Skip to main content
Quantum Computing

Return to Answer

added 24 characters in body
Source Link
Adam Zalcman
Adam Zalcman
  • 24.8k
  • 3
  • 38
  • 92

In the absence of the constraint on the marginal, it is true that there exists $U_{ABC}$ such that $U_{ABC}|\rho_{ABC}\rangle = |\sigma_{ABC}\rangle$. Indeed, extend $|\rho_{ABC}\rangle = |\rho_{ABC}^{(1)}\rangle$ to an orthonormal basis $|\rho_{ABC}^{(k)}\rangle$ and $|\sigma_{ABC}\rangle = |\sigma_{ABC}^{(1)}\rangle$ to an orthonormal basis $|\sigma_{ABC}^{(k)}\rangle$ and define $U_{ABC} =\sum_k |\sigma_{ABC}^{(k)}\rangle\langle\rho_{ABC}^{(k)}|$. This construction makes it clear that $U_{ABC}$ is not unique.

Now, enforcing the constraint on the marginal, it is true that for any $\rho_{AB}$ and $\sigma_{AB}$ with $\mathrm{tr}_B(\rho_{AB}) = \mathrm{tr}_B(\sigma_{AB})$ there exists $U_{BC}$ that achieves $(I_A\otimes U_{BC})|\rho_{ABC}\rangle = |\sigma_{ABC}\rangle$ where $|\rho_{ABC}\rangle$ and $|\sigma_{ABC}\rangle$ denote the respective purifications.

To see this, consider the Schmidt decompositions of the two pure states relative to the partitioning of $ABC$ into $A$ and $BC$

$$ |\rho_{ABC}\rangle = \sum_i \lambda_i |i_A\rangle|i_{BC}\rangle \\ |\sigma_{ABC}\rangle = \sum_i \mu_i |i_A'\rangle|i_{BC}'\rangle $$

where $|i_A\rangle$, $|i_A'\rangle$, $|i_{BC}\rangle$ and $|i_{BC}'\rangle$ are orthonormal sets of states (not necessarily full bases) and $\lambda_i$ and $\mu_i$ are positive real numbers. Note that

$$ \mathrm{tr}_B(\rho_{AB}) = \mathrm{tr}_{BC}(|\rho_{ABC}\rangle\langle\rho_{ABC}|) = \sum_i\lambda_i^2|i_A\rangle\langle i_A| \\ \mathrm{tr}_B(\sigma_{AB}) = \mathrm{tr}_{BC}(|\sigma_{ABC}\rangle\langle\sigma_{ABC}|) = \sum_i\mu_i^2|i_A'\rangle\langle i_A'| $$

and hence $\lambda_i = \mu_i$. Moreover, if $\lambda_i$ are distinct then the eigendecomposition above is unique and hence $|i_A\rangle = |i_A'\rangle$. If they are not, this is not guaranteed, but we can choose $|i_A\rangle = |i_A'\rangle$. In any case, we can write

$$ |\rho_{ABC}\rangle = \sum_i \lambda_i |i_A\rangle|i_{BC}\rangle \\ |\sigma_{ABC}\rangle = \sum_i \lambda_i |i_A\rangle|i_{BC}'\rangle. $$

Finally, set $U_{BC} = \sum_i |i_{BC}'\rangle\langle i_{BC}|$ and observe that

$$ (I_A\otimes U_{BC})|\rho_{ABC}\rangle = \sum_i \lambda_i |i_A\rangle U_{BC}|i_{BC}\rangle = \sum_i \lambda_i |i_A\rangle|i_{BC}'\rangle = |\sigma_{ABC}\rangle $$

as desired.

I don't know what the most general form of a marginal-preserving unitary is. It is certainly more general than $U = I\otimes U_{BC}$$I\otimes U_{BC}$ or even $U_A\otimes U_{BC}$, because there are examples of entangling unitaries that satisfy the constraint. For example, let $A$ and $B$ denote two qubits and $\rho_{AB} = |00\rangle\langle 00|$ and $\sigma_{AB} = |01\rangle\langle 01|$. Then $\mathrm{tr}_B\rho_{AB} = \mathrm{tr}_B\sigma_{AB} = |0\rangle\langle 0|$, but $\text{CNOT}\,\rho_{AB}\,\text{CNOT}^\dagger = \sigma_{AB}$.

In the absence of the constraint on the marginal, it is true that there exists $U_{ABC}$ such that $U_{ABC}|\rho_{ABC}\rangle = |\sigma_{ABC}\rangle$. Indeed, extend $|\rho_{ABC}\rangle = |\rho_{ABC}^{(1)}\rangle$ to an orthonormal basis $|\rho_{ABC}^{(k)}\rangle$ and $|\sigma_{ABC}\rangle = |\sigma_{ABC}^{(1)}\rangle$ to an orthonormal basis $|\sigma_{ABC}^{(k)}\rangle$ and define $U_{ABC} =\sum_k |\sigma_{ABC}^{(k)}\rangle\langle\rho_{ABC}^{(k)}|$. This construction makes it clear that $U_{ABC}$ is not unique.

Now, enforcing the constraint on the marginal, it is true that for any $\rho_{AB}$ and $\sigma_{AB}$ with $\mathrm{tr}_B(\rho_{AB}) = \mathrm{tr}_B(\sigma_{AB})$ there exists $U_{BC}$ that achieves $(I_A\otimes U_{BC})|\rho_{ABC}\rangle = |\sigma_{ABC}\rangle$ where $|\rho_{ABC}\rangle$ and $|\sigma_{ABC}\rangle$ denote the respective purifications.

To see this, consider the Schmidt decompositions of the two pure states relative to the partitioning of $ABC$ into $A$ and $BC$

$$ |\rho_{ABC}\rangle = \sum_i \lambda_i |i_A\rangle|i_{BC}\rangle \\ |\sigma_{ABC}\rangle = \sum_i \mu_i |i_A'\rangle|i_{BC}'\rangle $$

where $|i_A\rangle$, $|i_A'\rangle$, $|i_{BC}\rangle$ and $|i_{BC}'\rangle$ are orthonormal sets of states (not necessarily full bases) and $\lambda_i$ and $\mu_i$ are positive real numbers. Note that

$$ \mathrm{tr}_B(\rho_{AB}) = \mathrm{tr}_{BC}(|\rho_{ABC}\rangle\langle\rho_{ABC}|) = \sum_i\lambda_i^2|i_A\rangle\langle i_A| \\ \mathrm{tr}_B(\sigma_{AB}) = \mathrm{tr}_{BC}(|\sigma_{ABC}\rangle\langle\sigma_{ABC}|) = \sum_i\mu_i^2|i_A'\rangle\langle i_A'| $$

and hence $\lambda_i = \mu_i$. Moreover, if $\lambda_i$ are distinct then the eigendecomposition above is unique and hence $|i_A\rangle = |i_A'\rangle$. If they are not, this is not guaranteed, but we can choose $|i_A\rangle = |i_A'\rangle$. In any case, we can write

$$ |\rho_{ABC}\rangle = \sum_i \lambda_i |i_A\rangle|i_{BC}\rangle \\ |\sigma_{ABC}\rangle = \sum_i \lambda_i |i_A\rangle|i_{BC}'\rangle. $$

Finally, set $U_{BC} = \sum_i |i_{BC}'\rangle\langle i_{BC}|$ and observe that

$$ (I_A\otimes U_{BC})|\rho_{ABC}\rangle = \sum_i \lambda_i |i_A\rangle U_{BC}|i_{BC}\rangle = \sum_i \lambda_i |i_A\rangle|i_{BC}'\rangle = |\sigma_{ABC}\rangle $$

as desired.

I don't know what the most general form of a marginal-preserving unitary is. It is certainly more general than $U = I\otimes U_{BC}$, because there are examples of entangling unitaries that satisfy the constraint. For example, let $A$ and $B$ denote two qubits and $\rho_{AB} = |00\rangle\langle 00|$ and $\sigma_{AB} = |01\rangle\langle 01|$. Then $\mathrm{tr}_B\rho_{AB} = \mathrm{tr}_B\sigma_{AB} = |0\rangle\langle 0|$, but $\text{CNOT}\,\rho_{AB}\,\text{CNOT}^\dagger = \sigma_{AB}$.

In the absence of the constraint on the marginal, it is true that there exists $U_{ABC}$ such that $U_{ABC}|\rho_{ABC}\rangle = |\sigma_{ABC}\rangle$. Indeed, extend $|\rho_{ABC}\rangle = |\rho_{ABC}^{(1)}\rangle$ to an orthonormal basis $|\rho_{ABC}^{(k)}\rangle$ and $|\sigma_{ABC}\rangle = |\sigma_{ABC}^{(1)}\rangle$ to an orthonormal basis $|\sigma_{ABC}^{(k)}\rangle$ and define $U_{ABC} =\sum_k |\sigma_{ABC}^{(k)}\rangle\langle\rho_{ABC}^{(k)}|$. This construction makes it clear that $U_{ABC}$ is not unique.

Now, enforcing the constraint on the marginal, it is true that for any $\rho_{AB}$ and $\sigma_{AB}$ with $\mathrm{tr}_B(\rho_{AB}) = \mathrm{tr}_B(\sigma_{AB})$ there exists $U_{BC}$ that achieves $(I_A\otimes U_{BC})|\rho_{ABC}\rangle = |\sigma_{ABC}\rangle$ where $|\rho_{ABC}\rangle$ and $|\sigma_{ABC}\rangle$ denote the respective purifications.

To see this, consider the Schmidt decompositions of the two pure states relative to the partitioning of $ABC$ into $A$ and $BC$

$$ |\rho_{ABC}\rangle = \sum_i \lambda_i |i_A\rangle|i_{BC}\rangle \\ |\sigma_{ABC}\rangle = \sum_i \mu_i |i_A'\rangle|i_{BC}'\rangle $$

where $|i_A\rangle$, $|i_A'\rangle$, $|i_{BC}\rangle$ and $|i_{BC}'\rangle$ are orthonormal sets of states (not necessarily full bases) and $\lambda_i$ and $\mu_i$ are positive real numbers. Note that

$$ \mathrm{tr}_B(\rho_{AB}) = \mathrm{tr}_{BC}(|\rho_{ABC}\rangle\langle\rho_{ABC}|) = \sum_i\lambda_i^2|i_A\rangle\langle i_A| \\ \mathrm{tr}_B(\sigma_{AB}) = \mathrm{tr}_{BC}(|\sigma_{ABC}\rangle\langle\sigma_{ABC}|) = \sum_i\mu_i^2|i_A'\rangle\langle i_A'| $$

and hence $\lambda_i = \mu_i$. Moreover, if $\lambda_i$ are distinct then the eigendecomposition above is unique and hence $|i_A\rangle = |i_A'\rangle$. If they are not, this is not guaranteed, but we can choose $|i_A\rangle = |i_A'\rangle$. In any case, we can write

$$ |\rho_{ABC}\rangle = \sum_i \lambda_i |i_A\rangle|i_{BC}\rangle \\ |\sigma_{ABC}\rangle = \sum_i \lambda_i |i_A\rangle|i_{BC}'\rangle. $$

Finally, set $U_{BC} = \sum_i |i_{BC}'\rangle\langle i_{BC}|$ and observe that

$$ (I_A\otimes U_{BC})|\rho_{ABC}\rangle = \sum_i \lambda_i |i_A\rangle U_{BC}|i_{BC}\rangle = \sum_i \lambda_i |i_A\rangle|i_{BC}'\rangle = |\sigma_{ABC}\rangle $$

as desired.

I don't know what the most general form of a marginal-preserving unitary is. It is certainly more general than $I\otimes U_{BC}$ or even $U_A\otimes U_{BC}$, because there are examples of entangling unitaries that satisfy the constraint. For example, let $A$ and $B$ denote two qubits and $\rho_{AB} = |00\rangle\langle 00|$ and $\sigma_{AB} = |01\rangle\langle 01|$. Then $\mathrm{tr}_B\rho_{AB} = \mathrm{tr}_B\sigma_{AB} = |0\rangle\langle 0|$, but $\text{CNOT}\,\rho_{AB}\,\text{CNOT}^\dagger = \sigma_{AB}$.

Source Link
Adam Zalcman
Adam Zalcman
  • 24.8k
  • 3
  • 38
  • 92

In the absence of the constraint on the marginal, it is true that there exists $U_{ABC}$ such that $U_{ABC}|\rho_{ABC}\rangle = |\sigma_{ABC}\rangle$. Indeed, extend $|\rho_{ABC}\rangle = |\rho_{ABC}^{(1)}\rangle$ to an orthonormal basis $|\rho_{ABC}^{(k)}\rangle$ and $|\sigma_{ABC}\rangle = |\sigma_{ABC}^{(1)}\rangle$ to an orthonormal basis $|\sigma_{ABC}^{(k)}\rangle$ and define $U_{ABC} =\sum_k |\sigma_{ABC}^{(k)}\rangle\langle\rho_{ABC}^{(k)}|$. This construction makes it clear that $U_{ABC}$ is not unique.

Now, enforcing the constraint on the marginal, it is true that for any $\rho_{AB}$ and $\sigma_{AB}$ with $\mathrm{tr}_B(\rho_{AB}) = \mathrm{tr}_B(\sigma_{AB})$ there exists $U_{BC}$ that achieves $(I_A\otimes U_{BC})|\rho_{ABC}\rangle = |\sigma_{ABC}\rangle$ where $|\rho_{ABC}\rangle$ and $|\sigma_{ABC}\rangle$ denote the respective purifications.

To see this, consider the Schmidt decompositions of the two pure states relative to the partitioning of $ABC$ into $A$ and $BC$

$$ |\rho_{ABC}\rangle = \sum_i \lambda_i |i_A\rangle|i_{BC}\rangle \\ |\sigma_{ABC}\rangle = \sum_i \mu_i |i_A'\rangle|i_{BC}'\rangle $$

where $|i_A\rangle$, $|i_A'\rangle$, $|i_{BC}\rangle$ and $|i_{BC}'\rangle$ are orthonormal sets of states (not necessarily full bases) and $\lambda_i$ and $\mu_i$ are positive real numbers. Note that

$$ \mathrm{tr}_B(\rho_{AB}) = \mathrm{tr}_{BC}(|\rho_{ABC}\rangle\langle\rho_{ABC}|) = \sum_i\lambda_i^2|i_A\rangle\langle i_A| \\ \mathrm{tr}_B(\sigma_{AB}) = \mathrm{tr}_{BC}(|\sigma_{ABC}\rangle\langle\sigma_{ABC}|) = \sum_i\mu_i^2|i_A'\rangle\langle i_A'| $$

and hence $\lambda_i = \mu_i$. Moreover, if $\lambda_i$ are distinct then the eigendecomposition above is unique and hence $|i_A\rangle = |i_A'\rangle$. If they are not, this is not guaranteed, but we can choose $|i_A\rangle = |i_A'\rangle$. In any case, we can write

$$ |\rho_{ABC}\rangle = \sum_i \lambda_i |i_A\rangle|i_{BC}\rangle \\ |\sigma_{ABC}\rangle = \sum_i \lambda_i |i_A\rangle|i_{BC}'\rangle. $$

Finally, set $U_{BC} = \sum_i |i_{BC}'\rangle\langle i_{BC}|$ and observe that

$$ (I_A\otimes U_{BC})|\rho_{ABC}\rangle = \sum_i \lambda_i |i_A\rangle U_{BC}|i_{BC}\rangle = \sum_i \lambda_i |i_A\rangle|i_{BC}'\rangle = |\sigma_{ABC}\rangle $$

as desired.

I don't know what the most general form of a marginal-preserving unitary is. It is certainly more general than $U = I\otimes U_{BC}$, because there are examples of entangling unitaries that satisfy the constraint. For example, let $A$ and $B$ denote two qubits and $\rho_{AB} = |00\rangle\langle 00|$ and $\sigma_{AB} = |01\rangle\langle 01|$. Then $\mathrm{tr}_B\rho_{AB} = \mathrm{tr}_B\sigma_{AB} = |0\rangle\langle 0|$, but $\text{CNOT}\,\rho_{AB}\,\text{CNOT}^\dagger = \sigma_{AB}$.