# Physical Interpretation of Non-Trace Preserving Quantum Operations

In Chapter 8 of Nielsen and Chuang's Quantum Computation and Quantum Information, a mathematical framework is developed to describe the dynamics of open quantum systems. Suppose the initial state of principal system is $$\rho$$ and the initial state of the environment is $$\rho_{env}$$. Let the principal-system-environment system start out in the product state $$\rho \otimes \rho_{env}$$. Let's say after some unitary evolution $$U$$, we want to extract the "evolved" state of the principal system. We do this by tracing over the environment i.e. averaging over the states of environment as shown below $$\mathcal{E}(\rho) = tr_{env}(U(\rho \otimes \rho_{env})U^{\dagger})$$ Let $${|e_k\rangle}$$ be some orthonormal basis of the state space of the environment and $$\rho_{env} = |e_0\rangle\langle e_0|$$. Then we have

Now in the text, it is said that $$tr(\mathcal{E}(p))$$ can be less than 1 or $$\sum_k E_kE^{\dagger}_k . But how can this be true? Is the basis $${|e_k\rangle}$$ not complete?

In the text, non-trace preserving that follow the above properties are said to "not provide a complete description of the process that may occur in the system". An example is used to illustrate this :

Granted $$\mathcal{E}_0(\rho)$$ and $$\mathcal{E}_1(\rho)$$ don't provide full description of the system but that is only because the basis for these operations are not complete. So are non trace preserving quantum operations those that don't account for a complete orthonormal basis of environment? If not, is there a better example or physical intuition of understanding this "non trace preserving" property?

Let's consider a simple example to try and clarify the issue.

Consider a one-qubit state $$|\psi\rangle=\alpha|0\rangle+\beta|1\rangle$$ measured in the computational basis. You can describe the results of such a measurement via the map $$\Phi$$ acting as $$\newcommand{\ketbra}[1]{|#1\rangle\!\langle #1|}\Phi(|i\rangle\!\langle j|)=\delta_{ij}\ketbra i,$$ so that $$\Phi(\ketbra\psi)=|\alpha|^2\ketbra0+|\beta|^2\ketbra1.$$ You can equivalently write this map as $$\Phi(\rho)=\operatorname{Tr}_E[U(\rho\otimes\ketbra0)U^\dagger]$$ with $$U$$ such that $$U|0,0\rangle\equiv|0,0\rangle$$ and $$U|1,0\rangle=|1,1\rangle$$ (more generally, any unitary $$U$$ such that $$U|0,0\rangle=|0,u\rangle$$ and $$U|1,0\rangle=|1,u_\perp\rangle$$ with $$\langle u|u_\perp\rangle=0$$ does the job). The corresponding Kraus operators are then $$E_0=\ketbra0$$ and $$E_1=\ketbra1$$.

Now, what happens if I consider another map $$\tilde\Phi(\rho)=E_0\rho E_0^\dagger$$? Clearly this is not trace-preserving, and the physical interpretation of this is that you are not describing the full array of possible outputs. Because any physical process always gives some output state (by the definition itself of what "process" means here), there must always be a trace-preserving map describing it.

Another example is $$\Phi(\rho)=\ketbra0$$, which corresponds to $$E_0=|0\rangle\!\langle0|$$ and $$E_1=|0\rangle\!\langle 1|$$. Now a corresponding non-trace-preserving version could be $$\tilde\Phi(\rho)=E_0\rho E_0^\dagger$$. This is again non-trace-preserving, but the interpretation is slightly different than before. Rather than ignoring possible outputs, we are considering only one of the output environment states. You could think of this as focusing on what happens on $$\rho$$ when the environment has been found in the state $$|0\rangle$$, that is, as the output state of the system post-selected over an environment state.

More generally, a non-trace-preserving map can be thought of as describing the output state post-selected on a subset of the possible environment states. I should mention that it can be debatable whether such an interpretation of a map as generated by measuring the environment is sensible. In some cases, e.g. when the map describes a measurement, whether you accept such an interpretation as sensible boils down to whether you choose to interpret collapse as due to environmental decoherence or something else. Nevertheless, mathematically things can always be described in this fashion.

A next question would be: when is it useful to use non-trace-preserving maps to describe what happens in a physical scenario? I don't think I ever encountered such a situation. If anyone can think of any please let me know.

• Why do you delete the space between the ket and the bra in outer products? – tparker May 22 at 23:38
• @tparker uh, I just think $|1\rangle\!\langle2|$ looks much better than $|1\rangle\langle2|$. It's fine for something like $|\psi\rangle\langle1|2\rangle$ though – glS May 23 at 7:04
• Interesting. I personally think the opposite. – tparker May 23 at 13:28
• @tparker fwiw, the latex physics package also defines \ketbra without the middle space, so it's at least not just me – glS May 23 at 13:44
• Interestingly, the angle brackets render differently on different devices. On my laptop and tablet, $\rangle \! \langle$ gets merged into a single contiguous symbol that looks like an $X$. But on my phone screen, there's still a small space between them, and a noticeably larger space between the brackets in $\rangle \langle$ than on the larger devices. – tparker May 23 at 17:24

The basis $$| e_k\rangle$$ is complete, and quantum operations that take the form in your question are indeed always trace-preserving. Non-trace-preserving quantum operations are associated with measurements, which cannot be straightforwardly described by the unitary time evolution assumed in your question.

In the Copenhagen interpretation, it's just postulated that measurement processes aren't unitary, so the unitary time evolution assumed in your question doesn't apply. In the many-worlds interpretation, all time evolution is unitary, and any apparent non-unitarity from measurement processes just means that you aren't considering the full wave function of the universe, which allows superpositions large enough to encompass the observer. In this interpretation, if you need to use a non-trace-preserving quantum operation to describe the time evolution of your system, then you didn't actually purify your environment Hilbert space "large enough" to capture the rest of the universe, including yourself - because if you had, then you'd get time evolution of the form described in your question, which as you point out must be trace-preserving.