# the density matrix in QFT on a cylinder

My question regards the density matrix in quantum field theory on a cylinder.

The partition function is given by $Z=\text{Tr} e^{-\beta H}$. The elements of this thermal density matrix become \begin{align} \rho(\{\varphi(x_j)\};\{\varphi(x_i)\})&\equiv\left\langle\prod_j\{\varphi(x_j)\}\vert\hat{\rho}\vert\prod_i\{\varphi(x_i)\}\right\rangle\\ &=Z(\beta)^{-1}\left\langle\prod_j\{\varphi(x_j)\}\vert e^{-\beta H}\vert\prod_i\{\varphi(x_i)\}\right\rangle, \end{align} represented as a path integral \begin{align} \rho=Z^{-1}\int D\varphi(x,\tau)\prod_{i}\delta(\varphi(x,0)-\varphi(x_i))\prod_{j}\delta(\varphi(x,\beta)-\varphi(x_j))e^{-S}. \end{align} Now, from what i can read, it is the trace of the partition function that sews together boundaries $\varphi(x,0)$ and $\varphi(x,\beta)$. Can someone please explain how the trace does this?

• Wouldn't it be a bit easier for you to start from a single-particle (i.e. single coordinate) path integral? It's rather obvious why the trace does what you want in that case, and then it's easy to generalise to the many-body context. (Hint: the path integral gives you a transition amplitude $\langle x_f \rvert \mathrm{e}^{-\mathrm{i} Ht} \lvert x_i\rangle$. In the trace, you sum over terms of the form $\langle x_i \rvert \mathrm{e}^{- H\tau} \lvert x_i\rangle$.) – Mark Mitchison Mar 14 '16 at 12:44
• Thanks, that helps a bit. However, i'm still not entirely sure when the two boundaries are glued to a cylinder. Is it just that in the trace one requires that the initial and final states are the same? Can you please give me some more hints or somewhere i can look it up? – JungleR Mar 14 '16 at 13:37
• Yeah, the point is just that the "initial" and "final" states are the same. If you carefully derive the path integral representation of the partition function then you will see that the fields must have periodic boundary conditions. – Mark Mitchison Mar 14 '16 at 14:14
• Okay, i'll give it a try. Cheers. – JungleR Mar 14 '16 at 19:24