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From a proof that 2D Wightman CFT leads to a vertex algebra [1]:

Let $$ Y(a,z):=\frac{1}{(1+z)^{2\Delta_a}}\Phi_a\left(i\frac{1-z}{1+z}\right),\quad\text{with}\quad |z|<1. $$

Here $\Delta_a\ge 0$ is conformal weight of $\Phi_a$, and $\Phi_a$ is the scalar field satisfying the usual Wightman axioms (Poincare covariance, stable vacuum and positive spectrum of momentum, completeness, i.e. that polynomials of smeared fields applied on vacuum are dense in the Hilbert space, and locality (p. 6 of [1])) and in addition, satisfying conformal covariance $$ U(q,\Lambda,b)\Phi_a(x)U(q,\Lambda,b)^{-1}=(1+2x\cdot b+|x|^2 |b|^2)^{-\Delta_a}\Phi_a((q,\Lambda,b)\cdot x) $$ with $(q,\Lambda,b)\mapsto U(q,\Lambda,b)$ being a unitary representation of the conformal group, fixing the vacuum vector $|0\rangle$.

On p. 11 of [1] the author writes:

We expand a chiral field $Y(a,z)$ in a Fourier series: $$ Y(a,z)=\sum_n a_{(n)}z^{-n-1},\quad\quad(\star) $$ where $a_{(n)}\in\mathrm{End}\; \mathcal{D}$.

Here $\mathcal{D}$ is the span of polynomials in the fields applied on the vacuum and is dense by completeness assumption.

I am not certain if this expansion is justified. Especially because it allows the author to prove that we indeed get a vertex algebra and hence we have an operator product expansion of the fields. Moreover, I found in another book [2] a similar proof on p. 190 and there the author writes

The operator product expansion (Axiom 6 on p. 168) of the primary fields allows to understand the fields $\Phi_a$ as fields $$ \Phi_a(z)=\sum a_{(n)} z^{-n-1} \in \mathrm{End} D [[z,z^{-1}]] $$ in the sense of vertex algebras.

Here $D$ is some dense domain in Hilbert space which is left invariant by fields, i.e. in [2] completeness is not assumed.

In other words, the author of [2] needs the existance of operator product expansion to write such a Fourier expansion.

My question:

$$ \text{is }(\star)\text{ true?} $$


References:

[1] V. Kac. Vertex Algebras for Beginners, 1998.

[2] M. Schottenloher. A Mathematical Introduction to Conformal Field Theory, 2008.

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  • 1
    $\begingroup$ It depends on which family of test functions $\Phi_a$ is defined. In principle up to a phase, $a_n$ is obtained from $\Phi_a$ by smearing with the "test function" $f_n(x)=\frac{ (1+ix)^{n+\Delta-1}}{(1-ix)^{n+1-\Delta}}$, so if this lies in the domain of test functions, the Fourier mode is well-defined. If your Wightman functions are only defined on $\mathcal S(\mathbb R)$ you have show that you can extend this domain. $\endgroup$ – Marcel Bischoff Nov 28 '15 at 23:24
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Using Marcel Bischoff's comment: $$ Y(a,z)=\sum_n a_{(n)} z^{-n-1} \iff a_{(m)}=\frac{1}{2\pi i} \oint\limits_{|z|=1} Y(a,z)z^{m} \,\mathrm{d}z. $$

Since the author of [1] starts with the Schwartz space $\mathscr{S}(\mathbb{R})$, we need to show that $Y(a,z)$ is analytic in $|z|<1$.

By definition, $$ Y(a,z)=\frac{1}{(1+z)^{2\Delta_a}}\Phi\left(i\frac{1-z}{1+z}\right)\quad\text{with}\quad |z|<1. $$ Clearly, $$ \frac{1}{(1+z)^{2\Delta_a}} $$ is analytic in the domain $|z|<1$ since it is holomorphic. Letting $$ t:=i\frac{1-z}{1+z} $$ we note that $|z|<1$ gets mapped to $\mathrm{Im}\, t>0$, so we need to show that $\Phi_a(t)$ can be analytically continued to $\mathrm{Im}\, t>0$. We now reintroduce the second coordinate, since the joint spectrum of momentum operators is positive. From [1] $$ t:=x^0-x^1,\quad \bar{t}:=x^0+x^1\quad\text{and}\quad x:=(x^0,x^1) $$ with $t,\bar{t}$ the lightcone coordinates. Since the polynomials in fields applied to the vacuum are dense by completeness assumption, it suffices to prove that any $$ \Phi_{a_1}(x_1)\ldots\Phi_{a_n}(x_n)|0\rangle $$ can be analytically continued to the domain $$ \{\mathrm{Im}\, x^0_1>0\}\times\dots\times\{\mathrm{Im}\, x^0_n>0\}\text{ such that } \mathrm{Im}\,x^0_i\ne \mathrm{Im}\,x^0_j \text{ for } i\ne j. $$ The fact that coinciding points do not matter follows from locality [1] $$ (z-w)^N[Y(a,z),Y(b,w)]=0 \text{ for } N\gg0. $$ We now follow Section 2 of [3]. Let $$ \Psi(x_1,\ldots,x_n):=\Phi(x_1)\ldots\Phi(x_n)|0\rangle $$ with $\Phi$ being a Hermitian scalar field. Generalizations to fields of arbitrary spin are given in Section 9 of [3] (in 4 dimensions, here 2D). Note that $\Psi$ (smeared with a test function) belongs to the Hilbert space $\mathcal{H}$ by completeness. By positive spectrum $$ \Psi(x_1,\ldots,x_n)=\int d^2 p\; d^2 q_1\ldots d^2 q_{n-1} \tilde{\Psi}(p,q_1,\ldots,q_{n-1})e^{i\left(p x_1+\sum q_j (x_{j+1} -x_j)\right)}. $$ Here $\tilde{\Psi}$ is non-zero only if $p^0\ge0$ and all $q_i^0\ge 0$. Thus, $\Psi$ can be analytically continued to a vector-valued analytic function $\Psi\in\mathcal{H}$, i.e. we have \begin{align} \Psi(z_1,\dots,z_n)\quad\text{of}\quad z_k=x_k+i y_k\quad & \text{defined and holomorphic for}\\ &y_1\in V_+ \text{ and } y_j-y_i\in V_+ \text{ if } j>i\\ &(V_+ \text{ is the open forward lightcone}). \end{align} In [4] it was shown that locality and the edge of the wedge theorem allows to extend the domain of definition and analyticity even further:

Claim. $\Psi(z_1,\ldots,z_n)$ belongs to $\mathcal{H}$ and is analytic in the $z_k$ in a connected domain which includes the Euclidean points with $z_k=(i y^0_k, x_k)$ such that $y^0_k>0$ for all $k$, and $z_i^0\ne z_j^0$ if $i\ne j$.

Proof of Claim. Fix $\pi$ to be any permutation of $(1,\ldots,n)$ and $$ \Psi^{\pi} (z_1,\ldots,z_n):=\Psi(z_{\pi(1)},\ldots,z_{\pi(n)}),\quad z_k=x_k+i y_k. $$ By the above, $\Psi^{\pi}$ is well-defined and holomorphic in a domain containing the Euclidean points with $0<y^0_{\pi(1)}<\ldots<y^0_{\pi(n)}$. Furthermore, by locality $$ \Psi^{\pi}(x_1,\ldots,x_n)=\Psi(x_1,\ldots,x_n)\quad\text{if}\quad x_k\in\mathbb{R}:(x_i-x_j)^2<0 \quad\forall i\ne j, $$ i.e. all $\Psi^{\pi}$'s are equal on a real neighborhood. Now the Edge of the Wedge Theorem [5] shows that they are analytic continuations of one and the same analytic function. Moreover, the domain of analyticity of this function must contain the domains of analyticity each $\Psi^{\pi}$. Q.E.D.


Remark. I think $\Psi(x_1,\ldots,x_n)=\Phi(x_1)\ldots\Phi(x_n)|0\rangle$ generalizes to arbitrary bosonic fields trivially and for fermionic fields we would pick up some minus signs.


References:

[1] V. Kac. Vertex Algebras for Beginners, 1998.

[2] M. Schottenloher. A Mathematical Introduction to Conformal Field Theory, 2008.

[3] Lüscher, M. and Mack, G. Global conformal invariance in quantum field theory. Communications in Mathematical Physics, Springer-Verlag, 1975, 41, 203-234.

[4] Glaser, V. On the equivalence of the Euclidean and Wightman formulation of field theory. Comm. Math. Phys., 1974, 37, 257-272.

[5] Streater, R. F. and Wightman, A. S. PCT, Spin and Statistics, and All That, W. A. Benjamin, Inc., 1964.

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