In quantum mechanics it is usually the case that when degrees of freedom in a system are traced out (i.e. ignored), the evolution of the remaining system is no longer unitary and this is formally described as the entropy of the reduced density matrix ($S = Tr(\rho\ln{\rho}$)) attaining a nonzero value.

Why is it then that there exist certain quantum mechanical systems which have unitary evolution, but for which we clearly do not keep track of all degrees of freedom? Practically any non-relativistic quantum system such as atomic structure, spin chains, any implementation of a quantum computer etc. are clearly just effective theories which ignore the massive number of degrees of freedom underlying the more fundamental physics, i.e. quantum field theory. But somehow we are able to trace away all of this underlying physics into an effective potential/interaction term in a Hamiltonian and thus retain unitarity.

Now this does not work entirely for all systems, particularly it is impossible to hide away the full coupling of an atom to the EM field because you have spontaneous emission, a non-unitary effect. But still a large part of the electron/proton interaction (which is really mediated through the EM field) can be captured by the Coulomb potential, which ignores the EM field yet is still unitary. On the other hand, some systems, such as certain implementations of quantum computing/simulation claim to be able to achieve perfect unitarity and are only limited by imperfections in the system. At least, I have never heard of anyone talking about intrinsic unitarity limitations to a quantum computer.

My questions are:

-Under what conditions can underlying degrees of freedom be hidden away into a unitary interaction?

-Are there intrinsic limits to unitarity of any quantum system (such as an atom's unitarity being limited by spon. emission), assuming that you will always trace out a portion of the underlying field theory?

  • $\begingroup$ Here is a tangentially related post on the difficulty of creating large entangled states. $\endgroup$
    – DanielSank
    Aug 23, 2016 at 5:47
  • $\begingroup$ @DanielSank, any chance of an answer? Your hint about adiabatic elimination gave me something to go on, but you seemed so enthusiastic about answering that I'm sure you could give a fantastic response that I would be very happy to read. $\endgroup$ Sep 13, 2016 at 5:18
  • $\begingroup$ I thought about the question for a while but didn't come up with a good illustrative example. Maybe I'll try some more. $\endgroup$
    – DanielSank
    Sep 13, 2016 at 12:11

2 Answers 2


The answer to your question is a bit subtle and has to do with the various ways we can ignore degrees of freedom in physics. One way, as you mentioned, is if you have a system interacting with its environment but you don't care about the state of the environment. Then you can perform a partial trace over environmental states and obtain a reduced density matrix describing the quantum state of the system. This reduced density matrix will $not$ evolve unitarily.

However, there is another way to remove degrees of freedom in order to simplify your problem. This is called integrating out variables/fields. Suppose you have a theory that works to arbitrarily high energy, but you only care about low energy excitations of your system. You can "integrate out" the high energy modes of your system and obtain a low-energy effective field theory that is still unitary (as long as you remain at low energy). If you are familiar with QFT, a common example of this is the Four Fermi theory obtained by integrating out the very heavy W and Z bosons of the standard model.

The reason you can integrate out the W and Z bosons at low energy is that they are extremely massive compared to the other particles in the theory. Thus, at low energy they can never be produced. So when you integrate out the W and Z, you are not actually ignoring anything $physical$ as you are if you perform a partial trace over an environment. (If you were working at high enough energies to produce W and Z bosons then you would be ignoring physical particles, so your theory would no longer be unitary).

To use an example you mentioned: why can you not integrate out EM fields to get an effective theory for hydrogen atoms? The reason is that photons are massless, so no matter what energies you are working at, photons can always be created. Thus, integrating out photons would involve ignoring physical objects and would result in non-unitary behavior.

To recap: there are two ways of ignoring degrees of freedom:

(1) If you ignore physical things eg though partial traces, then you will not have unitary behavior.

(2) If you are working at low energy but simplify your theory by ignoring high energy physics, then you can obtain a low energy effective theory that is unitary. However if you push this low energy theory to high energy, it will (usually) fail to be unitary.


Mathematically it is quite simple: If your overall system unitary $U_{tot}$ can be expressed as $U_{tot} = U_1 \otimes U_2$, then the subsystems 1 and 2 have unitary evolution even when you trace out the partner.

Parameter counting can give you some very rough idea of how easy or hard this is. For instance, in a 2-qubit system the general unitary operator has 16 free parameters. A 1-qubit unitary has 4 parameters. So the total dimensionality of the separable space of unitaries is 8 out of the total 16 dimensions. As your system get bigger, the separable unitary gets to be a smaller and smaller sub-space of the total system. In the real world of course this is a gross oversimplification. Some of those parameters are automatically small or zero, others are very hard to control.

However, there is not any known intrinsic limit to how accurately a subsystem can be made unitary. Even your example of an atom in an excited state is not so limited. Spontaneous emission is a unitary and reversible process, so we can control it. By placing an atom in a cavity it is possible to suppress or enhance the spontaneous emission rate. This is known as the Purcell Effect. The ultimate limit of the atomic lifetime is then given by how well the cavity can be engineered.

  • $\begingroup$ I understand that if a unitary can be written as a product, then the corresponding subsystems can be separated. But this is just a trivial case where the two subsystems don't interact at all. I am thinking more about three subsystems: A, B, and a mediating field (e.g. two magntic dipoles interacting through the EM field). Here $U_{tot}$ cannot be decomposed into A, B, and field parts else there is no interaction at all. Yet we are still sometimes able to ignore the dynamics of the field and write a unitary operator for A,B alone (e.g. dipole potential). $\endgroup$ Aug 23, 2016 at 18:32
  • $\begingroup$ @user2640461 You are correct to question that. The essential quality is whether or not these extra degrees of freedom are adiabatic throughout the evolution. The criteria given by Evan are sufficient but not necessary. I think you're asking why the mediating field can be eliminated such that the remaining unitary is separable as Evan has written. Do I understand your question correctly? $\endgroup$
    – DanielSank
    Aug 23, 2016 at 20:03
  • $\begingroup$ More like: Can the mediating field be eliminated such that the evolution of the remaining density matrix (of A,B) is unitary? That the remaining unitary is seperable (into A and B) would be too strong, as that would again give no interaction between A and B. I have heard of the concept of "adiabatic elimination" before, are you suggesting this is what is at play? $\endgroup$ Aug 23, 2016 at 20:09
  • $\begingroup$ @user2640461 Ok got it. Yes adiabatic elimination is a related concept. I was trying to cook up a good example problem where we can solve the whole thing and see explicitly why the reduced dynamics of A and B are unitary, but I haven't come up with a complete example yet. I'll keep trying later though. This is an excellent Physics.SE post and deserves a thorough answer. $\endgroup$
    – DanielSank
    Aug 23, 2016 at 20:12

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