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$\Psi(x,t) = \sum_{n} a_n \psi(x)_n e^{-i E_n t / \hbar}$ Yes, if $\psi(x)_n$ and $E_n$ are the eigenfunctions and eigenvalues. When you take the absolute square, you must keep all the terms: $$|\Psi(x,t)|^2 = \sum_{k}\sum_{n} a_k^*a_n \psi^*_k(x)\psi_n(x) e^{-i (E_n-E_j) t / \hbar}$$ The probability density is $|\Psi(x,t)|^2 \textrm{d}x$, and it is time ...

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There's two different concepts that you are mixing up (perhaps partly because you use the same letter $\psi$ to refer to both concepts). There is the initial wave function, which I'll call $f$, so $\Psi(x,0)=f(x)$. Then there are the energy eigenfunctions, $u_n(x)$. It is true that \Psi(x,t)=\sum_n a_n u_n(x) e^{-i \omega_n t} ...

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Some history here might be useful; one of the oldest cosmological theories that we have was developed by Ionian philosophers which was an atomic theory; in that theory uncertainty as in random motion was taken as something fundamental (they called it the clinamen which is usually translated as swerve). This shows that pure determinism, physically, is ...

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The Heisenberg Uncertainty Principle (HUP) holds for special observables, as energy and time, space and momentum, .. To every observable there corresponds a quantum mechanical operator. Quantum mechanical operators either commute or not commute, and are seen in the commutation relationships. Observables that do not commute are what the HUP is about. It is ...

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First and foremost we have to understand that if we are deriving laws of nature then our primary assumption is that the natural phenomenons are not random. If they are random, it would be impossible to say anything about them. Now let's come back to Quantum Mechanics. There is nothing random about the motion of electrons or any other subatomic particles ...

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There is something going on. Which is that certain observables are fundamentally incompatible. That means firstly that you can do an experiment for one observable or for another observable, but you can't do an experiment for both observables at the same time. And what's worse if you did an experiment for A then one for A again and then one for B the two ...

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I think this is saying much the same as Gennaro, but I'd phrase things slightly differently. In config $\mathbf{1}$ your system has some eigenfunctions $\psi$, and you start with the system in one of these eigenfunctions $\psi_n$. When you change to config $\mathbf{2}$ the function $\psi_n$ is no longer an eigenfunction because your system now has a set of ...

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A quantum system is described by a set of self-adjoint operators $(A_1\ldots A_n, H)$ and a Hilbert space $\mathcal{H}$. The mentioned operators represent the observables that you can experimentally measure and their eigenvalues the possible outcomes. Among them there is a special one, the Hamiltonian $H$, describing the time evolution of the system. A state ...

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Do electrons always have a probability of being somewhere [in] the same way as when they surround a nucleus? Yes. Of course they don't have a probability a of being somewhere when surrounding a nucleus, they have a frequency of being found somewhere if measured, which is different. You can get a full probability too, but only if you specify even more ...

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Do electrons always have a probability of being somewhere? It's lies to Pies again I'm afraid. We make electrons (and positrons) out of light waves in pair production. We can diffract electrons because of the wave nature of matter. And waves are not point particles. They are extended entities. They do not exist at one point only. So all that ...

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The trick to exponentiate the $2\times 2$ matrix is to diagonalise it. A diagonal matrix $D$ exists such that $H= C^{t}DC$, with $C$ being orthogonal due to the fact that $H$ is hermitian. Using the orthogonality of $C$ and the definition of the exponential as power series leads you to prove that $$e^{-iHt} = e^{-i(C^{t}DC)t} = C^{t}\,e^{-i(D)t}\,C$$ and ...

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Another such argument was made by Jaume Garriga and Alexander Vilenkin, see here: A generic prediction of inflation is that the thermalized region we inhabit is spatially infinite. Thus, it contains an infinite number of regions of the same size as our observable universe, which we shall denote as O-regions. We argue that the number of possible histories ...

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A particle that is rolling up and down the walls of a well is nowhere near the ground state potential: the ground state would correspond to the particle being at the bottom of the well, and not moving. In quantum mechanics, you can't both be "at the bottom of the well" (well defined position) and "not moving" (well defined momentum). So the ground state is ...

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In classical physics there are classical forces and also you observe a particle somewhere because it is located there. In quantum physics there is no reason to think that a particle is located some particular place, and even if it were then there is still no reason to think an observation reveals that position rather than creating a new position. And even ...

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You could use a Wigner distribution which does generate positive numbers for regions of phase space that are large enough. But when operators don't commute it also means there is no experimental setup that can evolve them into a state that would give the same result repeatedly if performed again and again. So you've jumped over the whole fact that a so ...

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I have found a set of four amplitudes that seem to work. You have to assume the particle has two states. L Move left (amplitude = +1/2) L' Move left and flip polarisation (amplitude = +1/2) R Move right (amplitude = +1/2) R' Move right and flip polarisation (amplitude = -1/2) I've tested it up to t=3 and it seems to work. Although I haven't proved it ...

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Let $\psi(k,t)$ be the amplitude to locate the particle on site $k$ at time $t$. Also let $U_{j,k}$ be the matrix describing your process, such that $$\psi(j, t+1) = \sum_k{U_{j,k}\psi(k,t)}$$ Then the process you proposed is described by $$U_{j,k} = \frac{1}{\sqrt{2}}\left( \delta_{j,k-1} + \delta_{j, k+1}\right)$$ It can be verified immediately that ...

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Do you see an error in my steps? Yes, I see a conceptual error. So I'll talk about that. $\dfrac{\partial \rho}{\partial t} = \dfrac{\partial}{\partial t} (\Psi^* \Psi) = \dfrac{\partial \Psi^*}{\partial t} \Psi + \Psi^* \dfrac{\partial \Psi}{\partial t}$ should be $\dfrac{\partial \rho}{\partial t} = \dfrac{\partial}{\partial t} (\Psi^* \Psi) = ... 1 Now I just begin with double-slit experiment and then conclude that there is a probability wave for electron This is wrong. Firstly, the wave function is defined on configuration space, it isn't a wave in space. And secondly, it isn't using probability the way a mathematics textbook on probability uses the word probability. So it isn't a wave in space, ... 3 The question might well be too broad. One may, of course, reconfigure the variables x and p into different ones, and integrate w.r.t. the "irrelevant" one , e.g. the angular variable on phase space, to produce a marginal quasi-probability distribution in the other, e.g. the angular variable squared,$(x^2+p^2)/2$which happens to be the (rescaled variables') ... 1 The probabilistic interpretation of the Wigner function is already flawed at the level of the Kolmogorov axioms, even for unremittingly positive values. That is to say, two points in phase space within a distance less than$\hbar\$ are not mutually exclusive sample space contingencies. In physics parlance, these two points are not distinguishable in any ...

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