# Partition function in the non-interacting limit

Let's consider the partition function $$Z(\lambda)=Tr (e^{-\beta H})=Tr (e^{-\beta (H_1+\lambda H_2)})$$ for a quantum system with the Hamiltonian $$H=H_1+\lambda H_2$$ where $$H_1$$ is the free part of the hamiltonian and $$\lambda H_2$$ is the interacting part. I'm wondering if it is true that $$\lim_{\lambda \rightarrow 0} Z(\lambda)=Z(0)=Tr(e^{-\beta H_1})$$ Can there be some subtilty which makes $$Z$$ discontinous at $$\lambda=0$$?

• I don't get something. The Trace is simply the sum of diagonal element (in eigenstate basis or not). So $Tr'$ is the same as $Tr$ since $H_1$ and $H_1+\lambda H_2$ are matrices of the same size. – E. Bellec Feb 11 at 13:53
• Ok, I've edited my question. The problem is if the partition function Z(λ) is continous at λ=0 – Hodor Feb 13 at 11:25
• I did not find a reason for the partition function to be discontinuous. But it's derivative can be discontinuous in case of symmetry breaking. For example, take a 1D Ising ferromagnet under applied magnetic field $h$ at $T = 0$. If you take a look at the derivative (with respect to $h$) of the analytical solution to the partition function, it will be discontinuous jumping from $+\infty$ to $-\infty$ as you change the sign of $h$. – E. Bellec Feb 13 at 14:15
• Haha, found something finally. A jump in the partition function (or free energy) could be called a zeroth-order phase transition. Here, I found an article about it link.springer.com/article/10.1023/B:MATN.0000049669.32515.f0 (you can find more by googling "zeroth order phase transition"). – E. Bellec 2 days ago
• There 's only a small part of the full article here. But you can find it on scihub using this DOI 10.1023/B:MATN.0000049669.32515.f0 – E. Bellec 2 days ago