Expression of density operator States in Quantum Mechanics can be thought of as density operators, i.e., positive semi-definite, normalized trace class operators on a Hilbert Space $\mathcal{H}$.
In the case $\mathcal{H}=\mathbb{C}^{2}$ we have that a generic state $\rho$ can be expressed as:
$$
\rho=\frac{1}{2}\left(\mathbb{I} + x^{1}\sigma_{1} + x^{2}\sigma_{2} + x^{3}\sigma_{3}\right)
$$ 
where $\mathbb{I}$ is the identity matrix, the $\sigma_{i}$'s are the Pauli matrices and $||\,\vec{x}\,||\leq1$.
This can be easily shown because $\mathcal{H}$ is finite-dimensional, and thus the set $\mathcal{B}(\mathcal{H})$ of linear operators is finite-dimensional.
I would like to know if there is a similar expression in the case in which $\mathcal{H}$ is infinite-dimensional, specifically, when $\mathcal{H}=\left(\mathcal{L}^{2}(\mathbb{R}^{n})\,;dl\right)$, where $dl$ is the Lebesgue measure.
EDIT
In order to be more specific, in the case $\mathcal{H}=\left(\mathcal{L}^{2}(\mathbb{R}^{n})\,;dl\right)$, I would like to find an expression for a generic $\rho$ suitable for the calculation of $tr(A\rho)$, where $A\in\mathcal{B}(\mathcal{H})$.
EDIT 2
In the setting I have in mind the operator $A$ is unknown.
Moreover its definition depends on its action on states, in the sense that it is given as an assumption on the behaviour of $tr(A\rho)$, for example $tr(A\rho)>0$.
Therefore, a generic expression for $\rho$ is needed in order to obtain informations on $A$.
 A: In case $\mathcal{H}=L^{2}(\mathbb{R}^{n},d\mathbf{x})$ $\rho $ is a
positive semi-definite trace class operator and can be expressed as
$$
\rho =\sum_{k}\lambda _{k}|u_{k}><u_{k}|
$$
where $\{u_{k}\}$ is a basis for $\mathcal{H}$ and $\lambda _{k}$ is
non-negative with
$$
\sum_{k}\lambda _{k}=1.
$$
Addition in response to a question by Norbert Schuch.
This is common knowledge within the C*-algebra formulation of quantum
mechanics. In the Hilbert space ($\mathcal{H}$) situation observables are
bounded operators on $\mathcal{H}$ and states are positive linear
functionals of the algebra of observables. One such class, the density
operators, is contained in the trace class, the so-called pre-dual of the
set of bounded linear operators $\mathcal{B}(\mathcal{H})$ as a Banach
space. But, in the infinite-dimensional situation, there are also states in
the dual of $\mathcal{B(H)}$ not contained in the trace class.
As an example consider a  free particle in a box. Then the Hamiltonian $H$
has discrete spectrum bounded from below and the canonical equilibrium state
is well-defined, $\sim \exp [-\beta H]$. But if the linear dimensions of the
box tend to infinity $H$ becomes the kinetic energy operator $\mathbf{p}%
^{2}/(2m)$ which has purely continuous spectrum and $\exp [-\beta \mathbf{p}%
^{2}/(2m)]$ is no longer a trace class (density) operator. For a gas of free
particles in a box we have a similar but more complicared situation in the
thermodynamic limit. Also in certain scattering situations we encounter such
a case.
I hope that this gives an idea of what is going on. In practice, for
instance when only expectation values are involved, the introduction of such
non-normal states can be avoided.
A: The trace class operators form a Banach space. There is a concept of (countable) basis for Banach spaces that is called Schauder basis.
Not every Banach space has a Schauder basis, but it is true e.g. for the space of compact operators (the case of $\mathcal{K}(l^2)$ is given explicitly in the wikipedia article).
Since the trace class operators are imbedded in the compact operators you have also a Schauder basis for them (and that basis may be further specialized to converge in the trace norm I suppose).
Let $\{\rho_n\}_{n\in\mathbb{N}}$ be a Schauder basis for $\mathcal{L}^1(L^2(\mathbb{R}^d))$ (trace class operators on $L^2$), then for any $\rho\in\mathcal{L}^1$ there exist $\{\alpha_n\}$ complex numbers such that
$$\rho=\sum_{n=0}^\infty \alpha_n\rho_n$$
and that sum converges in trace norm.
Also, if the operator $A$ is self-adjoint, then you may write it in the spectral decomposition
$$A=\int_{\mathbb{R}}\lambda dP_\lambda$$
where $dP_\lambda$ is the projection valued measure. Then you may write
$$Tr(A\rho)=\sum_n \alpha_n\int \lambda Tr(\rho_n dP_\lambda)\; .$$
If in addition the spectrum of $A$ is purely discrete (i.e. $A$ compact or with compact resolvent) you get
$$Tr(A\rho)=\sum_{n,m=0}^\infty \alpha_n\lambda_m Tr(\rho_n \lvert\psi_m\rangle\langle\psi_m\rvert)$$
where $\psi_m$ is the eigenfunction and $\lambda_m$ is the corresponding eigenvalue.
A: Every density matrix on $\mathbb C^2$ is of that form because $\rho$ must be Hermitian. This property is clearly satisfied because of the form and the algebra of Pauli matrices. This can then be generalised to higher dimensional Hilbert spaces by introducing a basis for Hermitian matrices which satisfy the same algebra. So the problem is then equivalent to whether one can find a basis $\{\sigma_i\}$ for the self-adjoint operators on $\mathcal H$ such that
$$\{\sigma_i,\sigma_j\} = 2\delta_{ij}I,$$
i.e a real Clifford algebra, plus the trace-class condition.
