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).

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    $\begingroup$ $\sum_i E_i^A \langle \phi_i |_A \rho |\phi_i \rangle_A \ne \mathrm{Tr}_A \rho$. There is an extra $E^A$ inside the sum. $\endgroup$ Jan 11, 2021 at 12:23
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    $\begingroup$ Relevant Wiki: Open quantum systems $\endgroup$
    – J. Murray
    Jan 11, 2021 at 12:36
  • $\begingroup$ @BySymmetry Cant you define $| \tilde{\phi}_i \rangle_A := \sqrt{E_i^A}| \phi_i \rangle_A$ so that $\sum_i E_i^A \langle \phi_i |_A \rho |\phi_i \rangle_A = \sum_i \langle \tilde{\phi}_i |_A \rho | \tilde{\phi}_i \rangle_A = \text{Tr}_A \rho$? $\endgroup$ Jan 11, 2021 at 12:37
  • $\begingroup$ ^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? $\endgroup$
    – J. Murray
    Jan 11, 2021 at 12:41
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    $\begingroup$ 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. $\endgroup$
    – J. Murray
    Jan 11, 2021 at 13:04

1 Answer 1


As was discussed in the comments, the problem with what you have written is there is no change of basis such that $$ \sum_i E_i^A \langle\phi_i|_A \rho |\phi_i\rangle_A = \mathrm{Tr}_A \rho $$

More generally the problem with the notion of a reduced Hamiltonian is that the reduced system at a given moment in time typically does not contain enough information to reproduce the future dynamics of the system. If I don't know what the part I have traced out is doing, I can't predict what effect it is going to have on what's left.

The simplest consequence this can have a loss on unitary in the time evolution. As information about the system is lost to its environment the reduced density matrix generally gets more mixed and less pure over time. This cannot be modeled with a true Hamiltonian, but can be modeled with formalism such as the Lindblad master equation (in which the Louvillian plays a role analogous to the Hamiltonian) or non-Hermitian quantum mechanics. Both these approaches, however, rely on what is known as the Markov approximation. When information about the system is passed to it's surroundings it is either returned immediately or simply lost. We can never send out a 'signal' and some time later hear back an 'echo'. This loss of information leads to the non-unitary dynamics.

There are some methods that aim to move beyond the Markov approximation, such as the Nakajima-Zwanzig equation or the Feynmann-Vernon influence functional approach. These methods generally involve not only the instantaneous density matrix, but some form of integral over the past history of the system. If we know the initial state of the environment and all of our past interactions with it, we can reconstruct its current state and so calculate, exactly, the future dynamics. Since the equations of motion in these formalisms are time non-local there is really nothing i that resembles a Hamiltonian that closely.

  • $\begingroup$ 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! $\endgroup$ Jan 11, 2021 at 19:04

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