In order to distinguish the new definition of $ {\mu_{n}}(T) $ from the old one, let us call it $ {\mu^{\text{New}}_{n}}(T) $.
We shall assume throughout this discussion that $ \displaystyle \sum_{n=1}^{\infty} {\mu_{n}}(T) < \infty $.
By the Divergence Test from calculus (it’s hard to believe that something so simple can crop up here!), we have $ \displaystyle \lim_{n \to \infty} {\mu_{n}}(T) = 0 $. Hence, for any $ \epsilon > 0 $, there exists an $ n \in \mathbb{N} $ sufficiently large so that $ {\mu_{n}}(T) < \epsilon $, which means that we can find an $ F \in B(\mathcal{H}) $ of rank $ \leq n $ such that $ \| T - F \|_{B(\mathcal{H})} < \epsilon $. Therefore, $ T $ can be approximated in the operator norm by bounded operators of finite rank, making it a compact operator.
Recall that $ {\mu_{n}}(T) $ is defined as the $ n $-th term of the null sequence that is formed by listing the eigenvalues of the positive compact operator $ |T| $ in decreasing order, taking multiplicity into account. The Minimax Principle then says that $ {\mu^{\text{New}}_{n}}(T) = {\mu_{n}}(T) $ (please click here to access a set of notes on trace-class operators that contains a proof of this result; see Lemma 12). Therefore, ‘trace-class’ in the new sense is the same as ‘trace-class’ in the old sense, and so
$$
\text{Tr}(|T|) \stackrel{\text{def}}{=} \sum_{n=1}^{\infty} {\mu_{n}}(T) = \sum_{n=1}^{\infty} {\mu^{\text{New}}_{n}}(T).
$$
For any $ T \in K(\mathcal{H}) $, one can find orthonormal sequences $ (\mathbf{v}_{n})_{n \in \mathbb{N}} $ and $ (\mathbf{w}_{n})_{n \in \mathbb{N}} $, not necessarily complete, such that
$$
T = \sum_{n=1}^{\infty} {\mu_{n}}(T) \langle \mathbf{v}_{n},\bullet \rangle_{\mathcal{H}} \cdot \mathbf{w}_{n}.
$$
We thus obtain a more explicit approximation of $ T $ by bounded operators of finite rank. This is a standard result in the theory of compact operators; please refer to the Wikipedia article on Compact Operator or to Corollary 4 of the notes mentioned above.
Given any $T\in L(\mathbb C^n)$, you can write
$$
T=\frac{T+T^*}2+i\,\frac {T-T^*}{2i}
$$
so $T$ is a linear combination of selfadjoints. And for each selfadjoint you have the Spectral Theorem saying that they are a linear combination of rank-one projections (which are positive semidefinite of trace one).
The result is even true when $\mathcal X$ is an infinite-dimensional Hilbert space, although it is not trivial in that case.
Best Answer
The definition is the same as in the finite-dimensional case.
Note that the general density matrix (an operator) is given by \begin{align} \rho = \sum_n \lambda_n |\psi_n\rangle \langle \psi_n| \end{align} where $\lambda_n\geq 0$ and $\sum_n\lambda_n = 1$.
We say that $\rho$ is positive semidefinite if \begin{align} \langle \Psi\mid \rho\mid\Psi\rangle \geq 0 \end{align} for all $\Psi$. Observe \begin{align} \langle \Psi\mid \rho\mid\Psi\rangle = \sum_n \lambda_n \langle \Psi \mid\psi_n\rangle \langle \psi_n\mid \Psi\rangle = \sum_n \lambda_n |\langle \psi_n \mid\Psi\rangle|^2\geq 0. \end{align}