# Understanding the partial trace and deriving $\langle l|R_{B}|k\rangle = \text{Tr}((\mathbb{I}_{A} \otimes |k\rangle \langle l|)(R_{AB}))$

By definition according to the notes I am looking through:

The partial trace $\text{Tr}_A:L(H_A \otimes H_B) \rightarrow L(H_B)$ is the unique map that satisfies: $$\text{Tr}(L_B \cdot \text{Tr}_A(R_{AB})) = \text{Tr}((\mathbb{I} \otimes L_B)R_{AB})$$ for all $L_B \in L(H_B)$ and $R_{AB} \in L(H_A \otimes H_B)$.

Now, according to the Wikipedia definition,$$T \in L(V \otimes W) \mapsto \text{Tr}_W(T) \in L(V).$$

Now if, in trying to relate the two definitions I let $W = H_A$, and $T = R_{AB}$ in the second definition, then I recover $\text{Tr}_A(R_{AB})$ However in the first definition I have the outer trace and $L_B$ - the first part which is $\text{Tr}(L_B \cdots$

Why are these different (at least seemingly)? $\leftarrow$ Question (1)

Now I am trying to understand why, as in the subject line: $$\langle l|R_{B}|k\rangle = \text{Tr}((\mathbb{I}_{A} \otimes |k\rangle \langle l|)(R_{AB}))$$ First, I know that $\langle l|R_{B}|k\rangle = \text{Tr}(|k\rangle \langle l|R_{B})$, but I don't see how to get to the result. I think some of the confusion lies in that I don't know how to operate on $(\mathbb{I}_{A} \otimes |k\rangle \langle l|)(R_{AB})$. Since $R_{AB}$ is a matrix I don't know what to do with it in regards to the tensor product (I couldn't find or recognize a similar example with distributing a matrix over a tensor product).

Question 2: How do we get $\langle l|R_{B}|k\rangle = \text{Tr}((\mathbb{I}_{A} \otimes |k\rangle \langle l|)(R_{AB}))$?

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This notation is very hard for me to understand. I have seen a much nicer explanation of (I think) the partial trace in the first chapter in John Bell's book 'Speakable and Unspeakable in Quantum Mechanics'. If you read that carefully you might be able to work backwards on figuring this out. – dj_mummy Aug 23 '13 at 16:11
Practically, for a multi-particle systems, the density matrix is : $\rho^I_J= \rho^{i_1i_2....i_n}_{j_1j_2...j_n}$.You get the partial density matrix for the particles $1..m$, with $(\rho')^{I'}_{J'}= (\rho')^{i'_1 i'_2....i'_m}_{j'_1j'_2...j'_m}= \sum_{k_m+1,k_m+2,..,k_n}^{}\rho^{i'_1i'_2....i'_mk_{m+1}k_{m+2}...k_{n}}_{j'_1j‌​_2'...j'_mk_{m+1}k_{m+2}...k_{n}}$ – Trimok Aug 23 '13 at 16:54

I quite like your characterization of the partial trace!

I think you perceive a conflict with the Wikipedia definition because you are only taking part of the latter: given an operator $T\in L(V\otimes W)$, the requirement that its partial trace obey $$\text{Tr}_W(T)\in L(V)$$ simply says that the partial trace over $W$ be an operator on $V$, but that doesn't say which operator. (The specification of that is done in a more concrete, basis-dependent way.)

To obtain the first result that confuses you, $\langle k |R|l\rangle=\text{Tr}(|l\rangle\langle k|R)$ for some operator $R$, simply take the trace in the same orthogonal basis where $|k\rangle$ and $|l\rangle$ came from: $$\text{Tr}(|l\rangle\langle k|R)=\sum_j \langle j|l\rangle\langle k|R|j\rangle =\sum_j \langle k|R|j\rangle\langle j|l\rangle =\langle k|R|l\rangle.$$

Now, if you take $R=R_B=\text{Tr}_A(R_{AB})$, the matrix elements of this partial trace in the $B$ basis are, from the above, $$\langle k|R_B|l\rangle=\text{Tr}_B(|l\rangle\langle k|R_B)=\text{Tr}_{AB}((\mathbb I\otimes|l\rangle\langle k|)R_{AB}),$$ where the second equality is simply the fundamental definition of the partial trace, as you formulated it.

Now, I can understand it if all this simply looks complicated and does not provide any insight into what is going on - though that simply means that you need to look more closely into what your fundamental definition is saying.

Say I have a bipartite system $A\leftrightarrow B$, which may be initially entangled, and then I completely forget about the $A$ part of the system. Thus, I need to trade my full (possibly entangled) density matrix $\rho_{AB}$ for one I can deal with locally: a density matrix $\rho_B$ which acts only on the $B$ side, which I can act on with operators in $L(H_B)$, and which I can take the $B$ trace on. That is, I need to be able to speak of the object $$\text{Tr}(L_B\rho_B),$$ and that object embodies all I need in order to make predictions.

However, in terms of the full system, the state is $\rho_{AB}$, when I operate on it I am really using the operator $\mathbb I\otimes L_B$, and when I take the trace I am really taking the full trace $\text{Tr}_{AB}$ over the full space.

Since both viewpoints must match, these objects must obey $$\text{Tr}(L_B\rho_B)=\text{Tr}_{AB}((\mathbb I\otimes L_B)\rho_{AB}), \tag{1}$$ and this equation is simply a requirement on the only free object we have, $\rho_B$, which we call the partial trace $\rho_B:=\text{Tr}_A(\rho_{AB})$. As it happens, requiring $\text{Tr}_A$ to obey this for all $L_B\in L(H_B)$ and $\rho_{AB}\in L(H_A\otimes H_B)$* is enough to uniquely determine it, so that requirement can act as a definition (though, of course, you can have simpler definitions based on explicit basis-dependent formulae).

* Note that I am taking $\rho_{AB}$ to be a general operator, instead of only a density matrix, since we want to be able to act on $\rho_{AB}$ using entangling or correlated measurements before we forget about $B$. However, requiring (1) for all $L_B\in L(H_B)$ and only those $\rho_{AB}\in L(H_A\otimes H_B)$ such that $\rho_{AB}\geq 0$ and $\text{Tr}_{AB}(\rho_{AB})=1$ is enough to determine $\text{Tr}_A$ uniquely by linearity, as any operator $R=R_{AB}$ can be decomposed into positive-definite, trace-one operators $R_k$ as $R=r_1 R_1+ir_2R_2-r_3R_3-ir_4R_4$, with each $r_k\geq0$, by taking positive and negative parts of its hermitian and antihermtian parts.

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@user1922184: $\langle j |l\rangle$ and $\langle k| R|j\rangle$ commute because they are both numbers; after that I use the resolution of identity $\sum_j|j\rangle\langle j|=1$. I left it in because I thought it's clever (and doesn't depend on what basis you use), but it's a lot easier to enforce $\langle j|l\rangle=\delta_{jl}$ and kill the sum right away. – Emilio Pisanty Aug 26 '13 at 22:27
For any vector space $H$, in this answer $L(H)$ is the set of all linear operators $L:H\rightarrow H$, which given a basis of size $n$ for $H$ is isomorphic to the set of all $n$ by $n$ complex matrices. (Mathematicians call this $\text{End}(H)$, for endomorphism, and notation depends on who is writing.) Many of the things you'd put in for $L$ in $\text{Tr}(L\rho)$ have extra structure, but in finite dimension it's defined for all linear maps $L$. – Emilio Pisanty Aug 26 '13 at 22:57
The interpretation of $\text{Tr}(L\rho)$ depends on $L$. If it is an observable, the trace is its expectation value after measurement. If it is a projector (i.e. $L=L^\dagger=L^2$, and there exists some $M$ such that $L+M=1$) then $\text{Tr}(L\rho)$ is the probability you'll get the outcome $l$ instead of $m$ when measuring $lL+mM$. – Emilio Pisanty Aug 26 '13 at 23:01
You can break apart $|l\rangle \langle k|$ because once they are sandwiched in $\langle j|$ and $R|j\rangle$ they become pure numbers, and those do commute. You can't usually shuffle stuff around unless you know it's a number. You won't have seen things like $R|l\rangle\langle k|=|l\rangle \langle k|$ because it isn't true, but it does hold inside a trace: $$\text{Tr}(R|l\rangle\langle k|)=\text{Tr}(|l\rangle \langle k|).$$ – Emilio Pisanty Aug 26 '13 at 23:19
This is known as the cyclic property of the trace, $\text{Tr}(ABC)=\text{Tr}(CAB)$ ($\leftarrow$ prove it!), which is essentially all your question is about (and holds even when $A$, $B$ and $C$ have different dimensions, such as $A=\langle k|$, $B=R$ and $C=|l\rangle$), but that's probably pushing it way too far. In general, until you're comfortable with it,simplify! – Emilio Pisanty Aug 26 '13 at 23:19