# Is it correct, according to quantum mechanics, to say that an observer can get entangled with the measured system?

Entanglement applies to any physical system, observers included. However, what one typically means by the observer being entangled with the measured system seems not to comply with the requirements for such a phenomenon to take place. For example, splitting a spin-zero diatomic molecule will produce a pair of spin-entangled atoms as a result of angular momentum conservation. On any axis the spins of the atoms will be found to be anticorrelated. On the other hand, when an atom passes through a Stern–Gerlach device and is detected on a screen we need to apply momentum conservation to the entire experimental device, not only to the photons that ultimately arrive on the observer's retina. Therefore it seems to me that there can be no entanglement between the measured atom and the observer.

• If the observer does not perform a measurement, it is not clear why it should be called "observer". If it does perform a measurement, then there is no entanglement. It is confusing indeed; and it is known as the measurement problem. – Stéphane Rollandin Jul 5 '19 at 11:58
• @StéphaneRollandin: The statements you're making in the comment seem to assume the Copenhagen interpretation. – user4552 Jul 5 '19 at 12:05
• @Ben Crowell. No, I am not advocating any interpretation. The measurement problem is just an open question, and that actually is precisely the reason why we have different QM interpretations, each trying to solve it in its own way (BTW this is stated in the first sentence of the wikipedia page I linked to). Usually, it is when people say there is no measurement problem that they assume a specific interpretation. – Stéphane Rollandin Jul 5 '19 at 12:12

## 2 Answers

If a quantum mechanical particle's state is completely indeterminate - that is, if its probability of being found in state A is 50% and its probability of being found in state B is also 50% - then the observer, his/her measurement apparatus, and everything else that changes depending on the outcome of a measurement of the particle's state, becomes entangled with the particle when a measurement is performed.

From the Many Worlds perspective, the measurement causes the universal wave function to split into two independent components corresponding to the two possible outcomes of the measurement.

However "entanglement" of the observer and the particle is not something the observer detects. It would have to be detected by a second observer who can perform subsequent measurements on both the first observer and the measured particle, and can repeat his measurements multiple times on identically generated particles and (first) observers.

Edit: Though it's effectively impossible to create multiple observers all in the same quantum state, it might be possible to design an indirect experiment that can be done using a single observer to demonstrate that the observer is entangled with measured particles. I think this would boil down to a variation on Schroedinger's Cat, with the cat being the observer.

• S. McGrew, I agree that what you are saying is generally accepted. However, I see no justification for this in the quantum formalism. Entanglement appears as a result of a physical interaction between systems, such is the case with two atoms in a molecule or two electrons in an atom. It is possible for a particle to interact with the observer's retina and, if the particle is released, there will be entanglement between that particle and the observer. Such an interaction, however, does not convey any spin-related information to the observer. - to continue – Andrei Jul 5 '19 at 7:22
• -cont: A proper spin measurement, however, does not seem to result in an entangled state. If it would, one could be able to perform Bell-theorem tests using particle-observer pairs. I do not think such an experiment is possible. The correlation observer-particle seems to be a trivial classical correlation, not an entangled state. – Andrei Jul 5 '19 at 7:27
• The difficulty, of course, is in forming a lot of particle-observer pairs. A practical experiment to test the principle might need to use relatively simple multi-state molecules as proxies for observers. – S. McGrew Jul 5 '19 at 14:43
• Why is it difficult to prepare observer - particle pairs? You just ask the desired number of observers to measure the particles. – Andrei Jul 6 '19 at 5:37
• It is difficult because the observers must be identical. – S. McGrew Jul 6 '19 at 6:02

Let's say that we have a qubit. Lets also say we have a measuring device (or 'observer') which will measure the qubit in the computational basis. Our measuring device obeys quantum mechanics and returns a result $$M$$.

If we initialise our qubit in the state $$\left|0\right\rangle$$ and then measure it with our measuring device, the state of the measuring device becomes $$\left|M= 0\right\rangle$$. Similarly, if we initialise our qubit in the $$\left|1\right\rangle$$ state and meausre, our measuring device will end up in state $$\left|M=1\right\rangle$$. So far what we have described is just the definition of a measuring device which obeys quantum mechanics. If our measuring device behaved any differently to what we have just described, then either it does not obey quantum mechanics, or it is not a measuring device (eg. if, when we measured the $$\left|0\right\rangle$$ state, our device sometimes returns $$\left|M=1\right\rangle$$, it would not be a proper measurement device as it gives us wrong information).

Lets say that the measuring device is initialised to a state $$\left|I\right\rangle$$. We can therefore write the effect of the measurement process as follows:

$$\left|0\right\rangle\left|I\right\rangle \rightarrow \left|0\right\rangle \left|M=0\right\rangle$$

$$\left|1\right\rangle\left|I\right\rangle \rightarrow \left|1\right\rangle \left|M=1\right\rangle$$

Now, since quantum mechanics is linear, we can also have a qubit in a superposition state $$\left|\psi\right\rangle = a\left|0\right\rangle + b\left|1\right\rangle$$, where $$|a|^2 + |b|^2 = 1$$. What happens when we do a measurement on this state? Well, since quantum mechanics is linear our measuring device enters into a superposition:

$$(a\left|0\right\rangle + b\left|1\right\rangle)\left|I\right\rangle \rightarrow a\left|0\right\rangle \left|M=0\right\rangle + b\left|1\right\rangle\left|M=1\right\rangle$$

Notice now that the joint state of the qubit and measuring device is an entangled state. There is plenty of discussion by physicists about what qualifies as a measurement, but this is what people mean when they say a system becomes entangled with the measurement device. To satisfy your actual question, you can replace 'measurement device' with 'observer' without affecting the argument.

• asph, thanks for your detailed answer! There are still some points that I find difficult to accept. First, the states of the measurement device are not, as far as I can tell, "true" quantum states. There is no Hamiltonian written down, no Schrodinger equation, no solution of that equation being calculated. Some classical states (say pointer-left/pointer-right) corresponding to the macroscopic appearance of the instrument are assumed to be quantum states so that the rules of QM could be applied to them. Is this procedure a mathematically justified one? - to cont – Andrei Jul 12 '19 at 6:29
• -cont. Another issue I have is related to the way superposition states are prepared. If the spin of the particle is measured along X we have a spin-superposition on Z. As long as the instrument is not modified, the spin on Z being either +1/2 or -1/2 cannot generate any conflict with any possible observation due to the uncertainty principle (UP). A macroscopic superposition, such as pointer-right/pointer-left is far beyond the limits imposed by UP due to the large mass of the pointer. This would generate conflict with a measurement of the gravitational field for example. – Andrei Jul 12 '19 at 6:42
• Hi Andrei, your claim 'the states of the measurement device are not, as far as I can tell, "true" quantum states' is a controversial one! The way that many people view it is as follows: your measuring device is made out of protons/electrons etc which all obey quantum mechanics, therefore if you are claiming that the device itself does not obey QM, the onus is on you to say why. Until we have a reason to describe measuring equipment as 'non-quantum', we have to assume that they obey quantum mechanics. Also, bear in mind, that a measurement device may be something as simple as a photon. – asph Jul 12 '19 at 11:40
• Your comments about macroscopic superpositions are currently being investigated by physicists. There are many proposals to resolve the problem 'why do we not see macroscopic superpositions in everyday life?'. These include: decoherence theory (interaction with an environment makes the quantum state behave 'classically'), the many-worlds interpretation (macroscopic superpositions do exist, but the different 'branches' exist in different universes) and spontaneous collapse models (non-linear additions to the Schrodinger equation which cause superpostions to collapse when they get big enough). – asph Jul 12 '19 at 11:44
• asph, I did not intend to say that the instrument does not obey QM. An instrument in "ready" state is not unique, like the ground state of the hydrogen for example. It is a generic name for a potentially infinite number of states that have the macroscopic appearance of an instrument capable of performing the measurement. I agree that each of these states obeys QM. The same goes for the instrument+particle composite system that can be in any of a large number of "pointer-right" states or "pointer-left" states. - to cont – Andrei Jul 13 '19 at 6:44