Decoherence in Everett quantum mechanics

Take an initial state and its environment, $E$, as follows, $$|\psi\rangle_i = |0\rangle |E\rangle + \sqrt{2}|1\rangle |E\rangle.$$ Suppose that I've written it already in the basis in which the state de-coheres, such that after de-coherence, the wave-function is $$|\psi\rangle_f = |0\rangle |E_0\rangle + \sqrt{2}|1\rangle |E_1\rangle.$$ where the environment states are orthogonal. In the many-worlds interpretation, if an observer has equal chance to be in each branch, he won't see a Born rule; each outcome is equiprobable.

But suppose that when the state decoheres, it decoheres such that, $$|\psi\rangle_f = |0\rangle |E_0\rangle + |1\rangle |E_2\rangle + |1\rangle |E_3\rangle.$$ This can be obtained with a unitary transformation, $$U = 1 \otimes \left[\left(\frac{1}{\sqrt{2}}|E_2\rangle + |E_3\rangle\right)\langle E_1| + |E_0\rangle\langle E_0|\right]$$

If the wave-function decoheres in the second way, as usual each branch is equi-probable, but this time it results in a Born rule!

What are the problems with positing that the wave-function always decoheres and branches in such a way that, if you assign equal probabilites to each branch, the results are what you would have got from the Born rule?

I find your question a little unclear, but the following is my best understanding of your position. According to the MWI an observer will exist in multiple versions after a measurement. So then it is equally possible for him to be in either state and he should assign equal probability to each: let's call this the equality rule. It is important to note first that the equality rule is not the Born rule even if it happens to give the same result under some circumstances.

The equality rule leads to inconsistencies. Suppose for example, that you prepare the state $$\tfrac{1}{\sqrt{3}}|0\rangle+\sqrt{\tfrac{2}{3}}|1\rangle$$ and measure it. According to your proposed rule the probability of each outcome is 1/2. This measurement gives the state $$\tfrac{1}{\sqrt{3}}|0\rangle|0\rangle_M+\sqrt{\tfrac{2}{3}}|1\rangle|1\rangle_M.$$ Suppose you then get another system and set it up so that if the measurement apparatus is in the state $|1\rangle_M$ then the new system is in the state $$\tfrac{1}{\sqrt{2}}(|1\rangle_N+|2\rangle_N)$$ and otherwise it is in the state $|0\rangle_N$. Then you have $$\tfrac{1}{\sqrt{3}}(|0\rangle|0\rangle_M|0\rangle_N+|1\rangle|1\rangle_M|1\rangle_N+|1\rangle|1\rangle_M|2\rangle_N).$$ So then the probability of each outcome by your rule is 1/3. But then you have a problem for the probability that the original system had the state $|1\rangle$ is now 2/3 where formerly it was 1/2. Since the equality rule leads to inconsistencies, it is not a viable candidate for a probability rule.

There are two proposals I am aware of for how to explain the Born rule from quantum mechanics without collapse. One involves decision theory, see

http://arxiv.org/abs/0906.2718

and explains why other candidates for probability rules don't make sense. The other is Zurek's envariance argument:

Zurek fudges the issue of whether other universes exist for some reason best known to himself, but the explanation would work in the Everett interpretation if it is correct.

Update: I misinterpreted the question. The question was "Is it possible that the state always decoheres in a particular basis for the environment such that the equi-probability rule matches the Born rule?" I don't think this is possible because it is possible to prepare an atom or photon in a non-equiprobability superposition and then measure it. You could prepare a photon polarised at 30 degrees to the horizontal and then measure it with a horizontal polariser in front of a suitable detector. The resulting branches of the wavefunction would not be equiprobable.

You could say the photon is not the environment, but then I think you begin to get into setting up a position that hangs on the terminological issue of what you call the environment. Is the measuring apparatus the environment or only part of it? What about the first electron the photon interacts with in the measuring device? I don't see that this solves any problems. For different purposes it might be reasonable to draw the boundary in different ways. For example, if you have a detector that exhibits quantum coherence during some part of the detection process and you can reverse the detection, then maybe it shouldn't be included in the environment since it need not cause decoherence if you set up the experiment correctly. If you don't have such a detector maybe it should be included in the environment.

And why should the environment-system boundary be drawn in such a way as to make the equi-probability postulate true even if that is possible?

• Thanks, but I am positing that the state always decoheres in a particular basis for the environment such that the equi-probability rule matches the Born rule. Is that possible? – innisfree Jul 11 '14 at 12:27
• I don't fully understand your objection. With ordinary de-coherence, don't you always have to draw a distinction between the state being measured and the environment (including the measurement apparatus)? such that the Hilbert space is factorized into state times environment? – innisfree Jul 14 '14 at 14:55
• In my case the physics doesn't depend on where I draw the line. All that matters is that there is a line and there is stuff outside that line from which the information required to maintain coherence doesn't return. In your case the physics does depend on where you draw the line. After the interaction with the environment the state is equiprobability but before that interaction it may not be. So that line is where the probability changes from non-equal probability to equal probability. So the physics changes depending on the location of the line. – alanf Jul 14 '14 at 15:35
• Great point. I'll think it over. – innisfree Jul 15 '14 at 9:14