Struggles In Physics

1. Motivation In non-relativistic quantum mechanics, there exists in Cartesian coordinates the famous commutation relation between the position $X^i$ and momentum operator $P_i$ , namely,

$\left[ X^i, P_j \right] \equiv X^i P_j - P_j X^i = i \delta^i_{\phantom{i}j}.$

What is the corresponding commutator in relativistic — i.e., Minkowski-spacetime — quantum field theory (QFT)? For a scalar field $\varphi$ , say, we often state, in terms of Cartesian coordinates that parametrize an inertial frame,

(1.1): $\left. \left[ \varphi[t,\vec{x}], \frac{\partial \mathcal{L}}{\partial (\partial_{0} \varphi)}[t',\vec{x}'] \right] \right\vert_{t=t'} = i \delta^{(d-1)}\left[ \vec{x}-\vec{x}' \right],$

where $\mathcal{L}$ is the Lagrangian density and hence $\partial$ $\mathcal{L}$ / $\partial (\partial_{0} \varphi)$ is the conjugate momentum to $\varphi$ .

However, equation (1.1) — with its explicit dependence on the time-derivative of $\varphi$ — makes opaque the underlying Lorentz symmetry enjoyed by the field theory itself: namely, why isn’t there an “equal” treatment of space and time? This issue becomes more pressing when quantizing field theories in curved spacetimes, where the need for a generally covariant description of the field theory does not sit well with the need, in “canonical quantization”, for imposing a quantization condition at “equal times”. I believe it was Peierls who first advocated, through the brackets now named after him, a different route to quantization that allows the process to begin from a generally covariant stand-point. Peierls’ brackets have since been promoted repeatedly by Bryce DeWitt — see, for instance, his two-volume text on Quantum Field Theory. Here, I will follow the discussion in Rafael Sorkin’s lecture notes, to quantize a linear scalar field in curved spacetimes using the difference between its generally covariant classical retarded and advanced Green’s functions. In spacetimes where one can find a global timelike vector field $u^\mu$ , which we shall normalize to unity, i.e., $u^2 \equiv u_\mu u^\mu = 1$ , I will then show how to recover the conventional “equal-time” commutation relations.

2. Peierls’ bracket for a linear scalar field Let us begin with a scalar field $\varphi$ subject to the following (Heisenberg-picture) equations-of-motion:

(2.1): $\left( \Box + \xi \mathcal{R} \right) \varphi = 0$ ,

where $\Box \varphi = \partial_\mu \left( \sqrt{|g|} g^{\mu\nu} \partial_\nu \varphi \right)/\sqrt{|g|}$ is the wave operator; $g_{\mu\nu}$ is the spacetime metric; $|g|$ is the absolute value of its determinant; $\xi$ is some arbitrary real constant; and $\mathcal{R}$ is the Ricci scalar of the same geometry described by $g_{\mu\nu}$ .

To quantize this theory in a curved spacetime we shall postulate that the field operator $\varphi$ obeys the generally covariant commutation relation

(2.2): $\Big[ \varphi[x], \varphi[x'] \Big] = -i \Delta[x,x'],$

where $x$ and $x'$ are now shorthand for spacetime coordinates; and $\Delta[x,x']$ is defined as the difference between the retarded $G^\text{(ret)}$ and advanced $G^\text{(adv)}$ Green’s functions:

(2.3): $\Delta[x,x'] \equiv G^\text{(ret)}[x,x'] - G^\text{(adv)}[x,x'],$

with (cf. eq. (2.1))

(2.4): $\left( \Box + \xi \mathcal{R} \right) G^\text{(ret)} = \left( \Box + \xi \mathcal{R} \right) G^\text{(adv)} = \frac{\delta^{(d)}\left[x-x'\right]}{\sqrt[4]{g[x] g[x']}}.$

The left hand side of eq. (2.4) holds with respect to both $x$ and $x'$ . Moreover, the retarded Green’s function $G^\text{(ret)}[x,x']$ is characterized by the condition that, for a fixed source spacetime position $x'$ , it is non-zero only when the observer spacetime location $x$ lies to the future of all appropriately defined constant-time surfaces containing $x'$ . Similarly, the advanced Greens function $G^\text{(adv)}[x,x']$ is characterized by the condition that, for a fixed $x'$ , it is non-zero only when $x$ lies to the past of all appropriately defined constant-time surfaces containing $x'$ .

Anti-symmetry Because $G^{\text{(ret)}}[x',x]=G^{\text{(adv})}[x,x']$ , we may see from eq. (2.3):

(2.5): $\Delta[x,x'] = -\Delta[x',x].$

3. Klein-Gordon product When dealing with QFT in some $d$ -dimensional curved spacetime, one would meet the indefinite “inner product” of Klein-Gordon (KG). For a pair of scalar functions $f$ and $g$ , their KG product on some constant-time hypersurface $\Sigma[t]$ , which we also assume is orthogonal to some unit norm future-directed timelike vector $u^\mu$ , is defined as:

(3.1): $\left( f \odot g \right)[t] \equiv \int_{\Sigma[t]} d^D \Sigma_\mu \left( (\nabla^\mu f) g - f (\nabla^\mu g) \right).$

Here, if one may endow the $\Sigma[t]$ surface with $D \equiv d-1$ spatial coordinates $\vec{y}$ such that the induced metric on the former is $h_{ij}[\vec{y}]$ , the directed volume/area element is then

$d^D \Sigma_\mu \equiv d^D\vec{y} \sqrt{\det h_{ij}[\vec{y}]} u_\mu[t,\vec{y}],$

where $u_\mu$ $[t,\vec{y}]$ means the timelike vector $u_\mu$ is evaluated at the time $t$ , corresponding to the constant-time hypersurface $\Sigma[t]$ , and the spatial position $\vec{y}$ .

4. A lemma To recover equal-time commutation relations familiar from canonical quantization procedures, we first need the following lemma involving the KG product of $\Delta$ in eq. (2.3) and some arbitrary scalar function $f$ .

Let $\Delta[x,x']$ and $f[x']$ vanish at the (spatial) boundaries of $\Sigma[t]$ for all relevant times $t$ ; i.e.,

$\Delta\left[ x,x' \in \partial \Sigma[t] \right] = f\left[ x' \in \partial \Sigma[t] \right] = 0, \qquad \forall t$ .

Furthermore, let $f$ obey the classical Klein-Gordon equation (cf. eq. (2.1)),

(4.1): $\left( \Box + \xi \mathcal{R} \right) f = 0$ .

Then,

(4.2): $\left( \Delta \odot f\right)_x[t] \equiv \int_{\Sigma[t]} d^D \Sigma_{\mu'} \left( \nabla^{\mu'} \Delta[x,x'] f[x'] - \Delta[x,x'] \nabla^{\mu'} f[x'] \right) = -f[x].$

(The prime on the spacetime indices indicates the associated spacetime coordinate is $x'$ ; for e.g., $\nabla^{\mu'} \equiv \nabla^{x'^\mu}$ .)

Proof To see this, we first suppose $x$ lies to the future of $\Sigma[t]$ . Then, only the retarded portion $G^{\text{(ret)}}$ of $\Delta$ in eq. (2.3) contributes to the integral, namely

$\left( \Delta \odot f \right)_x[t] = \int_{\Sigma[t]} d^D \Sigma_{\mu'} \left( \nabla^{\mu'} G^{\text{(ret)}}[x,x'] f[x'] - G^{\text{(ret)}}[x,x'] \nabla^{\mu'} f[x'] \right).$

We may convert this integral into a “closed-surface” one by introducing another constant-time hypersurface $\Sigma[t_>]$ that lies to the future of $x$ . For, since the retarded Green’s function $G^{\text{(ret)}}[x,x'] = 0$ whenever $x'$ lies on $\Sigma[t_>]$ ,

$\left( \Delta \odot f\right)_x[t] = \left(\int_{\Sigma[t]} - \int_{\Sigma[t_>]}\right) d^D \Sigma_{\mu'} \left( \nabla^{\mu'} G^{\text{(ret)}}[x,x'] f[x'] - G^{\text{(ret)}}[x,x'] \nabla^{\mu'} f[x']\right).$

That this is now a “closed-surface” integral is because we have assumed $\Delta[x,x']$ (viewed as a function of $x'$ ) and $f[x']$ both vanish on the spatial boundaries of $\Sigma[t]$ for all times $t$ . Since, by assumption, $u^\mu$ is “pointing” towards the future, we have Gauss’ theorem informing us that

$\left( \Delta \odot f\right)_x[t] = - \int_{t}^{t_>} d^d x' \sqrt{|g[x']|} \nabla_{\mu'} \left( \nabla^{\mu'} G^{\text{(ret)}}[x,x'] f[x'] - G^{\text{(ret)}}[x,x'] \nabla^{\mu'} f[x'] \right),$

where the limits $t$ and $t_>$ remind us the domain of the spacetime-volume integral $\int d^d x' \sqrt{|g[x']|}$ is bounded by the constant-time hypersurfaces $\Sigma[t]$ and $\Sigma[t_>]$ . At this point, by expanding out the covariant derivatives, one should find

$\left( \Delta \odot f\right)_x[t] = - \int_{t}^{t_>} d^d x' \sqrt{|g[x']|} \left( \Box_{x'} G^{\text{(ret)}}[x,x'] f[x'] - G^{\text{(ret)}}[x,x'] \Box_{x'} f[x'] \right).$

By adding and subtracting $\xi \mathcal{R}[x'] G^{\text{(ret)}}[x,x'] f[x']$ within the integrand, followed by using the equations-of-motion equations (2.4) and (4.1), one would find

$\left( \Delta \odot f\right)_x[t] = - \int_{t}^{t_>} d^d x' \sqrt{|g[x']|} \frac{\delta^{(d)}[x-x']}{\sqrt[4]{g[x]g[x']}} f[x'].$

Since $x$ always lies within the integration domain $\int_{t}^{t_>} d^d x' \sqrt{|g[x']|}$ we have arrived at eq. (4.2).

For the case where $x$ lies to the past of $\Sigma[t]$ , only the advanced Green’s function now contributes to the integral $(\Delta \odot f)_x[t]$ . A similar line of reasoning — introducing a hypersurface $\Sigma[t_<]$ lying to the past of $x$ — can be used to convert $(\Delta \odot f)_x[t]$ into a “closed-surface” integral, from which eq. (4.2) would again be established once equations (2.4) and (4.1) are employed.

5. Equal-time commutators Let the above constant-time hypersurface $\Sigma[t]$ be parametrized by $D \equiv d-1$ spatial coordinates and let $\vec{y}$ and $\vec{y}'$ denote two distinct locations on $\Sigma[t]$ ; also denote the “time-derivative” of $\varphi$ as

$\dot{\varphi} \equiv u^\mu \nabla_\mu \varphi.$

The following ‘equal-time’ commutation results hold on the hypersurface $\Sigma[t]$ :

(I) The curved spacetime QFT analog of the quantum mechanical $[X^i, P_j] = i\delta^i_j$ reads

(5.1): $\Big[ \varphi[t,\vec{y}], \dot{\varphi}[t,\vec{y}'] \Big] = i \frac{ \delta^{(D)}[\vec{y}-\vec{y}'] }{\sqrt[4]{\det h_{ab}[\vec{y}] \det h_{a'b'}[\vec{y}']}} ;$

(II) while the “equal-time” auto-commutator of the fields and their time derivatives are zero, namely

(5.2): $\Big[ \varphi[t,\vec{y}], \varphi[t,\vec{y}'] \Big] = \Big[ \dot{\varphi}[t,\vec{y}], \dot{\varphi}[t,\vec{y}'] \Big] = 0.$

Proof Let $f$ and $g$ satisfy the KG equation (4.1), but are otherwise arbitrary. (Since they obey second-order-in-time differential equations, by arbitrary we mean they have arbitrary initial conditions $(f,\dot{f})$ and $(g,\dot{g})$ on $\Sigma[t]$ .) From the lemma in eq. (4.2), we may note that

$\left( f \odot \Delta \odot g\right) = -\left( f \odot g\right).$

(One can check that the KG product is associative, in that $f \odot \Delta \odot g = (f \odot \Delta) \odot g = f \odot (\Delta \odot g)$ .) This can be written as

(5.3): $\left( f \odot \Delta \odot g\right)[t] \\ = -\int_{\Sigma[t]} d^D\vec{y}' \sqrt{\det h_{a'b'}} d^D\vec{y}'' \sqrt{\det h_{a''b''}} \frac{\delta^{(D)}[\vec{y}'-\vec{y}'']}{\sqrt[4]{\det h_{i'j'}[\vec{y}'] \det h_{i''j''}[\vec{y}'']}} \\ \qquad\qquad \times \left( \dot{f}[x'] g[x''] - f[x'] \dot{g}[x''] \right),$

where $x' \equiv (t,\vec{y}')$ and $x'' \equiv (t,\vec{y}'')$ . On the other hand, via a direct calculation,

(5.4): $\left( f \odot \Delta \odot g \right)[t] \\= \int_{\Sigma[t]} d^D\vec{y}' \sqrt{\det h_{a'b'}} d^D\vec{y}'' \sqrt{\det h_{a''b''}} \left( \dot{f}[x'] u^{\mu''} \nabla_{\mu''} \Delta[x',x''] - f[x'] u^{\alpha'} u^{\mu''} \nabla_{\alpha'} \nabla_{\mu''} \Delta[x',x''] \right) g[x''] \\ - \int_{\Sigma[t]} d^D\vec{y}' \sqrt{\det h_{a'b'}} d^D\vec{y}'' \sqrt{\det h_{a''b''}} \left( \dot{f}[x'] \Delta[x',x''] - f[x'] u^{\alpha'} \nabla_{\alpha'} \Delta[x',x''] \right) \dot{g}[x''].$

Since the initial data $(f, \dot{f})$ and $(g, \dot{g})$ are arbitrary, we may now proceed to choose $\dot{f} = g = 0$ , and equate the coefficients of $f$ and $\dot{g}$ in equations (5.3) and (5.4) to deduce

$-u^{\mu''} \partial_{\mu''} \Delta[x'=(t,\vec{y}'),x''=(t,\vec{y}'')] = \frac{\delta^{(D)}[\vec{y}'-\vec{y}'']}{\sqrt[4]{\det h_{i'j'}[\vec{y}'] \det h_{i''j''}[\vec{y}'']}},$

where we have further employed the anti-symmetric property of $\Delta$ in eq. (2.5). Multiply both sides of the above equation with $i$ , and utilize Peierls’ bracket in eq. (2.2) to replace $-i\Delta[x',x'']$ with $[\varphi[x'],\varphi[x'']]$ . This brings us to eq. (5.1).

Eq. (5.2), in turn, can be verified by choosing the initial conditions $f = g = 0$ and $\dot{f} = \dot{g} = 0$ .

A side note: this analysis indicates more general commutation relations can likely be derived by choosing profiles for the initial data $(f,\dot{f},g,\dot{g})$ more sophisticated than the ones considered here.

6. Example: Minkowski spacetime As a simple application of this formalism, let us quantize the massless scalar field $\varphi$ in Minkowski spacetime $g_{\mu\nu} = \eta_{\mu\nu} \equiv \text{diag}[1,-1,\dots,-1]$ using Cartesian coordinates $x^\mu \equiv (t,\vec{x})$ and $x'^\mu \equiv (t',\vec{x}')$ .

The Heisenberg equations-of-motion of $\varphi$ is

(6.1): $\partial^2 \varphi \equiv \eta^{\mu\nu} \partial_\mu \partial_\nu \varphi = 0.$

Denoting the step function as $\Theta$ , the retarded Green’s function $1/\partial^2$ has the following Fourier representation:

$G^{\text{(ret)}}[x-x'] = i \Theta[t-t'] \int_{\mathbb{R}^D} \frac{d^D \vec{k}}{(2\pi)^D} \left( e^{-i|\vec{k}|(t-t')} - e^{+i|\vec{k}|(t-t')} \right) \frac{e^{i\vec{k}\cdot(\vec{x}-\vec{x}')}}{2|\vec{k}|}.$

The advanced Green’s function $1/\partial^2$ , on the other hand, has the following Fourier representation:

$G^{\text{(adv)}}[x-x'] = -i \Theta[t'-t] \int_{\mathbb{R}^D} \frac{d^D \vec{k}}{(2\pi)^D} \left( e^{-i|\vec{k}|(t-t')} - e^{+i|\vec{k}|(t-t')} \right) \frac{e^{i\vec{k}\cdot(\vec{x}-\vec{x}')}}{2|\vec{k}|}.$

Taking their difference thus yields the following version of the Peierls’ quantization condition in eq. (2.2):

(6.2): $\Big[ \varphi[x], \varphi[x'] \Big] = \int_{\mathbb{R}^D} \frac{d^D \vec{k}}{(2\pi)^D} \left( e^{-i|\vec{k}|(t-t')} - e^{+i|\vec{k}|(t-t')} \right) \frac{e^{i\vec{k}\cdot(\vec{x}-\vec{x}')}}{2|\vec{k}|},$

where we have used $\Theta[t-t'] + \Theta[t'-t] = 1$ .

The timelike vector field which we may employ to define “time-derivatives” is simply $\partial_0$ itself,

$u^\mu \partial_\mu = \partial_0$ .

It is now not difficult to check — by direct calculation — that the equal time relations

$\Big[ \varphi[t,\vec{x}], \dot{\varphi}[t,\vec{x}'] \Big] = i \delta^{(D)}[\vec{x}-\vec{x}']$

and

$\Big[ \varphi[t,\vec{x}], \varphi[t,\vec{x}'] \Big] = \Big[ \dot{\varphi}[t,\vec{x}], \dot{\varphi}[t,\vec{x}'] \Big] = 0$

are consequences of eq. (6.2), thereby yielding consistency with the general arguments made to derive equations (5.1) and (5.2).

Particle interpretation From eq. (6.1), we may deduce the general solution

(6.3): $\varphi[t,\vec{x}] = \int_{\mathbb{R}^D} \frac{d^D \vec{k}}{(2\pi)^D \sqrt{2 |\vec{k}|} } \left( a_{\vec{k}} e^{-i|\vec{k}|t} e^{i\vec{k}\cdot\vec{x}} + a_{\vec{k}}^\dagger e^{+i|\vec{k}|t} e^{-i\vec{k}\cdot\vec{x}} \right),$

where $a_{\vec{k}}$ and its adjoint $a_{\vec{k}}^\dagger$ are linear operators. (The $1/\sqrt{2|\vec{k}|}$ normalization was chosen for technical convenience.)

Plug eq. (6.3) into the left hand side of eq. (6.2), and proceed to equate from both sides of the ensuing equation the coefficients of the basis solutions $\{ e^{\pm i |\vec{k}|t} e^{i\vec{k}\cdot\vec{x}} \}$ and $\{ e^{\pm i |\vec{k}'|t'} e^{i\vec{k}'\cdot\vec{x}'} \}$ . This yields the famous Fourier-space/ladder-operator commutation relations

$\left[ a_{\vec{k}}, a_{\vec{k}'}^\dagger \right] = (2\pi)^D \delta^{(D)}[\vec{k}-\vec{k}']$

and

$\left[ a_{\vec{k}}, a_{\vec{k}'} \right] = \left[ a_{\vec{k}}^\dagger, a_{\vec{k}'}^\dagger \right] = 0.$

These are continuum (i.e., $\vec{k}$ -space) analogs of the commutation relations obeyed by the ladder operators in the quantum mechanical simple harmonic oscillator system, and is key to building and interpreting the $n$ -particle states of the QFT.

7. References

R.E. Peierls, “The Commutation laws of relativistic field theory,” Proc. Roy. Soc. Lond. A 214, 143 (1952). doi:10.1098/rspa.1952.0158
B. S. DeWitt, “The global approach to quantum field theory. Vol. 1, 2,” Int. Ser. Monogr. Phys. 114, 1 (2003).
R.D. Sorkin, “From Green Function to Quantum Field,” Int. J. Geom. Meth. Mod. Phys. 14, no. 08, 1740007 (2017). doi:10.1142/S0219887817400072 [arXiv:1703.00610 [gr-qc]].

Struggles In Physics

Peierls’ Brackets, Green’s Functions, and Equal-Time Commutators

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