Finding the trace of a system explicitly Consider that we are working with a joint system composed of system A with basis $|\alpha_j\rangle$ and system B with basis $|\beta_j\rangle$.
In my notes the density operator is denoted as follows:
$$\space\space\rho = \sum_{j,k,l,m} \langle\alpha_j| \langle\beta_k |\rho |\alpha_l\rangle |\beta_m\rangle  |\alpha_j\rangle |\beta_k\rangle \langle\alpha_l| \langle \beta_m|$$
whereby my notes state that $$ \rho_{jklm}  = \langle\alpha_j| \langle\beta_k |\rho |\alpha_l\rangle |\beta_m\rangle $$
They also state the following equations for the Trace of A and the trace of B:
$$\rho_\beta = Tr_\alpha(\rho) = \sum_{l,m}(\sum_{j} \rho_{j,l,j,m}) |\beta_l\rangle \langle\beta_m| $$
$$\rho_\alpha = Tr_\beta(\rho) = \sum_{j,k}(\sum_{l} \rho_{j,l,k,l}) |\alpha_j\rangle \langle\alpha_k| $$
My main question is how would one write out $\rho_{j,l,k,l}$ and $\rho_{j,l,j,m}$ explcitly as what I get do not seem to agree with a worked example in my book and so I am quite confused.
Thanks
 A: 
Well because if I were to do it myself I would write it as follows: $\rho_{jlkl} =\langle \alpha_j|\langle \beta_l| \rho |\alpha_k\rangle |\beta_l\rangle $ However I am unsure because the worked examples I have seen suggest the following $\rho_{jlkl} =\langle \alpha_j|\langle \beta_l| \rho |\beta_l\rangle |\alpha_k\rangle $.

It seems you are misunderstanding the idea of a tensor product of states, so I'll review that briefly.  Let $\mathcal H_A$ and $\mathcal H_B$ be Hilbert spaces, and let $\alpha \in \mathcal H_A$ and $\beta \in \mathcal H_B$.  The tensor product of $\alpha$ and $\beta$ is the ordered pair $(\alpha,\beta)$ which has the following properties:

*

*$(\alpha,\beta+\gamma)=(\alpha,\beta)+(\alpha,\gamma)$ for all $\alpha\in\mathcal H_A, \beta,\gamma \in \mathcal H_B$

*$(\alpha+\delta,\beta)=(\alpha,\beta)+(\delta,\beta)$ for all $\alpha,\delta \in \mathcal H_A, \beta \in \mathcal H_B$

*$\lambda (\alpha,\beta) = (\lambda \alpha,\beta) = (\alpha,\lambda \beta)$ for all $\lambda \in \mathbb C, \alpha\in\mathcal H_A, \beta \in \mathcal H_B$
Rather than write $(\alpha,\beta)$ for the tensor product, it is standard notation to write $\alpha \otimes \beta$.

The tensor product of Hilbert spaces $\mathcal H_A$ and $\mathcal H_B$ is the space of all tensor products of the form $\alpha\otimes \beta$ with $\alpha\in\mathcal H_A$ and $\beta \in \mathcal H_B$, and all linear combinations thereof.  The inner product on this space is taken to be
$$\bigg< (\alpha,\beta), (\gamma,\delta)\bigg>_{\mathcal H_A\otimes \mathcal H_B} := \left<\alpha,\gamma\right>_{\mathcal H_A} \cdot \left<\mathcal \beta ,\mathcal \delta\right>_{\mathcal H_B}$$
Therefore, an element $\psi \in \mathcal H_A \otimes \mathcal H_B$ might look like
$$\psi= \alpha\otimes \beta + 3\gamma \otimes \delta$$
It is clear from the definition that $\alpha$ and $\gamma$ belong to $\mathcal H_A$ while $\beta$ and $\delta$ belong to $\mathcal H_B$.  Again per standard convention, we reuse the symbol $\otimes$ and denote the tensor product of Hilbert spaces by $\mathcal H_A \otimes \mathcal H_B$.

If you'd like to work with Dirac notation, then you can write something like $|\psi\rangle = |\alpha\rangle \otimes |\beta \rangle$.  The corresponding bra would be $\langle \psi| = \langle \alpha| \otimes \langle \beta |$.  If we let $|\phi\rangle = |\gamma\rangle \otimes |\delta \rangle$, then
$$\langle \psi|\phi\rangle = \bigg(\langle \alpha| \otimes \langle \beta|\bigg) \bigg( |\gamma \rangle \otimes |\delta \rangle\bigg) = \langle \alpha|\gamma\rangle \cdot \langle \beta|\delta\rangle$$
The convention is that whether you're talking about a bra or a ket, the first quantity in the tensor product belongs to $\mathcal H_A$ (or its dual space) and the second belongs to $\mathcal H_B$ (or its dual space).

With all that being said, your expression
$$\rho_{j,l,k,l} = \langle\alpha_j| \langle\beta_l |\rho |\beta_l\rangle |\alpha_k\rangle$$
doesn't make sense to me, because the tensor product ket on the right is in the wrong order.
A: First of all, it should be noted that the way you understand $\rho_{ijk\ell}$ is first and foremost a matter of convention. That said, some conventions are certainly more "natural" than others.
One way to think about it is that the matrix components of $\rho$ in a composite space $\mathcal H\equiv \mathcal X\otimes\mathcal Y$ are nothing but that: matrix components in some space. If you use the indices $I,J$ to label the elements of a basis of $\mathcal H$, you can write the matrix components as
$$\rho_{I,J}\equiv \langle I|\rho|J\rangle, \qquad |I\rangle,|J\rangle\in\mathcal H.$$
However, this notation does not take into account the bipartite structure of $\mathcal H$. To do this, we observe that we can always find a basis of $\mathcal H$ that is built from bases of $\mathcal X$ and $\mathcal Y$. We can thus label the basis elements of $\mathcal H$ using two indices, denoting the corresponding basis elements of $\mathcal X$ and $\mathcal Y$. In other words, we can write
$$\mathcal H = \mathrm{span}(\{|i,j\rangle\equiv|i\rangle\otimes|j\rangle : \quad |i\rangle\in\mathcal X, \,\,|j\rangle\in\mathcal Y\}).$$
Then, instead of an index $I$, we use a pair of indices, say $(i,j)$. The matrix elements of $\rho$ then become
$$\rho_{(i,j),(k,\ell)} \equiv \langle i,j|\rho|k,\ell\rangle \equiv (\langle i|\otimes\langle j|)\rho(|k\rangle\otimes |\ell\rangle),$$
where I'm including different equivalent ways to write the expression.
Note that I wrote the "input" and "output" indices of $\rho$ using pairs $(i,j)$ and $(k,\ell)$ here, to stress the different roles the indices have.
For brevity, one does not usually do this, and simply writes $\rho_{ijk\ell}$ to mean $\rho_{(i,j),(k,\ell)}$.
Now, you can also decide to use $\rho_{ijk\ell}$ to mean something like $\langle \ell,j|\rho|k,i\rangle$. That would be quite an awkward notation though.
