Partial trace over a product of matrices - one factor is in tensor product form
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$$Tr(rho^{AB} (sigma^A otimes I/d)) = Tr(rho^A sigma^A)$$
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, $$rho^{AB} = sum_{mnop} rho_{mnop} |morangle langle np|$$
Then $$rho^{A} = sum_{mno} rho_{mnoo} |mrangle langle n|$$
Let $$sigma^A = sum_{ij} sigma_{ij} |irangle langle j|$$
And $$I/d = sum_{xy} [I/d]_{xy} |xrangle langle y|$$
RHS
$$Tr(rho^A sigma^A)\
= Tr(sum_{mno} rho_{mnoo} |mrangle langle n|sum_{ij} sigma_{ij} |irangle langle j|)\
= Tr(sum_{mnoj} rho_{mnoo} sigma_{nj} | m rangle langle j|)\
= sum_{mno} rho_{mnoo} sigma_{nm}$$
LHS
$$Tr(rho^{AB} (sigma^A otimes I/d)\
= Tr(sum_{mnop} rho_{mnop} |morangle langle np| sum_{ijxy} sigma_{ij} [I/d]_{xy} |ixrangle langle jy|)\
= Tr(sum_{mnoxjy}rho_{mnox} sigma_{nj} [I/d]_{xy} | mo rangle langle jy |)\
= sum_{mnyx} rho_{nm}[I/d]_{xy}\
= (1/d)sum_{mny} rho_{mnyy} sigma_{nm}$$
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?
quantum-information density-matrix tensor-product nielsen-and-chuang partial-trace
$endgroup$
|
show 2 more comments
$begingroup$
$$Tr(rho^{AB} (sigma^A otimes I/d)) = Tr(rho^A sigma^A)$$
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, $$rho^{AB} = sum_{mnop} rho_{mnop} |morangle langle np|$$
Then $$rho^{A} = sum_{mno} rho_{mnoo} |mrangle langle n|$$
Let $$sigma^A = sum_{ij} sigma_{ij} |irangle langle j|$$
And $$I/d = sum_{xy} [I/d]_{xy} |xrangle langle y|$$
RHS
$$Tr(rho^A sigma^A)\
= Tr(sum_{mno} rho_{mnoo} |mrangle langle n|sum_{ij} sigma_{ij} |irangle langle j|)\
= Tr(sum_{mnoj} rho_{mnoo} sigma_{nj} | m rangle langle j|)\
= sum_{mno} rho_{mnoo} sigma_{nm}$$
LHS
$$Tr(rho^{AB} (sigma^A otimes I/d)\
= Tr(sum_{mnop} rho_{mnop} |morangle langle np| sum_{ijxy} sigma_{ij} [I/d]_{xy} |ixrangle langle jy|)\
= Tr(sum_{mnoxjy}rho_{mnox} sigma_{nj} [I/d]_{xy} | mo rangle langle jy |)\
= sum_{mnyx} rho_{nm}[I/d]_{xy}\
= (1/d)sum_{mny} rho_{mnyy} sigma_{nm}$$
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?
quantum-information density-matrix tensor-product nielsen-and-chuang partial-trace
$endgroup$
2
$begingroup$
How do you define $rho_a$ in terms of $rho_{ab}$?
$endgroup$
– DaftWullie
Dec 25 '18 at 8:34
$begingroup$
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.
$endgroup$
– Blue♦
Dec 25 '18 at 9:55
$begingroup$
@Blue, edited title, check now?
$endgroup$
– Mahathi Vempati
Dec 25 '18 at 11:05
$begingroup$
@DaftWullie, Edited the question
$endgroup$
– Mahathi Vempati
Dec 25 '18 at 11:06
1
$begingroup$
@MahathiVempati Thanks. Looks good!
$endgroup$
– Blue♦
Dec 25 '18 at 13:09
|
show 2 more comments
$begingroup$
$$Tr(rho^{AB} (sigma^A otimes I/d)) = Tr(rho^A sigma^A)$$
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, $$rho^{AB} = sum_{mnop} rho_{mnop} |morangle langle np|$$
Then $$rho^{A} = sum_{mno} rho_{mnoo} |mrangle langle n|$$
Let $$sigma^A = sum_{ij} sigma_{ij} |irangle langle j|$$
And $$I/d = sum_{xy} [I/d]_{xy} |xrangle langle y|$$
RHS
$$Tr(rho^A sigma^A)\
= Tr(sum_{mno} rho_{mnoo} |mrangle langle n|sum_{ij} sigma_{ij} |irangle langle j|)\
= Tr(sum_{mnoj} rho_{mnoo} sigma_{nj} | m rangle langle j|)\
= sum_{mno} rho_{mnoo} sigma_{nm}$$
LHS
$$Tr(rho^{AB} (sigma^A otimes I/d)\
= Tr(sum_{mnop} rho_{mnop} |morangle langle np| sum_{ijxy} sigma_{ij} [I/d]_{xy} |ixrangle langle jy|)\
= Tr(sum_{mnoxjy}rho_{mnox} sigma_{nj} [I/d]_{xy} | mo rangle langle jy |)\
= sum_{mnyx} rho_{nm}[I/d]_{xy}\
= (1/d)sum_{mny} rho_{mnyy} sigma_{nm}$$
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?
quantum-information density-matrix tensor-product nielsen-and-chuang partial-trace
$endgroup$
$$Tr(rho^{AB} (sigma^A otimes I/d)) = Tr(rho^A sigma^A)$$
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, $$rho^{AB} = sum_{mnop} rho_{mnop} |morangle langle np|$$
Then $$rho^{A} = sum_{mno} rho_{mnoo} |mrangle langle n|$$
Let $$sigma^A = sum_{ij} sigma_{ij} |irangle langle j|$$
And $$I/d = sum_{xy} [I/d]_{xy} |xrangle langle y|$$
RHS
$$Tr(rho^A sigma^A)\
= Tr(sum_{mno} rho_{mnoo} |mrangle langle n|sum_{ij} sigma_{ij} |irangle langle j|)\
= Tr(sum_{mnoj} rho_{mnoo} sigma_{nj} | m rangle langle j|)\
= sum_{mno} rho_{mnoo} sigma_{nm}$$
LHS
$$Tr(rho^{AB} (sigma^A otimes I/d)\
= Tr(sum_{mnop} rho_{mnop} |morangle langle np| sum_{ijxy} sigma_{ij} [I/d]_{xy} |ixrangle langle jy|)\
= Tr(sum_{mnoxjy}rho_{mnox} sigma_{nj} [I/d]_{xy} | mo rangle langle jy |)\
= sum_{mnyx} rho_{nm}[I/d]_{xy}\
= (1/d)sum_{mny} rho_{mnyy} sigma_{nm}$$
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?
quantum-information density-matrix tensor-product nielsen-and-chuang partial-trace
quantum-information density-matrix tensor-product nielsen-and-chuang partial-trace
edited Dec 31 '18 at 23:43
Blue♦
6,63141556
6,63141556
asked Dec 25 '18 at 8:15
Mahathi VempatiMahathi Vempati
4908
4908
2
$begingroup$
How do you define $rho_a$ in terms of $rho_{ab}$?
$endgroup$
– DaftWullie
Dec 25 '18 at 8:34
$begingroup$
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.
$endgroup$
– Blue♦
Dec 25 '18 at 9:55
$begingroup$
@Blue, edited title, check now?
$endgroup$
– Mahathi Vempati
Dec 25 '18 at 11:05
$begingroup$
@DaftWullie, Edited the question
$endgroup$
– Mahathi Vempati
Dec 25 '18 at 11:06
1
$begingroup$
@MahathiVempati Thanks. Looks good!
$endgroup$
– Blue♦
Dec 25 '18 at 13:09
|
show 2 more comments
2
$begingroup$
How do you define $rho_a$ in terms of $rho_{ab}$?
$endgroup$
– DaftWullie
Dec 25 '18 at 8:34
$begingroup$
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.
$endgroup$
– Blue♦
Dec 25 '18 at 9:55
$begingroup$
@Blue, edited title, check now?
$endgroup$
– Mahathi Vempati
Dec 25 '18 at 11:05
$begingroup$
@DaftWullie, Edited the question
$endgroup$
– Mahathi Vempati
Dec 25 '18 at 11:06
1
$begingroup$
@MahathiVempati Thanks. Looks good!
$endgroup$
– Blue♦
Dec 25 '18 at 13:09
2
2
$begingroup$
How do you define $rho_a$ in terms of $rho_{ab}$?
$endgroup$
– DaftWullie
Dec 25 '18 at 8:34
$begingroup$
How do you define $rho_a$ in terms of $rho_{ab}$?
$endgroup$
– DaftWullie
Dec 25 '18 at 8:34
$begingroup$
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.
$endgroup$
– Blue♦
Dec 25 '18 at 9:55
$begingroup$
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.
$endgroup$
– Blue♦
Dec 25 '18 at 9:55
$begingroup$
@Blue, edited title, check now?
$endgroup$
– Mahathi Vempati
Dec 25 '18 at 11:05
$begingroup$
@Blue, edited title, check now?
$endgroup$
– Mahathi Vempati
Dec 25 '18 at 11:05
$begingroup$
@DaftWullie, Edited the question
$endgroup$
– Mahathi Vempati
Dec 25 '18 at 11:06
$begingroup$
@DaftWullie, Edited the question
$endgroup$
– Mahathi Vempati
Dec 25 '18 at 11:06
1
1
$begingroup$
@MahathiVempati Thanks. Looks good!
$endgroup$
– Blue♦
Dec 25 '18 at 13:09
$begingroup$
@MahathiVempati Thanks. Looks good!
$endgroup$
– Blue♦
Dec 25 '18 at 13:09
|
show 2 more comments
2 Answers
2
active
oldest
votes
$begingroup$
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:
$$
text{Tr}bigl(rho^{AB} bigl(sigma^A otimes Ibigr)bigr) = text{Tr}bigl(rho^A sigma^Abigr).
$$
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
$$
text{Tr}bigl(bigl(rho^A otimesrho^Bbigr) bigl(sigma^A otimes Ibigr)bigr) = text{Tr}bigl(rho^A sigma^Abigr)text{Tr}bigl(rho^Bbigr) = text{Tr}bigl(rho^A sigma^Abigr),
$$
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
$$
text{Tr}bigl(rho^{AB} bigl(sigma^A otimes Ibigr)bigr);;text{and};;text{Tr}bigl(rho^A sigma^Abigr)
$$
depend linearly on $rho^{AB}$, and the set of all product states $rho^Aotimesrho^B$ spans the vector space of all operators acting on $H_Aotimes H_B$.
Alternatively, we have
$$
text{Tr}((Xotimes Y)(Zotimes I)) = text{Tr}(XZ)text{Tr}(Y) = text{Tr}bigl(text{Tr}_B(Xotimes Y), Zbigr)
$$
for all operators $X$ and $Z$ acting on $H_A$ and all $Y$ acting on $H_B$, irrespective of their traces, and therefore
$$
text{Tr}(W(Zotimes I)) = text{Tr}bigl(text{Tr}_B(W) Zbigr)
$$
for all operators $W$ acting on $H_Aotimes H_B$ by linearity.
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add a comment |
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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.
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Hey, thank you very much for answering! But I don't understand the Einstein notation very well. Can you explain what the rhs means?
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– Mahathi Vempati
Dec 25 '18 at 12:46
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@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 $A_{ij} x_j$
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– Алексей Уваров
Dec 25 '18 at 13:27
$begingroup$
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.
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– Алексей Уваров
Dec 25 '18 at 13:33
1
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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: $rho = sum_{i,j,k,l} rho_{ijkl} (| irangle langle j | otimes | krangle langle l |)$.
$endgroup$
– Алексей Уваров
Dec 26 '18 at 8:18
1
$begingroup$
@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
$endgroup$
– Алексей Уваров
Dec 28 '18 at 8:53
|
show 2 more comments
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2 Answers
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2 Answers
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$begingroup$
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:
$$
text{Tr}bigl(rho^{AB} bigl(sigma^A otimes Ibigr)bigr) = text{Tr}bigl(rho^A sigma^Abigr).
$$
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
$$
text{Tr}bigl(bigl(rho^A otimesrho^Bbigr) bigl(sigma^A otimes Ibigr)bigr) = text{Tr}bigl(rho^A sigma^Abigr)text{Tr}bigl(rho^Bbigr) = text{Tr}bigl(rho^A sigma^Abigr),
$$
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
$$
text{Tr}bigl(rho^{AB} bigl(sigma^A otimes Ibigr)bigr);;text{and};;text{Tr}bigl(rho^A sigma^Abigr)
$$
depend linearly on $rho^{AB}$, and the set of all product states $rho^Aotimesrho^B$ spans the vector space of all operators acting on $H_Aotimes H_B$.
Alternatively, we have
$$
text{Tr}((Xotimes Y)(Zotimes I)) = text{Tr}(XZ)text{Tr}(Y) = text{Tr}bigl(text{Tr}_B(Xotimes Y), Zbigr)
$$
for all operators $X$ and $Z$ acting on $H_A$ and all $Y$ acting on $H_B$, irrespective of their traces, and therefore
$$
text{Tr}(W(Zotimes I)) = text{Tr}bigl(text{Tr}_B(W) Zbigr)
$$
for all operators $W$ acting on $H_Aotimes H_B$ by linearity.
$endgroup$
add a comment |
$begingroup$
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:
$$
text{Tr}bigl(rho^{AB} bigl(sigma^A otimes Ibigr)bigr) = text{Tr}bigl(rho^A sigma^Abigr).
$$
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
$$
text{Tr}bigl(bigl(rho^A otimesrho^Bbigr) bigl(sigma^A otimes Ibigr)bigr) = text{Tr}bigl(rho^A sigma^Abigr)text{Tr}bigl(rho^Bbigr) = text{Tr}bigl(rho^A sigma^Abigr),
$$
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
$$
text{Tr}bigl(rho^{AB} bigl(sigma^A otimes Ibigr)bigr);;text{and};;text{Tr}bigl(rho^A sigma^Abigr)
$$
depend linearly on $rho^{AB}$, and the set of all product states $rho^Aotimesrho^B$ spans the vector space of all operators acting on $H_Aotimes H_B$.
Alternatively, we have
$$
text{Tr}((Xotimes Y)(Zotimes I)) = text{Tr}(XZ)text{Tr}(Y) = text{Tr}bigl(text{Tr}_B(Xotimes Y), Zbigr)
$$
for all operators $X$ and $Z$ acting on $H_A$ and all $Y$ acting on $H_B$, irrespective of their traces, and therefore
$$
text{Tr}(W(Zotimes I)) = text{Tr}bigl(text{Tr}_B(W) Zbigr)
$$
for all operators $W$ acting on $H_Aotimes H_B$ by linearity.
$endgroup$
add a comment |
$begingroup$
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:
$$
text{Tr}bigl(rho^{AB} bigl(sigma^A otimes Ibigr)bigr) = text{Tr}bigl(rho^A sigma^Abigr).
$$
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
$$
text{Tr}bigl(bigl(rho^A otimesrho^Bbigr) bigl(sigma^A otimes Ibigr)bigr) = text{Tr}bigl(rho^A sigma^Abigr)text{Tr}bigl(rho^Bbigr) = text{Tr}bigl(rho^A sigma^Abigr),
$$
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
$$
text{Tr}bigl(rho^{AB} bigl(sigma^A otimes Ibigr)bigr);;text{and};;text{Tr}bigl(rho^A sigma^Abigr)
$$
depend linearly on $rho^{AB}$, and the set of all product states $rho^Aotimesrho^B$ spans the vector space of all operators acting on $H_Aotimes H_B$.
Alternatively, we have
$$
text{Tr}((Xotimes Y)(Zotimes I)) = text{Tr}(XZ)text{Tr}(Y) = text{Tr}bigl(text{Tr}_B(Xotimes Y), Zbigr)
$$
for all operators $X$ and $Z$ acting on $H_A$ and all $Y$ acting on $H_B$, irrespective of their traces, and therefore
$$
text{Tr}(W(Zotimes I)) = text{Tr}bigl(text{Tr}_B(W) Zbigr)
$$
for all operators $W$ acting on $H_Aotimes H_B$ by linearity.
$endgroup$
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:
$$
text{Tr}bigl(rho^{AB} bigl(sigma^A otimes Ibigr)bigr) = text{Tr}bigl(rho^A sigma^Abigr).
$$
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
$$
text{Tr}bigl(bigl(rho^A otimesrho^Bbigr) bigl(sigma^A otimes Ibigr)bigr) = text{Tr}bigl(rho^A sigma^Abigr)text{Tr}bigl(rho^Bbigr) = text{Tr}bigl(rho^A sigma^Abigr),
$$
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
$$
text{Tr}bigl(rho^{AB} bigl(sigma^A otimes Ibigr)bigr);;text{and};;text{Tr}bigl(rho^A sigma^Abigr)
$$
depend linearly on $rho^{AB}$, and the set of all product states $rho^Aotimesrho^B$ spans the vector space of all operators acting on $H_Aotimes H_B$.
Alternatively, we have
$$
text{Tr}((Xotimes Y)(Zotimes I)) = text{Tr}(XZ)text{Tr}(Y) = text{Tr}bigl(text{Tr}_B(Xotimes Y), Zbigr)
$$
for all operators $X$ and $Z$ acting on $H_A$ and all $Y$ acting on $H_B$, irrespective of their traces, and therefore
$$
text{Tr}(W(Zotimes I)) = text{Tr}bigl(text{Tr}_B(W) Zbigr)
$$
for all operators $W$ acting on $H_Aotimes H_B$ by linearity.
answered Dec 28 '18 at 13:36
John WatrousJohn Watrous
1,91828
1,91828
add a comment |
add a comment |
$begingroup$
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.
$endgroup$
$begingroup$
Hey, thank you very much for answering! But I don't understand the Einstein notation very well. Can you explain what the rhs means?
$endgroup$
– Mahathi Vempati
Dec 25 '18 at 12:46
$begingroup$
@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 $A_{ij} x_j$
$endgroup$
– Алексей Уваров
Dec 25 '18 at 13:27
$begingroup$
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.
$endgroup$
– Алексей Уваров
Dec 25 '18 at 13:33
1
$begingroup$
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: $rho = sum_{i,j,k,l} rho_{ijkl} (| irangle langle j | otimes | krangle langle l |)$.
$endgroup$
– Алексей Уваров
Dec 26 '18 at 8:18
1
$begingroup$
@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
$endgroup$
– Алексей Уваров
Dec 28 '18 at 8:53
|
show 2 more comments
$begingroup$
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.
$endgroup$
$begingroup$
Hey, thank you very much for answering! But I don't understand the Einstein notation very well. Can you explain what the rhs means?
$endgroup$
– Mahathi Vempati
Dec 25 '18 at 12:46
$begingroup$
@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 $A_{ij} x_j$
$endgroup$
– Алексей Уваров
Dec 25 '18 at 13:27
$begingroup$
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.
$endgroup$
– Алексей Уваров
Dec 25 '18 at 13:33
1
$begingroup$
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: $rho = sum_{i,j,k,l} rho_{ijkl} (| irangle langle j | otimes | krangle langle l |)$.
$endgroup$
– Алексей Уваров
Dec 26 '18 at 8:18
1
$begingroup$
@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
$endgroup$
– Алексей Уваров
Dec 28 '18 at 8:53
|
show 2 more comments
$begingroup$
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.
$endgroup$
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.
edited Dec 28 '18 at 9:06
answered Dec 25 '18 at 12:42
Алексей УваровАлексей Уваров
1737
1737
$begingroup$
Hey, thank you very much for answering! But I don't understand the Einstein notation very well. Can you explain what the rhs means?
$endgroup$
– Mahathi Vempati
Dec 25 '18 at 12:46
$begingroup$
@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 $A_{ij} x_j$
$endgroup$
– Алексей Уваров
Dec 25 '18 at 13:27
$begingroup$
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.
$endgroup$
– Алексей Уваров
Dec 25 '18 at 13:33
1
$begingroup$
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: $rho = sum_{i,j,k,l} rho_{ijkl} (| irangle langle j | otimes | krangle langle l |)$.
$endgroup$
– Алексей Уваров
Dec 26 '18 at 8:18
1
$begingroup$
@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
$endgroup$
– Алексей Уваров
Dec 28 '18 at 8:53
|
show 2 more comments
$begingroup$
Hey, thank you very much for answering! But I don't understand the Einstein notation very well. Can you explain what the rhs means?
$endgroup$
– Mahathi Vempati
Dec 25 '18 at 12:46
$begingroup$
@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 $A_{ij} x_j$
$endgroup$
– Алексей Уваров
Dec 25 '18 at 13:27
$begingroup$
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.
$endgroup$
– Алексей Уваров
Dec 25 '18 at 13:33
1
$begingroup$
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: $rho = sum_{i,j,k,l} rho_{ijkl} (| irangle langle j | otimes | krangle langle l |)$.
$endgroup$
– Алексей Уваров
Dec 26 '18 at 8:18
1
$begingroup$
@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
$endgroup$
– Алексей Уваров
Dec 28 '18 at 8:53
$begingroup$
Hey, thank you very much for answering! But I don't understand the Einstein notation very well. Can you explain what the rhs means?
$endgroup$
– Mahathi Vempati
Dec 25 '18 at 12:46
$begingroup$
Hey, thank you very much for answering! But I don't understand the Einstein notation very well. Can you explain what the rhs means?
$endgroup$
– Mahathi Vempati
Dec 25 '18 at 12:46
$begingroup$
@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 $A_{ij} x_j$
$endgroup$
– Алексей Уваров
Dec 25 '18 at 13:27
$begingroup$
@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 $A_{ij} x_j$
$endgroup$
– Алексей Уваров
Dec 25 '18 at 13:27
$begingroup$
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.
$endgroup$
– Алексей Уваров
Dec 25 '18 at 13:33
$begingroup$
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.
$endgroup$
– Алексей Уваров
Dec 25 '18 at 13:33
1
1
$begingroup$
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: $rho = sum_{i,j,k,l} rho_{ijkl} (| irangle langle j | otimes | krangle langle l |)$.
$endgroup$
– Алексей Уваров
Dec 26 '18 at 8:18
$begingroup$
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: $rho = sum_{i,j,k,l} rho_{ijkl} (| irangle langle j | otimes | krangle langle l |)$.
$endgroup$
– Алексей Уваров
Dec 26 '18 at 8:18
1
1
$begingroup$
@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
$endgroup$
– Алексей Уваров
Dec 28 '18 at 8:53
$begingroup$
@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
$endgroup$
– Алексей Уваров
Dec 28 '18 at 8:53
|
show 2 more comments
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2
$begingroup$
How do you define $rho_a$ in terms of $rho_{ab}$?
$endgroup$
– DaftWullie
Dec 25 '18 at 8:34
$begingroup$
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.
$endgroup$
– Blue♦
Dec 25 '18 at 9:55
$begingroup$
@Blue, edited title, check now?
$endgroup$
– Mahathi Vempati
Dec 25 '18 at 11:05
$begingroup$
@DaftWullie, Edited the question
$endgroup$
– Mahathi Vempati
Dec 25 '18 at 11:06
1
$begingroup$
@MahathiVempati Thanks. Looks good!
$endgroup$
– Blue♦
Dec 25 '18 at 13:09