I was reading the paper No Universal Flipper for Quantum States. In this paper they have tried to prove by contradiction that a universal flipping machine cannot exist. By flipping I mean if I have a qubit $\alpha|0 \rangle +\beta|1\rangle$, then the flipped qubit will be $\beta^*|0 \rangle -\alpha^*|1\rangle$. In the paper they take an entangled system between two parties Alice and Bob, and apply an operation on Bob's side. But they say that the operation is not necessarily unitary ( its just linear ) but it does preserve the trace of Alice's reduced density operator before and after the operation.

But I thought on closed quantum mechanical systems only unitary operations could be applied. Am I missing something?

  • $\begingroup$ Evolution operators have to be unitary because you want the scalar product to be conserved by evolution. This is not true for e.g. observables, that are surely admitted in QM as operators (they are the "building blocks" of the theory), but are self-adjoint and not unitary. $\endgroup$ – yuggib Mar 25 '15 at 9:13
  • $\begingroup$ by observables do yo mean measurement operators ? $\endgroup$ – sashas Mar 25 '15 at 9:40
  • $\begingroup$ I mean e.g. energy, momentum, angular momentum, spin operators. Experimentally, when you perform a measurement of one of these quantities and obtain a result, you are projecting on a spectral subspace of those operators, and that again is not a unitary operation. $\endgroup$ – yuggib Mar 25 '15 at 9:49
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    $\begingroup$ the example of flipping you gave is implemented by an antiunitary operator, so my feeling is that somehow this is related to the question whether antilinear operator should be considered or not. Unfortunately I'm not an expert of quantum computing/information so I don't really know the answer to this last question. $\endgroup$ – Phoenix87 Mar 25 '15 at 14:51

Correct, the evolution of a quantum system does not necessarily need to be unitary. However, if the system is closed, it needs to be unitary. The latter is due to the fact that we assume the system evolves due to the Schrödinger equation with a self-adjoint Hamiltonian, which induces a unitary evolution operator via Stone's theorem.

For non-closed systems, the idea is the following: Let's suppose you have your system (denote it by a state $\rho$) and maybe some other system, the environment (denoted by a state $\rho_E$). Then you start out with the state $\rho\otimes \rho_E$ and you can perform any unitary operation you want, i.e. $\tilde{\rho}:=U(\rho\otimes \rho_E)U^*$. Now so far, we have dealt with what you learn in your first quantum theory course. The idea is that in our case, we are simply not interested in (or can't keep track of) the evolution of the environment. Hence the state we have at the end is the reduced density matrix of $\tilde{\rho}$ restricted to our system, i.e:

$$ \rho_{final}=\operatorname{tr}_E(U(\rho\otimes \rho_E)U^*) $$ with the partial trace over the environment. Now it turns out that this is (more or less) the most general form of what is called a completely positive map and hence we usually say that those are the maps governing quantum mechanics in closed as well as open systems. In order to keep the normalization of a state, these need to be trace-preserving, hence we say that a general quantum evolution is governed by a completely positive trace-preserving map (CPTP) or quantum channel.

Note that the only point to argue is about $\rho\otimes \rho_E$: It is not so clear that we should have to start with an uncorrelated systems, but there are good reasons to do so.

Now, maybe you don't even know how the environment looks like. These systems are called "open quantum systems" and one way to describe their evolution is by as master equation as in statistical equation (for quantum mechanics, the corresponding type is also often known as Lindblad-equation). However, the corresponding map on density matrices is also a quantum channel and thus of the form outlined above.

Another characterization of quantum channels might be helpful for your understanding: They can be axiomatized (in finite dimensions for simplicity) via:

  • A quantum channel is a linear map from matrices to matrices (we need linearity physically; see below).
  • It is a positive map, i.e. it sends positive semidefinite matrices to positive semidefinite matrices (states are positive semidefinite and our map better send states to states).
  • Moreover, if $T$ is the map, then also $T\otimes \operatorname{id}_n$, where $\operatorname{id}_n$ is the identity is also a positive map for all $n$ (this is what is meant by "complete" positivity: If our map acts just on a subsystem, the natural extension to the whole system should definitely be a valid map).
  • It preserves the trace of the matrices.

Except for maybe the third property, it should be immediately clear that the other properties should be true for any quantum evolution starting of any system - and these should also be the properties the authors use for their proof.

Linearity: Finally, let's talk about antiunitary evolutions: It turns out that a map as proposed is not linear (it's antilinear), so we have to exclude antiunitary maps, too, because they would violate the linearity of quantum mechanics. This linearity however is established by many experiments and lies at the heart of quantum theory - there is no good reason why we should abandon it.


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