I saw the following definition for the partial trace operator:

$\rho_A=\sum_k \langle e_k|\rho_{AB}|e_k\rangle$, where $e_k$ is basis for the state space of system $B$.

From what I know, in the Dirac notation, the meaning of $\langle v|A|u\rangle$ is the inner product of the vectors $|v\rangle$ and $A|u\rangle$, so I have two problems with this notation of the partial trace.

First, how can an inner product be an operator? An inner product should be a complex number, so I guess that inner product represents an operator somhow.

Second, what is the meaning of $\rho_{AB}|e_k\rangle$? $\rho_{AB}$ is a mapping on the space $A\otimes B$, but $e_k$ is a vector from the space $B$. So, I don't really know how to interpret the meaning of this notation.

## Best Answer

I believe that an example will help clarify your confusion about notation (as examples usually do). Consider a system of two qubits, $A$ and $B$, with Hilbert spaces $V_A$ and $V_B$ spanned by two orthonormal eigenbasis of $\sigma_z$, $|0\rangle_A$ and $|1\rangle_A$; and $|0\rangle_B$ and $|1\rangle_B$. Now suppose that we have a Bell state, $$|\Psi\rangle_{AB} = \frac{1}{\sqrt{2}} (|0\rangle_A \otimes |0\rangle_B + |1\rangle_A \otimes |1\rangle_B).$$ This state corresponds to a density matrix, $$\rho_{AB}=|\Psi\rangle_{AB}\langle\Psi|_{AB}$$ $$=\frac{1}{2}(|0\rangle_A \otimes |0\rangle_B \langle0|_A \otimes \langle0|_B+|0\rangle_A \otimes |0\rangle_B\langle1|_A \otimes \langle1|_B$$ $$+|1\rangle_A \otimes |1\rangle_B\langle0|_A \otimes \langle0|_B+|1\rangle_A \otimes |1\rangle_B\langle1|_A \otimes \langle1|_B).$$ Now suppose that we wish to get the reduced density matrix for system A. We use your definition for the partial trace over system $B$ with $|e_1 \rangle=|0\rangle_B$ and $|e_1 \rangle=|1\rangle_B$, together with the fact that $\langle \phi' |_B(|\psi \rangle_A \otimes |\phi\rangle_B)=(\langle \phi' |_B|\phi\rangle_B)|\psi \rangle_A $ (which is just the inner product of $|\phi\rangle_B$ and $|\phi'\rangle_B$, a number, times $|\psi \rangle_A$) as well as orthonormality, to get, $$\rho_{A}=\frac{1}{2}(|0\rangle_A\langle0|_A+|1\rangle_A\langle1|_A), $$ a completely mixed state.

Incidentally, this appearance of a completely mixed state is the reason there is no FTL signalling in Bell experiments - a mixed state is complete ignorance about what is going with $B$ if we only study $A$ locally.