I read that quantum decoherence is subjective, in the sense that two observers may not have the same "environment" and after each one has traced over those degrees of freedom they will end up with a different density matrix for the couple system-apparatus.

I kind of understand the idea, but has an example of this calculation already been done? With the same system looked at by two different observers, and both density matrices calculated/compared.


There are two observers and one qubit in the state $| 0 \rangle$ (known to both of them).

Then the qubit is randomly with equal probabilities either intact $U=\mathbb{I}$ or reversed $U=\sigma_x$ (i.e. $| 0 \rangle \mapsto | 1 \rangle$), but only the first observer knowns which action was applied.

Consequently, the first observer has one of the following states $$\rho = \left[ \begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array} \right] \quad \text{or} \quad \rho = \left[ \begin{array}{cc} 0 & 0 \\ 0 & 1 \end{array} \right]$$ while the second $$ \rho = \left[ \begin{array}{cc} \tfrac{1}{2} & 0 \\ 0 & \tfrac{1}{2} \end{array} \right]. $$

The random transformation $U$ can either be classical (e.g. switching a magnetic field) or quantum. For example, lets have the following state $$\left( \frac{|0\rangle + |1\rangle}{\sqrt{2}}\right) \otimes |0\rangle,$$ where the first particle is accessible only to the first observer, while the second - to both of them. After controlled-not gate we get $$\frac{|0\rangle \otimes |0\rangle + |1\rangle \otimes |1\rangle}{\sqrt{2}}.$$ The first observer can measure the first particle (so knows the state of the second), while as the second observer cannot - for him the final state is completely mixed.


Here's the simplest example I can think of. Consider a machine that can, at the flip of a switch, prepare the an electron in the spin state $|{\uparrow} \rangle$ or $|{\downarrow}\rangle$. We agree that I will flip a coin and then prepare the electron in one state or the other, without telling you which.

I flip the coin and it comes up heads, so I set the machine to prepare the electron in the state $|{\uparrow} \rangle$. For me the electron is in a pure spin state, represented by the density matrix $|{\uparrow} \rangle \langle {\uparrow}|$. From this I can calculate that if I measure the spin along the same axis that it was prepared, it will be up with 100% probability.

However, you don't know whether my coin came up heads or tails, so for you the electron could be in the state $|{\uparrow} \rangle$ or $|{\downarrow}\rangle$ with equal probability. Therefore you have to represent it with the density matrix $\frac{1}{2}|{\uparrow} \rangle \langle {\uparrow}| + \frac{1}{2}|{\downarrow} \rangle \langle {\downarrow}|$, which tells you that you have an equal probability of measuring an up or down spin, no matter which axis you measure it on.

edit: here's a second example, which is similar to the first but involves entanglement rather than a classical coin flip.

We prepare a pair of electrons in the singlet state $\frac{1}{\sqrt{2}}|{\uparrow}{\downarrow}\rangle-\frac{1}{\sqrt{2}}|{\downarrow}{\uparrow}\rangle$. (This is a maximally entangled pure state.) Now you go off into another room and measure one of the electrons' spins in the $y$ direction. Let's say it comes out to be up. Now you know that if you measure the other electron on the same axis its spin must come out to be down, so for you the other electron is in the pure state $|{\downarrow}\rangle$, represented by the density matrix $|{\downarrow} \rangle \langle {\downarrow}|$. But I don't know what the outcome of your measurement was, so for me the remaining electron is still in the mixed state $\frac{1}{2}|{\uparrow} \rangle \langle {\uparrow}| + \frac{1}{2}|{\downarrow} \rangle \langle {\downarrow}|$.

  • $\begingroup$ Thank you for this explanation, but I don't think it really answers my question (or at least I don't see it) I was asking specifically about the process of decoherence, how can it be subjective? I know that two observers who don't share the same information about the preparation of the same system will describe it with two different density operators, but if they are both looking at the same all big general density operator (system apparatus environment), how can they disagree on the environment degrees of freedom to trace over, and finally have different reduced density operators? $\endgroup$ – toot Apr 10 '12 at 16:02
  • $\begingroup$ I see. I'd sort-of missed the point about decoherence, even though it's in your title. I have a rough idea of how to answer it, but it would involve a bit of conjecture on my part. (As in, I can see how something like what you describe could happen, but I've no idea if what I'm thinking of is the same as what you've read about.) I'll edit my answer to cover it once I've had a bit of a think about it, if nobody else gives a good answer first. $\endgroup$ – Nathaniel Apr 10 '12 at 18:52

When you do a measurement, the information of the phase of the resulting projected eigenstate with the former full state is lost to you, but it is still available to other observers.

This is because of the nature of entanglement; when two systems initially unentangled interact, assuming one of them is largely classical and the other is quantum (in superposition of multiple non-degenerate states), the final state is largely described by a sum of entangled pairs, of quantum eigenstate $\times$ a classical system coupled to that eigenstate. This is the Everett picture of the measurement.

What eigenstate basis? it depends on the hamiltonian of the measurement apparatus, different apparatus will couple to the quantum system differently, making the entanglement along different basis.

If you think on the sums of the resulting state after measurement, the i-th classical observer that sees the i-th eigenstate, did lose all phase information, and worse any physical way to retrieve it back, even in principle. After a quantum system has been projected to your apparatus, there is nothing you can do to retrieve the non-projected component of the wavefunction. Measurements are inherently non-unitary steps of a system evolution.

But this is not agreed by all observers; other observers (let's call them Meta) that have not interacted directly with either the system or the observer (described by the classical component) will still see both systems to interact and evolve unitarily at all times; if he is clever enough and make the appropiate measurements he can confirm that the measurements done by the classic observer has not erased any phase information, but for all practical purposes, the resulting phase of the coupled pair is a complex result of both systems interaction and initial phases, so it would in general average out as random noise.

So in short, unitarity as a history of the evolution of systems is not an universal property agreed by all observers, it will depend on what systems they are entangled and what measurements (intentional or not) have they done. Note however, that this is not the same as "subjectivity", but more akin to how different observers in relativity might see different quantities like elapsed time, curvature, etc.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.