Question about the Dirac notation for partial trace

Question

I saw the following definition for the partial trace operator:

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

From what I know, in the Dirac notation, the meaning of $langle v|A|urangle$ is the inner product of the vectors $|vrangle$ and $A|urangle$, 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 $rho_{AB}|e_krangle$? $rho_{AB}$ is a mapping on the space $Aotimes B$, but $e_k$ is a vector from the space $B$. So, I don't really know how to interpret the meaning of this notation.

Bubble · Accepted 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 $sigma_z$, $|0rangle_A$ and $|1rangle_A$; and $|0rangle_B$ and $|1rangle_B$. 
Now suppose that we have a Bell state,
$$|Psirangle_{AB} = frac{1}{sqrt{2}} (|0rangle_A otimes |0rangle_B + |1rangle_A otimes |1rangle_B).$$
This state corresponds to a density matrix,
$$rho_{AB}=|Psirangle_{AB}langlePsi|_{AB}$$
$$=frac{1}{2}(|0rangle_A otimes |0rangle_B langle0|_A otimes langle0|_B+|0rangle_A otimes |0rangle_Blangle1|_A otimes langle1|_B$$
$$+|1rangle_A otimes |1rangle_Blangle0|_A otimes langle0|_B+|1rangle_A otimes |1rangle_Blangle1|_A otimes langle1|_B).$$
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 $|e_1 rangle=|0rangle_B$ and $|e_1 rangle=|1rangle_B$, together with the fact that $langle phi' |_B(|psi rangle_A otimes |phirangle_B)=(langle phi' |_B|phirangle_B)|psi rangle_A $ (which is just the inner product of $|phirangle_B$ and $|phi'rangle_B$, a number, times $|psi rangle_A$) as well as orthonormality, to get, 
$$rho_{A}=frac{1}{2}(|0rangle_Alangle0|_A+|1rangle_Alangle1|_A), $$ 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.

Trimok · Answer

A basis for the Hilbert space $A otimes B$ could be written :

$$|e_{i_1 i_2}rangle = |e_{i_1}rangle otimes |e_{i_2}rangle tag{1}$$

We may user the notation  $I$ representing a composite index : $ I = (i_1 i_2)$, so that $|e_{I}rangle = |e_{i_1}rangle otimes |e_{i_2}rangle $

The density matrix could be written :

$$rho_{I'I} = rho_{ large (i'_1 i'_2)(i_1 i_2)} tag{2}$$

For instance, the action of the density matrix on a state $|psi rangle = sum_{i_1,i_2} psi_{i_1 i_2} |e_{i_1 i_2}rangle$ is $|psi' rangle = rho |psi rangle$, that is ${psi'}_I{'} = sum_{I} rho_{I'I} psi_I$ ,or, in a more detailed way :

$${psi'}_{i'_1 i'_2}  = sum_{i_1,i_2} rho_{ large (i'_1 i'_2)(i_1 i_2)} psi_{i_1 i_2} tag{3}$$

The partial trace on system $(2)$, that we called $rho_1$, is defined by :

$$(rho_1)_{i_1' i_1} = sum_{i_2}rho_{ large (i'_1 i_2)(i_1 i_2)} tag{4}$$

Now, we may write $rho_{I'I} = langle e_{I'}|rho|e_{I}rangle$, that is  : $$rho_{ large (i'_1 i'_2)(i_1 i_2)} = langle e_{i'_1 i'_2}|rho |e_{i_1 i_2}rangle = ( langle e_{i'_1}| otimes langle e_{i'_2}|) ~~rho ~~(|e_{i_1}rangle otimes |e_{i_2}rangle) tag{5}$$

So, finally, we may write :

$$(rho_1)_{i_1' i_1} = sum_{i_2} langle e_{i'_1 i_2}|rho |e_{i_1 i_2}rangle= sum_{i_2} ( langle e_{i'_1}| otimes langle e_{i_2}|) ~~rho ~~(|e_{i_1}rangle otimes |e_{i_2}rangle)tag{6}$$

Emilio Pisanty · Answer

This is indeed slightly confusing.
It is probably a little bit easier to understand starting with the bra. If $langle varphi|$ is a bra in the $A$ side of an $mathcal H_Aotimes mathcal H_B$ tensor product, then it is a linear transformation from $mathcal H_A$ into $mathbb C$:
$$langle varphi|:mathcal H_Arightarrow mathbb C.$$
Similarly to operators $hat V_A:mathcal H_Arightarrow mathcal H_A$, when you want to talk about their action on the tensor product $mathcal H_Aotimes mathcal H_B$, you should always tensor-product them with an identity on the $B$ side. Thus, on the composite system, the usual convention is that
$$langle varphi|text{ is shorthand for }langlevarphi|otimes mathbb I_B:mathcal H_Aotimesmathcal H_Brightarrow mathbb Cotimesmathcal H_B=mathcal H_B,$$
in the same way that $hat V_A$ is shorthand for$hat V_A otimes mathbb I_B:mathcal H_Aotimesmathcal H_Brightarrow mathcal H_Aotimesmathcal H_B$.
In an analogous way, vectors in a single tensor factor must also be tensored with an identity on the rest (unless, of course, they're part of a proper tensor product with vectors on all factors, like $$|varphi_Arangleotimes|varphi_Branglein mathcal H_Aotimesmathcal H_B.)$$
Thus, when doing a partial trace, the ket $|varphirangle$ on the right is shorthand for $|varphirangleotimes mathbb I_B$. It is slightly unnatural to see this as an operator but you can see it as going from $mathcal H_B=mathbb Cotimes mathcal H_B$ into $mathcal H_Aotimesmathcal H_B$.
Thus, for a density matrix $rho_{AB}:mathcal H_Aotimesmathcal H_Brightarrowmathcal H_A otimes mathcal H_B$, the partial trace over $A$ is a sum of terms of the form
$$
rho_Bsim langlevarphi|rho_{AB}|varphirangle
text{ which is shorthand for }
(langlevarphi|otimesmathbb I_B)rho_{AB}(|varphirangleotimesmathbb I_B)
,$$
and is an operator on $mathcal H_B$.

Question about the Dirac notation for partial trace

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