# 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. Commented Jan 29, 2015 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. Commented Jan 30, 2015 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. Commented Jan 31, 2015 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".