Partial trace over a product of matrices - prove that ${\rm Tr}(\rho^{AB}(\sigma^A\otimes I))={\rm Tr}(\rho^A\sigma^A)$

Question

$$Tr(\rho^{AB} (\sigma^A \otimes I/d)) = Tr(\rho^A \sigma^A)$$

I came across the above, but I'm not sure how it's true. I figured they first partial traced out the B subsystem, and then trace A, but I don't see how you are allowed to partial trace out B from both the factors in the arguments. A proof or any intuition on this would be appreciated.

Edit 1:

The notation

$\rho^{AB}$ is a state in Hilbert space $H_A \otimes H_B$

$\sigma^A$ is a state in Hilbert space $H_A$

$\rho^A$ is $\rho^{AB}$ with $B$ subsystem traced out.

$I/d$ is the maximally mixed state in Hilbert space $B$.

I saw this being used in Nielsen and Chuang, section 11.3.4, in the proof of subadditivity of entropy.

Edit 2:

So, I tried to write an answer based on DaftWullie's comment and Алексей Уваров's answer, but I am stuck again.

So, $$\rho^{AB} = \sum_{mnop} \rho_{mnop} |mo\rangle \langle np|$$

Then $$\rho^{A} = \sum_{mno} \rho_{mnoo} |m\rangle \langle n|$$

Let $$\sigma^A = \sum_{ij} \sigma_{ij} |i\rangle \langle j|$$

And $$I/d = \sum_{xy} [I/d]_{xy} |x\rangle \langle y|$$

RHS

$$Tr(\rho^A \sigma^A)\\ = Tr(\sum_{mno} \rho_{mnoo} |m\rangle \langle n|\sum_{ij} \sigma_{ij} |i\rangle \langle j|)\\ = Tr(\sum_{mnoj} \rho_{mnoo} \sigma_{nj} | m \rangle \langle j|)\\ = \sum_{mno} \rho_{mnoo} \sigma_{nm}$$

LHS

$$Tr(\rho^{AB} (\sigma^A \otimes I/d)\\ = Tr(\sum_{mnop} \rho_{mnop} |mo\rangle \langle np| \sum_{ijxy} \sigma_{ij} [I/d]_{xy} |ix\rangle \langle jy|)\\ = Tr(\sum_{mnoxjy}\rho_{mnox} \sigma_{nj} [I/d]_{xy} | mo \rangle \langle jy |)\\ = \sum_{mnyx} \rho_{nm}[I/d]_{xy}\\ = (1/d)\sum_{mny} \rho_{mnyy} \sigma_{nm}$$

Which is the same as the RHS, but there's an extra $1/d$ factor?

Also, am I thinking about this the wrong way? Is there a simpler way to look at this?

Answer 1

The equation at the top of the question is not correct: there is a missing factor of $1/d$ on the right-hand side. Let's eliminate this factor from the left-hand side to make it simpler, so that the equation we want is this: $$ \text{Tr}\bigl(\rho^{AB} \bigl(\sigma^A \otimes I\bigr)\bigr) = \text{Tr}\bigl(\rho^A \sigma^A\bigr). $$

To see why this is true, it helps to start with an easy special case, which is that $\rho^{AB}$ is a product state: $$ \rho^{AB} = \rho^A \otimes \rho^B. $$ In this case we have $$ \text{Tr}\bigl(\bigl(\rho^A \otimes\rho^B\bigr) \bigl(\sigma^A \otimes I\bigr)\bigr) = \text{Tr}\bigl(\rho^A \sigma^A\bigr)\text{Tr}\bigl(\rho^B\bigr) = \text{Tr}\bigl(\rho^A \sigma^A\bigr), $$ using just elementary properties of tensor products and their traces.

Now, given that the equation is true in the special case, it has to be true in general because the expressions $$ \text{Tr}\bigl(\rho^{AB} \bigl(\sigma^A \otimes I\bigr)\bigr)\;\;\text{and}\;\;\text{Tr}\bigl(\rho^A \sigma^A\bigr) $$ depend linearly on $\rho^{AB}$, and the set of all product states $\rho^A\otimes\rho^B$ spans the vector space of all operators acting on $H_A\otimes H_B$.

Alternatively, we have $$ \text{Tr}((X\otimes Y)(Z\otimes I)) = \text{Tr}(XZ)\text{Tr}(Y) = \text{Tr}\bigl(\text{Tr}_B(X\otimes Y)\, Z\bigr) $$ for all operators $X$ and $Z$ acting on $H_A$ and all $Y$ acting on $H_B$, irrespective of their traces, and therefore $$ \text{Tr}(W(Z\otimes I)) = \text{Tr}\bigl(\text{Tr}_B(W) Z\bigr) $$ for all operators $W$ acting on $H_A\otimes H_B$ by linearity.

Answer 2

Here the important fact is that the maximally mixed state is in fact an identity matrix.

Let me rewrite the expression on the left in index notation (the summation sign is omitted according to the Einstein convention):

$$ Tr(\rho^{AB} (\sigma^A \otimes I/d)) = [\rho^{AB}]_{ijkl} [\sigma^A]_{ji} [I/d]_{lk} $$

But $[I/d]_{lk} = \frac1d \delta_{lk}$, therefore $[\rho^{AB}]_{ijkl} [\sigma^A]_{ji} [I/d]_{lk} = \frac1d [\rho^{AB}]_{ijkk} [\sigma^A]_{ji}$, which is exactly what happens if you first trace out the subsystem $B$ (UPD: up to the prefactor of $1/d$ apparently).

The physical intuition would be as follows. This expression is basically an expected value of a Hermitian operator $\frac1d \sigma^A \otimes I$ over a state $\rho$. This operator only acts nontrivially on the first subsystem, thus we can safely trace out the rest.

EDIT: Also, this contraction problem can be understood better if you use tensor network notation. Learning it requires some time, but if you do, I suggest starting here and here.

Answer 3

We need to prove that $Tr\Big(\rho^{AB}(\sigma^A\otimes I)\Big)=Tr(\rho^A\sigma^A)$

Let's write $\rho^{AB}=\sum_{a,a',b,b'}\rho^{AB}_{a,a',b,b'}|a\rangle\langle a'|\otimes|b\rangle\langle b'|$ and $\sigma^A=\sum_{a,a'}\sigma_{a,a'}^A|a\rangle\langle a'|\implies \langle a'|\sigma^A|a\rangle=\sigma^A_{a',a}$ where $\{|a\rangle\}$, $\{|b\rangle\}$ are orthonormal basis, we can also write $\rho^A=\sum_{a,a'}\rho^A_{a,a'}|a\rangle\langle a'|$

\begin{align} Tr(\rho^A\sigma^A)&=\sum_{a,a'}\rho^A_{a,a'}Tr(|a\rangle\langle a'|\sigma^A)\\ &=\sum_{a,a'}\rho^A_{a,a'}\langle a'|\sigma^A|a\rangle\\ &=\sum_{a,a'}\rho^A_{a,a'}\sigma^A_{a',a}\\ \end{align}

\begin{align} Tr\Big(\rho^{AB}(\sigma^A\otimes I)\Big)&=\sum_{a,a',b,b'}\rho^{AB}_{a,a',b,b'}Tr\Big(|a\rangle\langle a'|\sigma^A\otimes|b\rangle\langle b'|\Big)\\ &=\sum_{a,a',b,b'}\rho^{AB}_{a,a',b,b'}Tr\Big(|a\rangle\langle a'|\sigma^A\Big)Tr\Big(|b\rangle\langle b'|\Big)\\ &=\sum_{a,a',b,b'}\rho^{AB}_{a,a',b,b'}\langle a'|\sigma^A|a\rangle\langle b'|b\rangle\\ &=\sum_{a,a',b,b'}\rho^{AB}_{a,a',b,b'}\langle a'|\sigma^A|a\rangle.\delta_{b',b}\\ &=\sum_{a,a',b}\rho^{AB}_{a,a',b,b}\langle a'|\sigma^A|a\rangle\\ &=\sum_{a,a',b}\rho^{AB}_{a,a',b,b}\sigma^A_{a',a}\\ \end{align}

\begin{align} \rho^A=Tr_B(\rho^{AB})&=\sum_{b''}(I\otimes\langle b''|)\rho^{AB}(I\otimes |b''\rangle)\\ &=\sum_{b''}(I\otimes\langle b''|)\Big(\sum_{a,a',b,b'}\rho^{AB}_{a,a',b,b'}|a\rangle\langle a'|\otimes|b\rangle\langle b'|\Big)(I\otimes |b''\rangle)\\ &=\sum_{a,a',b,b',b''}\rho^{AB}_{a,a',b,b'}|a\rangle\langle a'|\otimes \langle b''|b\rangle\langle b'|b''\rangle\\ &=\sum_{a,a',b,b',b''}\rho^{AB}_{a,a',b,b'}|a\rangle\langle a'|\otimes \delta_{b'',b}\delta_{b',b''}\\ &=\sum_{a,a',b}\rho^{AB}_{a,a',b,b}|a\rangle\langle a'|\\ \end{align} $\implies \rho^A_{a,a'}=\langle a|\rho^A|a'\rangle=\sum_b\rho^{AB}_{a,a',b,b}$ Therefore,

\begin{align} Tr\Big(\rho^{AB}(\sigma^A\otimes I)\Big)&=\sum_{a,a',b}\rho^{AB}_{a,a',b,b}\sigma^A_{a',a}\\ &=\sum_{a,a'}(\sum_b\rho^{AB}_{a,a',b,b})\sigma^A_{a',a}\\ &=\sum_{a,a'}\rho^{A}_{a,a'}\sigma^A_{a',a}\\ &=Tr(\rho^A\sigma^A) \end{align}