# Is the Casimir energy in CFT an observable?

We know that if we transform a 2d conformal field theory from a plane to a cylinder with perimeter $$L$$, the ground state energy will be shifted by $$E = -\frac{c}{24L}$$ due to the Schwarzian derivative term in the transformation of stress energy tensor.

This energy is the difference of a theory on a cylinder and the same theory on a plane. How can we compare the ground state energy of two theories on different spacetime? Therefore I would like to know is this energy a physical observable? And if not, why is it important?

## 1 Answer

Of course, the free energy on the cylinder is not a measurable observable if you're given the theory on the infinite plane. But one can measure other observables which are proportional to the central charge, such as the two-point function of the stress-energy tensor.

There are situations where that expression is an observable. If you have a one-dimensional quantum system with periodic boundary conditions that flows to a (1+1)-dimensional CFT, then its ground state energy will generically be given by the formula $$E = E_1 L + E_0 - \frac{\pi v c}{6L} + \cdots,$$ where the higher-order terms are lower-order in $$L$$. (See below about the mismatch between our expressions.) Here, $$E_1$$, $$E_0$$, and $$v$$ are non-universal constants ($$v$$ is the velocity of excitations at low-energy, usually called the "speed of light" in a field theory textbook). Then it is possible to "measure" the central charge term. For example, say you do some Monte-Carlo simulations to obtain the velocity $$v$$ of excitations, and then numerically calculate the ground state energy for several (large) values of $$L$$ and match it to the above equation. This lets you determine $$c$$.

In practice, it is much easier to extract central charge from the entanglement entropy. In particular, for an open one-dimensional quantum system, the entropy associated with tracing out half of the system is $$S = (c/6) \log L$$.

As a side-note, I think that what you are calling $$L$$ is really the radius of the cylinder, which is related to the perimeter by a factor of $$2 \pi$$. Finally, you are only considering the holomorphic sector, and above I'm everywhere considering also the antiholomorphic sector with an identical central charge. So that's why my expression is off by $$4 \pi$$ compared to yours.

• Thank you so much for the detailed answer. My aim is not to determine the central charge. I am actually confused by a more basic question: How would you measure (the shift of) the ground state energy? I would usually think that the ground state energy is defined. Nov 23 '19 at 3:19
• In a quantum CFT in one spatial dimension, one typically defines the ground state energy to vanish in an infinite volume, since in this limit there is no energy scale in the problem. Given this definition, one can ask what the ground state energy is in a periodic system with length L. The answer is $-\pi v c/6L$. This $L$-dependent energy shift can be "measured" in appropriate one-dimensional quantum systems as I describe in my answer. Nov 23 '19 at 4:02