# Modeling a measurement as unitary

1. Jun 19, 2014

### Staff: Mentor

This question is inspired by reading Arnold Neumaier's article on Everett's measurement theory:

http://www.mat.univie.ac.at/~neum/physfaq/topics/everett

Suppose I have a system whose state lies in a two-dimensional Hilbert space, and I want to model a measurement of its state as a unitary evolution. As I understand it, the standard way to do this is to first find a basis for the system's Hilbert space; let's call the two basis states $\vert \psi \rangle$ and $\vert \chi \rangle$. Then we have to model the system doing the measurement; this system will have to have a three-dimensional Hilbert space with basis states $\vert R \rangle$, $\vert \Phi \rangle$, and $\vert X \rangle$, where $\vert R \rangle$ is the "ready" state and $\vert \Phi \rangle$ and $\vert X \rangle$ are the states that indicate that $\vert \phi \rangle$ and $\vert \chi \rangle$, respectively, have been measured.

Then the measurement process is modeled by assuming that the Hamiltonian of the total system, which I won't write down explicitly, induces the following:

First, for $\vert \Phi \rangle$ and $\vert X \rangle$ to be proper "indicator" states, we must have

$$\vert \phi \rangle \vert \Phi \rangle \rightarrow \vert \phi \rangle \vert \Phi \rangle$$

$$\vert \chi \rangle \vert X \rangle \rightarrow \vert \chi \rangle \vert X \rangle$$

i.e., both of these states are left invariant by the Hamiltonian (because the measuring system already indicates the "correct" state for the measured system). Then, for the measuring system to actually make a measurement, we must have

$$\vert \phi \rangle \vert R \rangle \rightarrow \vert \phi \rangle \vert \Phi \rangle$$

$$\vert \chi \rangle \vert R \rangle \rightarrow \vert \chi \rangle \vert X \rangle$$

i.e., the "ready" state gets changed to the appropriate "indicator" state by the measurement.

However, it seems like there is a problem with the above evolution: it can't be unitary, because it's not invertible! Note that the states $\vert \phi \rangle \vert \Phi \rangle$ and $\vert \chi \rangle \vert X \rangle$ appear on the RHS of *two* separate evolution rules, with *different* states on the LHS. That means that, for example, if we find the system in state $\vert \phi \rangle \vert \Phi \rangle$, if the above rules are correct, we can't tell whether that state arose from $\vert \phi \rangle \vert \Phi \rangle$ in the past, i.e., because the system was just sitting there unchanged, or whether it arose from $\vert \phi \rangle \vert R \rangle$ in the past, i.e., because a measurement took place.

So it seems like there's no way to model what we usually think of as a "measurement" by a unitary operator. Am I missing something?

Last edited: Jun 19, 2014
2. Jun 19, 2014

### Staff: Mentor

Its the decoherence thing. Obviously observation is not unitary - that has been known for ages.

The system, environment, and observational apparatus all evolve unitarily. But the entangled system via tracing over the environment doesn't have to - which is how a superposition is transformed to a mixed state.

Thanks
Bill

3. Jun 19, 2014

### Staff: Mentor

But the non-unitarity is not supposed to be fundamental, correct? It's supposed to be possible, at least in principle, to write down a unitary dynamics for the entire system; things only appear non-unitary when you restrict attention to subsystems. I'm trying to understand how, in principle, we write down the unitary dynamics for the entire system.

I'm still not clear on how we would write this down in a simple case like the one in the OP. That is, I don't see how adding the environment to the mix changes anything. What you're saying here basically amounts to saying that the non-unitary evolution rules I wrote down in the OP can be obtained from the correct unitary rules by tracing over the environment; but how would that work?

4. Jun 19, 2014

### atyy

If there is a single outcome for each measurement trial, then measurement cannot be unitary.

What a description involving the environment, measurement apparatus and system can do is to evolve unitarily, so that the system comes close to a state in which there is a unique basis in which the system is diagonal, thus picking out the measurement outcomes. If the environment is not included, then the apparatus and system evolve unitarily, but there isn't a unique basis indicating the measurement outcomes. Compare Eq 2.2, 2.3 with Eq 3.5 in http://arxiv.org/abs/quantph/0312059.

If one does not insist on a single outcome, ie. all outcomes occur, then there is some hope for unitary evolution to be preserved via a many-worlds interpretation. I'm not sure all problems with many-worlds have been solved, but one that seems pretty convincing is Deutsch-Wallace's version http://arxiv.org/abs/0906.2718.

5. Jun 19, 2014

### Staff: Mentor

I am not sure I understand your issue.

But the situation is this.

When a system becomes entangled with the environment there are other factors influencing its evolution than its Hamiltonian - that's why its possible for its evolution to not be unitary. Technically you do a partial trace of the system and environment (see Lubos's answer):
http://physics.stackexchange.com/qu...ake-the-partial-trace-to-describe-a-subsystem

Since we are only interested in observation of the system, not the system and environment, it is now described by the state after it has been traced over the environment. That's how non unitary evolution enters into it.

I really cant explain it any better, and if it hasn't resolved your issue I may have to leave it to someone else.

Thanks
Bill

6. Jun 19, 2014

### Staff: Mentor

This is the Achilles Heal of decoherence as the explanation of measurement - it looks like a proper mixed state - but isn't really. Basically you assume it is ie place the von Neumann cut right there. That is permissible - but the collapse is still there, hidden, and non unitary 'collapse' has occurred.

MW does indeed help solve the issue since no collapse has occurred, and Consistent histories as well, since that doesn't even have an observation being the stochastic theory of histories, but the ignorance ensemble interpretation stands powerless - a collapse has occurred - somehow.

Thanks
Bill

7. Jun 19, 2014

### Staff: Mentor

I think these comments capture at least part of what I was getting at. But I'm still not sure they capture all of it. I'll see if I can think of a better way to formulate the issue I'm trying to get at.

8. Jun 19, 2014

### Staff: Mentor

If its that measurements cant be described unitarily - they sure cant - despite what some people who misinterpret decoherence may tell you - I know - I nearly fell into that trap myself, but there were those around here that cured me of it.

However you may find chapter 8 and 9 of Ballentine helpful.

Thanks
Bill

9. Jun 20, 2014

### kith

I don't think that the issue with unitarity here is the problem of definite outcomes (or collapse) because Peter's problem already occurs if the system has well-defined properties (with respect to the measurement basis).

Peter's argument is concerned only with the physical interaction between the system and the apparatus which is described by an appropriate Hamiltonian. He notes that two different initial states with well-defined system properties may be mapped to the same final state. This implies that the time evolution operator is not invertible so it can't be unitary.

Another formulation of the problem would be "if a system in an eigenstate doesn't change it's state, how can it ever get there in a fundamentally reversible way?". This sounds much like a thermodynamical question and I think in both cases, the answer is the same: we have to include the environment.

So I would say that the state of the environment changes in such a way that the time evolution of system+apparatus+environment is reversible and unitary. The time evolution of system+apparatus alone is not unitary and may be approximated by something like the Lindblad equation.

10. Jun 20, 2014

### Staff: Mentor

Yes, this is a good brief statement of the problem. Collapse doesn't come into it, because I was only talking in the OP about eigenstates of the object (the system being measured).

In other words, you're saying that the evolution rules I gave in the OP should be modified to look something like this:

$$\vert \phi \rangle \vert R \rangle \vert E \rangle \rightarrow \vert \phi \rangle \vert \Phi \rangle \vert E_{1A} \rangle$$

$$\vert \chi \rangle \vert R \rangle \vert E \rangle \rightarrow \vert \chi \rangle \vert X \rangle \vert E_{2A} \rangle$$

$$\vert \phi \rangle \vert \Phi \rangle \vert E \rangle \rightarrow \vert \phi \rangle \vert \Phi \rangle \vert E_{1B} \rangle$$

$$\vert \chi \rangle \vert X \rangle \vert E \rangle \rightarrow \vert \chi \rangle \vert X \rangle \vert E_{2B} \rangle$$

where the $E$ states are the various environment states. In other words, by looking at the state of the environment after the interaction, I can in principle tell, for example, whether the object-detector state $\vert \phi \rangle \vert \Phi \rangle$ came from starting state $\vert \phi \rangle \vert R \rangle$, i.e., from a measurement, or from starting state $\vert \phi \rangle \vert \Phi \rangle$, i.e., just from the fact that it is an eigenstate of the object-detector subsystem. In practice I can't do this because I can't actually measure the state of the environment that precisely, but in principle I could.

11. Jun 20, 2014

### Demystifier

You CAN know that. If you know the Hamiltonian, or equivalently if you know the unitary operator responsible for this evolution, then you know whether the system was just sitting there or whether it arose from another state.

12. Jun 20, 2014

### stevendaryl

Staff Emeritus
I think that there are two different things going on in measurement, and I'm not completely sure how to "disentangle" them. First, there is an entanglement, where you really have at least three correlated systems: (1) The particle that you're measuring. (2) The device you're using to measure it. (3) The environment (everything else in the universe that isn't covered by (1) and (2)). Because of (3), you can't model the situation as a product state of the form $|\psi \rangle | \Psi \rangle$ where $|\psi \rangle$ is the state of the particle, and $|\Psi \rangle$ is the state of the detector. If you want to ignore the environment, you have to trace over the states of the environment, and get a density matrix, rather than a pure state.

The second problem, which I think might be different, or might be the same--it's not clear to me--is irreversibiliy. In a detection event, a small cause, a tiny particle, causes an irreversible change to a macroscopic object, the detector. I don't know how you can model irreversible processes quantum mechanically. Microscopically, things are always reversible, so irreversibility must be some kind of law of large numbers thing, but I don't really understand how it comes about, in detail.

13. Jun 20, 2014

### Staff: Mentor

But in practice I don't know the full Hamiltonian. I know the object-detector part of it, but I don't know the environment part.

Also, even if I knew the environment part of the full Hamiltonian, how would that help me if I couldn't tell which state the environment was in? If I don't have any way to tell whether the environment is, for example, in state $\vert E_{1A} \rangle$ or state $\vert E_{1B} \rangle$, then even if I know the object-detector subsystem is in state $\vert \phi \rangle \vert \Phi \rangle$, I can't tell which prior state this state came from, because to know that I have to know the environment's state.

14. Jun 20, 2014

### stevendaryl

Staff Emeritus
I might not be understanding the concern, but it seems to me that you PREPARE your measurement device so that it is in the "ready" state $|R\rangle$. If you later find it in the state $|\Phi \rangle$, then you know that it is because it detected a particle in the $|\phi\rangle$ state.

15. Jun 20, 2014

### Staff: Mentor

Yes, that was basically what I was getting at in the OP.

I think irreversibility, as far as the question I asked in my OP goes, is simple: the evolution rule in my OP looks irreversible, but that's because I left out the environment states; putting them back in makes it reversible. In practice, we can't tell apart the different environment states in the reversible evolution rule, so the rule looks to us like the irreversible rule I gave in the OP; but the true rule, if the microphysics has to be reversible, must be the rule I gave in my later post, with the environment states included.

I think there are other issues that come under the heading of "irreversibility", but those have more to do with initial conditions than evolution rules.

16. Jun 20, 2014

### Staff: Mentor

No, you don't, because just by looking at the state of the measuring device, you don't know if it was prepared previously or not. If it helps, pretend that I'm the one in charge of the measuring device, but I'm secretive and don't tell anybody whether I actually prepared it or not; so when you come along and look at its state, you have no way of knowing whether it's in that state because it just measured the object, or whether it has just been sitting there, and I didn't touch anything at all.

The key is that $R$ and $\Phi$ are both eigenstates of the measuring device subsystem, and eigenstates are left invariant by the Hamiltonian, so if all I know is the state of the subsystem, and it's in an eigenstate, I can't tell how it originally got into that eigenstate. I have to look at the full system, including the environment, to know for sure what happened, because only the evolution of the full system is reversible.

17. Jun 20, 2014

### stevendaryl

Staff Emeritus
I guess I'm still not comprehending the problem. To me, a detector can only function as a detector if I have some control over its initial state. For example, I can use photographic film to detect whether an object has been exposed to light. If it has been exposed, then the film will be dark, and otherwise, it will be light. But that only works if I know that INITIALLY the film was in the state of being unexposed. If someone manufactured it so that it started out dark, then I certainly can't use its darkness to detect whether it's been exposed to light.

So, to me, it only makes sense to talk about a device being a detector (of whatever) if I have reasonable assurance that the device started out in a "nothing detected" state. If I have no control whatsoever over the initial state, then I can't learn anything by looking at its state now.

18. Jun 20, 2014

### The_Duck

I don't see why you want to enforce these conditions. If you drop them your problem disappears, and they are not necessary to model a realistic measurement. Consider probing the state of an atom by firing a photon at it. The post-measurement state "atom + scattered photon" is not invariant under time evolution.

19. Jun 20, 2014

### Staff: Mentor

In a practical sense, yes, this is true, and the oversimplified model of measurement I've been using in this thread does not account for that, because there's no evolution rule that gets the detector into the $\vert R \rangle$ state from any other state. A more realistic model would have to include that.

But doing so would raise the same kinds of issues, because in order to reset the detector into the "ready" state, the evolution rule as restricted to the detector subsystem has to be able to map multiple initial states, like $\vert \Phi \rangle$ and $\vert X \rangle$, to the same final state, $\vert R \rangle$. This is obviously not invertible as it stands, so again we have to add the environment states in order to make the full evolution rule unitary. (I'm assuming that resetting the detector does not change the state of the object being measured at all.)

20. Jun 20, 2014

### Staff: Mentor

See my response to stevendaryl.

True, but there are other measurements that don't work like this. For example, in a Stern-Gerlach measurement on an atom, the post-measurement state has the atom's spin correlated with its momentum, and is (at least roughly speaking) invariant under time evolution (because spin eigenstates and momentum eigenstates both are).

21. Jun 20, 2014

### stevendaryl

Staff Emeritus
Well, to continue with my analogy of photographic plates to detect the presence of light: Obviously, the way that we produce unexposed photographic plates is not by reversing the process that darkens them in the first place. The process for putting it in the initial (unexposed) state is completely different from the process that the process that takes it from unexposed to exposed.

I think that's pretty much true in general for detectors. We have a process to reset the detector in its initial state, but that process is not anything like reversing the detection process. To give another example: Jiggling a set mouse trap will cause it to snap shut, but the way you set the mouse trap in the first place is NOT by jiggling it in a time-reversed manner.

So, even if the process of detection is reversible, at a microscopic level (as it must be, since the fundamental laws are reversible), that isn't particularly relevant to the process of resetting the detector in its initial state.

22. Jun 20, 2014

### Strilanc

Related: in quantum computing, measurement is basically equivalent to a controlled-not gate. You flip the value of an otherwise unused wire in the parts of the superposition where the "measured" wire is true. Controlled nots are unitary.

There's also the deferred measurement principle. Delaying measurement until the end of a circuit doesn't change the resulting probabilities. The delayed-measurement circuit would be harder to implement, since you have to isolate more stuff to run it, but gives the same answer.

Last edited: Jun 20, 2014
23. Jun 20, 2014

### The_Duck

A Stern-Gerlach apparatus is just the same as my example of scattering a photon. The post-measurement state of the Stern-Gerlach apparatus is *not* invariant under time evolution, and it can't be, for the reasons you've spelled out in this thread. I don't see why you want to say that it's "roughly" time-invariant. It's only "roughly" time-invariant in the sense that a wide wave packet can take a long time to cross its own width. But in fact the Stern-Gerlach measurement takes at least one packet-width-crossing-time to complete, so on the relevant time scale the wave packet is not time-invariant.

24. Jun 20, 2014

### Staff: Mentor

That's because photographic plates are not reusable. But a reusable detector should be capable of being modeled (in an oversimplified way) as I described. I agree that a different kind of model would need to be constructed for non-reusable detectors.

It doesn't have to be. All it has to do is map multiple initial states (the "indicator" states of the detector) to one final state (the "ready" state), and for any reusable detector, resetting it *has* to do that, otherwise it's not a reusable detector. As I said above, I agree that non-reusable detectors require a different model; but that just means the current discussion needs to be restricted to reusable detectors.

But if the microscopic laws are reversible, then resetting a reusable detector must also be reversible, just as the detection process is. (So must the process of constructing and initializing a non-reusable detector, of course; but as above, that requires a different model than the one we're currently discussing.)

25. Jun 20, 2014

### Staff: Mentor

I didn't say the post-measurement state of the *apparatus* was invariant under time evolution; I said the post-measurement state of the *observed atom* was. The post-measurement state of the observed atom is just the tensor product of a spin eigenstate and a momentum eigenstate (at least, in the oversimplified model I'm using--see below), and such a state is invariant under time evolution.

Because in any real experiment, the post-measurement state of the observed object won't be an exact eigenstate of momentum. But in principle we could collimate the output beams sufficiently to make them eigenstates of momentum.