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 $\langle l|R_{B}|k\rangle = \text{Tr}((\mathbb{I}_{A} \otimes |k\rangle \langle l|)(R_{AB}))$?

  • $\begingroup$ 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 $(\rho')^{I'}_{J'}= (\rho')^{i'_1 i'_2....i'_m}_{j'_1j'_2...j'_m}= \sum_{k_m+1,k_m+2,..,k_n}^{}\rho^{i'_1i'_2....i'_mk_{m+1}k_{m+2}...k_{n}}_{j'_1j_2'...j'_mk_{m+1}k_{m+2}...k_{n}}$ $\endgroup$ – 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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    $\begingroup$ @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. $\endgroup$ – Emilio Pisanty Aug 26 '13 at 22:27
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    $\begingroup$ 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 $L$. $\endgroup$ – Emilio Pisanty Aug 26 '13 at 22:57
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    $\begingroup$ 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 $lL+mM$. $\endgroup$ – Emilio Pisanty Aug 26 '13 at 23:01
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    $\begingroup$ 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: $$\text{Tr}(R|l\rangle\langle k|)=\text{Tr}(|l\rangle \langle k|).$$ $\endgroup$ – Emilio Pisanty Aug 26 '13 at 23:19
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    $\begingroup$ 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! $\endgroup$ – Emilio Pisanty Aug 26 '13 at 23:19

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