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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 $\lambda=0$?

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    $\begingroup$ 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. $\endgroup$ – E. Bellec Feb 11 at 13:53
  • $\begingroup$ Ok, I've edited my question. The problem is if the partition function Z(λ) is continous at λ=0 $\endgroup$ – Hodor Feb 13 at 11:25
  • $\begingroup$ 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 $h$. $\endgroup$ – E. Bellec Feb 13 at 14:15
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    $\begingroup$ 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"). $\endgroup$ – E. Bellec 2 days ago
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    $\begingroup$ 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 $\endgroup$ – E. Bellec 2 days ago

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