4
$\begingroup$
  1. Is state-operator correspondence exclusive in radial quantization?

  2. Usually we say that a state in CFT is corresponding to an operator in radial quantization. If I do a map (not necessarily a global conformal transformation) to some other geometry, and then choose a slice to do quantization, then will I still have this state-operator correspondence?

  3. I have a lot of confusions about this problem. Like how does the states in the Hilbert space transform?

  4. Originally we classify the states in the Hilbert space using the eigenvalues of the states under the dilation operator. Now if I do a transformation, what operator should I use to classify the states?

$\endgroup$
3
$\begingroup$
  1. Radial quantization is necessary neither for the state-operator correspondence, nor for CFT in general. However, radial quantization provides an intuitive, geometric explanation of why there is a state-operator correspondence, by showing that a sphere (where a state lives) can be collapsed to a point (where an operator lives) by a dilatation.

  2. Yes. Your operators are intrinsic objects, their fundamental properties are not affected by a choice of quantization, which is just one way of constructing the CFT.

  3. I am not sure I understand the question. A priori, states do not depend on space, and the action of conformal transformations of states may not be defined without further assumptions.

  4. States belong to representations of the conformal algebra. The axiom that there is a state-operator correspondence that commutes with the action of that algebra, defines an action of that algebra at each point of space. Calling $T$ an element of the algebra, $v$ a state, $O_v$ the corresponding operator, and $x$ a point, we have $$ T^{(x)}O_v(x) = O_{Tv}(x) $$ It is always the same algebra, with the same generators (including the generator of dilatations), that acts on the space of states. On the other hand, that algebra has representations $T^{(x)}$ depending on $x$, that describe its action on operators.

| cite | improve this answer | |
$\endgroup$
2
$\begingroup$

In a general quantum field theory, with a field operator $\phi$ with a vacuum vector $|0\rangle$, an incoming state can be defined as,

$$|\phi_{\mathrm{in}}\rangle := U(0, -\infty) \phi|0\rangle$$

where $U(t_1,t_2)$ is the time-evolution operator. It is now a general statement that for a conformal field theory, the assignment, $\phi \mapsto |\phi_{\mathrm{in}}\rangle$ is a bijection. In less formal terms, it means the map going the other way is unique, that is, we can for each state find some local operator representing it.

It is simply that for particular geometries, there is a neat, geometrical interpretation of this, like the case you noted of radial quantisation. However, it is not exclusively dependent on this.

| cite | improve this answer | |
$\endgroup$
2
$\begingroup$

Let's consider a Quantum Field Theory (QFT) defined on a cylinder with coordinates $(\tau,\sigma)$. In QFT, it is defined a chronological product (time-ordering product):

\begin{equation} T\left(\phi(t,\vec{x})\psi(l,\vec{y}\right)):= \begin{cases} \phi(t,\vec{x})\psi(l,\vec{y})\quad \text{if }l<t \\ \psi(l,\vec{y})\phi(t,\vec{x})\quad \text{if }l>t \end{cases} \end{equation} The time on the cylinder is $\tau$, which is indentified with the flat direction, and $\sigma$ is the spatial direction, indetified with the "circular" direction. Then: $$(\tau,\sigma)\in\mathbb{R}\times[0,L]$$ where $L$ is the length (perimeter) of the circular sections of the cylinder. Now, we can map the cylinder in the complex plane, by means of the following identification: $$z=e^{i\sigma+\tau}$$ This means that if you have a space translation in the cylinder $\sigma\to\sigma'=\sigma+\eta$ and time translation $\tau\to\tau'=\tau+\theta$, in the complex coordinates: $$z\to z'_{space-tr}=e^{i\sigma'+\tau}=e^{i\sigma+i\eta+\tau}=e^{i\eta}e^{i\sigma+\tau}=e^{i\eta}z$$ $$z\to z'_{time-tr}=e^{i\sigma+\tau'}=e^{i\sigma+\tau+\theta}=e^{\theta}e^{i\sigma+\tau}=e^{\theta}z$$which means that: $$(\text{space translations on the cylinder})\iff(\text{rotations in complex plane})$$ $$(\text{time translations on the cylinder})\iff(\text{dilatation in complex plane})$$ In other words, paths in the cylinder with a fixed time $\tau_0,\tau_1,...,\tau_n$ become concentric circles in the complex plane since the temporal part is the radial amplitude in the complex number $e^{i\sigma+\tau}$.

In turn the time-ordering product $T$ in the complex plane, becomes a Radial Ordering: \begin{equation} R(\phi(z)\psi(z')):= \begin{cases} \phi(z)\psi(z')\quad \text{if }|z'|<|z| \\ \psi(z')\phi(z)\quad \text{if }|z'|>|z| \end{cases} \end{equation}

Now, if we define in the Hilbert space $\mathscr{H}$ a vacuum state $|0\rangle$, and it is well defined in every reference frame system since we have the Poincaré invariance given by the conformal invariance (we couldn't base our theory if this state wouldn't be the same in every reference frame, in fact we would slip into 'problems' like Hawking Radiation).

Now, in Scattering theory, one defines the asymptotic 'in'-states and 'out'-states as limits for $t\to-\infty$ and $t\to+\infty$ respectively. In this case, since our remote past is $z=0$ and our remote future is $z=\infty$ (In general we compactify the $\mathbb{C}$ plane into the Riemann sphere), we define: $$|\phi_{in}\rangle:=\lim_{z,\bar{z}\to 0}\phi_{\Delta,\bar{\Delta}}(z,\bar{z})|0\rangle$$ where $\phi_{\Delta,\bar{\Delta}}(z,\bar{z})$ is a primary field with conformal dimensions $(\Delta,\bar{\Delta}).$ After defining this state, one can define its adjoint (the asymptotic out state) $$\langle\phi_{out}|=(|\phi_{in}\rangle)^\dagger$$ where the adjoint for a primary field is deifned as: $$\phi_{\Delta,\bar{\Delta}}(z,\bar{z})^\dagger:=z^{-2\bar{\Delta}}\bar{z}^{-2\Delta}\phi\left(\frac{1}{\bar{z}},\frac{1}{z}\right)$$ Thanks to this definition, the inner product: $$\langle\phi_{out}|\phi_{in}\rangle=1$$

Claerly the discourse is much more longer, but in turn, it is in this sense that we have a correspondence between states in the Hilbert space and Operators.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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