# Operator that takes us from one density matrix to another?

Let's say we have two systems A and B. Each system is described by a density matrix $\rho_A$ and $\rho_B$. I'm wondering about the formal notation to write down the expectation value of an operator that takes us from system A to system B or vice versa.

Normally the expectation value of an operator M acting on a system X is just:

$<M> = Tr(\rho_X M)$.

But how would I write this for two systems? Would it be a tensor product of the two density matrices: $\rho_A \otimes \rho_B \rightarrow$ $<M> = Tr(\rho_A \otimes \rho_B M)$ ?

If I would use state vectors I would just write: $<M_{ij}> = <i|M|j>$ and get my expectation value for a specific M or for all M if I do that for all possible states (in this case pure states for simplification). Any idea how to express this in the density matrix trace formalism? How would it look for pure and mixed states?

• I don't understand what you mean in most of this question: you seem to be confused about a few things. 1) What do you mean by "an operator that takes us from system A to system B"? Do you mean an operator that acts on both $A$ and $B$? 2) The state of two systems can only be written $\rho_A\otimes\rho_B$ if the two systems are uncorrelated. In the vast majority of cases this is not so. Oct 10 '13 at 20:49
• 3) Your expression $\langle M_{ij} \rangle = \langle i |M_{ij} | j\rangle$ doesn't make sense. You could write $M_{ij} = \langle i |M | j\rangle$ for the matrix element of the operator $M$, but this is not an expectation value unless $i = j$. Oct 10 '13 at 20:51
• Hello Mark, thanks for your comment! For 3): Yes, that was a typo, I meant $M_{ij} = <i|M|j>$. For 1) and 2): I admit, I am confused. Let's say we have a STM tip with a single atom at the end and a sample that has a single atom on the surface. Then these two atoms have a spin. If you bring them close enough they can interact (e.g. through the Kondo effect) while at the same time electrons "flow" through the STM from the tip atom to the sample atom. Im struggling to find a way to describe how the electron goes from one state into the other.I will probably write more details in my question tmrw.
– Mike
Oct 10 '13 at 22:11
• Sounds like you need a bit more thought to define what your system is first mathematically, and then exactly what expectation value you want to evaluate. Doesn't sound like you can model this with just two atoms. The tip atom should be in contact with a reservoir of billions of electrons inside the STM tip. This means you would need to treat the two atoms as an open system. The coupling between the two atoms could be described by a Hamiltonian like $a^{\dagger}_1a_2 + h.c.$, which describes hopping of an electron from atom $1$ to atom $2$. Oct 10 '13 at 22:19
• The expectation value of this operator would give you the correlation energy between the two atoms, which I'm not sure if you want, since it's usually not very important if they're weakly coupled. What you might want is something like a current operator, which will look something like $i a_1^{\dagger}a_2 - h.c.$. You can derive its precise form by considering the local continuity equation for the electron density $\sim a^{\dagger}a$. Its expectation value gives you the current. However, unless you take the STM electron reservoir into account explicitly I don't see how you'd get any current. Oct 10 '13 at 22:23

I would agree with the comments that you seem to be confusing a few notions here, but let me give it a shot: If you say "an operator, which takes us from system A to system B", you are talking about a map $T$ that (linearly, because we do quantum mechanics) sends density matrices to density matrices, i.e.:
Let $\mathcal{H}$ denote some Hilbert space, then a density matrix $\rho\in\mathcal{B}(\mathcal{H})$ (bounded linear operators on the Hilbert space - in finite dimensions those are just matrices) is a positive semidefinite matrix with unit trace ($\operatorname{tr}(\rho)=1$). Then "an operator which takes us from system A to system B" is a linear map $$T:\mathcal{B}(\mathcal{H}_A)\to\mathcal{B}(\mathcal{H}_B);~~~\rho_A\mapsto \rho_B:=T(\rho)$$ In order for that map to be a map that sends density matrices to density matrices, you must require that it is trace-preserving and completely positive (positive in the sense that it sends positive semidefinite matrices to positive semidefinite matrices, trace-preserving in that $\operatorname{tr}(T(\rho))=\operatorname{tr}(\rho)$ such that the density matrices are normalized and I'm going to skip the "complete" in "completely positive").
Now you talk about "expectation values". Those occur, if you measure a system (i.e. you measure an observable). A measurement doesn't take you from system A to system B, it takes the state and gives you a (real) value from some probability distribution. Granted, you can ask, how the state looks like after the measurement - and this then defines a map as above, but that is a different question. More precisely, given an observable $M$ and a state $\rho$, its expectation value is given by $$\langle M \rangle_{\rho}=\operatorname{tr}(\rho M)$$ as you write above. If you have a larger system that consists of multiple tensor factors, e.g. your system is $\mathcal{B}(\mathcal{H}_A)\otimes\mathcal{B}(\mathcal{H}_B)$ then your density matrices are positive, normalized elements $\rho\in\mathcal{B}(\mathcal{H}_A)\otimes\mathcal{B}(\mathcal{H}_B)$ and the above formula still holds. It might of course happen, that your state is a product state, i.e. $\rho=\rho_A\otimes \rho_B$, but then, the formula still holds and it just reads: $\operatorname{tr}(\rho M)=\operatorname{tr}(\rho_A\otimes \rho_B M)$. There is no difference if you have multiple tensor factors.
Let's now answer the question posed above: You have a state in system A, you send it to B, where you measure an observable $M\in\mathcal{B}(\mathcal{H}_B)$ - how would the expectation look like? Well, if $T:\mathcal{B}(\mathcal{H}_A)\to\mathcal{B}(\mathcal{H}_B)$ is the map that sends us from A to B and our state is $\rho\in\mathcal{B}(\mathcal{H}_A)$, then $$\langle M\rangle_{T(\rho)}=\operatorname{tr}(T(\rho)M)$$ Maybe that already clears up enough for you?