How do we know that, if only $\rho_A$ evolves, then the evolution of $\rho_{AB}$ is given by $(\mathcal{L}_A \otimes 1)(\rho_{AB})$?

Question

I am currently learning about quantum maps, ie maps that transform a density matrix into another one.

Assume we are in the Hilbert space: $H_A \otimes H_B$. I call the quantum map on the density matrix $\rho_A$ living in $ H_A$: $\mathcal{L}_A$.

The postulates are the following :

• "convex" linearity

$$\mathcal{L}_A(p\rho^1_A+q\rho^2_A)=p\mathcal{L}_A (\rho^1_A)+q\mathcal{L}_A(\rho^2_A)$$ where $p+q=1$

• Conservation of hermiticity

$$\mathcal{L}_A(\rho_A)^{\dagger}=\mathcal{L}_A(\rho_A)$$

• Conservation of trace

$$Tr(\mathcal{L}_A(\rho_A))=1$$

• Positivity

$$ \forall |\phi^{A}\rangle : \langle \phi^{A} | \mathcal{L}_A(\rho_A) | \phi^{A} \rangle \geq 0 $$

Those postulates ensure us that $\mathcal{L}_A(\rho_A)$ is a density matrix of $H_A$.

But there is an extra postulate that is :

$\forall \rho_{AB}$ density matrix of $H_A \otimes H_B$, we have :

$$ \forall |\phi^{AB}\rangle : \langle \phi^{AB} | (\mathcal{L}_A \otimes 1)\rho_{AB} | \phi^{AB} \rangle \geq 0$$

I understand this postulate as :

If I imagine a transformation of $\rho_A=Tr_B(\rho_{AB})$ that does'nt affect $\rho_B = Tr_A(\rho_{AB})$, then the evolution of $\rho_{AB}$ is written $(\mathcal{L}_A \otimes 1)(\rho_{AB})$, and we want this last matrix to be positive (to keep having a density matrix).

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My question is:

How do we know that the evolution of $\rho_{AB}$ will be given by $(\mathcal{L}_A \otimes 1)(\rho_{AB})$ under the assumption that only $\rho_A$ evolve?

Indeed, for this we would need to have :

We have : $\rho_{AB}$ evolve, thus :

$$ \rho_{AB}' = \mathcal{L}(\rho_{AB}) $$

The constraint are :

• $\rho_A$ evolve under $\mathcal{L}_A$ :

$$ \rho_A'=\mathcal{L}_A(\rho_A) $$

• $\rho_B$ doesn't evolve :

$$ \rho_B'=\rho_B $$

How from these two last constraints we can prove that actually :

$$ \mathcal{L}=\mathcal{L}_A \otimes 1 $$

For me, this is not at all obvious.

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[edit] : I tried to look at the trick proposed by Luzanne in the comment but I don't find a solution.

So I fix $\mathcal{L}_A$ and I wonder what will be $\mathcal{L}$.

I know that for density matrices in the form $\rho_{AB}=\rho_A \otimes \rho_B$, I have :

$$ \mathcal{L}(\rho_{AB})=\mathcal{L}_A(\rho_A) \otimes \rho_B $$

I try to use those particular cases to show that $\mathcal{L}=\mathcal{L}_A \otimes 1$.

$$ \rho_{AB}=\sum_{ijkl} a_{ij} b_{kl} |u_i\rangle \langle u_j| \otimes |v_k\rangle \langle v_l| $$

Thus :

$$ \rho_{AB}=\sum_{ijkl} a_{ij} b_{kl} \mathcal{L}(|u_i\rangle \langle u_j| \otimes |v_k\rangle \langle v_l|)= \rho_{AB}=\sum_{ijkl} a_{ij} b_{kl} (\mathcal{L_A} \otimes 1)(|u_i\rangle \langle u_j| \otimes |v_k\rangle \langle v_l|)$$

To show the two linear maps are equal I have to check on every vector of the basis, but I must have $\rho_A$ and $\rho_B$ density matrices here.

So by taking $\rho_A=|u_i \rangle \langle u_i |$ and $\rho_B=|v_k \rangle \langle u_k |$, I can have :

$$\mathcal{L}(|u_i v_k \rangle \langle u_i v_k|)= (\mathcal{L}_A \otimes 1)(|u_i v_k \rangle \langle u_i v_k|)$$

But I don't see how to prove it as well for the non diagonal elements of the basis which is also necessary here...

Answer 1

So using the reference you provided (specifically the Appendix B where the heavy lifting is done), we can extend $\mathcal{L}$ and $\mathcal{L}_A$ as real-linear maps on the space of Hermitian matrices on $\mathcal{H} := \mathcal{H}_A \otimes \mathcal{H}_B$, resp. $\mathcal{H}_A$ (said reference then goes on to define complex-linear maps on the space of all matrices but I won't need that).

Special case: $\rho_A \otimes |\psi_B\rangle\langle\psi_B|$ with $\rho_A$ a density matrix

First, let $\rho_A$ be a density matrix over $\mathcal{H}_A$ and let $\psi_B \in \mathcal{H}_B$ with $\|\psi_B\|=1$. Defining $\rho'_{AB} := \mathcal{L}(\rho_A \otimes |\psi_B\rangle\langle\psi_B|)$, we have: $$\text{Tr}_A(\rho'_{AB}) = \text{Tr}_A(\rho_A \otimes |\psi_B\rangle\langle\psi_B|) = |\psi_B\rangle\langle\psi_B|.$$ Since $\rho'_{AB}$ is a density matrix there exist reals $p_k \in ]0,1]$ with $\sum_k p_k = 1$ and unit vectors $\Psi_k \in \mathcal{H}$ such that: $$\rho'_{AB} = \sum_k p_k |\Psi_k\rangle\langle\Psi_k|,$$ so defining the orthogonal projector $\Pi := \mathbb{1} \otimes |\psi_B\rangle\langle\psi_B|$ and using the properties of the partial trace we have: $$\sum_k p_k \langle\Psi_k|\Pi|\Psi_k\rangle = \text{Tr} (\rho'_{AB} \Pi) = 1 = \sum_k p_k \langle\Psi_k|\Psi_k\rangle.$$ Using that all $p_k$ are positive with $\langle\Psi_k|\Pi|\Psi_k\rangle \leq \langle\Psi_k|\Psi_k\rangle$, we deduce $\langle\Psi_k|\Pi|\Psi_k\rangle = \langle\Psi_k|\Psi_k\rangle$ and therefore $|\Psi_k\rangle = \Pi|\Psi_k\rangle$. In other words there exist unit vectors $\phi_k \in \mathcal{H}_A$ such that $\Psi_k = \phi_k \otimes \psi_B$. Defining the density matrix $\rho'_A := \sum_k p_k |\phi_k\rangle\langle\phi_k|$, we thus have: $$\rho'_{AB} = \rho'_A \otimes |\psi_B\rangle\langle\psi_B|,$$ and, since $\rho'_A = \text{Tr}_B \rho'_{AB} = \mathcal{L}_A(\rho_A)$, we conclude: $$\mathcal{L}(\rho_A \otimes |\psi_B\rangle\langle\psi_B|) = (\mathcal{L}_A \otimes \mathbb{1})(\rho_A \otimes |\psi_B\rangle\langle\psi_B|).$$

Extending by linearity to $\sigma_A \otimes |\psi_B\rangle\langle\psi_B|$ with $\sigma_A$ an arbitrary Hermitian matrix

We can then extend this result by linearity to arbitrary Hermitian matrices $\sigma_A$ on $\mathcal{H}_A$ (for any such Hermitian matrix can be written as a linear combination of density matrices over $\mathcal{H}_A$: specifically as $\sigma_A = r^+ \rho_A^+ - r^- \rho_A^-$ with $r^+,r^-$ non-negative reals and $\rho_A^+,\rho_A^-$ density matrices; see the above-mentioned reference).

Extending by linearity to general arbitrary Hermitian matrices $\sigma_{AB}$

Now, let $\sigma_{AB}$ be a general Hermitian matrix on $\mathcal{H}$ and let $\left(e_i \right)_i$ be an orthonormal basis of $\mathcal{H}_B$. We have: $$\sigma_{AB} = \sum_{i,j} \tau_A^{ij} \otimes |e_i\rangle\langle e_j|,$$ with $\tau_A^{ji} = \left(\tau_A^{ij}\right)^{\dagger}$. Reorganizing terms this becomes: $$\sigma_{AB} = \sum_{i} \tau_A^{ii} \otimes |e_i\rangle\langle e_i| + \sum_{i<j} \frac{\tau_A^{ij} + \tau_A^{ji}}{2} \otimes (|e_i\rangle\langle e_j| + |e_j\rangle\langle e_i|) + \frac{i\tau_A^{ij} - i\tau_A^{ji}}{2} \otimes (|e_i\rangle\langle ie_j| + |ie_j\rangle\langle e_i|).$$ And we also have: $$|e_i\rangle\langle e_j| + |e_j\rangle\langle e_i| = \frac{|e_i+e_j\rangle}{\sqrt{2}}\frac{\langle e_i+e_j|}{\sqrt{2}} - \frac{|e_i - e_j\rangle}{\sqrt{2}}\frac{\langle e_i - e_j|}{\sqrt{2}},$$ as well as a similar formula for $|e_i\rangle\langle ie_j| + |ie_j\rangle\langle e_i|$. Putting everything together there exist Hermitian matrices $\sigma_A^k$ and unit vectors $\psi_B^k$ such that: $$\sigma_{AB} = \sum_k \sigma_A^k \otimes |\psi_B^k\rangle\langle\psi_B^k|,$$ so by linearity the general result $\mathcal{L}(\sigma_{AB}) = (\mathcal{L}_A \otimes \mathbb{1})(\sigma_{AB})$ follows from the previous case.

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Note 1: An alternative proof for the last part would be to use successively that any density matrix $\rho_B$ is a linear combination of $|\psi_B\rangle\langle\psi_B|$'s, any Hermitian matrix $\sigma_B$ is a linear combination of $\rho_B$'s, and any Hermitian matrix $\sigma_{AB}$ is a linear combination of $\sigma_A \otimes \sigma_B$'s. In a way the proof above just make this decomposition explicit. I like that it shows better what happens to the non-diagonal terms, namely that they can be made diagonal in an overcomplete (non-orthogonal) basis.

Note 2: Vice-versa, rather than invoking the linked reference to extend the special case result from $\rho_A \otimes |\psi_B\rangle\langle\psi_B|$ (with $\rho_A$ density matrix) to $\sigma_A \otimes |\psi_B\rangle\langle\psi_B|$ (with $\sigma_A$ Hermitian matrix), we could have used such an explicit decomposition of $\sigma_A$ (it would have looked very similar to the formulas from the last part, except with simple complex coefficients $\lambda^{ij}$ instead of $\otimes$-multiplied matrices $\tau^{ij}$).

Note 3: Many questions about density matrices have analogues in terms of classical probability densities, where we may have more intuition. The analogue problem here would be, given a linear transformation of the joint probability: $$p'_{AB}(a,b) = \int \!da' db'\, K(a,b;a',b') \, p_{AB}(a',b'),$$ which, for any $p_{AB}$, transforms the marginal probability for $A$ as: $$p'_A(a) = \int \!da'\, K_A(a;a') \, p_A(a'),$$ and leaves the marginal probability for $B$ unchanged, what is the form of the kernel $K$? A way to solve this classical problem would to get rid of the complexity of the full joint probability, by looking first at what happens if the state of $B$ is certain, ie. $p_{AB}(a,b) = p_A(a) \delta(b-b_o)$. Then, the marginal probability for $B$ after the transformation will still be $\delta(b-b_o)$, ie. the state of $B$ will still be certain, and therefore the joint probability will have the form $p'_{AB}(a,b) = p'_A(a) \delta(b-b_o)$, yielding $K(a,b;a',b_o) = K_A(a;a') \delta(b-b_o)$ (or, in linear operator notation: $K = K_A \otimes \mathbb{1}$). Once we understand this classical case, we can try and adapt the proof to the quantum problem, replacing $\delta$ by a pure state for $B$. Of course, there are complications in the quantum case (in particular, we need to use Hermitian matrices that are not density matrices in intermediary steps, while the classical case could be done entirely acting on probability densities only and using exclusively convex linearity), but the spirit is the same.