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3

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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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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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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"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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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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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 ...


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Are you thinking of illustrations like this? If so, the sign just means that (in this case) $\Psi(-x) = -\Psi(x)$. There is no absolute sense in which one half of the orbital is negative and the other positive.


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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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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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