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Can an isometry leave entropy invariant?
Proof of an Holevo information inequalityRelating min-entropy with conditional entropyControlling high-dimensional Hilbert spaces with a single qubitKraus operator of dephasing channelPartial Trace over a complicated looking statePartial trace over a product of matrices - one factor is in tensor product formApplication of improved compatibilityBurnside Decomposition in Kuperberg's Hidden ShiftHow to show a density matrix is in a pure/mixed state?Encoding bosonic degrees of freedom
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Consider two finite dimensional Hilbert spaces $A$ and $B$. If I have an isometry $V:Arightarrow Aotimes B$, under what condition can I find a unitary $U:Aotimes Bto Aotimes B$ such that $$Urho_Aotimes |psiranglelangle psi|_BU^dagger=Vrho_AV^dagger$$
for some state $|psirangle_B$? The motivation is I have an isometry like $V$ and I would like it very much for a state to satisfy $S(A)_rho=S(AB)_Vrho V^dagger$ where $S$ is quantum entropy. Is this possible?
quantum-information mathematics entropy
asked 3 hours ago
user2723984user2723984
1625
$endgroup$
add a comment |
$begingroup$
Consider two finite dimensional Hilbert spaces $A$ and $B$. If I have an isometry $V:Arightarrow Aotimes B$, under what condition can I find a unitary $U:Aotimes Bto Aotimes B$ such that $$Urho_Aotimes |psiranglelangle psi|_BU^dagger=Vrho_AV^dagger$$
for some state $|psirangle_B$? The motivation is I have an isometry like $V$ and I would like it very much for a state to satisfy $S(A)_rho=S(AB)_Vrho V^dagger$ where $S$ is quantum entropy. Is this possible?
quantum-information mathematics entropy
asked 3 hours ago
user2723984user2723984
1625
$endgroup$
add a comment |
$begingroup$
Consider two finite dimensional Hilbert spaces $A$ and $B$. If I have an isometry $V:Arightarrow Aotimes B$, under what condition can I find a unitary $U:Aotimes Bto Aotimes B$ such that $$Urho_Aotimes |psiranglelangle psi|_BU^dagger=Vrho_AV^dagger$$
for some state $|psirangle_B$? The motivation is I have an isometry like $V$ and I would like it very much for a state to satisfy $S(A)_rho=S(AB)_Vrho V^dagger$ where $S$ is quantum entropy. Is this possible?
quantum-information mathematics entropy
asked 3 hours ago
user2723984user2723984
1625
$endgroup$
Consider two finite dimensional Hilbert spaces $A$ and $B$. If I have an isometry $V:Arightarrow Aotimes B$, under what condition can I find a unitary $U:Aotimes Bto Aotimes B$ such that $$Urho_Aotimes |psiranglelangle psi|_BU^dagger=Vrho_AV^dagger$$
for some state $|psirangle_B$? The motivation is I have an isometry like $V$ and I would like it very much for a state to satisfy $S(A)_rho=S(AB)_Vrho V^dagger$ where $S$ is quantum entropy. Is this possible?
quantum-information mathematics entropy
quantum-information mathematics entropy
asked 3 hours ago
user2723984user2723984
1625
asked 3 hours ago
user2723984user2723984
1625
edited 3 hours ago
user2723984
asked 3 hours ago
user2723984user2723984
1625
asked 3 hours ago
user2723984user2723984
1625
asked 3 hours ago
user2723984user2723984
1625
1625
add a comment |
add a comment |
2 Answers
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$begingroup$
You don't need any additional conditions beyond those already stated in the question. That is, for any isometry $V: A rightarrow Aotimes B$ and any unit vector $|psirangle_B$, there will always be a unitary $U$ satisfying the equation in the question (simultaneously for every choice of $rho_A$).
One way to see this is to first pick any orthonormal basis $1rangle,ldots,$ for $A$, and then consider the two sets $1rangleotimes$ and $V$. These are orthonormal sets (using the fact that $V$ is an isometry in the second case), so you can extend them both to be orthonormal bases of $Aotimes B$ and choose a unitary $U$ that maps the first completed basis to the second. Of course there are many ways to do this in general, but in particular you can choose $U$ so that it maps $|krangleotimes|psirangle$ to $V|krangle$ for each $k in 1,ldots,n$. This implies that the equation in the question is satisfied.
Note that if what you really want is $S(A)_rho = S(AB)_Vrho V^dagger$, then it is simpler to conclude that this is always true from the observation that $rho$ and $V rho V^dagger$ must agree on their nonzero eigenvalues.
answered 1 hour ago
John WatrousJohn Watrous
2,03328
$endgroup$
add a comment |
$begingroup$
You can always do that.
Subspaces $textIm(V)$ and $Aotimes |0rangle$ have the same dimension, so there must be some unitary that translates one subspace to another. That is, $exists W in textUnitary(Aotimes B), W(textIm(V)) = Aotimes |0rangle$. Now $WV$ translates $A$ to $Aotimes |0rangle$. Since $V$ is isometry $WV$ is also isometry, hence there exists unitary $Y$ on $A$ such that $WV|phirangle = Y|phirangleotimes |0rangle$.
Now the unitary you are looking is $U = W^-1cdot Yotimes I$.
answered 1 hour ago
Danylo YDanylo Y
76516
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
You don't need any additional conditions beyond those already stated in the question. That is, for any isometry $V: A rightarrow Aotimes B$ and any unit vector $|psirangle_B$, there will always be a unitary $U$ satisfying the equation in the question (simultaneously for every choice of $rho_A$).
One way to see this is to first pick any orthonormal basis $1rangle,ldots,$ for $A$, and then consider the two sets $1rangleotimes$ and $V$. These are orthonormal sets (using the fact that $V$ is an isometry in the second case), so you can extend them both to be orthonormal bases of $Aotimes B$ and choose a unitary $U$ that maps the first completed basis to the second. Of course there are many ways to do this in general, but in particular you can choose $U$ so that it maps $|krangleotimes|psirangle$ to $V|krangle$ for each $k in 1,ldots,n$. This implies that the equation in the question is satisfied.
Note that if what you really want is $S(A)_rho = S(AB)_Vrho V^dagger$, then it is simpler to conclude that this is always true from the observation that $rho$ and $V rho V^dagger$ must agree on their nonzero eigenvalues.
answered 1 hour ago
John WatrousJohn Watrous
2,03328
$endgroup$
add a comment |
$begingroup$
You don't need any additional conditions beyond those already stated in the question. That is, for any isometry $V: A rightarrow Aotimes B$ and any unit vector $|psirangle_B$, there will always be a unitary $U$ satisfying the equation in the question (simultaneously for every choice of $rho_A$).
One way to see this is to first pick any orthonormal basis $1rangle,ldots,$ for $A$, and then consider the two sets $1rangleotimes$ and $V$. These are orthonormal sets (using the fact that $V$ is an isometry in the second case), so you can extend them both to be orthonormal bases of $Aotimes B$ and choose a unitary $U$ that maps the first completed basis to the second. Of course there are many ways to do this in general, but in particular you can choose $U$ so that it maps $|krangleotimes|psirangle$ to $V|krangle$ for each $k in 1,ldots,n$. This implies that the equation in the question is satisfied.
Note that if what you really want is $S(A)_rho = S(AB)_Vrho V^dagger$, then it is simpler to conclude that this is always true from the observation that $rho$ and $V rho V^dagger$ must agree on their nonzero eigenvalues.
answered 1 hour ago
John WatrousJohn Watrous
2,03328
$endgroup$
add a comment |
$begingroup$
You don't need any additional conditions beyond those already stated in the question. That is, for any isometry $V: A rightarrow Aotimes B$ and any unit vector $|psirangle_B$, there will always be a unitary $U$ satisfying the equation in the question (simultaneously for every choice of $rho_A$).
One way to see this is to first pick any orthonormal basis $1rangle,ldots,$ for $A$, and then consider the two sets $1rangleotimes$ and $V$. These are orthonormal sets (using the fact that $V$ is an isometry in the second case), so you can extend them both to be orthonormal bases of $Aotimes B$ and choose a unitary $U$ that maps the first completed basis to the second. Of course there are many ways to do this in general, but in particular you can choose $U$ so that it maps $|krangleotimes|psirangle$ to $V|krangle$ for each $k in 1,ldots,n$. This implies that the equation in the question is satisfied.
Note that if what you really want is $S(A)_rho = S(AB)_Vrho V^dagger$, then it is simpler to conclude that this is always true from the observation that $rho$ and $V rho V^dagger$ must agree on their nonzero eigenvalues.
answered 1 hour ago
John WatrousJohn Watrous
2,03328
$endgroup$
You don't need any additional conditions beyond those already stated in the question. That is, for any isometry $V: A rightarrow Aotimes B$ and any unit vector $|psirangle_B$, there will always be a unitary $U$ satisfying the equation in the question (simultaneously for every choice of $rho_A$).
One way to see this is to first pick any orthonormal basis $1rangle,ldots,$ for $A$, and then consider the two sets $1rangleotimes$ and $V$. These are orthonormal sets (using the fact that $V$ is an isometry in the second case), so you can extend them both to be orthonormal bases of $Aotimes B$ and choose a unitary $U$ that maps the first completed basis to the second. Of course there are many ways to do this in general, but in particular you can choose $U$ so that it maps $|krangleotimes|psirangle$ to $V|krangle$ for each $k in 1,ldots,n$. This implies that the equation in the question is satisfied.
Note that if what you really want is $S(A)_rho = S(AB)_Vrho V^dagger$, then it is simpler to conclude that this is always true from the observation that $rho$ and $V rho V^dagger$ must agree on their nonzero eigenvalues.
answered 1 hour ago
John WatrousJohn Watrous
2,03328
answered 1 hour ago
John WatrousJohn Watrous
2,03328
answered 1 hour ago
John WatrousJohn Watrous
2,03328
answered 1 hour ago
John WatrousJohn Watrous
2,03328
2,03328
add a comment |
add a comment |
$begingroup$
You can always do that.
Subspaces $textIm(V)$ and $Aotimes |0rangle$ have the same dimension, so there must be some unitary that translates one subspace to another. That is, $exists W in textUnitary(Aotimes B), W(textIm(V)) = Aotimes |0rangle$. Now $WV$ translates $A$ to $Aotimes |0rangle$. Since $V$ is isometry $WV$ is also isometry, hence there exists unitary $Y$ on $A$ such that $WV|phirangle = Y|phirangleotimes |0rangle$.
Now the unitary you are looking is $U = W^-1cdot Yotimes I$.
answered 1 hour ago
Danylo YDanylo Y
76516
$endgroup$
add a comment |
$begingroup$
You can always do that.
Subspaces $textIm(V)$ and $Aotimes |0rangle$ have the same dimension, so there must be some unitary that translates one subspace to another. That is, $exists W in textUnitary(Aotimes B), W(textIm(V)) = Aotimes |0rangle$. Now $WV$ translates $A$ to $Aotimes |0rangle$. Since $V$ is isometry $WV$ is also isometry, hence there exists unitary $Y$ on $A$ such that $WV|phirangle = Y|phirangleotimes |0rangle$.
Now the unitary you are looking is $U = W^-1cdot Yotimes I$.
answered 1 hour ago
Danylo YDanylo Y
76516
$endgroup$
add a comment |
$begingroup$
You can always do that.
Subspaces $textIm(V)$ and $Aotimes |0rangle$ have the same dimension, so there must be some unitary that translates one subspace to another. That is, $exists W in textUnitary(Aotimes B), W(textIm(V)) = Aotimes |0rangle$. Now $WV$ translates $A$ to $Aotimes |0rangle$. Since $V$ is isometry $WV$ is also isometry, hence there exists unitary $Y$ on $A$ such that $WV|phirangle = Y|phirangleotimes |0rangle$.
Now the unitary you are looking is $U = W^-1cdot Yotimes I$.
answered 1 hour ago
Danylo YDanylo Y
76516
$endgroup$
You can always do that.
Subspaces $textIm(V)$ and $Aotimes |0rangle$ have the same dimension, so there must be some unitary that translates one subspace to another. That is, $exists W in textUnitary(Aotimes B), W(textIm(V)) = Aotimes |0rangle$. Now $WV$ translates $A$ to $Aotimes |0rangle$. Since $V$ is isometry $WV$ is also isometry, hence there exists unitary $Y$ on $A$ such that $WV|phirangle = Y|phirangleotimes |0rangle$.
Now the unitary you are looking is $U = W^-1cdot Yotimes I$.
answered 1 hour ago
Danylo YDanylo Y
76516
answered 1 hour ago
Danylo YDanylo Y
76516
answered 1 hour ago
Danylo YDanylo Y
76516
answered 1 hour ago
Danylo YDanylo Y
76516
76516
add a comment |
add a comment |
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