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6

I would rephrase the correct answer by Thatpotatoisaspy as: when there are no boundary conditions there are the plane wave solutions of the quantum mechanical equations. Plane waves go from -infinity to + infinity so they are not good for modeling the probability of finding the electrons in our experiment. But, there exist wave packets, which can model ...

2

I think it's important to point out that if a problem admits bound states, then there is always a boundary condition at infinity, which comes from demanding that the wave function go to zero as $x \to \pm \infty$. It's interesting to note that for a particle in a box, the quantisation condition that leads to discrete solutions comes from demanding that the ...

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The quantity $P(t)=|\langle \Psi|\exp(-iHt)\Psi\rangle|^2$ is the probability of survival of your state, and not the probability of survival per unit time! It is a dimensionless quantity. For a vast number of quantum systems, exponentially decaying it is something like $e^{-t/\tau}$, for a (mean) lifetime τ. It is then apparent how the lifetime is gotten ...

1

If $X$ is a random variable, which is distributed according to a Poissonian distribution with rate parameter $\lambda_1$, we write $X\sim \textrm{Pois}(\lambda_1)$. Now, if $Y\sim \textrm{Pois}(\lambda_2)$ then $$Z=X+Y\sim \textrm{Pois}(\lambda_1) + \textrm{Pois}(\lambda_2) = \textrm{Pois}(\lambda_1 + \lambda_2)$$ In your case you have the same rate ...

1

Yes. A good example is a wave packet for the free particle. The ordinary free particle solution is not normalisable, but you can construct another solution which is, the wave packet.

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