# How does the wavefunction "know" the observables?

1. Dec 8, 2015

### Coherence

The thread title is probably confusing, but I couldn't really think of a better one. There is a basic feature of quantum mechanics that I have been puzzled by for a long time. My guess is that my issue stems from some fundamental misunderstanding of the theory, so I would appreciate any efforts to help clear things up!

Let's prepare a simple quantum system, like a particle in a one-dimensional infinite square well. If we now measure the energy of the particle at time t = 0, we get some value E_n and its state "collapses" into an energy eigenstate |E_n>, which can be represented in the position basis as a standing wave in the well. Energy eigenstates (equivalently, eigenstates of the Hamiltonian) are stationary states, meaning that we can measure the energy of the particle again at a later time t = t1 and obtain the same energy value E_n. We can then repeat the energy measurement at an even later time t = t2 and should still obtain the same energy value E_n.

Now, let's imagine that instead of measuring the energy again at time t = t1, we now measure the position of the particle at t = t1. We get some value x, and the state "collapses" from the energy eigenstate |E_n> into a position eigenstate |x>, which can be represented in the position basis as a delta function. |x> is obviously not an eigenstate of the Hamiltonian and therefore not stationary. However, we can express |x> as a super-position of energy eigenstates, each one of which has unitary evolution. At time t = t2, the state of the particle is no longer a position eigenstate, but can still be expressed as a super-position of energy eigenstates. If we now measure the energy of the particle at t = t2, there is some probability that we will once again measure value E_n and "collapse" the state back to |E_n>. However, there is also now a probability that we will measure a different energy E_m and "collapse" into a different state |E_m>.

Nothing that I've described so far is mysterious to me from the mathematical formalism of quantum mechanics (i.e., non-commuting operators representing observables that do not have simultaneous eigenstates). My problem is this: how does the wavefunction "know" which observable was measured? In the first example that I described, the energy was measured three times (at t = 0, t = t1, and t = t2), always yielding the result E_n. In the second example that I described, the energy was measured at t = 0, yielding E_n. Then the position was measured at t = t1. And then the energy was measured again at t = t2, generally yielding a different energy E_m. The position measurement at t = t1 clearly has an effect on the result of the energy measurement at t = t2. But how is the wavefunction "aware" of the quantity that is being measured? Certainly we can devise various experimental apparatuses to measure position, and various experimental apparatuses to measure energy. How does the wavefunction "learn" that the arbitrary apparatus we were using was designed to measure position and not energy, and vice versa, causing it to "adjust" its behavior appropriately in the future?

I realize that this whole analysis might be flawed, because in any real experiment, the measurement of the particle energy or position would destroy or significantly perturb the particle (i.e., observer effect), so unitary evolution with the Hamiltonian corresponding to the infinite square well is no longer valid. But it still seems like the example I described is consistent with the kinds of problems you might see in an undergraduate text book, so there has to be some practical reality to it... or no?

2. Dec 8, 2015

### Staff: Mentor

The wave function is simply a description of the state of the system. The wave function doesn't "know" about the measurement, the wave function is different before and after the measurement. In other words, the act of measurement changes the state of the quantum system.

3. Dec 8, 2015

### bhobba

Just to elaborate further see post 137:

Thanks
Bill

4. Dec 8, 2015

### Demystifier

Measurement is nothing but a special case of interaction. The interaction always affects the wave function, so in this sense the wave function "knows" what is the interaction. Hence it is the interaction which determines what observable (if any) is measured.

For more details try to google about decoherence and preferred basis problem.

5. Dec 9, 2015

### Coherence

Thanks for the suggested readings, I will look at this.

I completely understand from the mathematical framework of quantum mechanics that the measurement changes the wave function, generally from a super-position state to an eigenstate of that observable. And I also understand that measurement is ultimately just an interaction with some measurement apparatus (which can be a detector, or another particle, etc.).

But what specifically about the interaction, in an actual laboratory experiment, forces the wavefunction into an eigenstate of a certain observable, rather than another one? (I realize that the collapse mechanism is not a part of the quantum mechanical framework, but that is irrelevant to my question). Certainly there are a lot of different types of apparatuses to measure position. And there are a lot of different types of apparatuses to measure energy. The interaction with the wavefunction is unique to each one of these apparatuses. And yet, no matter which combination of apparatuses I choose to use for position and energy, the situation that I described in my original post will happen. Starting with an eigenstate of energy, measurement of energy followed by another measurement of energy yields very different results than measurement of position followed by measurement of energy. Quantum mechanics succeeds in making the correct statistical predictions no matter what specific apparatuses I choose to use. Specifically, the energy measurement after the position measurement will generally yield a different result than the original energy measurement, with a probability distribution that can be computed from the evolution of a position eigenstate. So how does the wavefunction "know" that the arbitrary apparatus I am using is "trying" to extract the position, rather than the energy, or vice versa? How does it "know" to collapse into the eigenstate of that observable?

6. Dec 9, 2015

### Staff: Mentor

As of today... We don't know.

You're asking if there is a deeper theory that explains why quantum mechanics works the way it does (sort of the way that the atomic theory is needed to explain why stoichiometry works the way it does - always exactly two moles of hydrogen for one of oxygen, never a bit more or a bit less). So far no one has been able to come up with such a theory, or even a plausible conjecture. That doesn't mean it doesn't exist, just that if it does we haven't found it yet.

You may find this a bit disconcerting, but consider that the theory of gravity was in the same situation for centuries. Newton's $F=Gm_1m_2/r^2$ accurately described the force between the earth and the sun, but offered no explanation for how that force came to be. How does each atom in the earth "know" that there is a solar-sized mass 93 million miles away so that it can move accordingly? That doesn't stop us from using Newton's theory.

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7. Dec 9, 2015

### Demystifier

Typically, if the interaction Hamiltonian is proportional to an observable $O$, then the wave function splits into (approximate) eigenstates of the observable $O$, provided that the interaction Hamiltonian is "much larger" than the free Hamiltonian, and that the "interaction" is an interaction with a system ("measuring apparatus") with a large number of degrees of freedom.

This, indeed, is a basis for the theory of decoherence.

8. Dec 9, 2015

### bhobba

Its a logical consequence of the axiom I gave in my link:
An observation/measurement with possible outcomes i = 1, 2, 3 ..... is described by a POVM Ei such that the probability of outcome i is determined by Ei, and only by Ei, in particular it does not depend on what POVM it is part of.

So your question is why that axiom.

The answer is we don't know - its simply how nature works.

One can use other axioms that are more intuitively transparent eg:
http://arxiv.org/pdf/quant-ph/0101012.pdf

That shows the key assumption is the necessity of continuous transformations between pure states. That is very intuitive because we expect if a system is in a state and a second later in another state it went through some state at half a second.

Very intuitively pleasing - but all you have done is replace something that's elegant but opaque with something intuitive and clear - but the question remains - why is nature like that. The answer is the same - we don't know.

But in pondering this bear in mind the nature of explanation - every theory - every single one - assumes things that are assumed true and not explained by the theory. You just have to accept them.

Thanks
Bill

9. Dec 9, 2015

### gjonesy

I think the biggest problem with most of the popular material on these subjects is the simple English wording they use in the descriptions. The etymology of the word "know" and all of its common usage and definitions greatly differ from the context of how its used here. How does the wavefunction "know" (implying consciousness awareness and thought) when its actually a description of change through interaction. I think DrClaude describes it much easier and in a more understandable way.

10. Dec 9, 2015

### kith

The "forces the wavefunction into an eigenstate"-part of your question is controversial and a matter of interpretational debate. The attitudes range from "we need a more fundamental theory" over "this question can't be answered by science" to exotic ontologies like "we need parallel worlds". I won't go into this but only address the "of a certain observable, rather than another one"-part of your question.

The general mechanism for this is the following: the interaction between the apparatus and the system leads to entanglement between the two. Since the apparatus is a macroscopic object, the interaction is irreversible in the sense of statistical mechanics (for example, if a particle leads to a dot on a phosphor screen, it is incredibly unlikely that the dot will disappear and produce the initial particle again). So we have a stable final state which is entangled.

Since the state of the whole is entangled, you cannot assign a state vector to the system alone anymore. The state of the system is now represented by a so-called reduced density matrix.

In our case, this matrix has to be diagonal. Because if it is, the eigenvectors of the matrix constitute a preferred basis in the following sense: if you look at the observable which is built from these eigenvectors and use the Born rule to calculate the probabilities for the possible measurement outcomes, you don't get interference anymore. If the matrix wasn't diagonal, you would get interference so it couldn't describe the outcomes of a measurement.

So what happens is this: the mathematical form of the interaction between the system and the apparatus leads to a (unitary) time evolution which results in a reduced density matrix which is diagonal. So the mathematical form of the interaction singles out a preferred basis. This determines the observable which your apparatus measures. (Zurek somehow related the Coulomb interaction to the position basis but this is only half knowledge on my part. Important key words here are "decoherence" and "einselection".)

As noted in the introduction, an important caveat: You don't get a definite pure state as final state, only the interference in a certain basis goes away. How exactly the final state should be interpreted is a matter of debate.

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11. Dec 9, 2015

### bhobba

Just to elaborate on what Kith said this is the opposite view to that of the post I linked to that used Gleason and is favoured by Zurek.

Gleason starts with observations and develops the conept of state. Zurek starts with state and develops the concept of observation (he calls it Quantum Darwinism):
http://arxiv.org/pdf/0903.5082v1.pdf

As I said, you can start from different places and arrive at the same destination. But all require assumptions.

Thanks
Bill

Last edited: Dec 9, 2015
12. Dec 9, 2015

### bhobba

13. Dec 9, 2015

### kith

Yes. I just want to state explicitly that the question "how does the measurement apparatus interact with the system?" can be addressed also from the Copenhagen-like viewpoint which starts with observables. You can always try to treat the system together with the apparatus as a combined system of investigation.

Thanks, I had forgotten about this. The argument is that the position operator commutes with the Coulomb interaction term (which is proportional to 1/r). I just don't see right away how this leads to the preferred basis. It's been quite a while since I dealt with these things.

For reference, Schlosshauer talks about this in 2.8.4 and the whole section 2.8 (Environment-Induced Superselection) is very relevant to the discussion. It also ties in with what Demystifier wrote in post #7.