I have a feeling this is a very basic question. I apologize if it is.
Using Dirac's notation, can every (mixed) density operator $\rho_A$ of system $A$ be written as the ket-bra (outer) product $|a_1 \rangle \langle a_2|$ for some vectors $|a_1\rangle , |a_2\rangle \ \in A$ ?
I ask this because in the book "Quantum Computation and Quantum Information" by Nielsen & Chuang, the definition of reduced density operator of operator $\rho^{AB}$ of systems $A$ and $B$ is given as so:
$$\rho^{A} =\mathrm{tr}_B(\rho^{AB}) = \mathrm{tr}_B \left( |a_1 \rangle \langle a_2| \otimes |b_1 \rangle \langle b_2| \right) = | a_1 \rangle \langle a_2 | \ \ \mathrm{tr} (|b_1 \rangle \langle b_2| )$$
which — to me at least– is implicitly asserting two things:
- Every operator $\rho^{AB} \in A \otimes B $ can be expressed as a tensor product $\rho_A \otimes \rho_B$, for some $\rho_A \in A, \ \rho_B \in B$, which I though was only true for separable $\rho^{AB}$ ,
- Every $A$ can be expressed as a product $|a_1 \rangle \langle a_2|$, and similarly for $B$.
So this second point is the one I'm asking.
Edit
Thanks to @NorbertSchuch for pointing out my mistake.
In the above question I wrongly merged two definitions of Nielsen and Chuang's book into one equation.
The first one is the definition of the reduced density operator
$$\rho^{A} =\mathrm{tr}_B(\rho^{AB})$$
The second one is the definition of the partial trace, which is defined (completely independently of the above definition) as
$$\mathrm{tr}_B \left( |a_1 \rangle \langle a_2| \otimes |b_1 \rangle \langle b_2| \right) = | a_1 \rangle \langle a_2 | \ \ \mathrm{tr} (|b_1 \rangle \langle b_2| )$$
What the book doesn't say –which is what I was after– is what the general definition of the partial trace is in terms of bra-ket notation.
I found the general definition of the partial trace here.
So, thanks to this and @CraigGidney, @WetSavannaAnimalakaRodVance, @ACuriousMind answers I now know that the answer is no.
Best Answer
Not every tensor is a simple tensor. The simple tensors of the form $a\otimes b\in A\otimes B$ span the tensor product, which means that a general element $t$ of the tensor product is a linear combination of these simple tensors, i.e. $$ t = \sum c_{ij} a_i\otimes b_j$$ for some basis $a_i,b_j$ of $A$ and $B$ respectively. If $A$ itself is a space of operators on a finite-dimensional Hilbert space $H_A$, then apply this property again to $A = H_A\otimes H_A^\ast$ to get a decomposition of such an operator in terms of bras and kets. Additionally, the equation you have written down doesn't make any sense - $\rho^A$ is supposed to be an operator on $A$, but $\langle a_1 \vert a_2 \rangle$ and $\operatorname{tr}(\lvert b_1 \rangle\langle b_2 \rvert)$ seem to be both numbers. You can find the proper definition of the partial trace in the Wikipedia article.