# Different kinds of trace for statistical ensembles

In the chapter 7 of the book "A Modern Course in Statiscal Physics" by L. Reichl, we found $Tr[\hat{\rho}]=1$ for microcanonical ensembles and $Tr_N[\hat{\rho}]=1$ for canonical and grandcanonical ones. I looked for the meaning of $Tr_N$ in the book but I didn't find it. It seems that it is related to number of states with a given energy E, but I don't know how this relation looks like. Thus

-What does $Tr_N$ mean?

or

-What is the difference between $Tr_N$ and $Tr$?

A consequence of this difference in the book is

$Tr\left[e^{\left(\frac{\alpha_0}{k_B}-1\right)\hat{I}}\right]=1$, (1a)

$e^{\left(\frac{\alpha_0}{k_B}-1\right)}N=1$, (1b)

for microcanonical ensemble, and

$Tr_N\left[e^{\left(\frac{\alpha_0}{k_B}-1\right)\hat{I}+\frac{\alpha_E}{k_B}\hat{H}}\right]=1=e^{\frac{\alpha_0}{k_B}-1}Tr_N\left[e^{\frac{\alpha_E}{k_B}\hat{H}}\right]$, (2)

for canonical or grandcanonical, where $\hat{H}$ is the hamiltonian operator and $\alpha_0$, $\alpha_E$ and $k_B$ are constants, that I also doesn't understand these results.

About Eq. (2), it seems that, if $Tr_N$ has the property

$Tr_N\left[\hat{A}\hat{B}\right]\equiv\frac{1}{N}Tr[\hat{A}]Tr[\hat{B}]$, (3)

where $\hat{A}$ and $\hat{B}$ are diagonal matrices and $N$ is the dimension of $\hat{A}$ and $\hat{B}$, I can understand (2), but not even more Eqs. (1).

• I would be a lot easier to answer this if you could define $\text{Tr}_N$ explicitly. – DanielSank Jan 29 '15 at 21:02
• Adding new information in the question. Resuming to DanielSank: I didn't find in the book what $Tr_N$ exactly means. And thanks for your attention, Daniel. – Caetes Jan 30 '15 at 1:32
• @ErmsPereira Note that $e^{\alpha \hat{I}} = e^{\alpha}\hat{I}$, since $\hat{I}^n = \hat{I}$. That is why you can take the first term outside the trace in Eq. (2). You don't need to assume any special properties of the trace. – Mark Mitchison Jan 31 '15 at 15:37

Using the comment of @MarkMitchison, since $e^\hat{C}=\sum_{k=0}^\infty\frac{\hat{C}^k}{k!}$ and $e^x=\sum_{k=0}^\infty\frac{x^k}{k!}$ (as can be seen here), so $e^{\alpha\hat{I}}=e^\alpha\hat{I}$, and I "can take the first term outside the trace in Eq. (2). You don't need to assume any special properties of the trace".