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

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

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?

• How do you define $\rho_a$ in terms of $\rho_{ab}$? Dec 25, 2018 at 8:34
• Hi Maharthi. It would be better if you can edit to make the title more descriptive. Questions with math-only titles are not easily searchable. Dec 25, 2018 at 9:55
• @Blue, edited title, check now? Dec 25, 2018 at 11:05
• @DaftWullie, Edited the question Dec 25, 2018 at 11:06
• It wasn’t a question in terms of me not understanding what you mean, but for you to answer and set you on your way towards a proper answer. Dec 25, 2018 at 11:30

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.

• Tr((𝑋⊗𝑌)(𝑍⊗𝐼)) =T r(𝑋𝑍)Tr(𝑌) = Tr(Tr𝐵(𝑋⊗𝑌)𝑍). This statement does not hold true irrespective of traces, right? In particular, the Trace of Y has to be 1 for this to hold true? Jun 21, 2019 at 4:54
• No, it holds for all $X$, $Y$, and $Z$. Jun 21, 2019 at 12:48
• Just to clarify, when you say that "the set of all product states $\rho_A\otimes\rho_B$ spans $H_A\otimes H_B$", do you mean that the $\rho_A$ and $\rho_B$ here are arbitrary matrices? We cannot restrict them to those with unit trace or to be positive semidefinite, correct? Jun 15, 2020 at 3:31
• No, I mean that the span is the entire space, even if $\rho_A$ and $\rho_B$ range over all positive semidefinite, trace-one matrices. Jun 15, 2020 at 14:06

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.

• Hey, thank you very much for answering! But I don't understand the Einstein notation very well. Can you explain what the rhs means? Dec 25, 2018 at 12:46
• @MahathiVempati By $[M]_{ij}$ I mean the element of a matrix M with indices i, j. The square brackets are unnecessary in general, but I used them to separate the subsystem indices like A, B from the tensor indices. Now, whenever there are two coinciding indices, there is a summation over this index, e.g. an ordinary matrix-vector product would look like $A_{ij} x_j$ Dec 25, 2018 at 13:27
• Now, intuitively what's going on in the rhs is that the density matrix $\rho^{AB}$ is contracted with the density matrix $\sigma^A \otimes I/d$, and then all of this is traced. If they both were presented as regular matrices, it would just be a product, but in tensor form it might be somewhat more confusing. Dec 25, 2018 at 13:33
• Hmm, when I was answering the question, I thought about ij as the row and column for the first subsystem and kl as that for the second. I think it can be written as follows: $\rho = \sum_{i,j,k,l} \rho_{ijkl} (| i\rangle \langle j | \otimes | k\rangle \langle l |)$. Dec 26, 2018 at 8:18
• @MahathiVempati Hmm, looks like you're right about the $1/d$ prefactor. However, in my copy of Nielsen and Chuang I can't find this exact expression. They seem to prove subadditivity somewhat differently Dec 28, 2018 at 8:53

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}