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Usually, the textbook starts with a few assumptions of what density operator $\rho$ has.

One of them is $Tr(\rho) = 1$.

Why is that?

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    $\begingroup$ So the probabilities of measurement outcomes sum to $1$. $\endgroup$ – Rammus Apr 6 at 17:42
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Let $\rho$ be a density operator and let $\{M_x\}_x$ be a POVM. Then the probability we get outcome $x$ for the state $\rho$ is given by the Born rule as $$ p(x) = \mathrm{Tr}[\rho M_x]. $$ But then $$ \begin{aligned} \sum_x p(x) &= \sum_x \mathrm{Tr}[\rho M_x] \\ &= \mathrm{Tr}[\rho \sum_x M_x] \\ &= \mathrm{Tr}[\rho I] \\ &= \mathrm{Tr}[\rho] \\ &= 1 \end{aligned} $$ where we used the fact that for a measurement $\sum_x M_x = I$ where $I$ is the identity operator (also crucial for the probabilities to sum to $1$). Thus we see that in order for the probabilities of measurement outcomes to sum to one we require $\mathrm{Tr}[\rho] = 1$.

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The unit trace constraint on density matrix $\rho$ ensures that the probabilities of measurement outcomes sum to $1$ for every possible measurement performed on $\rho$.

A measurement can be described by a collection of positive definite operators $E_m$, one for each measurement outcome $m$, that satisfy the completness relation

$$ \sum_m E_m = I\tag1 $$

where the sum ranges over all measurement outcomes and $I$ stands for the identity operator. In this description of measurement, known as the POVM, the probability of outcome $m$ is given by the formula

$$ p(m|\rho) = \mathrm{tr}(\rho E_m)\tag2. $$

The unit trace constraint follows from the following simple calculation

$$ 1 = \sum_m p(m|\rho) = \sum_m \mathrm{tr}(\rho E_m) = \mathrm{tr}\left(\rho \sum_m E_m\right) = \mathrm{tr}(\rho I) = \mathrm{tr}(\rho) $$

where first we used the fact that the the sum of $p(m|\rho)$ over all measurement outcomes $m$ is $1$, then we used the Born rule $(2)$, then linearity of the trace and finally completeness relation $(1)$.

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Diagonal element of density matrix represents the probability that the system is in the correspondent basis state (in other words, population of the basis state). The sum of probabilities should be equal to 1, hence the normalization condition$$Tr(\rho)=1$$

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    $\begingroup$ +1 Very nice and simple explanation. We can derive the interpretation of the diagonal elements directly from the Born rule by noticing that $\rho_{ii} = \langle i|\rho|i\rangle = \mathrm{tr}(\rho|i\rangle\langle i|) = p(i|\rho)$. $\endgroup$ – Adam Zalcman Apr 7 at 2:16

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