How much can an observer know about its own quantum state?

Question

Let me state my background: I am familiar with the usual formalism of QM, of how measurements are modeled, measurement problem, decoherence, etc.

In thinking about the process of measurement, I was thinking about how an observer's knowledge is encoded in its state and how much the observer can know about its state. This is not so much a question, rather it is a request for relevant literature on the topic.

Here are more specific examples on my thoughts/things I've read elsewhere to give you the 'vibe' of what I'm looking for and the interpretation I'm leaning towards:

Ignore collapse for a second, imagine an idealized situation where we are able to prepare a human in a coherent superposition of two macroscopically distinct states. Could the observer even tell? How? My gut says it can't because whatever property of iself it tries to measure, it needs some outside reference, some interaction with another system. For example, the reason I know I am not in a superposition of two far away positions is... I don't really know since there's no absolute reference. What I know is that I sit inside a room and the room is inside a building and the building has those trees around it and so on. So what I know is how my position is correlated with the positions of other systems I interact with.
Again, let's try to ignore collapse, entertain the idea that dynamics are completely unitary. Say there's an isolated spin 1/2 particle prepared in the state $\frac{1}{\sqrt{2}}( | \uparrow \, \rangle + | \downarrow \, \rangle)$. Say we measure it with some kind of apparatus that has a pointer that goes to one side if its spin up and to another if its spin down. Then the observer looks at the pointer. The state after this kind of unitary evolution would be $\frac{1}{\sqrt{2}}( | \uparrow,\ pointer\ \uparrow,\ observer\ sees\ pointer \uparrow \rangle + | \downarrow,\ pointer\ \downarrow,\ observer\ sees\ pointer \downarrow \rangle)$. Now, the observer cannot possibly know/experience this whole state since there's nothing in the state that corresponds to that knowledge/experience. And it's hard for me to conceive of this as something other as two equally real 'quantum trajectories' for this observer.

The two examples stated above make me lean towards the Many-Worlds interpretation. And the reason is that while the usual formalism does include collapse, it doesn't strictly prohibit these kinds of unitary evolutions. And I have a hard time interpreting them in any way other than some kind of branching happening.

To be completely honest, I have tried to avoid "believing" in Many-Worlds interpretation for a long time, but these thought experiments have me stumped.

score 11 · Answer 1 · answered Feb 05 '25 at 11:06

While word observer is routinely used in quantum mechanics, it does not refer to an intelligent human being making observations, but rather to interaction with any classical object (see, e.g., When will a wave function collapse if the observer was only a camera and the video was watched later in time?.) In other words, the problem of observation and the problem of knowing/awareness are two distinct problems - one is physical, the other is philosophical.

Now, if we assume for a moment that we "believe" in quantum mechanics, so that we do not treat observer as an extraneous classical object, but rather as a quantum object behaving in classical limit. This object is the behaving classically because:

it was from the beginning in a mixed state (rather than a pure state describable by a wave function)
it constantly interacts with the environment (decoherence)

Although macroscopic objects can in principle be put in a quantum state, their complexity makes it difficult to achieve this. While we still can put in a quantum state a buckyball molecule, a large system such as a human is well in thermodynamic limit. In other words, probability of putting a human in a quantum state is like probability of parting waters of the Red Sea - technically not zero, but for all the practical purposes negligeable (we are talking about probabilities of order $e^{-N_A}$, where $N_A=10^{23}$.)

Another thing to consider is that biological processes (human thinking included) are irreversible thermodynamic processes - incessant decoherence is an essential part of being alive. In this sense, a person in a quantum state is not alive... or at least not capable of thinking/knowing.

score 7 · Answer 2 · answered Feb 05 '25 at 12:05

If someone (or something) is in a superposition of two states, say

$$ |\psi\rangle = \frac{1}{\sqrt 2}(|A\rangle + |B\rangle) $$

then linearity of the time evolution operator implies that it will always remain in such a superposition

$$ U(t)|\psi\rangle = \frac{1}{\sqrt 2}(U(t)|A\rangle + U(t)|B\rangle) \equiv \frac{1}{\sqrt 2}|A(t)\rangle + |B(t)\rangle $$

and if $|A\rangle$ and $|B\rangle$ are orthogonal then so will be $|A(t)\rangle$ and $|B(t)\rangle$. The future state $|A(t)\rangle$ is clearly independent of the other parts of the superposition and therefore cannot conatain any information about them.

alanf · Accepted Answer · 2025-02-05T15:07:17.080

There are not many papers on the topic of what an observer can know about his own quantum state, although I will cite some exceptions below.

One reason for this is that an observer in reality is going to be a complex physical system with multiple subsystems. As a result, there is a bit of fuzziness about what you choose to regard as part of the observer. An observer could look at one of his parts as a separate system and then measure it and understand the measurements as he would for any other system. This is one reason why trying to make an observer a fundamental part of physics doesn't make a lot of sense.

So then you would treat that measurement along the lines of standard quantum measurement theory. Standard measurement theory includes repeated, continuous and unsharp measurements and collapse doesn't work well for treating those

https://arxiv.org/abs/1604.05973

However, once you start treating the observer as a quantum system this opens up other possibilities for doing experiments. In Section 8 of his 1985 paper "Quantum theory as a universal physical theory" David Deutsch gives an account of an experiment where an AI implemented on a quantum computer could do an interference experiment on himself and know that he was in an unsharp state during that experiment:

https://boulderschool.yale.edu/sites/default/files/files/Deutsch.pdf

Deutsch interprets this as a test of the many worlds interpretation (MWI) versus the Copenhagen interpretation (CI) since the CI wouldn't allow an observer to undergo interference and the MWI would.

There are a lot of Wigner's friend type experiments on the quantum theory of observers such as the Frauchiger and Renner paper:

https://arxiv.org/abs/1604.07422

and there are other more recent papers along similar lines:

https://arxiv.org/abs/2209.06236

https://arxiv.org/abs/2407.06279

There is a large literature on this sort of thing and the papers above their citations and references to them on Google scholar might provide you with a way to start to look into it.

score 4 · Answer 4 · answered Feb 05 '25 at 13:04

The simple answer is that an "observer" in quantum mechanics is by definition a classical system or measuring device. As stated in other answers, there is no requirement that the "observer" needs to be conscious - however, the "observer" needs to behave as a classical system in order to produce a definite measurement. If the "observer" behaved as a quantum system then it would be in a superposition of states and could not produce a definite measurement or "observation" - unless it was in turn "observed" by some other system, which gives you an infinite regress.

Since an "observer" behaves as a classical system it does not have a quantum state (or, more precisely, its quantum state is the same as its classical state). There is therefore no problem with the "observer" being "aware" of or recording its own state. This is exactly what it does when it produces a measurement.

score 1 · Answer 5 · answered Feb 05 '25 at 13:55

Let $U$ be the state of the pointer-plus-observer system in which the pointer is up and the observer sees that it's up. Let $D$ be the state in which the pointer is down and the observer sees that it's down.

For purposes of this answer, call $U$ and $D$ the "definite" states of the system.

Now as the observer, ask yourself: "Did I see the system in a definite state"? Answering that question constitutes a measurement on the system, represented by some Hermitian operator $H$.

If the system is in either of the two definite states, that measurement presumably returns the answer "yes". In other words, both $U$ and $D$ are in the "yes" eigenspace of $H$.

It follows from the linearity of $H$ that any linear combination of $U$ and $D$ --- including the state of the form you've displayed in your point (2) --- is also in the "yes" eigenspace of $H$.

So an observer in that state, when asked "did you see the pointer in a definite state?" will answer "yes", although he did not in fact see the pointer in a definite state. He will answer this way when you query him, and he will answer this way when he queries himself.

This suggests that the observer is unable to distinguish between having seen, or having not seen, the pointer in a definite state.

Steven Sagona · Answer 6 · 2025-02-07T08:00:38.440

In short: For a state, such as $|\psi\rangle = \frac{1}{\sqrt{2}}(|A\rangle + |B\rangle)$ that has been already observed by an observer, that observer cannot measure the quantum state because it has collapsed in their perspective. You are correct that this is in contradiction with an outside observer who will see that the observer is in a superposition of having measured A and B, $ \frac{1}{\sqrt{2}}(|A\rangle |O_A\rangle + |B\rangle|O_B\rangle)$.

So much of what you are asking is similar to the Wigner's friend thought experiment. Everett (inventor of MWI) was a student of Wigner and these thought experiments led directly to the many-worlds interpretation. In my opinion the wigner's friend thought experiment is "proof" of the many-worlds.

To competely rephrase your problem in terms of the thought experiment:

Consider the Wigner's friend experiment. We have a Friend $|F\rangle$ who makes a measurement of a system in $\frac{1}{\sqrt{2}}(|A\rangle + |B\rangle)$. As we know, to Wigner who is outside the experiment, the friend is in a superposition of two outcomes:

$$ |F_A\rangle|A\rangle + |F_B\rangle|B\rangle $$

However to the friend, who has made the measurement, the outcome is, with certainty, either A or B. However to Wigner this is clearly in a superposition state.

In the Friend's perspective, the state IS A. The friend cannot measure that it is in a superposition state because it is NOT in a superposition state in his perspective. It is very much in the state A. And he knows with 100% certainty that it is in A.

You are correct that this is in direct contradiction with Wigner's perspective. In Wigner's prespective, there is a superpostion of the friend seeing state A, and the state being A, which is in superpostion with the friend seeing state B and the state being in B.

I discuss this in greater detail in the supplemental of this paper. Personally, I can only see this being resolved via the MWI, as did those who came up with the thought experiment.