frac{n\pi}{\omega}$.

One might otherwise think there is an inherent ‘fuzziness’ to quantum motion, but at half-periods, all the quantum aspects of the dynamics seem to make their appearance only in the Maslov index $1/i^n$ multiplying the Dirac delta-function on the right hand side of eq. (2). As we will now see, eq. (2) is a direct consequence of the exact invariance of the SHO Hamiltonian in eq. (1) under parity. If $P$ is the parity operator that implements $P X P^{-1} = -X$, then the commutator $[P, H_{\text{SHO}}]$ vanishes.  Therefore the SHO energy eigenstates must be simultaneous eigenstates of the parity operator; specifically, the $\{ \vert E_\ell \rangle \}$ in

(2′):     $H_{\text{SHO}} \vert E_\ell \rangle = \left(\ell + \frac{1}{2}\right) \omega \vert E_\ell \rangle$

obey

(3):     $P \vert E_\ell \rangle = (-)^\ell \vert E_\ell \rangle, \qquad \ell = 0,1,2,\dots$.

Now, by inserting a complete set of energy eigenstates, we may begin from the left hand side of eq. (2), to consider motion over a single half period $\Delta t_1 = \pi/\omega$:

(4):    $\langle x \vert \exp[-i H_{\text{SHO}} (\pi/\omega)] \vert x' \rangle = \sum_{\ell=0}^\infty \langle x \vert E_\ell \rangle \langle E_\ell \vert x' \rangle \exp[-i (\ell+1/2) \pi]$ .

Observe that $e^{-i\pi/2} = 1/i$; whereas $e^{-i\ell \pi} = (-)^\ell$. Invoking the parity property of the energy eigenstates in eq. (3) to say $e^{-i\ell \pi} \langle x \vert E_\ell \rangle = \langle -x \vert E_\ell \rangle$, we may thus write

(4′):    $\langle x \vert \exp[-i H_{\text{SHO}} (\pi/\omega)] \vert x' \rangle = \frac{1}{i} \sum_{\ell=0}^\infty \langle -x \vert E_\ell \rangle \langle E_\ell \vert x' \rangle$ .

The completeness relation in the position representation is $\sum_\ell \langle u \vert E_\ell \rangle \langle E_\ell \vert v \rangle = \delta[u-v]$; comparing this to eq. (4′) we arrive at

(5):    $\langle x \vert \exp[-i H_{\text{SHO}} (\pi/\omega)] \vert x' \rangle = \frac{1}{i} \delta[(-) x - x'] = \frac{1}{i} \delta[x - (-) x']$ .

By recognizing the unitary nature of the time-evolution operator $\exp[-i H_{\text{SHO}} t ]$, and applying the result in eq. (5) $n$ times to evolve a quantum particle from $x'$ to some other location $x$ over $n$ half periods, one will recover the primary result in eq. (2).

Reference

• P. A. Horvathy, “The Maslov correction in the semiclassical Feynman integral,” Central Eur. J. Phys. 9, 1 (2011) doi:10.2478/s11534-010-0055-3 [quant-ph/0702236].

## Synge World Function & Shapiro Delay

Geodesics     Consider all possible spacetime trajectories joining the points $x'$ and $x$. Suppose you found a trajectory $Z^\mu[0 \leq \lambda \leq 1]$ (obeying $Z[0]=x'$ and $Z[1]=x$) such that any perturbation away from it yields a slightly longer or shorter path — then such a path is said to extremize the distance between $x'$ and $x'$. In differential geometric/general relativity lingo, such a path is called a geodesic.

In a given spacetime metric $g_{\mu\nu}$ and given a geodesic trajectory $Z^\mu$ joining $x'$ to $x$, half the square of the geodesic distance between $x'$ and $x$ can be written as

(1):    $\sigma[x,x'] = \frac{1}{2} \int_0^1 g_{\mu\nu}[Z] \frac{d Z^\mu}{d\lambda} \frac{d Z^\nu}{d\lambda} d\lambda , \qquad \qquad Z[0]=x', \ Z[1]=x ;$

where $\sigma[x,x']$ is dubbed “Synge’s World Function” in the literature.

Conversely, if we view eq. (1) as a functional of the trajectory $Z$, and we demand it to be extremized, then it would yield the (affinely parametrized) geodesic equation

(1′):    $\frac{d^2 Z^\mu}{d \lambda^2} + \Gamma^\mu_{\phantom{\mu}\alpha\beta} \frac{d Z^\alpha}{d \lambda} \frac{d Z^\beta}{d \lambda} = 0 .$

To sum: Synge’s world function is the action principle for affinely parametrized geodesic motion; and when evaluated on a given geodesic $Z^\mu[0 \leq \lambda \leq 1]$ joining $x'$ to $x$, it hands us half the square of the geodesic distance between this pair of spacetime points.

Perturbation Theory     In a weakly curved spacetime of the form

(2):    $g_{\mu\nu} = \eta_{\mu\nu} + h_{\mu\nu},$

where the components of $h_{\mu\nu}$ are assumed to be much less than unity; the Synge’s world function may be used to find an $\mathcal{O}[h]$ accurate integral solution for geodesic distances merely from the geodesic solutions in flat spacetime — namely, a straight line $\bar{Z}^\mu$ joining $x'$ to $x$ — precisely because $\sigma[x,x']$ is the geodesic action principle. To see this, we first express the geodesic solution $Z^\mu$ of the geometry in eq. (2) as a perturbation away from a straight line:

(2′):    $Z^\mu = \bar{Z}^\mu + \delta Z^\mu,$

where the straight line itself is

(2.S):    $\bar{Z} = x' + \lambda (x-x') .$

Up to first order in $\delta Z$, Synge’s world function is

(2”):    $\sigma[x,x'] = \int_0^1 \left( \frac{1}{2} \left( \eta_{\mu\nu} + h_{\mu\nu}[\bar{Z}] \right) (x-x')^\mu (x-x')^\nu - \delta Z^\kappa \left( \eta_{\kappa\mu} + h_{\kappa\mu}[\bar{Z}] \right) \frac{D^2 \bar{Z}^\mu}{d \lambda^2} + \mathcal{O}\left[(\delta Z)^2\right] \right) d\lambda .$

Here, the boundary conditions $Z[0]=x', \ Z[1]=x \Leftrightarrow \delta Z[0]=0=\delta Z[1]$ were used to set to zero the boundary terms; we have used eq. (2.S) to infer $\dot{Z} = x-x'$; and, finally, the ‘geodesic operator’ $D^2/d \lambda^2$ reads

(2”’):    $\frac{D^2 Y^\mu}{d\lambda^2} \equiv \frac{d^2 Y^\mu}{d \lambda^2} + \Gamma^\mu_{\phantom{\mu}\alpha\beta}[g=\eta+h] \frac{d Y^\alpha}{d \lambda} \frac{d Y^\beta}{d \lambda} .$

Both $\delta Z$ and $D^2 \bar{Z}/d \lambda^2$ in eq. (2”) must scale as order $h$ or higher since they vanish in the limit as $h_{\mu\nu} \to 0$. (The arguments for $D^2 \bar{Z}/d \lambda^2 \sim \mathcal{O}[h]$ and $\delta Z \sim \mathcal{O}[h]$ can be made explicit by direct computation for the former; and, for the latter, by first converting the geodesic equation of eq. (1′) into an integral equation, followed by employing the Born-series-approximation iteration technique.) Therefore the $\delta Z (\dots) D^2 \bar{Z}/d \lambda^2$ group of terms on the right in eq. (2”) must scale as $\mathcal{O}[h^2]$ or higher and may thus be dropped if all we are seeking is a first order accurate expression.

To summarize: at first order in the metric perturbation, half the square of the geodesic distance between the pair of spacetime points $x'$ and $x$ in a weakly curved spacetime (cf. eq. (2)) is given by the integral

(2.Synge):    $\sigma[x,x'] = \frac{1}{2} (x-x')^\mu (x-x')^\nu \int_0^1 \left( \eta_{\mu\nu} + h_{\mu\nu}[\bar{Z}] \right) d\lambda + \mathcal{O}[h^2],$

with the straight line already given in eq. (2.S).

Linearized Einstein’s Equations     If we choose the de Donder gauge

(3):    $\partial^\sigma \bar{h}_{\sigma\mu} = 0 ,$

where we are moving indices with the flat Cartesian metric and

(3′):    $\bar{h}_{\mu\nu} = h_{\mu\nu} - \frac{\eta_{\mu\nu}}{2} \eta^{\sigma\rho} h_{\sigma\rho} ;$

Einstein’s field equations $G_{\alpha\beta} = 8\pi G_\text{N} T_{\alpha\beta}$ linearized about flat spacetime yields

(3”):    $\partial^2 \bar{h}_{\mu\nu} = -16\pi G_\text{N} \bar{T}_{\mu\nu} ,$

with $\partial^2$ denoting the flat spacetime wave operator and $\bar{T}_{\mu\nu}$ is the portion of the matter stress-energy tensor $T_{\mu\nu}$ that does not contain any metric perturbations. Now, in the non-relativistic limit, stress-energy-momentum is dominated by the energy density; if $v$ is some characteristic speed of the internal dynamics of the source (in its rest frame), we usually have $\bar{T}_{0i}/\bar{T}_{00} \sim \mathcal{O}[v]$ and $\bar{T}_{ij}/\bar{T}_{00} \sim \mathcal{O}[v^2]$. In such a scenario, we may parametrize the metric perturbation as a unit matrix proportional to the Newtonian potential

(3.N):    $h_{\mu\nu} = 2\Phi \delta_{\mu\nu}$

such that eq. (3”) is now dominated by the Poisson equation

(3”’):    $\vec{\nabla}^2 \Phi = \frac{16\pi G_\text{N}}{d} \bar{T}_{00}.$

($d$ is the spacetime dimension.) What’s crucial for the Shapiro delay discussion below is that this Newtonian potential is strictly negative — provided the energy density is positive ($\bar{T}_{00} \geq 0$) — since the Euclidean Green’s function $1/\vec{\nabla}^2$ is strictly negative:

(3.NS):    $\Phi[t,\vec{x}] = - \frac{16\pi G_\text{N}}{d} \int_{\mathbb{R}^{d-1}} d^{d-1}\vec{x}' \frac{\Gamma[\frac{d-3}{2}]}{4\pi^{\frac{d-1}{2}} |\vec{x}-\vec{x}'|^{d-3}} \overline{T}_{00}[t,\vec{x}'] .$

Shapiro Delay     Consider two observers at spatial locations $\vec{x}$ and $\vec{x}'$ sending signals to each other via null rays (e.g., high frequency electromagnetic waves). Suppose the null rays pass through a region of spacetime near an isolated non-relativistic matter source, we may compute the time-of-flight between emission at $(t',\vec{x}')$ to reception at $(t,\vec{x})$ using the Synge’s world function. Since null rays are involved, that means the Synge’s world function in eq. (2.Synge) is zero:

(4):    $0 = (x-x')^\mu (x-x')^\nu \int_0^1 \left( \eta_{\mu\nu} + 2\Phi[\bar{Z}] \delta_{\mu\nu} \right) d\lambda , \qquad \qquad x\equiv(t,\vec{x}), \ x'\equiv(t',\vec{x}') \\ = T^2 - R^2 + 2 (T^2 + R^2) \int_0^1 \Phi[\bar{Z}] d\lambda .$

(Remember, $\bar{Z}^\mu = x'+\lambda(x-x')$.) We have used the shorthand $T\equiv t-t'$ for the time elapsed; and $R\equiv |\vec{x}-\vec{x}'|$ for the Euclidean coordinate distance between the observers. We see that, if there were no matter source, so that $\Phi = 0$, the Minkowski light cone condition $T^2 = R^2$ would be recovered. That $T^2 = R^2(1 + \mathcal{O}[\Phi])$ in turn means the $T^2$ multiplying the $\Phi$-integral may be replaced with $R^2$, since the error incurred would be of second order. This leads us to deduce from eq. (4)

(4′):    $T^2 = R^2 \left( 1 - 4 \int_0^1 \Phi[\bar{Z}] d\lambda + \mathcal{O}[\Phi^2] \right).$

Taking the positive square root on both sides, we find that the time elapsed is

(4.Shapiro):    $t-t' = |\vec{x}-\vec{x}'| \left( 1 - 2 \int_0^1 \Phi[\bar{Z}] d\lambda + \mathcal{O}[\Phi^2] \right) .$

We have arrived at the main result: Shapiro time delay. If the matter source were absent, the spacetime would be completely flat and $T=R$. But now that $\Phi$ is non-trivial, we see it increases the time-of-flight because the $-2 \Phi$ in eq. (4.Shapiro) is positive; which in turn is due to the positive character of the energy density $\bar{T}_{00}$ in eq. (3.NS). In fact, if energy density were strictly negative, $\bar{T}_{00} < 0$, notice this would decrease the time-of-flight and the effective speed of light would be faster than that in flat spacetime!

References

• I.I. Shapiro, “Fourth Test of General Relativity,” Phys. Rev. Lett. 13, 789 (1964). doi:10.1103/PhysRevLett.13.789
• I.I. Shapiro, G.H. Pettengill, M.E. Ash, M.L. Stone, W.B. Smith, R.P. Ingalls and R.A. Brockelman, “Fourth Test of General Relativity: Preliminary Results,”
Phys. Rev. Lett. 20, 1265 (1968). doi:10.1103/PhysRevLett.20.1265
• I.I. Shapiro, M.E. Ash, R.P. Ingalls, W.B. Smith, D.B. Campbell, R.B. Dyce, R.F. Jurgens and G.H. Pettengill, “Fourth test of general relativity – new radar result,” Phys. Rev. Lett. 26, 1132 (1971). doi:10.1103/PhysRevLett.26.1132
• M.J. Pfenning and E. Poisson, “Scalar, electromagnetic, and gravitational selfforces in weakly curved space-times,” Phys. Rev. D 65, 084001 (2002) doi:10.1103/PhysRevD.65.084001 [gr-qc/0012057].
• Y. Z. Chu and G.D. Starkman, “Retarded Green’s Functions In Perturbed Spacetimes For Cosmology and Gravitational Physics,” Phys. Rev. D 84, 124020 (2011) doi:10.1103/PhysRevD.84.124020 [arXiv:1108.1825 [astro-ph.CO]].

## Simple Harmonic Oscillator Path Integral

If you have taken quantum mechanics up to graduate school, you’d certainly have learned about the following transition amplitude — i.e., the path integral — for the simple harmonic oscillator (SHO):

(1):    $\langle x \vert \exp[-i(t-t') H_{\text{SHO}}] \vert x' \rangle = \sqrt{ \frac{\omega}{2\pi i \sin[\omega(t-t')]} } \exp\left[ \frac{i\omega}{2\sin[\omega(t-t')]} \left((x^2+x'^2)\cos[\omega(t-t')]-2xx'\right) \right] ,$

where we assume $t-t' \geq 0$; and the Hamilton of the SHO is

(1′):    $H_{\text{SHO}} = \frac{1}{2} p^2 + \frac{\omega^2}{2} X^2 .$

I will focus on the 1-dimensional case for simplicity. The key point in this post is that the discussion found in textbooks is usually not complete. For one, what does the square root of $1/i$ mean in eq. (1)? And, what happens when $\omega(t-t')$ is an integer multiple of $\pi$ — the sine goes to zero, and does that imply the path integral becomes ill defined?

The full path integral — the quantum mechanical transition amplitude for the SHO particle to propagate from $x'$ to $x$ is actually as follows. When the time elapsed lie within the ‘first half-period’, namely $0 < t-t' < \pi/\omega$, we have

(2):    $\langle x \vert \exp[-i(t-t') H_{\text{SHO}}] \vert x' \rangle = e^{-i\frac{\pi}{4}} \sqrt{ \frac{\omega}{2\pi |\sin[\omega(t-t')]|} } \exp\left[ \frac{i\omega}{2\sin[\omega(t-t')]} \left((x^2+x'^2)\cos[\omega(t-t')]-2xx'\right) \right] ,$

where here and below the square roots of real quantities are the positive ones. When the time elapsed lie between the $n$th and the $(n+1)$th half-periods, $n \pi/\omega < t-t' < (n+1) \pi/\omega$, we have

(2′):    $\langle x \vert \exp[-i(t-t') H_{\text{SHO}}] \vert x' \rangle = \frac{e^{-i\frac{\pi}{4}}}{i^n} \sqrt{ \frac{\omega}{2\pi |\sin[\omega(t-t')]|} } \exp\left[ \frac{i\omega}{2\sin[\omega(t-t')]} \left((x^2+x'^2)\cos[\omega(t-t')]-2xx'\right) \right] .$

In more detail, if we define $K_0[t-t';x,x'] \equiv e^{-i\frac{\pi}{4}} \sqrt{ \frac{\omega}{2\pi |\sin[\omega(t-t')]|} } \exp\left[ \frac{i\omega}{2\sin[\omega(t-t')]} \left((x^2+x'^2)\cos[\omega(t-t')]-2xx'\right) \right]$,

(2′.1):    $\langle x \vert \exp[-i(t-t') H_{\text{SHO}}] \vert x' \rangle = K_0[t-t';x,x'], \qquad\qquad 0 < t-t' < \frac{\pi}{\omega},$

(2′.2):     $\langle x \vert \exp[-i(t-t') H_{\text{SHO}}] \vert x' \rangle = \frac{1}{i} K_0[t-t';x,x'], \qquad\qquad \frac{\pi}{\omega} < t-t' < \frac{2\pi}{\omega},$

(2′.3):     $\langle x \vert \exp[-i(t-t') H_{\text{SHO}}] \vert x' \rangle = \frac{1}{i^2} K_0[t-t';x,x'], \qquad\qquad \frac{2\pi}{\omega} < t-t' < \frac{3\pi}{\omega},$

(2′.4):     $\langle x \vert \exp[-i(t-t') H_{\text{SHO}}] \vert x' \rangle = \frac{1}{i^3} K_0[t-t';x,x'], \qquad\qquad \frac{3\pi}{\omega} < t-t' < \frac{4\pi}{\omega} \dots$

These $1/i$ jumps in the phase factor when transitioning across half-periods are known as Maslov indices, and come about because one additional eigenvalue of the operator $(d/dt)^2 + \omega^2$ flips sign whenever the time elapsed $t-t'$ crosses over from say the $n$th to $(n+1)$th half-period.$\,^\star$

What happens when the time elapsed is some multiple of $\pi/\omega$? When $t-t' = n\pi/\omega$, for integer $n$, the path integral collapses to a Dirac delta function:

(2”):    $\langle x \vert \exp[-i(t-t') H_{\text{SHO}}] \vert x' \rangle = \frac{1}{i^n} \delta\left[ x - (-)^n x' \right] .$

As I hope to show in an upcoming post, the $(-)^n$ occurring within the Dirac delta function in eq. (2”) is intimately related to the fact that the energy eigenstates of the SHO system are also eigenstates of the parity operator.

$\,^\star$   I learned about this subtle aspect of the path integral from a comment left on Peter Woit’s excellent blog. The Maslov index, which falls within the more general rubric of Morse theory, is a generic feature of path integrals. It also shows up in the description of geometric optics and caustics.

References

• P. A. Horvathy, “The Maslov correction in the semiclassical Feynman integral,” Central Eur. J. Phys. 9, 1 (2011) doi:10.2478/s11534-010-0055-3 [quant-ph/0702236].
• L. S. Schulman, “Techniques and Applications of Path Integration,” Dover Publications (December 27, 2005)
• H. Kleinert, “Path Integrals in Quantum Mechanics, Statistics, Polymer Physics, and Financial Markets.” World Scientific Publishing Company; 5 edition (May 18, 2009)

## Mentorship Irresponsibilities of Physics Graduate Schools

Why go to graduate school in theoretical physics? If you are thinking about doing so, and particularly if you reside in North America or Europe, my recommendation would be: don’t do it! Find a career where your quantitative/analytical-thinking/programming skills are in demand, so that — relative to your chances within academia — you’ll likely find much higher standards of professionalism and more plentiful opportunities to grow.

♦     While I was a postdoc at U of MN Duluth (UMD), which does not award PhD degrees, one of the top MS students of the faculty who hired me had written 5 papers by time he was applying to PhD programs. Knowing the competitive nature of the admission process, however, I encouraged him to visit his top choices to deliver talks on his research. Given his track record, caliber, and experience, I told him he should not shy away from directly and aggressively approaching potential advisors to start working with them as soon as possible. The reader may not be surprised to hear — for, otherwise I would not be writing about this! — my advice was not taken; and while this student did get into a decent program he hadn’t found an advisor nearly a year into his tenure as a PhD student. I did hear from his former UMD classmate, this student did eventually find a mentor, but not within his original field of interest.

♦    Another ex-student of my research supervisor at UMD, whom I had co-mentored rather closely, spent 2+ years working with him — writing a robust numerical Python code to solve a class of Euclidean Quantum Field Theories — only to end up falling out with his advisor. He did end up writing his MS Project on the same topic, but with a completely different faculty. As I understood it, his supervisor falsely claimed all the 2+ years’ of work by the student was “trivial”. The only plausible explanation I have been able to come up with is that, due to his well-known conflict with the only other research-active theorist in the Department and out of his own pride, my then-supervisor did not want his students to only complete an MS degree with a Project. Rather he required them to produce something publishable (the student did not produce enough for a paper) in order to graduate from his theory group.

♦    What happens when an advisor produces an ill conceived, misguided project for his PhD student? Answer: his student struggles mightily with it, displaying great competence and scrupulousness — with little substantive aid from the advisor himself — only to end up wasting 2+ years and facing poor job prospects. Zero consequences for the advisor, as far as I can tell, other than padding his CV and grant applications with the paper the student did still manage to produce despite the trying circumstances.

♦    If you work with a senior and/or famous faculty, you may notice they publish very frequently and easily. This is at least partly because of the social-political dynamics of theoretical physics — it often takes just several conversations or so (especially if it involves junior people) for senior/famous folks to get their names on papers. Moreover, if you are senior/famous, it is easy to find junior scientists to explain things to you… Whereas, as already alluded to, as a graduate student you may have to struggle on your own with the technical implementation of your advisor’s ideas — and the burden of failure often falls entirely upon your shoulders.

♦    I took a year off after obtaining my MS degree to properly search for a graduate school to pursue my interests in theoretical cosmology, gravitation and field theory. During my visits to various schools I ran into quite a number of unhappy graduate students. One of them brought me inside a conference room, closed the door, and proceeded to warn me: if I ever came here, I should make sure not to work with this student’s advisor! The advisor is the sort that would barely write a single paper with his graduate student; and then leave them to find their own projects to work on. In theoretical physics, finding your own project as a graduate student is usually not only very difficult, it is a very poor strategy if one wishes to mature as a scientist. Unfortunately, this student’s experience is not an uncommon phenomenon.

♦    I obtained my MS degree in Physics from Yale. One of my classmates there was an international scholar who had graduated valedictorian from a small but elite school in the US. Like my UMD research supervisor’s top student mentioned above, however, he struggled to find an advisor that matched his research interests. He had to quit Yale, took a break from school, and eventually returned to his undergraduate alma mater to complete his PhD degree. The last time I heard from him, he was a physics and math private tutor in Texas. I suppose it is not too unreasonable to assert: a person of his abilities does not need to travel across the Atlantic (he’s European) to waste his time getting the most advanced degree possible in theoretical physics just to become a private tutor in a land so distant from home. (As a side note: I do personally know of at least 1-2 other PhD classmates of mine who have gone on to become high school teachers — a respectable profession, of course; but not one requiring a doctorate in theoretical physics. This side remark is not one about poor mentorship per se, but about the dearth of job opportunities.)

♦    When I was at Yale I came to appreciate that high energy theory was in the decline due to the severe lack of guidance from experiment: as of this writing, there are still no discoveries in particle physics that can illuminate the way forward for theorists working to extend the Standard Model to account for Dark Matter and other mysteries. I decided to turn my attention towards cosmology. I was temporarily relieved that the high energy theory group had just hired a new cosmologist; and I quickly made sure to approach him to indicate my interest in working with him. The way things worked at Yale, I was supposed to find an internship of sorts with a potential advisor during my first summer there. But the cosmologist told me he would be away and thus cannot take me. It turned out he was around, and even took on an undergraduate student. I repeated my interest in working with him during the following Fall. He asked me what classes I was taking and I happily told him I was auditing a Numerical Methods course offered by the Astronomy Department. This cosmologist was a heavily numerical one, and I had taken the initiative to pick up some numerical/programming skills. He told me he would take me “over all other students at Yale” but also did remark that numerical analysis was a very “portable skill,” a puzzling comment that I only understood later on. As far as Life itself was concerned, I guess I was still in kindergarten then, not recognizing that when a faculty is repeatedly vague with you, that most likely indicates a “No”. After more than 1+ years of pinning my hopes upon that cosmologist, I finally learned he — as brand new faculty at Yale — was taking his one and only student from a completely different institution. Only during Spring of my second year at Yale did the cosmologist reveal he was not inclined to take Yale students because he was worried they have a tendency to become quants right after graduation. When discussing this with my white American classmate later on, he then informed me he had overheard conversations between faculty that they believed Chinese graduate students have a tendency to go into finance directly after obtaining their PhD.

I had thought that this student the cosmologist had taken from another institution must be extraordinary — well trained in cosmology and quantum field theory perhaps. During the student’s first year at Yale, however, he had to take a cosmology reading course with his advisor and quantum field theory with mine. And, he did not do spectacularly in the latter. (I know, because I was his TA; I am glad to report, on the other hand, the top student in that QFT course is now a faculty at an elite US institution.)

At Yale, if you do not find an advisor by the end of the second year of graduate school, you’re fucked your support will cease completely. After graduating from Kindergarten into Elementary School of Life, I proceeded to pursue the cosmologist’s next-door neighbor out of panic — despite hearing he had already taken a student. Now, a dating advice: you should be very wary of partners who are ready to find a new fling at the slightest temptation. (Did I mention advisor seeking is like dating?) My experiences with this advisor employer, whom I worked with for a whole year before taking a year off from school, can easily fill a blog post or two. Suffice to say, theoretical high energy physicists aren’t the nicest lot. A gravitation theorist once told me he chose not to do particle theory because of its hard-nosed culture; I find the latter to be an understatement.

These examples merely illustrate a single aspect of the deeply flawed reward structure that exists within the academia I am familiar with: the extraordinarily low standards of mentorship within the theoretical physics community. Unfortunately, nowadays, Western academia — including physics/astrophysics — enjoys emphasizing the need for “diversity,” but given how left-leaning illiberal/biased it has grown, however, I fear this just translates into an obsession with identity politics that is highly hypocritical and destructive.

## Green’s Functions in Translation Symmetric and Parity Invariant Spaces

Setup     In $D$ spatial dimensions, suppose we wish to solve the following Green’s function equation

(1)    $\left(\mathcal{W} - \vec{\nabla}^2_{\vec{x},D}\right) G_D = \left(\mathcal{W} - \vec{\nabla}^2_{\vec{x}',D}\right) G_D = M \cdot \delta^{(D)}[\vec{x}-\vec{x}'] .$

The $\mathcal{W}$ is some linear differential operator and $\vec{\nabla}^2_{\vec{x},D}$ and $\vec{\nabla}^2_{\vec{x}',D}$ are Laplacians in flat/Euclidean space with respect to some observer at $\vec{x}$ and point source at $\vec{x}'$. The $\delta^{(D)}[\vec{x}-\vec{x}']$ is the $D$ dimensional Dirac delta function in flat space. Note that we are not assuming anything about the geometry of spacetime itself at this point; but we shall assume that both $\mathcal{W}$ and $M$ do not depend on $D$ nor on the spatial coordinates $\vec{x}$ and $\vec{x}'$, so that the system is invariant under spatial translations

(2)    $\vec{x} \to \vec{x} + \vec{a} \qquad \text{and} \qquad \vec{x}' \to \vec{x}' + \vec{a}$

for constant $\vec{a}$; as well as parity flips

(3)    $\vec{x} \to -\vec{x} \qquad \text{and} \qquad \vec{x}' \to -\vec{x}' .$

We then expect that the Green’s function $G_D$ be a function of space solely through the Euclidean distance

(4)    $R \equiv |\vec{x}-\vec{x}'| .$

Result     The central assertion of this post is that

The Green’s function in odd spatial dimensions may be obtained from its 1-dimensional counterpart

(4′)    $G_{\text{odd }D \geq 1}[R] = \left(-\frac{1}{2\pi R} \frac{\partial}{\partial R}\right)^{\frac{D-1}{2}} G_1[R] .$

The Green’s function in even spatial dimensions may be obtained from its 2-dimensional counterpart

(4”)    $G_{\text{even }D \geq 2}[R] = \left(-\frac{1}{2\pi R} \frac{\partial}{\partial R}\right)^{\frac{D-2}{2}} G_2[R] .$

Proof     In $D+2$ spatial dimensions

(5)    $\left(\mathcal{W} - \vec{\nabla}^2_{\vec{x},D+2}\right) G_{D+2} = \left(\mathcal{W} - \vec{\nabla}^2_{\vec{x}',D+2}\right) G_{D+2} = M \cdot \delta^{(D+2)}[\vec{x}-\vec{x}'] .$

Let us consider applying the $D$ dimension version of the differential operator in eq. (1) onto the following integral involving the $G_{D+2}$:

(5′)     $\widehat{G}_{D} \equiv \int_{\mathbb{R}^2} d x'^{D+1} d x'^{D+2} G_{D+2} .$

By translation symmetry, note that we may also integrate with respect to $(x^{D+1},x^{D+2})$.

(5”)     $\widehat{G}_{D} = \int_{\mathbb{R}^2} d x^{D+1} d x^{D+2} G_{D+2} .$

We have

(6)     $\left( \mathcal{W} - \vec{\nabla}_{\vec{x},D} \right) \widehat{G}_D \\ = \left( \mathcal{W} - \vec{\nabla}_{\vec{x},D} - \partial_{x^{D+1}}^2 - \partial_{x^{D+2}}^2\right) \int_{\mathbb{R}^2} d x'^{D+1} d x'^{D+2} G_{D+2} ;$

where the $-\partial_{x^{D+1}}^2-\partial_{x^{D+2}}^2$ was inserted without any impact because, if we perform the integral of $(x'^{D+1},x'^{D+2})$ the result is independent of $(x^{D+1},x^{D+2})$. However, if we now interchange the order of integration and differentiation, followed by employing eq. (5),

(7)     $\left( \mathcal{W} - \vec{\nabla}_{\vec{x},D} \right) \widehat{G}_D \\ = M \int_{\mathbb{R}^2} d x'^{D+1} d x'^{D+2} \delta^{(D)}[\vec{x}-\vec{x}'] \delta[x^{D+1}-x'^{D+1}] \delta[x^{D+2}-x'^{D+2}] \\ = M \cdot \delta^{(D)}[\vec{x}-\vec{x}'].$

Similar arguments would demonstrate that

(8)     $\left( \mathcal{W} - \vec{\nabla}_{\vec{x}',D} \right) \widehat{G}_D = M \cdot \delta^{(D)}[\vec{x}-\vec{x}'] .$

In other words, eq. (5′) is in fact $G_D$ itself, i.e., we may drop the $\widehat{\cdot}$. Strictly speaking, if we were dealing with spacetime Green’s functions of wave equations, we would also have to argue that integrating retarded (advanced) Green’s functions $G_{D+2}$ would yield retarded (advanced) Green’s functions $G_D$. That this is true can be seen by viewing the integral over the 2 ‘extra dimensions’ as an instantaneous 2D sheet source — and the observer confined on the $D$ dimensional space receives a retarded (advanced) signal from the 2D sheet of uniform ‘charge density’.

By translation symmetry, we may write equations (5′) and (5”) as

(8)    $G_D[R] = \int_{\mathbb{R}^2} d (x-x')^{D+1} d (x-x')^{D+2} G_{D+2}[R] ,$

Note that this integral relation on its own does not require parity invariance, but the assumption that $G$ is a function of space through $R$ does. Now, on the right hand side of eq. (8),

(9)    $R \equiv R_{D+2} = \sqrt{R_{D} + \rho^2}$

where

(9′)    $R_D \equiv \sqrt{\sum_{i=1}^{D} ((x-x')^i)^2}, \qquad \qquad \rho \equiv \sqrt{((x-x')^{D+1})^2+((x-x')^{D+2})^2} .$

Switching the Cartesian integration variables of eq. (8) to 2D cylindrical coordinates $(\rho,\phi),$

(8)    $G_D[R] = 2\pi \int_{0}^\infty d \rho \cdot \rho G_{D+2}\left[ \sqrt{R^2+\rho^2} \right] = 2\pi \int_{R}^\infty d R' \cdot R' G_{D+2}\left[ R' \right] .$

At this point, differentiating both sides with respect to $R$ yields the differential recursion relation

(9)    $G_{D+2}[R] = -\frac{1}{2\pi R} \frac{\partial}{\partial R} G_D[R] .$

By applying the ‘raising operator’ $\mathcal{D}_R \equiv - (2\pi R)^{-1} \partial_R$ $n$ times on $G_D[R]$, we have

(10)    $G_{D+2n}[R] = \left(-\frac{1}{2\pi R} \frac{\partial}{\partial R}\right)^n G_D[R] .$

We arrive at equations (4′) and (4”) by setting, respectively, $D=1$ and $D=2$.

Application to Wave Equations     The fundamental solution of wave equations lie in their Green’s functions, which in turn can be thought of the field engendered by a spacetime-point source. The actual solution will then involve the superposition of all spacetime point sources weighted by the physical source in the setup in question. In spatial translation symmetric and parity invariant spacetimes, we would expect the Green’s functions of the wave operator in $d$ spacetime dimensions, namely $G_d \equiv 1/\Box$, to depend on space only through the function $R$. Thus, they should all obey, from equations (4′) and (4”),

(11)    $G_{\text{even }d}[t-t',R] = \left(-\frac{1}{2\pi R} \frac{\partial}{\partial R} \right)^{\frac{d-2}{2}} G_2[t-t',R]$

and

(11′)    $G_{\text{odd }d}[t-t',R] = \left(-\frac{1}{2\pi R} \frac{\partial}{\partial R} \right)^{\frac{d-3}{2}} G_3[t-t',R].$

These statements include the Minkowski and spatially flat Friedmann–Lemaître–Robertson–Walker cosmological spacetimes. The flat spacetime case, in particular, takes the form

(12)    $G^+_{\text{even }d \geq 4}[t-t',R] = \left(-\frac{1}{2\pi} \frac{\partial}{\partial R} \right)^{\frac{d-2}{2}} \left( \frac{\delta[t-t'-R]}{4\pi R} \right)$

and

(12′)    $G^+_{\text{odd }d \geq 3}[t-t',R] = \left(-\frac{1}{2\pi R} \frac{\partial}{\partial R} \right)^{\frac{d-3}{2}} \left( \frac{\Theta[t-t'-R]}{2\pi\sqrt{(t-t')^2-R^2}} \right).$

These are the retarded Green’s functions, where cause (i.e., the source) at $(t',\vec{x}')$ precedes the effect (i.e., the signal) at $(t,\vec{x})$; they satisfy

(12”)     $\left(\partial_0^2 - \partial_i \partial_i \right) G = \delta[t-t'] \delta^{(d-1)}[\vec{x}-\vec{x}'] .$

I have chosen to begin the even dimensional results at $d = 4$ to demonstrate the well known fact that waves associated with massless particles in even dimensional spacetimes higher than 2 propagate strictly on the light cone — as encoded by the Dirac delta function $\delta[t-t'-R]$ of eq. (12); whereas, according to eq. (12′), with $\Theta[z]=1$ when $z>0$ and $\Theta[z]=0$ when $z<0$, massless signals do in fact travel inside the null cone — tails develop — in all odd dimensional flat spacetimes.

Speculations     Can the ‘raising operator’ $-(2\pi R)^{-1} \partial_R$ be viewed as a genuine operator acting on a Hilbert space involving some form of union of Minkowski or spatially flat cosmological spaces of all even (or odd) dimensions? The reason for this speculation is that the presence of tails in eq. (12′) is likely due to the incomptability of strictly null propagation with spherical symmetry; in other words, in odd-dimensional Minkowski, there is no solution for an infinitesimally thin spherical shell propagating at the speed of light outward from some instantaneous point source at $(t',\vec{x}')$. Can this be elucidated more precisely — i.e., using equations — through such an operator approach?

References

• Y.-Z. Chu, “More On Cosmological Gravitational Waves And Their Memories,” Class. Quant. Grav. 34, no. 19, 194001 (2017) doi:10.1088/1361-6382/aa8392
[arXiv:1611.00018 [gr-qc]].
• Y.-Z. Chu, “Transverse traceless gravitational waves in a spatially flat FLRW universe: Causal structure from dimensional reduction,” Phys. Rev. D 92, no. 12, 124038 (2015) doi:10.1103/PhysRevD.92.124038 [arXiv:1504.06337 [gr-qc]].
• H. Soodak and M. S. Tiersten, Am. J. Phys. 61 (5), May 1993

## Integration Via Differential Equations: Two Bessel(-Trigonometric) Integrals

It is rather easy to find integrals that cannot be expressed in “closed form”, which in this context means functions whose properties we know a lot about. I, for one, am rather grateful for the resources — and the people who compiled them! — such as the Table of Integrals, Series, and Products, DLMF, AMS55 and Wolfram Math World that we may consult for calculus results and properties of “special functions”.

In this post I will discuss how to evaluate the following two integrals by solving the relevant differential equations they satisfy.

(I): $\frac{E_1[m]}{I_\nu[m]} \equiv \int_{-Z-\sqrt{Z^2-1}}^{-Z+\sqrt{Z^2-1}} \frac{d \rho'}{\sqrt{\rho'}} \frac{I_\nu[m \rho'] \cos\left[ m \sqrt{2 \bar{\sigma}} \right]}{I_\nu[m] \sqrt{2 \bar{\sigma}}} = \pi P_{\nu-\frac{1}{2}}\left[-Z\right]$

(II): $\frac{E_2[m]}{I_\nu[m]} \equiv \int_{-Z-\sqrt{Z^2-1}}^{-Z+\sqrt{Z^2-1}} \frac{d \rho'}{\rho'} \frac{I_{\nu}[m \rho']}{I_\nu[m]} J_0\left[m\sqrt{2\bar{\sigma}}\right] = \frac{1}{\nu} \left\{ \left( -Z + \sqrt{Z^2-1} \right)^\nu - \left( -Z - \sqrt{Z^2-1} \right)^\nu \right\}$

Here, $J_\nu$ is the Bessel function of the first kind; $I_\nu$ is the modified Bessel function of the first kind; and

(III): $\bar{\sigma} \equiv -\frac{1}{2} \left( \rho'^2 + 1 + 2 \rho' Z \right) \\ = -\frac{1}{2} \left\{ \rho' - \left( -Z - \sqrt{Z^2-1} \right) \right\} \left\{ \rho' - \left( -Z + \sqrt{Z^2-1} \right) \right\}$.

Motivation       Since there are infinitely many intractable integrals anyway, you may wonder why you should pay attention to this result. There is in fact a physical reason for doing so. I hope to start writing about it more, but in curved spacetimes, waves associated with massless particles in Nature — including light itself — do not in fact travel strictly on the null cone. Results (I) and (II) describe the inside-the-light cone (aka “tail”) portion of massive scalar waves in de Sitter spacetime, by viewing the latter as a hyperboloid situated in 1 higher dimensional Minkowski spacetime. The massless wave tails can in turn be obtained by setting $m=0$.

Derivation of I and II       By a direct calculation, you may readily verify that

(D1): $\mathcal{D}_m E[m] \equiv m^2 E''[m] + m E'[m] - (m^2+\nu^2) E[m] = 0$

where here $E[m]$ is either $E_1$ or $E_2$. Note that this is the ordinary differential equation (ODE) satisfied by $I_\nu[m]$ itself; i.e., $\mathcal{D}_m I_\nu[m] = 0$.

In more detail, you should find that applying $\mathcal{D}_m$ upon the left hand sides of equations (I) and (II) yields

(D2): $\mathcal{D}_m E_1[m] = -2m \int_{-Z-\sqrt{Z^2-1}}^{-Z+\sqrt{Z^2-1}} d\rho' \frac{\partial}{\partial \rho'} \left( \sqrt{\rho'} \sin\left[ m \sqrt{2\bar{\sigma}} \right] I_\nu[m\rho'] \right)$

and

(D3): $\mathcal{D}_m E_2[m] = -2m \int_{-Z-\sqrt{Z^2-1}}^{-Z+\sqrt{Z^2-1}} d\rho' \frac{\partial}{\partial \rho'} \left( \sqrt{2\bar{\sigma}} I_\nu[m\rho'] J_1\left[ m \sqrt{2\bar{\sigma}} \right] \right) .$

That is, acting $\mathcal{D}_m$ on $E_{1,2}$ converts the integrands into total derivatives, which then tells us the result is simply these integrands evaluated at the end points $\rho_\pm \equiv -Z \pm \sqrt{Z^2-1}$. But from eq. (III) we see these $\rho_\pm$ are precisely the zeroes of $\bar{\sigma}$ and hence of $\sin[m \sqrt{2\bar{\sigma}}]$ and $\sqrt{2\bar{\sigma}} J_1[m \sqrt{2 \bar{\sigma}}]$; which in turn means the integrals are zero. In other words, $\mathcal{D}_m E[m] = 0$. But as already alluded to, this is precisely the ODE satisfied by $I_\nu[m]$ itself. However, since there are two linearly independent solutions, we still need to show that our integrals satisfy $E[m] \propto I_\nu$. For non-integer $\nu$, note that $I_{\pm\nu}$ are linearly independent. Moreover, we may check from equations (I) and (II) that $E_{1,2}[m]$ are in fact power series in $m$ that begin with an overall $m^\nu$ pre-factor arising from the $I_\nu[m \rho']$. Since $I_{\pm\nu}[m]$ is $m^{\pm\nu}$ times a positive power series in $m^2$, this tells us there cannot be a $I_{-\nu}[m]$ term in our $E_{1,2}$. What remains is to figure out the $\chi_{1,2}$ in

(D4): $E_{1,2}[m] = \chi_{1,2} I_\nu[m] .$

To do so, notice $\chi_{1,2}$ cannot depend on $m$. We may therefore extract their values through the limits

(D5): $\chi_{1,2} = \lim_{m \to 0} \frac{E_{1,2}[m]}{I_\nu[m]} .$

From equations (I) and (II), and utilizing the Taylor series results

$I_\nu[z] = \frac{(z/2)^\nu}{\Gamma[\nu+1]} \left( 1 + \mathcal{O}[z^2] \right)$

and

$J_0[z] = 1 + \mathcal{O}[z^2]$

we have

(D6): $\lim_{m \to 0} \frac{E_1[m]}{I_\nu[m]} = \int_{-Z-\sqrt{Z^2-1}}^{-Z+\sqrt{Z^2-1}} d \rho' \frac{\rho'^{\nu-\frac{1}{2}}}{\sqrt{-\rho'^2 - 1 - 2\rho' Z}}$

and

(D7): $\lim_{m \to 0} \frac{E_2[m]}{I_\nu[m]} = \int_{-Z-\sqrt{Z^2-1}}^{-Z+\sqrt{Z^2-1}} d \rho' \rho'^{\nu-1} = \frac{1}{\nu} \left\{ \left( -Z + \sqrt{Z^2-1} \right)^\nu - \left( -Z - \sqrt{Z^2-1} \right)^\nu \right\} .$

From equations (D4), (D5) and (D7), we may see that equations (II) has been proven for non-integer $\nu$. What remains, therefore, is to tackle eq. (D6). If we put

(D8): $\rho' \equiv - Z + \cos[u] \sqrt{Z^2-1}, \qquad\qquad u \in [0,\pi]$

so that

$\frac{d\rho'}{\sqrt{2 \bar{\sigma}}} = \frac{d\rho'}{\sqrt{-\rho'^2 -1 - 2\rho' Z}} = \frac{d(\cos u) \sqrt{Z^2-1}}{\sin[u] \sqrt{Z^2-1}};$

(D9): $\lim_{m \to 0} \frac{E_1[m]}{I_\nu[m]} = \int_{u=0}^{u=\pi} d u \left( -Z + \sqrt{Z^2-1} \cos u\right)^{\nu-\frac{1}{2}}.$
Since the cosine is an even function we may extend the integration limit to $u = -\pi$, namely
(D9′): $\lim_{m \to 0} \frac{E_1[m]}{I_\nu[m]} = \frac{1}{2}\int_{u=-\pi}^{u=\pi} d u \left( -Z + \sqrt{Z^2-1} \cos u\right)^{\nu-\frac{1}{2}} .$
At this point, referring to the integral representation of the Legendre function $P_{\nu-1/2}[z]$ — see here, for instance — tells us we have arrived at the right-hand-side of eq. (I), at least for non-integer $\nu$. Our results are very likely true for integer $\nu$ as well, since I believe it is safe to assume they are continuous functions of $\nu$.