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we know that one property is that after measurement the probability of observing the qubit in either zero state or one state changes. This is not necessarily true. Consider the $\left|0\right\rangle$ or $\left|1\right\rangle$ state transforming under operators of the form $\begin{bmatrix}1 & 0\\ 0 & e^{i\phi}\end{bmatrix}$. The only requirement* ...

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I have found an interesting discussion about this question in the book (in french) "Mécanique quantique, Bases et applications" by Constantin Piron. He proves a Gleason-like theorem (A.2 Théorème fondamental, p.172) stating something like: if you would like to associate to each state (= vector) and proposition (= closed subspace of the Hilbert space) a real ...

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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 ...

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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 ...

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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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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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I'll try to give an answer that's as intuitive as possible. In probability you can often* replace any statement containing 'or' with addition. If you calculate the probability of throwing 1 or 2 with a dice you get $P(1\ \text{or}\ 2)=P(1\lor2)=P(1)+P(2)=1/6+1/6=1/3$. You can phrase the question 'what is the probability the atom has decayed after 3 half ...

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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 ...

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Googling on the list of names in the footnote turned up the paper https://arxiv.org/pdf/1405.6755.pdf , which led to Deutsch and the other references below. Edward Farhi, Jeffrey Goldstone, and Sam Gutmann, "How Probability Arises in Quantum Mechanics," Annals of Physics, 192 (1989) 368, June 1989. http://www.sciencedirect.com/science/article/pii/...

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