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When calculating expectation values, you need to know a few things: What is my random variable? What is my distribution function? What is my desired quantity in terms of the random variable? The general form in one dimension would look like this $$\langle G(x) \rangle = \frac{\int_{x_{min}}^{x_{max}} f(x) G(x) dx}{\int_{x_{min}}^{x_{max}} f(x) dx} \, ... 0 Aside from the obvious answer of randomness in probability distribution, each atom's decay event does depend on its overall energy, which cannot be measured individually by current technology. For example it may be contained in a gas where the density in one area of the gas is slightly higher than other areas and may be undergoing more "collisions" ... 0 It means, that the cosmological perturbations satisfy Gaussian initial conditions. This means, that the probability of the perturbation amplitude has a Gaussian shape about the mean value. Considering linear perturbation theory, the initial Gaussian probability distribution will remain Gaussian for all times, e.g. today. It means, that there is e.g. no ... 0 It's convention to normalize the wave function, but this is purely something done to make calculations easier. States multiplied by a c number(complex number) are completely equivalent. This is why we choose to normalize states (magnitude 1). It makes the calculations for things like the transitions probability between states easier. If you include your ... 1 The coefficients a_k quantity the projection of the state |\psi\rangle onto the k^\mathrm{th} basis state. So if you measure in that basis you would expect |a_k|^2 of the time to measure state |\phi_k\rangle. If you change the basis you will need to recalculate the projection coefficients. If you are in a non-orthogonal basis the projections are ... 29 The decay phenomenon is a purely quantum mechanical property. This problem is equivalent to a particle in a finite potential well, and a lower potential state that is available outside the well. Classically if the energy of the particle in the well is lower than the potential barrier - it will never get to the lower state. By quantum mechanics, the particle ... 2 It's basically got to do with the fact that nuclear decay is a quantum mechanical process, and quantum mechanical processes are not deterministic in the traditional sense, i.e. given a set of conditions you can't predict exactly what will happen in a particular process, only the probability of something occurring. In this case, nuclear decay occurs through ... 15 Speaking loosely, each individual atom has a desire to become stable, but that translates into a probability of decaying. This means, since there are billions and billions of atoms in a macroscopically significat chunk of material, that there are always going to be unlikely holdouts, and these holdouts are responsible for radiation that after the initial ... 2 All the atoms have the same chance to decay at any given moment. If you have more of them at the same place you will simply have bigger chance of them decaying. It's like dice, you have 1 in 6 chance of getting a 6. If you have 100 dices you will have 100 times more chances of getting a 6. Thous you'll have more 6-es. Unless you cheat ;) 1 For a classical Maxwell-Boltzmann gas, the partition function is given by$$Z=\sum_i g_i e^{-\beta \epsilon_i}$$where g_i is the degeneracy. And the probability for each level to be occupied by one particle is given by$$P_i=\frac{g_i e^{-\beta \epsilon_i}}{Z}$$The partition function function in the above expression is essentially a normalization factor ... 2 The partition function is not in general equal to one. It is a normalization constant, i.e. the probability of being in a configuration \{\sigma\} is$$ P(\{\sigma\}) = \frac{e^{-H(\{\sigma\})/k_B T}}{Z} $$where$$ Z(T) = \sum_{\{\sigma\}} e^{-H(\{\sigma\})/k_B T}  Since multiple configuration can have the same energy, you can define (with a little ...