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You can find derivation of these operators in most standard quantum mechanics textbooks. For your convenience, see https://en.wikipedia.org/wiki/Momentum_operator and https://en.wikipedia.org/wiki/Position_operator. For the second question, Paul Dirac said in his classic The Principles of Quantum Mechanics: A measurement always causes the system to jump ...


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The former and the latter are really the same: "$c_n=\psi(x)$". If you want to measure positions, then possible outcome states are $|x\rangle$, therefore you write $$ |\psi\rangle = \sum_x|x\rangle\langle x|\psi\rangle:=\sum_x\psi(x)|x\rangle :=\sum_xc_x|x\rangle $$ This tells you, the probability to find the particle at position $x$, i.e. to measure it in ...


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Here's a simple-minded answer: Let's just compute the momentum of a particle with a Bloch wave function $$\begin{eqnarray} \left.\langle x \right| \hat{p}\left|\Psi \rangle\right. &=& -i\hbar \left(\frac{d}{dx}\right) u_k(x) e^{i k x} \\ &=& -i \hbar \left( i k u_k(x) e^{ikx} + u_k'(x)e^{ikx}\right) \\ &=& \left( pu_k(x) - i\hbar ...


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If you solve the Schrodinger equation for a free particle the solutions are plane waves, and any sum of plane waves is also a solution. Since any wave packet profile can be constructed by summing plane waves then your equation with any $A(k)$ is also a solution of the Schrodinger equation. The $A(k)$ is not determined by the Schrodinger equation but rather ...


0

The explanation is really very simple to understand intuitively, and very beautiful. Imagine that a particle an uncertainity in its velocity $v$ of $\delta v$. Suppose at $t=0$ we have $x=x_{0}$. After $t=T$, the location of the particle will be given by the range $(x_{0}+Tv-T\delta v,x_{0}+Tv+T\delta v)$, because we dont know the exact velocity the ...


2

This wavefunction is an idealization of a wave function that has very very steep, but not discontinuous, behavior at $0$ and $a/2$. You are right, the wavefunction as written is not a proper wave function. That's a good observation. Often in physics the math is much easier if a real situation is modeled by one that is close to it, but mathematically more ...


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No $\hat\phi|0\rangle$ is not an eigenvector of $\hat\phi$. You can see this, for example, by writing out $\hat\phi$ in terms of creation and annihilation operators, then compare $\hat\phi|0\rangle$ against $\hat\phi^2|0\rangle$, and observe that one is not a scalar multiple of the other. So as you suspected, eq. 5 is not correct To obtain some analogy of ...


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You can utilize the construction of the $Q$ space, as described in Reed and Simon vol.2, page 228-230. Oversimplifying, you can make the analogy $\lvert \phi\rangle \sim \lvert x\rangle$, but the associated momentum is not $\hat{p}$, but $\hat{\pi}$ (the canonical conjugate momentum of the field $\hat{\phi}$). With slightly more precision: the Fock space ...


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First of all, it is indeed correct to model decoherence the system has to interact with what is called the "environment". Basically you have a joint CLOSED (unitary) evolution of system+environment, after which you discard the environment (technically called a partial trace), and you are left with the state of the system. Your "observer" can be taken as part ...


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All the definitions you've posted are correct, and they aren't in conflict with each other, although they are a bit imprecise. I'll try to explain in more detail what these concepts are. Probability I'll assume we don't need to attempt to define what probability is. :) I'll just note that it's formalized in mathematics under the aegis of measure theory. ...


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To understand the difference between probability and probability density consider the difference between mass and density. Density is the mass per unit volume, so to find the mass you multiply the density by the volume: $$ mass = density \times volume $$ In some cases the density will be a function of position and we have to write it as a function of the ...


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It has been many years since I have used this information (so please forgive my inaccuracies, but I wanted to get you PART of the way there with an answer (since no one has responded yet). The probability density is the integral (area under the curve) of the probability. I think the reason for the funky discrepancy between the two definitions is because.. ...


2

"Multiplying the wavefunctions" is a pretty nebulous term. Let's work with some definite vocabulary here, shall we? $(1)$ The states of one QM particle are elements of some Hilbert space $\mathcal{H}$. If we care only about position on a line as completely defining the state (which we can for a scalar boson), i.e. demand that the space be spanned by the ...


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What you refer to is probably to ground state of a infinite potential wall. If this is the case, then there, in the ground state, the particles are not localized. You can find the solution of the orthogonal ground-states here: https://en.wikipedia.org/wiki/Infinite_potential_well We can reduce the Problem to a one dimensional case, that doesn't change the ...


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1) It's basically just a description of what a particle's doing. This involves its energy level, the probability of where to "find it", its spin, etc. 2) I'd say it's somewhat analogous to describing a ball rolling down a hill in terms of its energy, where you describe its potential energy, kinetic energy, and rotational energy. But there's no "equivalence" ...


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The wavefunction, $\psi$, is the most complete possible description of a particle (or collection of particles). From the wavefunction, one can calculate a probability distribution for the the outcome of any measurement. Remember, quantum mechanics is probabilistic - one cannot make exact predictions for all measurable quantities, even if you know the ...


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I will give you an example from classical wave theory. Take Melde's experiment with a rope where you can have different modes in this rope. Those modes are discrete, like say a photon in a box. And you have a dispersion relation that relates the wave number $k$ to the frequency $\omega$. Therefore, if we try to make an analogy with a photon in a box, the ...


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$\newcommand{\spinup}{\vert{\uparrow\rangle}} \newcommand{\spindown}{\vert{\downarrow\rangle}}$ The "spin property" you mention is false. Since spin up and spin down by definition have different eigenvalues of the spin operator, they must be orthogonal states. Therefore for a general state, there is no relation between $\psi(r,s)$ and $\psi(r,-s)$ other ...


0

The point of your "Poincaré recurrence theorem analog in quantum mechanics" is that any sum of complex exponentials $$f(t) = \sum_{n=1}^N c_n e^{i \omega_n t}$$ with $\omega_n \in \mathbb{R}$ is almost periodic, i.e. $$\forall \epsilon > 0 ,\ \ \exists T \in \mathbb{R},\ \ \max_t |f(t) - f(t+T)| < \epsilon$$ The proof can be seen there : Almost ...


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One could look at the two slit experiment and the interference pattern as a 4 dimensional event. So if the event was run for say 5 minutes, it would be a single 5 minute 4 dimensional event. However, if we detected the photons passing through the slits, or detect them just before they reach the screen, then each of these detections is an event in itself, ...


0

An electron with total orbital angular momentum of $L^2 = \hbar^2 l(l+1)$ will experience a centrifugal force in addition to the Coulomb force from the nucleus. The result is that, in the frame rotating with the electron (don't read too much into this), the electron will see an effective potential energy given by: $$ V_l(r) = - \frac{e^2}{4 \pi \epsilon_o ...


3

You have fallen prey to the same confusion that many people have with regards to the wave/particle duality: The quantum objects that constitute our world are neither waves nor classical particles, and it is an error to believe that electrons/photons/whatever can "propagate as a wave" in one moment and "behave like a particle" in the next. The quantum ...



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