# Is there a notion of a "reduced" Hamiltonian?

Similar to how we can construct a density matrix $$\rho_A$$ that represents states in a subsystem $$A$$ by performing a partial trace on $$\rho$$ (the full density matrix of the whole subsystem (say $$AB$$)), is there a similar operation we can do on the hamiltonian $$H$$?

Consider $$i\frac{d}{dt} \rho_B = i\frac{d}{dt} \text{Tr}_A (\rho) = \text{Tr}_A ([H,\rho]),$$ where the last step is true by linearity. If for simplicity we are in a bipartite system such that $$H=H_A \otimes H_B$$ then we can further massage above expression \begin{align} i\frac{d}{dt} \rho_B & =\text{Tr}_A ([H,\rho]) \\ & = \sum_i (\langle \phi_i|_A\otimes 1_B)[H_A \otimes H_B,\rho](| \phi_i \rangle _A\otimes 1_B) \\ & =\sum_i E_i^A\underbrace{( 1_A \otimes H_B)}_\text{abuse of notation}(\langle \phi_i|_A\otimes 1_B)\rho(| \phi_i \rangle _A\otimes 1_B) \\ & \qquad -E_i^A(\langle \phi_i|_A\otimes 1_B)\rho(| \phi_i \rangle _A\otimes 1_B)\underbrace{( 1_A \otimes H_B)}_\text{abuse of notation} \\ & =\left[H_B, \sum_iE_i^A(\langle \phi_i|_A\otimes 1_B)\rho(| \phi_i \rangle _A\otimes 1_B)\right] \\ & = [H_B,\rho_B], \end{align} where the abuse of notation is that the identity now $$1_A:\mathcal{H}^* \rightarrow \mathcal{H}^*$$ and not $$1_A:\mathcal{H} \rightarrow \mathcal{H}$$, $$|\phi_i\rangle_A$$ is the eigenket of $$H_A$$ with eigenenergy $$E_i^A$$ and in the last line I redefined the basis of the partial trace.

This calculation, if correct implies that $$H_B$$ is the "reduced" Hamiltonian on $$B$$, is this correct?

To me this is not trivial as there might be interactions between $$A$$ and $$B$$ (nor that it matters why is not trivial to me).

• $\sum_i E_i^A \langle \phi_i |_A \rho |\phi_i \rangle_A \ne \mathrm{Tr}_A \rho$. There is an extra $E^A$ inside the sum. Jan 11, 2021 at 12:23
• Relevant Wiki: Open quantum systems Jan 11, 2021 at 12:36
• @BySymmetry Cant you define $| \tilde{\phi}_i \rangle_A := \sqrt{E_i^A}| \phi_i \rangle_A$ so that $\sum_i E_i^A \langle \phi_i |_A \rho |\phi_i \rangle_A = \sum_i \langle \tilde{\phi}_i |_A \rho | \tilde{\phi}_i \rangle_A = \text{Tr}_A \rho$? Jan 11, 2021 at 12:37
• ^The trace is taken with respect to an orthonormal basis, so no. The factors of $E_i$ make it clear that the sum you have written is not the trace of $A$, so you certainly shouldn't be able to sweep that under the rug with algebraic manipulation, right? Jan 11, 2021 at 12:41
• The trace is invariant, but that formula isn't. Note for a general change of basis $|\phi\rangle \mapsto P|\phi\rangle$ and $A\mapsto PAP^{-1}$, we have $\sum \langle \phi_i|A|\phi_i\rangle \mapsto \sum \langle \phi_i|P^\dagger \big(P A P^{-1}\big)P|\phi_i\rangle \neq \sum \langle\phi_i|A|\phi_i\rangle$ unless $P$ is unitary. I'm not comfortable enough with open quantum systems to give a good answer here, unfortunately, but in general the evolution of $\rho_B$ will be nonunitary and look very different - see the link I posted above. Jan 11, 2021 at 13:04

As was discussed in the comments, the problem with what you have written is there is no change of basis such that $$\sum_i E_i^A \langle\phi_i|_A \rho |\phi_i\rangle_A = \mathrm{Tr}_A \rho$$
• I now see the non-existence of such transformation. So all we can say, if any, is that $$i\frac{d}{dt} \rho_B =\left[H_B, \sum_iE_i^A(\langle \phi_i|_A\otimes 1_B)\rho(| \phi_i \rangle _A\otimes 1_B)\right],$$ whatever that might mean. I completly agree with your intuition behind that the lack of information implies that a "reduced" Hamiltonian does not exist in general (that´s why I was surprised and asked this question :) ). I will think about the rest of your answer and check the literature you mention. Thanks a lot for your time and patience! Jan 11, 2021 at 19:04