# Tag Info

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From the equations, $\phi$ is the operator acting on the variable/state $\xi$. It is important to notice also that in the factorization the $a_i$ numbers are real and the $c_i$ are complex. This factorization comes just from the mathematical fact that for a polynomial equation of degree $n$, such as the first equation, there are $n$ complex roots.

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This makes sense to me. $l$ and $m$ are two different summation indices. If you used the same index, say $l$, you would have only the $c^*_{lu}$ $c_{lu}$ terms, without the cross terms when $l\neq m$.

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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 might be interested in this question or this question (and some others I cannot track down now). The basic problem is this: It is not clear what we exactly mean by "deterministic". If you mean that we can in principle determine the future state of a system solely from initial conditions, then the time evolution given by the Schrödinger equation is ...

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To say the least, they are inseparable. The "indeterminacy" is meant to be a synonym of the "uncertainty" (original in German: Unschärfe oder Unbestimmtheit), e.g. the nonzero values of $\Delta x$ (uncertainty of position) and $\Delta p$ (uncertainty of momentum) that obey $$\Delta x \cdot \Delta p \geq \frac\hbar 2$$ This is a consequence of the nonzero ...

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So there is no wave particle duality. By particle we tend to use the classical macroscopic definition of a particle, which means a shape( a volume) a center of mass that describes absolutely the particle's position in (x,y,z,t). A billiard ball, for example. Electron is always a particle. But the location of electron is represented as wave, because ...

2

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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See this answer and the comments. There is an explanation how occures fringes due to the EM field from the surface of the slit or wire: https://commons.m.wikimedia.org/wiki/File:Moellenstedt_biprisma_voltage_shadow.JPG It shows the influence of an electrical field to fringes. ...

3

Isn't it just generally true that in the absence of any potential, the momentum eigenfunctions are also energy eigenfunctions? In other words, when there is no potential, (in the right units) $$H = p^2/2m + V(x) = p^2/2m$$ Since the Hamiltonian is proportional to the momentum operator squared, it's easy to see that any eigenket of the momentum operator ...

1

These relations are based on the fact that both the position and the momentum distributions are centred around zero, which is in turn due to the symmetry of the atom. Given that, the width of the position and momentum distributions ($\Delta x$ and $\Delta p$) is of the same order as a typical position or momentum within those distributions ($r$ and $p$).

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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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Saying that things "are both waves and particles" is a vestige of the 18th century way of thinking, and really ought to be done away with. Everything is described by a wavefunction. Period. What is a wavefunction? It is a complex-valued function. If you are interested in an electron's position, it is easy to think of it as a complex-valued function of ...

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An electron can interfere with itself. there have been experiment of interference with single electron. Saying that an electron is alway a particule is then wrong. The wave function is the "best" way we have find to describe electrons and other quantons. object at the quantum level are not wave or particles. They just follow quantums rules and waves or ...

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In the special case where $\ell^2$ or $s^2$ has eigenvalue zero, then $j^2$ is fixed. Otherwise you must know the projections $m_\ell,m_s$ to find $j$.

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The way this is justified is as follows: We start with the uncertainty principle, which can be roughly stated as $$\Delta x \Delta p \geq \hbar$$ For this rough estimate, we will ignore some factors of perhaps $2$ or $\pi$, but we're interested in some order of magnitude, not the exact result. Now, our second assumption will be that the ground state of the ...

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Perhaps you are starting by the wrong end. Your concern seems to be related in the first term with the totally misleading notation of integrals in quantum mechanics, and this is more related with the spectral theorem than with distributions itself. Distributions only appear in Quantum mechanics when certain operators has empty spectrum in the usual Hilber ...

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There seems to be no way to proceed unless more information is given. In fact, from what you have above (correcting the typo pointed out by Bernhard and ticster): $$\vec{j}=\vec{l}+\vec{s} \quad \Rightarrow \quad \vec{j}^2=\big(\vec{l}+\vec{s}\big)^2 = \vec{l}^2+\vec{s}^2 + 2\, \vec{l}\cdot\vec{s}\,,$$ meaning that knowledge of the eigenvalues of ...

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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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