Mixing
Partial Trace property of an operator in a tensor space
$begingroup$
I was studying Alexander Holevo's book on Quantum Systems, Channels and Information, which is a small but terse introduction to Quantum Information Theory. The theory of Tensor products is used almost everywhere so I decided to study the same topic from different books. There is one property that I find hard to prove and I hope that someone could offer me some assistance in proving it.
N.B: As math stackexchange people may not be comfortable with the physics style notations (a.k.a Dirac's bra and ket notation), I'll use notations from John Conway Functional Analysis. If requested, I'll rephrase the question in Dirac's notation.
Let $mathcal{H}_1$ and $mathcal{H}_2$ be two finite dimensional Hilbert spaces (over $mathbb{C}$) with inner products $langle.rangle_1$ and $langle.rangle_2$ respectively.
Let ${e_i}$ and ${f_j}$ be bases for $mathcal{H}_1$ and $mathcal{H}_2$ respectively.
Let $T in mathcal{L}(mathcal{H}_1 otimesmathcal{H}_2)$, where $otimes$ represents the tensor product. Then for any operator $S in mathcal{L}(mathcal{H}_2)$, show that
$$Tr_{mathcal{H}_2}(T(I_{H_1}otimes S)) = Tr_{mathcal{H}_2}((I_{H_1}otimes S)T)$$
where $Tr_{H_2}(A)$ for any operator $A in mathcal{L}(mathcal{H}_1 otimesmathcal{H}_2)$, called the partial trace of $A$ w.r.t $mathcal{H}_2$, is defined as that operator on $mathcal{H}_1$ that satisfies the following (this is as per Holevo's book)
$$langlephi,Tr_{H_2}(A) psirangle_1 = sum_{j}langlephi otimes f_j,A (psi otimes f_j)rangle$$
for every $phi,psi in mathcal{H}_1$. The inner product on RHS is the inner product for $mathcal{H}_1 otimes mathcal{H}_2$ and is given by
$$langle phi_1 otimes phi_2, psi_1 otimes psi_2 rangle = langle phi_1, psi_1 rangle_1 langle phi_2, psi_2 rangle_2$$
for any $phi_i, psi_i in mathcal{H}_i;$ $i=1,2$.
It is somewhat similar to showing $Tr(AB)=Tr(BA)$ for the usual trace. However I had a lot of trouble showing this. Also I wanted to know if it suffices to prove the result for $phi = psi$ as I think I've managed some kind of proof for that part.
hilbert-spaces linear-transformations tensor-products
asked Mar 22 '17 at 9:35
Gautam ShenoyGautam Shenoy
7,15911645
$endgroup$
add a comment |
$begingroup$
I was studying Alexander Holevo's book on Quantum Systems, Channels and Information, which is a small but terse introduction to Quantum Information Theory. The theory of Tensor products is used almost everywhere so I decided to study the same topic from different books. There is one property that I find hard to prove and I hope that someone could offer me some assistance in proving it.
N.B: As math stackexchange people may not be comfortable with the physics style notations (a.k.a Dirac's bra and ket notation), I'll use notations from John Conway Functional Analysis. If requested, I'll rephrase the question in Dirac's notation.
Let $mathcal{H}_1$ and $mathcal{H}_2$ be two finite dimensional Hilbert spaces (over $mathbb{C}$) with inner products $langle.rangle_1$ and $langle.rangle_2$ respectively.
Let ${e_i}$ and ${f_j}$ be bases for $mathcal{H}_1$ and $mathcal{H}_2$ respectively.
Let $T in mathcal{L}(mathcal{H}_1 otimesmathcal{H}_2)$, where $otimes$ represents the tensor product. Then for any operator $S in mathcal{L}(mathcal{H}_2)$, show that
$$Tr_{mathcal{H}_2}(T(I_{H_1}otimes S)) = Tr_{mathcal{H}_2}((I_{H_1}otimes S)T)$$
where $Tr_{H_2}(A)$ for any operator $A in mathcal{L}(mathcal{H}_1 otimesmathcal{H}_2)$, called the partial trace of $A$ w.r.t $mathcal{H}_2$, is defined as that operator on $mathcal{H}_1$ that satisfies the following (this is as per Holevo's book)
$$langlephi,Tr_{H_2}(A) psirangle_1 = sum_{j}langlephi otimes f_j,A (psi otimes f_j)rangle$$
for every $phi,psi in mathcal{H}_1$. The inner product on RHS is the inner product for $mathcal{H}_1 otimes mathcal{H}_2$ and is given by
$$langle phi_1 otimes phi_2, psi_1 otimes psi_2 rangle = langle phi_1, psi_1 rangle_1 langle phi_2, psi_2 rangle_2$$
for any $phi_i, psi_i in mathcal{H}_i;$ $i=1,2$.
It is somewhat similar to showing $Tr(AB)=Tr(BA)$ for the usual trace. However I had a lot of trouble showing this. Also I wanted to know if it suffices to prove the result for $phi = psi$ as I think I've managed some kind of proof for that part.
hilbert-spaces linear-transformations tensor-products
asked Mar 22 '17 at 9:35
Gautam ShenoyGautam Shenoy
7,15911645
$endgroup$
$begingroup$
It would not suffice to prove it for $phi=psi$. It would suffice to prove it for $phi=e_i$ and $psi=e_j$.
$endgroup$
– luftbahnfahrer
Mar 22 '17 at 21:35
add a comment |
$begingroup$
I was studying Alexander Holevo's book on Quantum Systems, Channels and Information, which is a small but terse introduction to Quantum Information Theory. The theory of Tensor products is used almost everywhere so I decided to study the same topic from different books. There is one property that I find hard to prove and I hope that someone could offer me some assistance in proving it.
N.B: As math stackexchange people may not be comfortable with the physics style notations (a.k.a Dirac's bra and ket notation), I'll use notations from John Conway Functional Analysis. If requested, I'll rephrase the question in Dirac's notation.
Let $mathcal{H}_1$ and $mathcal{H}_2$ be two finite dimensional Hilbert spaces (over $mathbb{C}$) with inner products $langle.rangle_1$ and $langle.rangle_2$ respectively.
Let ${e_i}$ and ${f_j}$ be bases for $mathcal{H}_1$ and $mathcal{H}_2$ respectively.
Let $T in mathcal{L}(mathcal{H}_1 otimesmathcal{H}_2)$, where $otimes$ represents the tensor product. Then for any operator $S in mathcal{L}(mathcal{H}_2)$, show that
$$Tr_{mathcal{H}_2}(T(I_{H_1}otimes S)) = Tr_{mathcal{H}_2}((I_{H_1}otimes S)T)$$
where $Tr_{H_2}(A)$ for any operator $A in mathcal{L}(mathcal{H}_1 otimesmathcal{H}_2)$, called the partial trace of $A$ w.r.t $mathcal{H}_2$, is defined as that operator on $mathcal{H}_1$ that satisfies the following (this is as per Holevo's book)
$$langlephi,Tr_{H_2}(A) psirangle_1 = sum_{j}langlephi otimes f_j,A (psi otimes f_j)rangle$$
for every $phi,psi in mathcal{H}_1$. The inner product on RHS is the inner product for $mathcal{H}_1 otimes mathcal{H}_2$ and is given by
$$langle phi_1 otimes phi_2, psi_1 otimes psi_2 rangle = langle phi_1, psi_1 rangle_1 langle phi_2, psi_2 rangle_2$$
for any $phi_i, psi_i in mathcal{H}_i;$ $i=1,2$.
It is somewhat similar to showing $Tr(AB)=Tr(BA)$ for the usual trace. However I had a lot of trouble showing this. Also I wanted to know if it suffices to prove the result for $phi = psi$ as I think I've managed some kind of proof for that part.
hilbert-spaces linear-transformations tensor-products
asked Mar 22 '17 at 9:35
Gautam ShenoyGautam Shenoy
7,15911645
$endgroup$
I was studying Alexander Holevo's book on Quantum Systems, Channels and Information, which is a small but terse introduction to Quantum Information Theory. The theory of Tensor products is used almost everywhere so I decided to study the same topic from different books. There is one property that I find hard to prove and I hope that someone could offer me some assistance in proving it.
N.B: As math stackexchange people may not be comfortable with the physics style notations (a.k.a Dirac's bra and ket notation), I'll use notations from John Conway Functional Analysis. If requested, I'll rephrase the question in Dirac's notation.
Let $mathcal{H}_1$ and $mathcal{H}_2$ be two finite dimensional Hilbert spaces (over $mathbb{C}$) with inner products $langle.rangle_1$ and $langle.rangle_2$ respectively.
Let ${e_i}$ and ${f_j}$ be bases for $mathcal{H}_1$ and $mathcal{H}_2$ respectively.
Let $T in mathcal{L}(mathcal{H}_1 otimesmathcal{H}_2)$, where $otimes$ represents the tensor product. Then for any operator $S in mathcal{L}(mathcal{H}_2)$, show that
$$Tr_{mathcal{H}_2}(T(I_{H_1}otimes S)) = Tr_{mathcal{H}_2}((I_{H_1}otimes S)T)$$
where $Tr_{H_2}(A)$ for any operator $A in mathcal{L}(mathcal{H}_1 otimesmathcal{H}_2)$, called the partial trace of $A$ w.r.t $mathcal{H}_2$, is defined as that operator on $mathcal{H}_1$ that satisfies the following (this is as per Holevo's book)
$$langlephi,Tr_{H_2}(A) psirangle_1 = sum_{j}langlephi otimes f_j,A (psi otimes f_j)rangle$$
for every $phi,psi in mathcal{H}_1$. The inner product on RHS is the inner product for $mathcal{H}_1 otimes mathcal{H}_2$ and is given by
$$langle phi_1 otimes phi_2, psi_1 otimes psi_2 rangle = langle phi_1, psi_1 rangle_1 langle phi_2, psi_2 rangle_2$$
for any $phi_i, psi_i in mathcal{H}_i;$ $i=1,2$.
It is somewhat similar to showing $Tr(AB)=Tr(BA)$ for the usual trace. However I had a lot of trouble showing this. Also I wanted to know if it suffices to prove the result for $phi = psi$ as I think I've managed some kind of proof for that part.
hilbert-spaces linear-transformations tensor-products
hilbert-spaces linear-transformations tensor-products
asked Mar 22 '17 at 9:35
Gautam ShenoyGautam Shenoy
7,15911645
asked Mar 22 '17 at 9:35
Gautam ShenoyGautam Shenoy
7,15911645
asked Mar 22 '17 at 9:35
Gautam ShenoyGautam Shenoy
7,15911645
asked Mar 22 '17 at 9:35
Gautam ShenoyGautam Shenoy
7,15911645
asked Mar 22 '17 at 9:35
Gautam ShenoyGautam Shenoy
7,15911645
7,15911645
$begingroup$
It would not suffice to prove it for $phi=psi$. It would suffice to prove it for $phi=e_i$ and $psi=e_j$.
$endgroup$
– luftbahnfahrer
Mar 22 '17 at 21:35
add a comment |
$begingroup$
It would not suffice to prove it for $phi=psi$. It would suffice to prove it for $phi=e_i$ and $psi=e_j$.
$endgroup$
– luftbahnfahrer
Mar 22 '17 at 21:35
$begingroup$
It would not suffice to prove it for $phi=psi$. It would suffice to prove it for $phi=e_i$ and $psi=e_j$.
$endgroup$
– luftbahnfahrer
Mar 22 '17 at 21:35
$begingroup$
It would not suffice to prove it for $phi=psi$. It would suffice to prove it for $phi=e_i$ and $psi=e_j$.
$endgroup$
– luftbahnfahrer
Mar 22 '17 at 21:35
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
I don't know how to prove it using your notation, but I will introduce some notation to prove this. The Hilbert-Schmidt inner product on the space $mathcal{L}(mathcal{H})$ of linear operators on a finite-dimensional Hilbert space $mathcal{H}$ is defined as
$$
langle A,Brangle = operatorname{Tr}(AB^*)
$$
for all $A,Binmathcal{L}(mathcal{H})$, where ${}^*$ denotes the adjoint operator (i.e. the conjugate transpose, or ${}^dagger$ in quantum mechanics notation). Note that
$$
A=B
quadLongleftrightarrowquad
langle A,Xrangle = langle B,Xrangle quad text{for all }Xinmathcal{L}(mathcal{H})
$$
for operators $A,Binmathcal{L}(mathcal{H})$. With this inner product, note that
$$
langle XY,Zrangle = langle X,ZY^*rangle = langle Y,X^*Zrangle
$$
for any $X,Y,Zinmathcal{L}(mathcal{H})$
Let $Ainmathcal{L}(mathcal{H}_1otimesmathcal{H}_2)$. The partial trace $operatorname{Tr}_{mathcal{H}_2}$ can then be defined as the unique operator such that
$$
langle operatorname{Tr}_{mathcal{H}_2}(A), Xrangle = langle A,Xotimes I_{mathcal{H}_2}rangle
quad text{for all }Xinmathcal{L}(mathcal{H_1}).
$$
We can then prove the following proposition. It relies on the fact that $Xotimes I_{mathcal{H}_2}$ commutes with $I_{mathcal{H_1}}otimes S^*$.
Proposition. Let $Tinmathcal{L}(mathcal{H}_1otimesmathcal{H}_2)$ and $Sinmathcal{L}(mathcal{H_2})$. Then $operatorname{Tr}_{mathcal{H_2}}(T(I_{mathcal{H}_1}otimes S)) = operatorname{Tr}_{mathcal{H_2}}((I_{mathcal{H}_1}otimes S)T)$.
Proof.
Let $Xinmathcal{L}(mathcal{H_1})$. Then
begin{align*}
langle operatorname{Tr}_{mathcal{H_2}}(T(I_{mathcal{H}_1}otimes S)) , Xrangle
& = langle T(I_{mathcal{H}_1}otimes S), Xotimes I_{mathcal{H_2}}rangle\
& = langle T, (Xotimes I_{mathcal{H_2}})(I_{mathcal{H}_1}otimes S)^*rangle\
%& = langle T, Xotimes S^*rangle\
& = langle T, (I_{mathcal{H}_1}otimes S)^*(Xotimes I_{mathcal{H_2}})rangle\
& = langle (I_{mathcal{H}_1}otimes S)T,Xotimes I_{mathcal{H_2}}rangle\
& = langle operatorname{Tr}_{mathcal{H_2}}((I_{mathcal{H}_1}otimes S)T) , Xrangle .
end{align*}
Since this holds for all $X$, the result follows.
answered Mar 22 '17 at 18:46
luftbahnfahrerluftbahnfahrer
657314
$endgroup$
$begingroup$
I believe the key statement is "$X otimes I_{mathcal{H}_2}$" commutes with... This answer is useful, so let me try again.
$endgroup$
– Gautam Shenoy
Mar 23 '17 at 5:22
$begingroup$
Kindly go over my answer. Open for comments.
$endgroup$
– Gautam Shenoy
Mar 23 '17 at 6:16
add a comment |
$begingroup$
I think I was able to solve it. This assumes that any operator $T in mathcal{L}(mathcal{H}_1 otimes mathcal{H}_2)$ can be written in terms of basis operators
$$T = sum_{i,j,k,l} alpha_{i,j,k,l} E_{ij}otimes F_{kl}$$
where (got this from wikipedia article on Partial Trace) given the basis as in the question, $E_{i,j} in mathcal{L}(mathcal{H}_1)$ and $F_{k,l}in mathcal{L}(mathcal{H}_2)$, such that
$$E_{ij}: e_{i} mapsto e_j$$ and
$$F_{kl}: f_{k} mapsto f_l $$
the other basis vectors being mapped to $0$.
We also need to use that (I've proved this before)
$$Tr_{mathcal{H}_2}(Aotimes B) = A. Tr(B)$$
where $A in mathcal{L}(mathcal{H}_1)$ and $Bin mathcal{L}(mathcal{H}_2)$.
Now it's easy. Just observe
$$Tr_{mathcal{H}_2}(T(I_{mathcal{H}_2}otimes S)) = Tr_{mathcal{H}_2}(sum_{i,j,k,l} alpha_{i,j,k,l} (E_{ij}otimes F_{kl}) (I_{mathcal{H}_2}otimes S)) \
= sum_{i,j,k,l} alpha_{i,j,k,l} Tr_{mathcal{H}_2}(E_{ij}otimes F_{kl} S)\
= sum_{i,j,k,l} alpha_{i,j,k,l} E_{ij}.Tr(F_{kl}S)$$
Similarly we get
$$Tr_{mathcal{H}_2}((I_{mathcal{H}_2}otimes S)T) = sum_{i,j,k,l} alpha_{i,j,k,l} E_{ij}.Tr(SF_{kl})$$
But $Tr(F_{kl}S) = Tr(SF_{kl})$. Hence proved.
Request everyone to check the steps.
Update: Found a simpler proof. This doesn't require the decomposition of operators (except identity) and uses the observation that $|aotimes brangle langle c otimes d| = |aranglelangle c| otimes |branglelangle d|$.
Let $e_k$ and $f_j$ be orthonormal basis for $mathcal{H}_1$ and $mathcal{H}_2$ respectively. Then we have,
begin{eqnarray}
&&sum_k langlephi otimes f_k~|(I_{H_1} otimes S)T|~psi otimes f_krangle \
&=& sum_k langlephi otimes f_k~|(I_{H_1} otimes S)(I_{H_1}otimes I_{H_2})T|~psi otimes f_krangle \
&=& sum_{k,i,j} langlephi otimes f_k~|(I_{H_1} otimes S)(|e_iranglelangle e_i| otimes |f_jranglelangle f_j|)T|~psi otimes f_krangle \
&=& sum_{k,i,j} langlephi otimes f_k~|(I_{H_1} otimes S)(|e_iotimes f_jrangle langle e_i otimes f_j |)T|~psi otimes f_krangle \
&=& sum_{k,i,j} langlephi otimes f_k~||e_iotimes Sf_jrangle langle e_i otimes f_j |T|~psi otimes f_krangle \
&=& sum_{k,i,j} langlephi |e_irangle langle f_k| Sf_jrangle langle e_i otimes f_j| T|~psi otimes f_krangle \
&=& sum_{k,i,j} langlephi |e_irangle langle e_i otimes f_j| T|~psi otimes f_krangle langle f_k| Sf_jrangle\
&=& sum_{k,i,j} langlephiotimes f_j |e_iotimes f_jrangle langle e_i otimes f_j| T|~psi otimes f_krangle left(frac{langle psi otimes f_k||psi otimes Sf_jrangle}{|psi|^2}right)\
&=& sum_{k,i,j} langlephiotimes f_j|(|e_iranglelangle e_i| otimes |f_jrangle langle f_j|) T~(|psiranglelanglepsi|otimes |f_kranglelangle f_k|)left(frac{|psi otimes Sf_jrangle}{|psi|^2}right) \
&=& sum_{j} langlephiotimes f_j|(I_{H_1} otimes |f_jrangle langle f_j|) T~(|psiranglelanglepsi|otimes I_{H_2})left(frac{(I_{H_1}otimes S)|psi otimes f_jrangle}{|psi|^2}right) \
&stackrel{(a)}{=}& sum_{j} langlephiotimes f_j| T (I_{H_1}otimes S)left(frac{(|psiranglelanglepsi|otimes I_{H_2})|psi otimes f_jrangle}{|psi|^2}right) \
&=& sum_{j} langlephiotimes f_j| T (I_{H_1}otimes S)|psi otimes f_jrangle \
end{eqnarray}
where $(a)$ follows from,
$$langle phiotimes f_j|(I_{H_1} otimes |f_jrangle langle f_j|) = langle(I_{H_1} otimes |f_jrangle langle f_j|)^*(phiotimes f_j)| = langle phiotimes f_j|$$
where we used the self adjointness of the operator. Also $langle f_j|f_k rangle = delta_{j,k}$. $blacksquare$
answered Mar 23 '17 at 6:15
Gautam ShenoyGautam Shenoy
7,15911645
$endgroup$
$begingroup$
Your proof looks good too.
$endgroup$
– luftbahnfahrer
Mar 23 '17 at 20:14
add a comment |
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2 Answers
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2 Answers
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$begingroup$
I don't know how to prove it using your notation, but I will introduce some notation to prove this. The Hilbert-Schmidt inner product on the space $mathcal{L}(mathcal{H})$ of linear operators on a finite-dimensional Hilbert space $mathcal{H}$ is defined as
$$
langle A,Brangle = operatorname{Tr}(AB^*)
$$
for all $A,Binmathcal{L}(mathcal{H})$, where ${}^*$ denotes the adjoint operator (i.e. the conjugate transpose, or ${}^dagger$ in quantum mechanics notation). Note that
$$
A=B
quadLongleftrightarrowquad
langle A,Xrangle = langle B,Xrangle quad text{for all }Xinmathcal{L}(mathcal{H})
$$
for operators $A,Binmathcal{L}(mathcal{H})$. With this inner product, note that
$$
langle XY,Zrangle = langle X,ZY^*rangle = langle Y,X^*Zrangle
$$
for any $X,Y,Zinmathcal{L}(mathcal{H})$
Let $Ainmathcal{L}(mathcal{H}_1otimesmathcal{H}_2)$. The partial trace $operatorname{Tr}_{mathcal{H}_2}$ can then be defined as the unique operator such that
$$
langle operatorname{Tr}_{mathcal{H}_2}(A), Xrangle = langle A,Xotimes I_{mathcal{H}_2}rangle
quad text{for all }Xinmathcal{L}(mathcal{H_1}).
$$
We can then prove the following proposition. It relies on the fact that $Xotimes I_{mathcal{H}_2}$ commutes with $I_{mathcal{H_1}}otimes S^*$.
Proposition. Let $Tinmathcal{L}(mathcal{H}_1otimesmathcal{H}_2)$ and $Sinmathcal{L}(mathcal{H_2})$. Then $operatorname{Tr}_{mathcal{H_2}}(T(I_{mathcal{H}_1}otimes S)) = operatorname{Tr}_{mathcal{H_2}}((I_{mathcal{H}_1}otimes S)T)$.
Proof.
Let $Xinmathcal{L}(mathcal{H_1})$. Then
begin{align*}
langle operatorname{Tr}_{mathcal{H_2}}(T(I_{mathcal{H}_1}otimes S)) , Xrangle
& = langle T(I_{mathcal{H}_1}otimes S), Xotimes I_{mathcal{H_2}}rangle\
& = langle T, (Xotimes I_{mathcal{H_2}})(I_{mathcal{H}_1}otimes S)^*rangle\
%& = langle T, Xotimes S^*rangle\
& = langle T, (I_{mathcal{H}_1}otimes S)^*(Xotimes I_{mathcal{H_2}})rangle\
& = langle (I_{mathcal{H}_1}otimes S)T,Xotimes I_{mathcal{H_2}}rangle\
& = langle operatorname{Tr}_{mathcal{H_2}}((I_{mathcal{H}_1}otimes S)T) , Xrangle .
end{align*}
Since this holds for all $X$, the result follows.
answered Mar 22 '17 at 18:46
luftbahnfahrerluftbahnfahrer
657314
$endgroup$
$begingroup$
I believe the key statement is "$X otimes I_{mathcal{H}_2}$" commutes with... This answer is useful, so let me try again.
$endgroup$
– Gautam Shenoy
Mar 23 '17 at 5:22
$begingroup$
Kindly go over my answer. Open for comments.
$endgroup$
– Gautam Shenoy
Mar 23 '17 at 6:16
add a comment |
$begingroup$
I don't know how to prove it using your notation, but I will introduce some notation to prove this. The Hilbert-Schmidt inner product on the space $mathcal{L}(mathcal{H})$ of linear operators on a finite-dimensional Hilbert space $mathcal{H}$ is defined as
$$
langle A,Brangle = operatorname{Tr}(AB^*)
$$
for all $A,Binmathcal{L}(mathcal{H})$, where ${}^*$ denotes the adjoint operator (i.e. the conjugate transpose, or ${}^dagger$ in quantum mechanics notation). Note that
$$
A=B
quadLongleftrightarrowquad
langle A,Xrangle = langle B,Xrangle quad text{for all }Xinmathcal{L}(mathcal{H})
$$
for operators $A,Binmathcal{L}(mathcal{H})$. With this inner product, note that
$$
langle XY,Zrangle = langle X,ZY^*rangle = langle Y,X^*Zrangle
$$
for any $X,Y,Zinmathcal{L}(mathcal{H})$
Let $Ainmathcal{L}(mathcal{H}_1otimesmathcal{H}_2)$. The partial trace $operatorname{Tr}_{mathcal{H}_2}$ can then be defined as the unique operator such that
$$
langle operatorname{Tr}_{mathcal{H}_2}(A), Xrangle = langle A,Xotimes I_{mathcal{H}_2}rangle
quad text{for all }Xinmathcal{L}(mathcal{H_1}).
$$
We can then prove the following proposition. It relies on the fact that $Xotimes I_{mathcal{H}_2}$ commutes with $I_{mathcal{H_1}}otimes S^*$.
Proposition. Let $Tinmathcal{L}(mathcal{H}_1otimesmathcal{H}_2)$ and $Sinmathcal{L}(mathcal{H_2})$. Then $operatorname{Tr}_{mathcal{H_2}}(T(I_{mathcal{H}_1}otimes S)) = operatorname{Tr}_{mathcal{H_2}}((I_{mathcal{H}_1}otimes S)T)$.
Proof.
Let $Xinmathcal{L}(mathcal{H_1})$. Then
begin{align*}
langle operatorname{Tr}_{mathcal{H_2}}(T(I_{mathcal{H}_1}otimes S)) , Xrangle
& = langle T(I_{mathcal{H}_1}otimes S), Xotimes I_{mathcal{H_2}}rangle\
& = langle T, (Xotimes I_{mathcal{H_2}})(I_{mathcal{H}_1}otimes S)^*rangle\
%& = langle T, Xotimes S^*rangle\
& = langle T, (I_{mathcal{H}_1}otimes S)^*(Xotimes I_{mathcal{H_2}})rangle\
& = langle (I_{mathcal{H}_1}otimes S)T,Xotimes I_{mathcal{H_2}}rangle\
& = langle operatorname{Tr}_{mathcal{H_2}}((I_{mathcal{H}_1}otimes S)T) , Xrangle .
end{align*}
Since this holds for all $X$, the result follows.
answered Mar 22 '17 at 18:46
luftbahnfahrerluftbahnfahrer
657314
$endgroup$
$begingroup$
I believe the key statement is "$X otimes I_{mathcal{H}_2}$" commutes with... This answer is useful, so let me try again.
$endgroup$
– Gautam Shenoy
Mar 23 '17 at 5:22
$begingroup$
Kindly go over my answer. Open for comments.
$endgroup$
– Gautam Shenoy
Mar 23 '17 at 6:16
add a comment |
$begingroup$
I don't know how to prove it using your notation, but I will introduce some notation to prove this. The Hilbert-Schmidt inner product on the space $mathcal{L}(mathcal{H})$ of linear operators on a finite-dimensional Hilbert space $mathcal{H}$ is defined as
$$
langle A,Brangle = operatorname{Tr}(AB^*)
$$
for all $A,Binmathcal{L}(mathcal{H})$, where ${}^*$ denotes the adjoint operator (i.e. the conjugate transpose, or ${}^dagger$ in quantum mechanics notation). Note that
$$
A=B
quadLongleftrightarrowquad
langle A,Xrangle = langle B,Xrangle quad text{for all }Xinmathcal{L}(mathcal{H})
$$
for operators $A,Binmathcal{L}(mathcal{H})$. With this inner product, note that
$$
langle XY,Zrangle = langle X,ZY^*rangle = langle Y,X^*Zrangle
$$
for any $X,Y,Zinmathcal{L}(mathcal{H})$
Let $Ainmathcal{L}(mathcal{H}_1otimesmathcal{H}_2)$. The partial trace $operatorname{Tr}_{mathcal{H}_2}$ can then be defined as the unique operator such that
$$
langle operatorname{Tr}_{mathcal{H}_2}(A), Xrangle = langle A,Xotimes I_{mathcal{H}_2}rangle
quad text{for all }Xinmathcal{L}(mathcal{H_1}).
$$
We can then prove the following proposition. It relies on the fact that $Xotimes I_{mathcal{H}_2}$ commutes with $I_{mathcal{H_1}}otimes S^*$.
Proposition. Let $Tinmathcal{L}(mathcal{H}_1otimesmathcal{H}_2)$ and $Sinmathcal{L}(mathcal{H_2})$. Then $operatorname{Tr}_{mathcal{H_2}}(T(I_{mathcal{H}_1}otimes S)) = operatorname{Tr}_{mathcal{H_2}}((I_{mathcal{H}_1}otimes S)T)$.
Proof.
Let $Xinmathcal{L}(mathcal{H_1})$. Then
begin{align*}
langle operatorname{Tr}_{mathcal{H_2}}(T(I_{mathcal{H}_1}otimes S)) , Xrangle
& = langle T(I_{mathcal{H}_1}otimes S), Xotimes I_{mathcal{H_2}}rangle\
& = langle T, (Xotimes I_{mathcal{H_2}})(I_{mathcal{H}_1}otimes S)^*rangle\
%& = langle T, Xotimes S^*rangle\
& = langle T, (I_{mathcal{H}_1}otimes S)^*(Xotimes I_{mathcal{H_2}})rangle\
& = langle (I_{mathcal{H}_1}otimes S)T,Xotimes I_{mathcal{H_2}}rangle\
& = langle operatorname{Tr}_{mathcal{H_2}}((I_{mathcal{H}_1}otimes S)T) , Xrangle .
end{align*}
Since this holds for all $X$, the result follows.
answered Mar 22 '17 at 18:46
luftbahnfahrerluftbahnfahrer
657314
$endgroup$
I don't know how to prove it using your notation, but I will introduce some notation to prove this. The Hilbert-Schmidt inner product on the space $mathcal{L}(mathcal{H})$ of linear operators on a finite-dimensional Hilbert space $mathcal{H}$ is defined as
$$
langle A,Brangle = operatorname{Tr}(AB^*)
$$
for all $A,Binmathcal{L}(mathcal{H})$, where ${}^*$ denotes the adjoint operator (i.e. the conjugate transpose, or ${}^dagger$ in quantum mechanics notation). Note that
$$
A=B
quadLongleftrightarrowquad
langle A,Xrangle = langle B,Xrangle quad text{for all }Xinmathcal{L}(mathcal{H})
$$
for operators $A,Binmathcal{L}(mathcal{H})$. With this inner product, note that
$$
langle XY,Zrangle = langle X,ZY^*rangle = langle Y,X^*Zrangle
$$
for any $X,Y,Zinmathcal{L}(mathcal{H})$
Let $Ainmathcal{L}(mathcal{H}_1otimesmathcal{H}_2)$. The partial trace $operatorname{Tr}_{mathcal{H}_2}$ can then be defined as the unique operator such that
$$
langle operatorname{Tr}_{mathcal{H}_2}(A), Xrangle = langle A,Xotimes I_{mathcal{H}_2}rangle
quad text{for all }Xinmathcal{L}(mathcal{H_1}).
$$
We can then prove the following proposition. It relies on the fact that $Xotimes I_{mathcal{H}_2}$ commutes with $I_{mathcal{H_1}}otimes S^*$.
Proposition. Let $Tinmathcal{L}(mathcal{H}_1otimesmathcal{H}_2)$ and $Sinmathcal{L}(mathcal{H_2})$. Then $operatorname{Tr}_{mathcal{H_2}}(T(I_{mathcal{H}_1}otimes S)) = operatorname{Tr}_{mathcal{H_2}}((I_{mathcal{H}_1}otimes S)T)$.
Proof.
Let $Xinmathcal{L}(mathcal{H_1})$. Then
begin{align*}
langle operatorname{Tr}_{mathcal{H_2}}(T(I_{mathcal{H}_1}otimes S)) , Xrangle
& = langle T(I_{mathcal{H}_1}otimes S), Xotimes I_{mathcal{H_2}}rangle\
& = langle T, (Xotimes I_{mathcal{H_2}})(I_{mathcal{H}_1}otimes S)^*rangle\
%& = langle T, Xotimes S^*rangle\
& = langle T, (I_{mathcal{H}_1}otimes S)^*(Xotimes I_{mathcal{H_2}})rangle\
& = langle (I_{mathcal{H}_1}otimes S)T,Xotimes I_{mathcal{H_2}}rangle\
& = langle operatorname{Tr}_{mathcal{H_2}}((I_{mathcal{H}_1}otimes S)T) , Xrangle .
end{align*}
Since this holds for all $X$, the result follows.
answered Mar 22 '17 at 18:46
luftbahnfahrerluftbahnfahrer
657314
edited Mar 22 '17 at 21:31
answered Mar 22 '17 at 18:46
luftbahnfahrerluftbahnfahrer
657314
answered Mar 22 '17 at 18:46
luftbahnfahrerluftbahnfahrer
657314
answered Mar 22 '17 at 18:46
luftbahnfahrerluftbahnfahrer
657314
657314
$begingroup$
I believe the key statement is "$X otimes I_{mathcal{H}_2}$" commutes with... This answer is useful, so let me try again.
$endgroup$
– Gautam Shenoy
Mar 23 '17 at 5:22
$begingroup$
Kindly go over my answer. Open for comments.
$endgroup$
– Gautam Shenoy
Mar 23 '17 at 6:16
add a comment |
$begingroup$
I believe the key statement is "$X otimes I_{mathcal{H}_2}$" commutes with... This answer is useful, so let me try again.
$endgroup$
– Gautam Shenoy
Mar 23 '17 at 5:22
$begingroup$
Kindly go over my answer. Open for comments.
$endgroup$
– Gautam Shenoy
Mar 23 '17 at 6:16
$begingroup$
I believe the key statement is "$X otimes I_{mathcal{H}_2}$" commutes with... This answer is useful, so let me try again.
$endgroup$
– Gautam Shenoy
Mar 23 '17 at 5:22
$begingroup$
I believe the key statement is "$X otimes I_{mathcal{H}_2}$" commutes with... This answer is useful, so let me try again.
$endgroup$
– Gautam Shenoy
Mar 23 '17 at 5:22
$begingroup$
Kindly go over my answer. Open for comments.
$endgroup$
– Gautam Shenoy
Mar 23 '17 at 6:16
$begingroup$
Kindly go over my answer. Open for comments.
$endgroup$
– Gautam Shenoy
Mar 23 '17 at 6:16
add a comment |
$begingroup$
I think I was able to solve it. This assumes that any operator $T in mathcal{L}(mathcal{H}_1 otimes mathcal{H}_2)$ can be written in terms of basis operators
$$T = sum_{i,j,k,l} alpha_{i,j,k,l} E_{ij}otimes F_{kl}$$
where (got this from wikipedia article on Partial Trace) given the basis as in the question, $E_{i,j} in mathcal{L}(mathcal{H}_1)$ and $F_{k,l}in mathcal{L}(mathcal{H}_2)$, such that
$$E_{ij}: e_{i} mapsto e_j$$ and
$$F_{kl}: f_{k} mapsto f_l $$
the other basis vectors being mapped to $0$.
We also need to use that (I've proved this before)
$$Tr_{mathcal{H}_2}(Aotimes B) = A. Tr(B)$$
where $A in mathcal{L}(mathcal{H}_1)$ and $Bin mathcal{L}(mathcal{H}_2)$.
Now it's easy. Just observe
$$Tr_{mathcal{H}_2}(T(I_{mathcal{H}_2}otimes S)) = Tr_{mathcal{H}_2}(sum_{i,j,k,l} alpha_{i,j,k,l} (E_{ij}otimes F_{kl}) (I_{mathcal{H}_2}otimes S)) \
= sum_{i,j,k,l} alpha_{i,j,k,l} Tr_{mathcal{H}_2}(E_{ij}otimes F_{kl} S)\
= sum_{i,j,k,l} alpha_{i,j,k,l} E_{ij}.Tr(F_{kl}S)$$
Similarly we get
$$Tr_{mathcal{H}_2}((I_{mathcal{H}_2}otimes S)T) = sum_{i,j,k,l} alpha_{i,j,k,l} E_{ij}.Tr(SF_{kl})$$
But $Tr(F_{kl}S) = Tr(SF_{kl})$. Hence proved.
Request everyone to check the steps.
Update: Found a simpler proof. This doesn't require the decomposition of operators (except identity) and uses the observation that $|aotimes brangle langle c otimes d| = |aranglelangle c| otimes |branglelangle d|$.
Let $e_k$ and $f_j$ be orthonormal basis for $mathcal{H}_1$ and $mathcal{H}_2$ respectively. Then we have,
begin{eqnarray}
&&sum_k langlephi otimes f_k~|(I_{H_1} otimes S)T|~psi otimes f_krangle \
&=& sum_k langlephi otimes f_k~|(I_{H_1} otimes S)(I_{H_1}otimes I_{H_2})T|~psi otimes f_krangle \
&=& sum_{k,i,j} langlephi otimes f_k~|(I_{H_1} otimes S)(|e_iranglelangle e_i| otimes |f_jranglelangle f_j|)T|~psi otimes f_krangle \
&=& sum_{k,i,j} langlephi otimes f_k~|(I_{H_1} otimes S)(|e_iotimes f_jrangle langle e_i otimes f_j |)T|~psi otimes f_krangle \
&=& sum_{k,i,j} langlephi otimes f_k~||e_iotimes Sf_jrangle langle e_i otimes f_j |T|~psi otimes f_krangle \
&=& sum_{k,i,j} langlephi |e_irangle langle f_k| Sf_jrangle langle e_i otimes f_j| T|~psi otimes f_krangle \
&=& sum_{k,i,j} langlephi |e_irangle langle e_i otimes f_j| T|~psi otimes f_krangle langle f_k| Sf_jrangle\
&=& sum_{k,i,j} langlephiotimes f_j |e_iotimes f_jrangle langle e_i otimes f_j| T|~psi otimes f_krangle left(frac{langle psi otimes f_k||psi otimes Sf_jrangle}{|psi|^2}right)\
&=& sum_{k,i,j} langlephiotimes f_j|(|e_iranglelangle e_i| otimes |f_jrangle langle f_j|) T~(|psiranglelanglepsi|otimes |f_kranglelangle f_k|)left(frac{|psi otimes Sf_jrangle}{|psi|^2}right) \
&=& sum_{j} langlephiotimes f_j|(I_{H_1} otimes |f_jrangle langle f_j|) T~(|psiranglelanglepsi|otimes I_{H_2})left(frac{(I_{H_1}otimes S)|psi otimes f_jrangle}{|psi|^2}right) \
&stackrel{(a)}{=}& sum_{j} langlephiotimes f_j| T (I_{H_1}otimes S)left(frac{(|psiranglelanglepsi|otimes I_{H_2})|psi otimes f_jrangle}{|psi|^2}right) \
&=& sum_{j} langlephiotimes f_j| T (I_{H_1}otimes S)|psi otimes f_jrangle \
end{eqnarray}
where $(a)$ follows from,
$$langle phiotimes f_j|(I_{H_1} otimes |f_jrangle langle f_j|) = langle(I_{H_1} otimes |f_jrangle langle f_j|)^*(phiotimes f_j)| = langle phiotimes f_j|$$
where we used the self adjointness of the operator. Also $langle f_j|f_k rangle = delta_{j,k}$. $blacksquare$
answered Mar 23 '17 at 6:15
Gautam ShenoyGautam Shenoy
7,15911645
$endgroup$
$begingroup$
Your proof looks good too.
$endgroup$
– luftbahnfahrer
Mar 23 '17 at 20:14
add a comment |
$begingroup$
I think I was able to solve it. This assumes that any operator $T in mathcal{L}(mathcal{H}_1 otimes mathcal{H}_2)$ can be written in terms of basis operators
$$T = sum_{i,j,k,l} alpha_{i,j,k,l} E_{ij}otimes F_{kl}$$
where (got this from wikipedia article on Partial Trace) given the basis as in the question, $E_{i,j} in mathcal{L}(mathcal{H}_1)$ and $F_{k,l}in mathcal{L}(mathcal{H}_2)$, such that
$$E_{ij}: e_{i} mapsto e_j$$ and
$$F_{kl}: f_{k} mapsto f_l $$
the other basis vectors being mapped to $0$.
We also need to use that (I've proved this before)
$$Tr_{mathcal{H}_2}(Aotimes B) = A. Tr(B)$$
where $A in mathcal{L}(mathcal{H}_1)$ and $Bin mathcal{L}(mathcal{H}_2)$.
Now it's easy. Just observe
$$Tr_{mathcal{H}_2}(T(I_{mathcal{H}_2}otimes S)) = Tr_{mathcal{H}_2}(sum_{i,j,k,l} alpha_{i,j,k,l} (E_{ij}otimes F_{kl}) (I_{mathcal{H}_2}otimes S)) \
= sum_{i,j,k,l} alpha_{i,j,k,l} Tr_{mathcal{H}_2}(E_{ij}otimes F_{kl} S)\
= sum_{i,j,k,l} alpha_{i,j,k,l} E_{ij}.Tr(F_{kl}S)$$
Similarly we get
$$Tr_{mathcal{H}_2}((I_{mathcal{H}_2}otimes S)T) = sum_{i,j,k,l} alpha_{i,j,k,l} E_{ij}.Tr(SF_{kl})$$
But $Tr(F_{kl}S) = Tr(SF_{kl})$. Hence proved.
Request everyone to check the steps.
Update: Found a simpler proof. This doesn't require the decomposition of operators (except identity) and uses the observation that $|aotimes brangle langle c otimes d| = |aranglelangle c| otimes |branglelangle d|$.
Let $e_k$ and $f_j$ be orthonormal basis for $mathcal{H}_1$ and $mathcal{H}_2$ respectively. Then we have,
begin{eqnarray}
&&sum_k langlephi otimes f_k~|(I_{H_1} otimes S)T|~psi otimes f_krangle \
&=& sum_k langlephi otimes f_k~|(I_{H_1} otimes S)(I_{H_1}otimes I_{H_2})T|~psi otimes f_krangle \
&=& sum_{k,i,j} langlephi otimes f_k~|(I_{H_1} otimes S)(|e_iranglelangle e_i| otimes |f_jranglelangle f_j|)T|~psi otimes f_krangle \
&=& sum_{k,i,j} langlephi otimes f_k~|(I_{H_1} otimes S)(|e_iotimes f_jrangle langle e_i otimes f_j |)T|~psi otimes f_krangle \
&=& sum_{k,i,j} langlephi otimes f_k~||e_iotimes Sf_jrangle langle e_i otimes f_j |T|~psi otimes f_krangle \
&=& sum_{k,i,j} langlephi |e_irangle langle f_k| Sf_jrangle langle e_i otimes f_j| T|~psi otimes f_krangle \
&=& sum_{k,i,j} langlephi |e_irangle langle e_i otimes f_j| T|~psi otimes f_krangle langle f_k| Sf_jrangle\
&=& sum_{k,i,j} langlephiotimes f_j |e_iotimes f_jrangle langle e_i otimes f_j| T|~psi otimes f_krangle left(frac{langle psi otimes f_k||psi otimes Sf_jrangle}{|psi|^2}right)\
&=& sum_{k,i,j} langlephiotimes f_j|(|e_iranglelangle e_i| otimes |f_jrangle langle f_j|) T~(|psiranglelanglepsi|otimes |f_kranglelangle f_k|)left(frac{|psi otimes Sf_jrangle}{|psi|^2}right) \
&=& sum_{j} langlephiotimes f_j|(I_{H_1} otimes |f_jrangle langle f_j|) T~(|psiranglelanglepsi|otimes I_{H_2})left(frac{(I_{H_1}otimes S)|psi otimes f_jrangle}{|psi|^2}right) \
&stackrel{(a)}{=}& sum_{j} langlephiotimes f_j| T (I_{H_1}otimes S)left(frac{(|psiranglelanglepsi|otimes I_{H_2})|psi otimes f_jrangle}{|psi|^2}right) \
&=& sum_{j} langlephiotimes f_j| T (I_{H_1}otimes S)|psi otimes f_jrangle \
end{eqnarray}
where $(a)$ follows from,
$$langle phiotimes f_j|(I_{H_1} otimes |f_jrangle langle f_j|) = langle(I_{H_1} otimes |f_jrangle langle f_j|)^*(phiotimes f_j)| = langle phiotimes f_j|$$
where we used the self adjointness of the operator. Also $langle f_j|f_k rangle = delta_{j,k}$. $blacksquare$
answered Mar 23 '17 at 6:15
Gautam ShenoyGautam Shenoy
7,15911645
$endgroup$
$begingroup$
Your proof looks good too.
$endgroup$
– luftbahnfahrer
Mar 23 '17 at 20:14
add a comment |
$begingroup$
I think I was able to solve it. This assumes that any operator $T in mathcal{L}(mathcal{H}_1 otimes mathcal{H}_2)$ can be written in terms of basis operators
$$T = sum_{i,j,k,l} alpha_{i,j,k,l} E_{ij}otimes F_{kl}$$
where (got this from wikipedia article on Partial Trace) given the basis as in the question, $E_{i,j} in mathcal{L}(mathcal{H}_1)$ and $F_{k,l}in mathcal{L}(mathcal{H}_2)$, such that
$$E_{ij}: e_{i} mapsto e_j$$ and
$$F_{kl}: f_{k} mapsto f_l $$
the other basis vectors being mapped to $0$.
We also need to use that (I've proved this before)
$$Tr_{mathcal{H}_2}(Aotimes B) = A. Tr(B)$$
where $A in mathcal{L}(mathcal{H}_1)$ and $Bin mathcal{L}(mathcal{H}_2)$.
Now it's easy. Just observe
$$Tr_{mathcal{H}_2}(T(I_{mathcal{H}_2}otimes S)) = Tr_{mathcal{H}_2}(sum_{i,j,k,l} alpha_{i,j,k,l} (E_{ij}otimes F_{kl}) (I_{mathcal{H}_2}otimes S)) \
= sum_{i,j,k,l} alpha_{i,j,k,l} Tr_{mathcal{H}_2}(E_{ij}otimes F_{kl} S)\
= sum_{i,j,k,l} alpha_{i,j,k,l} E_{ij}.Tr(F_{kl}S)$$
Similarly we get
$$Tr_{mathcal{H}_2}((I_{mathcal{H}_2}otimes S)T) = sum_{i,j,k,l} alpha_{i,j,k,l} E_{ij}.Tr(SF_{kl})$$
But $Tr(F_{kl}S) = Tr(SF_{kl})$. Hence proved.
Request everyone to check the steps.
Update: Found a simpler proof. This doesn't require the decomposition of operators (except identity) and uses the observation that $|aotimes brangle langle c otimes d| = |aranglelangle c| otimes |branglelangle d|$.
Let $e_k$ and $f_j$ be orthonormal basis for $mathcal{H}_1$ and $mathcal{H}_2$ respectively. Then we have,
begin{eqnarray}
&&sum_k langlephi otimes f_k~|(I_{H_1} otimes S)T|~psi otimes f_krangle \
&=& sum_k langlephi otimes f_k~|(I_{H_1} otimes S)(I_{H_1}otimes I_{H_2})T|~psi otimes f_krangle \
&=& sum_{k,i,j} langlephi otimes f_k~|(I_{H_1} otimes S)(|e_iranglelangle e_i| otimes |f_jranglelangle f_j|)T|~psi otimes f_krangle \
&=& sum_{k,i,j} langlephi otimes f_k~|(I_{H_1} otimes S)(|e_iotimes f_jrangle langle e_i otimes f_j |)T|~psi otimes f_krangle \
&=& sum_{k,i,j} langlephi otimes f_k~||e_iotimes Sf_jrangle langle e_i otimes f_j |T|~psi otimes f_krangle \
&=& sum_{k,i,j} langlephi |e_irangle langle f_k| Sf_jrangle langle e_i otimes f_j| T|~psi otimes f_krangle \
&=& sum_{k,i,j} langlephi |e_irangle langle e_i otimes f_j| T|~psi otimes f_krangle langle f_k| Sf_jrangle\
&=& sum_{k,i,j} langlephiotimes f_j |e_iotimes f_jrangle langle e_i otimes f_j| T|~psi otimes f_krangle left(frac{langle psi otimes f_k||psi otimes Sf_jrangle}{|psi|^2}right)\
&=& sum_{k,i,j} langlephiotimes f_j|(|e_iranglelangle e_i| otimes |f_jrangle langle f_j|) T~(|psiranglelanglepsi|otimes |f_kranglelangle f_k|)left(frac{|psi otimes Sf_jrangle}{|psi|^2}right) \
&=& sum_{j} langlephiotimes f_j|(I_{H_1} otimes |f_jrangle langle f_j|) T~(|psiranglelanglepsi|otimes I_{H_2})left(frac{(I_{H_1}otimes S)|psi otimes f_jrangle}{|psi|^2}right) \
&stackrel{(a)}{=}& sum_{j} langlephiotimes f_j| T (I_{H_1}otimes S)left(frac{(|psiranglelanglepsi|otimes I_{H_2})|psi otimes f_jrangle}{|psi|^2}right) \
&=& sum_{j} langlephiotimes f_j| T (I_{H_1}otimes S)|psi otimes f_jrangle \
end{eqnarray}
where $(a)$ follows from,
$$langle phiotimes f_j|(I_{H_1} otimes |f_jrangle langle f_j|) = langle(I_{H_1} otimes |f_jrangle langle f_j|)^*(phiotimes f_j)| = langle phiotimes f_j|$$
where we used the self adjointness of the operator. Also $langle f_j|f_k rangle = delta_{j,k}$. $blacksquare$
answered Mar 23 '17 at 6:15
Gautam ShenoyGautam Shenoy
7,15911645
$endgroup$
I think I was able to solve it. This assumes that any operator $T in mathcal{L}(mathcal{H}_1 otimes mathcal{H}_2)$ can be written in terms of basis operators
$$T = sum_{i,j,k,l} alpha_{i,j,k,l} E_{ij}otimes F_{kl}$$
where (got this from wikipedia article on Partial Trace) given the basis as in the question, $E_{i,j} in mathcal{L}(mathcal{H}_1)$ and $F_{k,l}in mathcal{L}(mathcal{H}_2)$, such that
$$E_{ij}: e_{i} mapsto e_j$$ and
$$F_{kl}: f_{k} mapsto f_l $$
the other basis vectors being mapped to $0$.
We also need to use that (I've proved this before)
$$Tr_{mathcal{H}_2}(Aotimes B) = A. Tr(B)$$
where $A in mathcal{L}(mathcal{H}_1)$ and $Bin mathcal{L}(mathcal{H}_2)$.
Now it's easy. Just observe
$$Tr_{mathcal{H}_2}(T(I_{mathcal{H}_2}otimes S)) = Tr_{mathcal{H}_2}(sum_{i,j,k,l} alpha_{i,j,k,l} (E_{ij}otimes F_{kl}) (I_{mathcal{H}_2}otimes S)) \
= sum_{i,j,k,l} alpha_{i,j,k,l} Tr_{mathcal{H}_2}(E_{ij}otimes F_{kl} S)\
= sum_{i,j,k,l} alpha_{i,j,k,l} E_{ij}.Tr(F_{kl}S)$$
Similarly we get
$$Tr_{mathcal{H}_2}((I_{mathcal{H}_2}otimes S)T) = sum_{i,j,k,l} alpha_{i,j,k,l} E_{ij}.Tr(SF_{kl})$$
But $Tr(F_{kl}S) = Tr(SF_{kl})$. Hence proved.
Request everyone to check the steps.
Update: Found a simpler proof. This doesn't require the decomposition of operators (except identity) and uses the observation that $|aotimes brangle langle c otimes d| = |aranglelangle c| otimes |branglelangle d|$.
Let $e_k$ and $f_j$ be orthonormal basis for $mathcal{H}_1$ and $mathcal{H}_2$ respectively. Then we have,
begin{eqnarray}
&&sum_k langlephi otimes f_k~|(I_{H_1} otimes S)T|~psi otimes f_krangle \
&=& sum_k langlephi otimes f_k~|(I_{H_1} otimes S)(I_{H_1}otimes I_{H_2})T|~psi otimes f_krangle \
&=& sum_{k,i,j} langlephi otimes f_k~|(I_{H_1} otimes S)(|e_iranglelangle e_i| otimes |f_jranglelangle f_j|)T|~psi otimes f_krangle \
&=& sum_{k,i,j} langlephi otimes f_k~|(I_{H_1} otimes S)(|e_iotimes f_jrangle langle e_i otimes f_j |)T|~psi otimes f_krangle \
&=& sum_{k,i,j} langlephi otimes f_k~||e_iotimes Sf_jrangle langle e_i otimes f_j |T|~psi otimes f_krangle \
&=& sum_{k,i,j} langlephi |e_irangle langle f_k| Sf_jrangle langle e_i otimes f_j| T|~psi otimes f_krangle \
&=& sum_{k,i,j} langlephi |e_irangle langle e_i otimes f_j| T|~psi otimes f_krangle langle f_k| Sf_jrangle\
&=& sum_{k,i,j} langlephiotimes f_j |e_iotimes f_jrangle langle e_i otimes f_j| T|~psi otimes f_krangle left(frac{langle psi otimes f_k||psi otimes Sf_jrangle}{|psi|^2}right)\
&=& sum_{k,i,j} langlephiotimes f_j|(|e_iranglelangle e_i| otimes |f_jrangle langle f_j|) T~(|psiranglelanglepsi|otimes |f_kranglelangle f_k|)left(frac{|psi otimes Sf_jrangle}{|psi|^2}right) \
&=& sum_{j} langlephiotimes f_j|(I_{H_1} otimes |f_jrangle langle f_j|) T~(|psiranglelanglepsi|otimes I_{H_2})left(frac{(I_{H_1}otimes S)|psi otimes f_jrangle}{|psi|^2}right) \
&stackrel{(a)}{=}& sum_{j} langlephiotimes f_j| T (I_{H_1}otimes S)left(frac{(|psiranglelanglepsi|otimes I_{H_2})|psi otimes f_jrangle}{|psi|^2}right) \
&=& sum_{j} langlephiotimes f_j| T (I_{H_1}otimes S)|psi otimes f_jrangle \
end{eqnarray}
where $(a)$ follows from,
$$langle phiotimes f_j|(I_{H_1} otimes |f_jrangle langle f_j|) = langle(I_{H_1} otimes |f_jrangle langle f_j|)^*(phiotimes f_j)| = langle phiotimes f_j|$$
where we used the self adjointness of the operator. Also $langle f_j|f_k rangle = delta_{j,k}$. $blacksquare$
answered Mar 23 '17 at 6:15
Gautam ShenoyGautam Shenoy
7,15911645
edited Jan 7 at 19:44
answered Mar 23 '17 at 6:15
Gautam ShenoyGautam Shenoy
7,15911645
answered Mar 23 '17 at 6:15
Gautam ShenoyGautam Shenoy
7,15911645
answered Mar 23 '17 at 6:15
Gautam ShenoyGautam Shenoy
7,15911645
7,15911645
$begingroup$
Your proof looks good too.
$endgroup$
– luftbahnfahrer
Mar 23 '17 at 20:14
add a comment |
$begingroup$
Your proof looks good too.
$endgroup$
– luftbahnfahrer
Mar 23 '17 at 20:14
$begingroup$
Your proof looks good too.
$endgroup$
– luftbahnfahrer
Mar 23 '17 at 20:14
$begingroup$
Your proof looks good too.
$endgroup$
– luftbahnfahrer
Mar 23 '17 at 20:14
add a comment |
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$begingroup$
It would not suffice to prove it for $phi=psi$. It would suffice to prove it for $phi=e_i$ and $psi=e_j$.
$endgroup$
– luftbahnfahrer
Mar 22 '17 at 21:35