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The central idea is that you can translate probabilities from a single-particle picture into fractions of particles in a many-particle picture (assuming no interaction). Consider $T$ for a moment for a single particle. Let's just, for ease of writing, say $T=.9$. So if you send in a single particle from the far left, 90% of the probability density ...


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There is a sense in which specifying an unknown probability distribution with known mean and variance by a gaussian with the corresponding mean and variance is the correct choice, and that is when those are the only things you are willing to say about the random variable. The gaussian distribution is the maximum entropy distribution with a given mean and ...


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The usual answer--and I think what the question is looking for--is that there is a difference, but you can't see it with a $z$-axis Stern-Gerlach apparatus. Imagine you took the output from the $z$-axis SGA, merged the two beams back together, and sent it through an $x$-axis SGA. Then one of them is still in the $| + x \rangle$ eigenstate, so will always ...


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E. T. Jaynes: Probability Theory: The Logic of Science http://omega.albany.edu:8008/JaynesBook.html The book has also printed form. Jaynes also published readable and revealing papers on probability, statistical physics and other physics. Here you can find them: http://bayes.wustl.edu/etj/node1.html


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It isn't quite clear to me precisely what is being asked so consider this to be simply a guess. First I assume the measurements are made one after the other. Immediately after the first measurement of the the observable A is made, the system is in the state $| \alpha_1\rangle$. Then a measurement of the observable B is made and there is a $\frac{9}{25}$ ...


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There's an ambiguity here, somewhat. I'm assuming that the states are intended to be non-degenerate. By that I mean, I'm assuming $| \alpha_1 \rangle$ and $| \alpha_2 \rangle$ correspond to two different states of the system with two different eigenvalues. Let's call their eigenvalues $a_1$ and $a_2$. What this means is that if you measure $A$ on $| \alpha_i ...


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The $\sigma_A$ on the LHS is defined as $$\sigma_A = \omega(A^2 - \omega(A)^2I),$$ where $\omega$ is a state (so $\sigma_A$ also depends on $\omega$!) and the way this is defined by von Neumann is by taking a statistically relevant ensemble of very same copies of the very same system in the very same state $\omega$ and repeat the measurement of $A$ and $A^2$ ...


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Maybe statistics and the HUP are conjoined in an inseparable way in QM, * Yes, they are, because a quantum system has the property that some observables are non-defined. The standard QM says that this property is not a flaw in our measurements, but an intrinsic property of the quantum system itself. Let $x$ and $p_x$ be the two non-commuting ...


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My reading of the question is that this final measurement occurs after the measurement of $S_x = + \hbar /2$ and $S_z = - \hbar/2$. If so, we are now in the $| -, z \rangle$ state which as you point out has .5 projection on both $x$ states. However, in a classical world, we would have measured $S_x = + \hbar/2$, then measured something unrelated, and now ...


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I believe that part c of the exercise is asking you to consider the scenario of a measurement of $S_z$ which gives a state with a definite $z$ component of the spin, followed by a measurement of $S_x$. A classical vector along the $z$ axis has no components along the $x$ and $y$ axes. Would you then expect to find $+\hbar/2$ for $S_x$ on a beam coming from a ...


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I think what you are asking whether the relationship $$ \mathrm{normalizable} \iff \mathrm{continuous}$$ holds, which is utterly wrong! The wave function has to be continuous*. Notwithstanding take $\psi(x)=H(x-1/2)-H(x+1/2)$, where $H(x)$ is the Heaviside step function. $$ \implies \int_{-\infty}^\infty \mathrm{d}x \,\, ||{\psi(x)}||^2 = 1 $$ (Area ...


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My question is, why am I doing this? Becase, by convention, we want the probability account for all possible outcomes to sum to unity. The fact that we will get some outcome at a measurement, is guaranteed, and with this normalization it reflects a total probability of 1.



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