Sir R. Penrose in his article (https://doi.org/10.1007/BF00668831) has shown that there are close similarities in various aspects of twistor theory and quantum mechanics. In twistor theory, one can consider Hilbert space defined as space of analytic (in suitable domains) twistor functions $f(Z^{\alpha})$ , equipped with a Hermitian norm $\langle,|,\rangle$. The scalar product was formally defined as $$\langle f|g\rangle =\int f(Z^{\alpha})g(W_{\alpha})(W_{\beta}Z^{\beta})^{2s-2}DZ\wedge DW$$, where f,g are homogeneous of degree -2s-2, s being the helicity of massless particles(Note: this definition works for $|s|=0,1/2$). The twistor operators $\hat{Z}^{\alpha}$ and $\hat{\bar{Z}}_{\alpha}$ acts on these analytic functions and follows the same commutation rule as position and momentum operators in quantum mechanics. This formalism of twistors can be used to compute scattering amplitudes in QFT's for massless fields (see: https://doi.org/10.1016/0370-1573(73)90008-2).

Given these close similarities, I want to know if there is an analogous version for path integral formulation in twistor theory. To be more precise, let's say $x$ and $y$ are two points in Minkowski space which are either time-like or null separated, I want to know if generating functions $F(Z,W)$ exist such that we can identify $$\langle f|g\rangle \sim e^{-F(Z,W)}\leftrightarrow \langle \psi_x|\psi_y\rangle_{QM} \sim e^{iS(x,y)}$$ where $Z^{\alpha}\in L_x$ (projective line corresponding to position $x$) and $W^{\alpha}\in L_y$


1 Answer 1


After giving it some thoughts, I have made few elementary identifications which I think is worth mentioning and should supplement this question. Bottom line here is that the twistor functions themselves can in a certain sense play the role of action, from which we can derive our classical equation of motion.

Consider the projective lines $L_x, \; L_y\subset \mathbb{PT}$ , where say $y=x+\delta x$ and that the points $y,x$ are causally separated in $\mathbb{M}$. The projective line $L_x$ is topologically a Riemann sphere, where each point on the sphere corresponds to a null ray $Z$ passing through $x$. Such family of rays are characterized by twistors $Z^{\alpha}=(ix^{AA'}\pi_{A'},\pi_{A'})$, where $\pi_{A'}$ is the tangent spinor of each such ray.

Now take any twistor function $f\in H^1(\mathbb{PT}^+;\mathcal{O}(\pm n-2))$. Let $W^{\alpha}\in L_y$. We can Taylor expand $f(W)$ about $x$ considering small $\delta x$:

$$f(i(x^{AA'}+\delta x^{AA'})\pi_{A'},\pi_{A'})=f(\underbrace{ix^{AA'}\pi_{A'},\pi_{A'}}_Z)+\underbrace{\partial_{AA'}f(Z)\delta x^{AA'}}_{\delta f}+\cdots$$

Geometrically, $\delta f$ merely provides the difference b/w $f$ measured on $L_y$ and $L_x$, thus $\delta f=0$ at the intersection point of $L_x$,$L_y$. Also in $\mathbb{M}$ picture, intersection point will correspond to the null ray which passes through both $x$ and $y$. Thus, we may consider $\delta x^a$ to be null, i.e. $\delta x^{AA'}=\epsilon \pi_{A'}\bar{\pi}_A$, for $|\epsilon|<<1$. Thus $\delta f=0 \implies \pi_{A'}\bar{\pi}_A\partial_{AA'}f=0$. However, this is precisely the mass-less field equation which one obtains from Penrose's transform.

Thus, $\delta f=0$ implies classical E.O.M. (similar to $\delta S=0$ in least action principle).

Similar arguments can be made for interactions with some background field, e.g. scattering in presence of YM field. For instance in twisted photon construction, we may identify the self dual potential $-ieA_a=f^{-1}\partial_{AA'}f$, where cocycle $f\in H^1(\mathbb{PT}^+;\mathcal{O}(0))$. The classical equation of motion describing the dynamics of a free field $\alpha$ in presence of background $f$ is

$$\pi^{A'}(\partial_{AA'}-ieA_{AA'})\alpha=0$$ But this can be written more compactly as: $\pi^{A'}\partial_{AA'}(f\alpha)=0$. This gives the intuition that the incoming free field $\alpha$ interacts with YM field $f$ and escapes as the outgoing "free" field $\beta\equiv f\alpha$. Thus $\delta\beta=0$ gives the above field equation. This also justifies the choice of Hamiltonian $H=f$ in equation (4.3) of (https://doi.org/10.1016/0370-1573(73)90008-2).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.