5 ms·

I am not sure what you mean by "experiments do not have results." A measurement in Everett is the environment becoming entangled with the measured system, and a

by tbabb 8y ago

I am not sure what you mean by "experiments do not have results." A measurement in Everett is the environment becoming entangled with the measured system, and a "result" is a particular basis of the combined wavefunction.

im3w1l 8y ago
If you square the probability amplitude and take the absolute value, that you get something that works very much like a probability. But there is nothing in many-worlds theory that says this quantity has any special significance.
- tbabb 8y ago
  I still either don't see why you say that, or don't follow what you are trying to say. The Born rule is a fundamental feature of all quantum theories, and I don't see why Everett is an exception. If you want an intuitive picture of what the squared amplitude means, it seems that it would suffice to think of it as the fraction of phase space which evolves into the configuration associated with that amplitude.
  - TheOtherHobbes 8y ago
    In fact you can't get to the Born rule from WMI without doing some (IMO) strange things. See e.g. https://arxiv.org/abs/1405.7907 https://arxiv.org/abs/1405.7907
effie 8y ago
> what you mean by "experiments do not have results." Let the experiment measure spin projection s_y of an atom previously prepared in spin state |s_x+>. It is known empirically that this experiment has two possible mutually exclusive results, +1/2 and -1/2, with probability of each being 1/2. Standard quantum theory accepts this and uses mathematics to predict those probabilities from data on the mutual arrangement of preparator and analyser of the spin. Such predictions are difficult to be made in Everettian theory, because that theory describes both the particle with spin and those forming the measuring apparatus in the same way. When that is done, the physical process of measurement can be mathematically described by the Schroedinger equation, which may seem like a win, but it has a big downside: all it gives is a ray in the big Hilbert space of particle plus apparatus at later times; it is not clear what this calculation result is good for. It is not the prediction of real experiment's result; it is not even clear how to use it to calculate probabilities of those results. Certainly not by Born's rule - because what is the eigenfunction to be used in the Born formula? Such eigenfunction is, in standard theory, determined by the classical settings of the apparatus, such as orientation of the SG magnet. But there is no such thing in Everettian theory, there is just psi function.
- tbabb 8y ago
  If I have two bases |O_up> and |O_down>, representing the observer in the state of having measured an up spin and a down spin respectively, what is wrong with taking <O_up|psi> and <O_down|psi> to find the fraction of Hilbert space occupied by either possibility?
  - effie 8y ago
    It does not work as written. How is |O_up> defined in terms of the basis of the global Hilbert space? There is infinity of directions in space which we can choose as the defining axis of the ket |O_up>. You could calculate the inner product for any of them. Does that mean that the inner product is typically zero and to get a non-zero probability, one has to talk not about probability of possible results of measurement, but about probability on the space of possible measurement setups? If so, why bother with the concept of quantum measurement at all? Why not just say any configuration of the world from some measurable set has probability int |psi|^2 dq? That would be the other, non-projection Born rule (or interpretation of psi), which doesn't have the problem of preferred basis and does not need it to be chosen by the classical apparatus.

by tbabb 8y ago

im3w1l 8y ago

If you square the probability amplitude and take the absolute value, that you get something that works very much like a probability. But there is nothing in many-worlds theory that says this quantity has any special significance.

tbabb 8y ago
I still either don't see why you say that, or don't follow what you are trying to say. The Born rule is a fundamental feature of all quantum theories, and I don't see why Everett is an exception. If you want an intuitive picture of what the squared amplitude means, it seems that it would suffice to think of it as the fraction of phase space which evolves into the configuration associated with that amplitude.
- TheOtherHobbes 8y ago
  In fact you can't get to the Born rule from WMI without doing some (IMO) strange things. See e.g. https://arxiv.org/abs/1405.7907 https://arxiv.org/abs/1405.7907

effie 8y ago

> what you mean by "experiments do not have results." Let the experiment measure spin projection s_y of an atom previously prepared in spin state |s_x+>. It is known empirically that this experiment has two possible mutually exclusive results, +1/2 and -1/2, with probability of each being 1/2. Standard quantum theory accepts this and uses mathematics to predict those probabilities from data on the mutual arrangement of preparator and analyser of the spin. Such predictions are difficult to be made in Everettian theory, because that theory describes both the particle with spin and those forming the measuring apparatus in the same way. When that is done, the physical process of measurement can be mathematically described by the Schroedinger equation, which may seem like a win, but it has a big downside: all it gives is a ray in the big Hilbert space of particle plus apparatus at later times; it is not clear what this calculation result is good for. It is not the prediction of real experiment's result; it is not even clear how to use it to calculate probabilities of those results. Certainly not by Born's rule - because what is the eigenfunction to be used in the Born formula? Such eigenfunction is, in standard theory, determined by the classical settings of the apparatus, such as orientation of the SG magnet. But there is no such thing in Everettian theory, there is just psi function.

tbabb 8y ago
If I have two bases |O_up> and |O_down>, representing the observer in the state of having measured an up spin and a down spin respectively, what is wrong with taking <O_up|psi> and <O_down|psi> to find the fraction of Hilbert space occupied by either possibility?
- effie 8y ago
  It does not work as written. How is |O_up> defined in terms of the basis of the global Hilbert space? There is infinity of directions in space which we can choose as the defining axis of the ket |O_up>. You could calculate the inner product for any of them. Does that mean that the inner product is typically zero and to get a non-zero probability, one has to talk not about probability of possible results of measurement, but about probability on the space of possible measurement setups? If so, why bother with the concept of quantum measurement at all? Why not just say any configuration of the world from some measurable set has probability int |psi|^2 dq? That would be the other, non-projection Born rule (or interpretation of psi), which doesn't have the problem of preferred basis and does not need it to be chosen by the classical apparatus.