# Decoherence, collpse, and WHEN does the collapse occur?

The idea that decoherence stands behind the so-called collapse (reduction) of the wave-function doesn't seem satisfactory. Consider a quantum particle whose wave-function is of the form

(1) $$\left|Ψ_1\right\rangle = C_a \left|Ψ_{1,a}\right\rangle + C_b\left|Ψ_{1,b}\right\rangle$$

where the subscripts mean: a = transmitted by a beam-splitter, b = reflected by the beam-splitter. Assume now that we entangle this particle with many others, i.e. particle 2, particle 3, etc., in the form

(2) $$\left|\Psi\right\rangle = C_a \left|\Psi_{1,a}\right\rangle\left|\Psi_{2,c}\right\rangle|Ψ_{3,e}\rangle... + C_b \left|Ψ_{1,b}\right\rangle \left|\Psi_{2,d}\right\rangle \left|\Psi_{2,f}\right\rangle... .$$

The single particle wave-function (1) was decohered, but no collapse occurred.

Assume however that the particle 1 flies to the observer Alice, the particle 2 to the observer Bob, and each one DETECTS on which path a or b, respectively c or d, is her/his particle. The COLLAPSE occurs. But WHICH ONE of the two measurements produced the collapse? Assume that Alice and Bob travel in opposite directions with respect to the Earth, and by a clock on the Earth they measure at the SAME TIME. However, by the clock in Alice's space-ship, Bob's measurement will occur later. Symmetrically for the clock on Bob's space-ship.

So, WHICH ONE of the two measurements collapsed the wave-function, Alice's measurement, or Bob's ?

(For completeness of the discussion on the difference between decoherence and collapse, see also the answers at Decoherence and collapse .)

• I do not see why you say "The single particle wave-function (1) was decohered, but no collapse occurred.". Inmy books as long as you have kets the phases are there. If it is a many body system ( a lasing laser for example) they may be modified in the ensemble , but still there if you think density matrix off diagonal elements. – anna v Nov 23 '14 at 11:58
• I say so because the phases of the single-particle wave-function are no more numbers, so, we cannot do, for instance single-particle interference. The decoherence indeed begins with decohering the initial wave-function of the given system by entangling with ne more particle, and another one, and so on. The collapse begins by decoherence. Just, decoherence is NOT ENOUGH to produce the collapse, there is SOMETHING MORE. And we have difficulty to say what exactly is this "something". Please tell me if my answer is clear. – Sofia Nov 23 '14 at 12:16
• I think you are confusing entanglement with decoherence. Decoherence loses phases, entanglement keeps them, maybe distorted but phases are there. In my opinion decoherence happens when h_bar no longer characterizes the interactions, distances are too large or energies, and h_bar can be considered zero. Then the phases no longer have a measurable effect and the system decoheres. ( with exceptions like lasing and crystals etc) – anna v Nov 23 '14 at 12:23
• No, Anna, it is matter of definitions. The definitions that I was taught are that when a particle possessing a well defined wave-function (which is a superposition of a couple of states) is entangled with another particle, the coefficients of the states in the single-particle wave-function are no more constants, they are vectors (in another Hilbert space - that of the second particle). The SINGLE-PARTICLE WAVE-FUNCTION is decohered, but the TWO-PARTICLE WAVE-FUNCTION has constant coefficients. We cannot do anymore single-particle interferometry, but we can do two-particle interferometry. – Sofia Nov 23 '14 at 17:18
• (continuation) However, when bringing the wave-function in contact with a macroscopic apparatus it is not clear whether THIS ONE has a wave-function at all. It is hard to say how many particles it has. It constantly exchanges particles with the environments (e.g. photons), it is an open system. So, the entanglement with such an apparatus is not clear if it can be described by a wave-function. BUT I see in your comment something new for me: you say that h_bar becomes almost zero. It is not clear to me WHERE in this problem appears h_bar? Do you want to say that the apparatus has NO HAMILTONIAN? – Sofia Nov 23 '14 at 17:44

## 1 Answer

Your question presupposes that collapse actually occurs. A third observer might view Bob and Alice to be entangled with the system. The combined state lives in a "larger" Hilbert space (I mean in the sense that it is a product space). Now, there are projections of this state into spaces where Alice particle 1 to have gone down path a, Bob particle 2 to have gone down path B, etc, and also vice versa. If you buy this (a lot of people don't), then there is no collapse from the third observer's point of view. Also, the choice of projection is largely a human construct (although motivated by things like pointer states), and so we don't need the theory tell us anything about human constructs.

Clarification:

Suppose we send two particles through a beam splitter (I don't really know why we need more than one particle, but ok). For a single particle, you would just have $$\left|\Psi_1 \right\rangle = \frac{1}{\sqrt{2}} \left( \left|\Psi_{1,left}\right\rangle + \left|\Psi_{1,right}\right\rangle \right)$$ But if we say that they are entangled, then we need something more general: $$\left|\Psi \right\rangle = c_{ll} \left|\Psi_{1,left}\right\rangle \otimes \left|\Psi_{2,left}\right\rangle + c_{lr} \left|\Psi_{1,left}\right\rangle \otimes \left|\Psi_{2,right}\right\rangle + \mathrm{2\,more\,terms}$$ I don't think this represents "decoherence" in any case. This setup can still produce interference patterns. Decoherence is when such interferences are washed out by having too many degrees of freedom that are out of phase. Now, suppose we have two detectors, called A and B, that detect particles. A can only detect particles that went left, and B can only detect particles that went right. When the two particles leave the splitter, the combined system is in the state $$\left|\Psi\right\rangle \otimes \left|A_\mathrm{ready}\right\rangle \otimes \left|B_\mathrm{ready}\right\rangle$$ An external observer picks a time slice later, that, in her reference frame, she considers that $A$ and $B$ have had enough time to perform detection. Remember, quantum states are not reference frame invariant. In that frame, the state looks more like $$c_{ll} \left|\Psi_{1,left}'\right\rangle \left|\Psi_{2,left}'\right\rangle \left|A_{click}\right\rangle \left|B_\mathrm{ready}\right\rangle + c_{lr} \left|\Psi_{1,left}'\right\rangle \left|\Psi_{2,right}'\right\rangle \left|A_{click}\right\rangle \left|B_\mathrm{click}\right\rangle + \dots$$

Now, this is a superposition of different possible outcomes. If you say that detector A registered a click, then you project onto $\left| A_\mathrm{click}\right\rangle$, and you are left with only the first two terms of the above state. From that point of view (human construct), B is in a superposition. But you can do that for detector B also, and find A to be in a superposition. The theory just predicts what the theory predicts. you are interpreting this projection business as collapse.

But to answer your question, can you detect this superposition. Well, in principle, if detectors A and B then go on to deflect a photon with some mirrors and things, you could recover an interference pattern when you do this over and over again. In practice, there are other parts of the system that we didn't write down, and those all contain their own messy phases, which never quite line up, which makes it impossible to detect. People have been trying with things like SQUIDs and the like, but it is difficult to prevent decoherence.

• I don't understand some parts of what you say: 1) Bob and Alice don't have to be present at the measurements, these can be done by apparatuses. So, indeed we put each one of the two particles in contact with a macroscopic apparatus and the Hilbert space becomes larger. And? What next? 2) What means the "vice-versa", what you want to say with it? You say "if you BUY THIS". Buy WHAT? I lost you here. 3) About "human construct", as I said, there are apparatuses, not human beings, that do the recording. – Sofia Nov 23 '14 at 12:26
• (continuation) How can I escape the assumption of collapse? I get only ONE answer, e.g. Alice's particle on the path b and Bob's particle on the path d. WHERE ARE the branches a respectively c for these two particles, in the total wave-function? Can I detect these branches, are they somehow detectable? And, if these branches DISAPPEARED (collapse) WHICH ONE of the two measurements, that done by Alice, or that done by Bob, collapsed the wave-function? – Sofia Nov 23 '14 at 12:43
• To Lionel: What you mean by too many degrees of freedom that are "out of phase"? What is out of phase? Do you mean destructive interference? A degree of freedom interferes ONLY with itself. And no matter how many degrees of freedom we have, why shouldn't we STILL have them represented by a wave-function? – Sofia Nov 23 '14 at 21:21
• To Lionel(continuation): you also say "From that point of view (human construct), B is in a superposition". Which superposition? A detector is a BIG body, like Schrodinger's cat. Did you see a cat in a superposition of "alive" and "dead"? How can a detector be in a superposition of "ready" and "click"? Since the detector EITHER clicks, OR remains silent, we interpret this as collapse. Which other choice we have? – Sofia Nov 23 '14 at 21:33
• Those are a lot of questions. First I'll ask you to keep an open mind, and stop typing in caps. 1) to see interference between two states, their inner product must be non-zero. For a state that is a product of many wave functions, this inner product is generally the product of many numbers less than 1, and tends to 0 in the absence of coherence. – lionelbrits Nov 23 '14 at 23:23