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15

This is a good question---from the logical point of view, a scalar is not an object of the same "type" as the one by one matrix you can make from it. (I have just spent hours trying to compile someone else's code which won't compile because of similar type errors.) From the physical point of view, there is no difference if they both can be used for the ...


15

The can be represented by matrices, yes. That doesn't mean they are matrices, though. Matrices aren't really very meaningful mathematical objects, they just provide a handy way of specifying linear operators on paper or in computer programs. In case of Bra-Ket, the situation is actually this: you have a Hilbert Space $\mathcal{H}$, which contains certain ...


5

This may seem naive, but as an engineer used to visualizing things, I always understand the multiplication of vectors and matrices as in this diagram: In order to multiply two matrices AB, A has to have as many columns as B has rows, and the result has as many rows as A, and as many columns as B. You get each element of the result by multiplying ...


4

The chain of reasoning runs something like this: Assume that the distributions $\mu_0(\lambda)$ and $\mu_+(\lambda)$ overlap in $\Delta\subset\Lambda$, so with probability $q$ preparing $|0⟩$ will produce a state $\lambda$ consistent with $|+⟩$ and vice versa. Assume that the two systems can be prepared independently. Therefore, with probability $q$ the ...


4

The full work-through goes as follows: \begin{align} ⟨H⟩ & = \int \Psi^* H \Psi dx \\ & = \int \left(\sum_m e^{-iE_m t}c_{m} \psi_{m}\right)^*H\left(\sum_n e^{-iE_n t}c_{n} \psi_{n}\right)\mathrm dx \\ & = \sum_m \sum_n e^{iE_m t}e^{-iE_n t}c_{m}^* c_{n}\int \psi_{m}^*H \psi_{n}\mathrm dx \\ & = \sum_m \sum_n e^{i(E_m-E_n) t}c_{m}^* ...


4

After an extended search through my copy of Pais' biography Niels Bohr's Times, I believe I have found what exactly what you are referring to. There were multiple times in his life when Bohr wrote about the definition of the word phenomenon. On pages 431-433 of my copy of Pais' book, Bohr's thoughts on this matter are summarized. I will now reproduce the ...


4

To talk about $\langle \psi_i \vert U(t) \vert \psi_i \rangle$ as the "expectation value of the time evolution operator" is probably the least insightful way to talk about this quantity. Since $U(t) = \exp(-\mathrm{i}Ht)$ for time-independent Hamiltonians, if you want to look at expectation values, you could as well look at that of $H$ directly. Note that, ...


4

The point is that your equation (1) does not make any sense. On the left side, you have operator acting on a vector in some abstract vector space. On the right side you have the position representation of momentum operator acting again on the same abstract vector. Those objects live in different spaces, you cannot just multiply them (actually you can if you ...


3

No, atoms have the same number of protons and electrons so they have no net charge. On the other hand ions (cations and anions) would be repelled or attracted depending on their net charge. Atoms are bound together in a molecules by different means like covalent bonding, ionic bonding (which can be easily explained in terms of electrostatic forces) or ...


3

The state is a vector in the Hilbert space of the Hamiltonian, which gives it a natural basis in terms of the eigenvectors; distinct eigenvalues then exist in distinct (orthogonal) subspaces - for degenerate values the subspaces are larger, but they are still distinct from all others. Clearly this situation gives many advantages in analysis. However, this ...


3

This is the hydrogen atom energy level solutions, as an easy example. The electron sits at the ground state, and can be kicked up to an excited state by the appropriate photon i.e. given that the photon has the quantized energy needed. For each energy level one can calculate using the solutions of the Schrodinger equation, the probability for the ...


3

Your statement on a pseudo-probability interpretation is slightly off, in that you most certainly did not write the Wigner quasi-probability distribution in your edit. Taking your two bases to be $x$ and $p$ for specificity, what you wrote, $ρ(x,p)$, up to a phase: $e^{(ixp/ħ)}$, is the "standard ordering prescription" for quasi-distribution functions, ...


3

In answer to the title question: no, you can't always decompose an $L_2$ function in terms of only the bound spectrum of hydrogen. This is because there are orthogonal functions to all bound states, which naturally represent the free states of the electron. The quickest examples are of course the Coulomb-wave eigenfunctions $|\chi_{E,l,m}⟩$ of the ...


3

Let's write the full Hamiltonian of the problem as $$ H = H_A + H_P + H_C + V,$$ where $H_A + H_P + H_C$ are the local Hamiltonians of $A$, $P$ and $C$, and their interaction is $V$. If you want the interaction to take place in arm $a$ of the interferometer, the interaction should be of the form $$ V = H_{AP} \otimes \lvert C_a \rangle \langle C_a\rvert.$$ ...


3

As you intuit, it is indeed pretty hard to define a sense of "direction" for an atomic electron within quantum mechanics when the electron doesn't have an orbit but it is instead some fuzzy ball of probability, but it is doable and in fact it is one of the central constructs in atomic physics. What ends up mattering is angular momentum, i.e. how much the ...


2

Write $d^3 q = dq q^2 d\theta d\phi \sin\theta$ and integrate over the angular variables. The only angular dependence in the integrand is in $e^{i \vec{q} \cdot ( \vec{x}-\vec{y}) } = e^{i q r \cos\theta}$ where I've defined $r = | \vec{x} - \vec{y} |$. Then, we have $$ \int_0^{2\pi} d\phi \int_0^\pi d\theta \sin\theta e^{i q r \cos\theta} $$ There is no ...


2

OP's ket first equation$^1$ $$\tag{1} \hat{p}|x\rangle ~=~+i\hbar\frac{\partial |x\rangle}{\partial x}$$ is explained in eq. (7) of my Phys.SE answer here. In short, eq. (1) is consistent with the corresponding bra equation $$\langle x |\hat{p} ~=~-i\hbar \frac{\partial \langle x |}{\partial x} $$ via Hermitian conjugation. The bra equation in turn is ...


2

Suppose $\left\{\left|e_i\right\rangle|i\in I\right\}$ is an orthonormal basis of a Hilbert space $\mathcal{H}$, viz. $\left\langle e_i |e_j\right\rangle =\delta_{ij}$. Then the identity operator from $\mathcal{H}$ to $\mathcal{H}$ can be written as an outer product $$\mathbb{I}=\Sigma_{i\in I}\left|e_i\right\rangle\left\langle ...


2

No. The general relation is given by $$v = \lambda\nu$$ Where $v$ is the velocity of the considered wave and $\lambda$ and $\nu$ its wavelength and frequency. Of course in the case of an electromagnetic wave which is traveling at the speed of light you gain $$c = \lambda\nu$$ If you're treating instead some massive particle, then thou have $$p = ...


2

Before quantum mechanics, we were familiar with particles of momentum $\mathbf{p}$ and plane waves $\propto \exp i\left(\mathbf{k}\cdot\mathbf{x}-\omega_\mathbf{k} t\right)$ with $\mathbf{k}$ the wavevector and $\omega_\mathbf{k}$ a $\mathbf{k}$-dependent angular, not linear, frequency. (The formula relating $\omega$ to $\mathbf{k}$ is called a dispersion ...


2

I don't know if I'm understanding your question right, but I think you are trying to pose a deeper question than it might seem at first sight... In ordinary quantum mechanics, when you study the hydrogen atom, you derive a set of solutions for the electron wavefunction using the Schrödinger equation (with different values of the energy). These are the ...


2

MWI is a minority position amongst interpretations in QM; famously Deutsch is an exponent. It gives the wave function of a QM ontological weight; with branching occurring at measurement, where typically, in the Copengagen-type interpretations, collapse occurs; I think Everrets original motivation was to remove this 'discontinuity'. But this can't be quite ...


2

The physical description of what Bell's inequality is about is as follows. Suppose that the outcomes of measurements are described by stochastic variables. That is, each quantity has a value that is some number picked at random in some way. And suppose that for each system the quantities influencing how the numbers are picked are determined locally. That is, ...


2

An eigenvector is simply a vector that is unaffected (to within a scalar value) by a transformation. Formally, an eigenvector is any vector $x$ such that for an operator $\Omega$, $\Omega x = \lambda x$ for some scalar constant $\lambda$. All operators of dimension $n$ have exactly $n$ eigenvectors/eigenvalues (though these are only all distinct if $\Omega$ ...


2

tl;dr: Your points 1 to 5 are misunderstandings. The answer to your question follows from a better understanding: Entanglement is no active link, hence there is no need for instantaneous action. However, if you want a certain kind of ontology, then you must accept a certain kind of instantaneous action/non-locality (which doesn't necessarily violate ...


2

I fear you are overthinking it. I am not sure why the original references, Kubo (1964) and Royer (1976) and their proofs are not adequate for you. Working in Fock space with creation and annihilation operators is a bit self-defeating, unless you were suitably adroit. Sticking to standard phase space operators $\hat{x}, \hat{p}$ and using the standard CBH ...


2

So, first states 1,2 and 3 are not the eigenstates of the Hamiltonian. To see this, try carrying out this operation $$ \hat{H} |1\rangle = E |1\rangle $$ What you will find is that this is NOT true. What you should find is that you get a combination of states 1 and 3 when you do this. So the question is what are the eigenstates? The best way to determine ...


2

Raman scattering is inelastic scattering from molecules. The photon interacts with the molecule and changes the molecules vibrational, rotational or electron energy. Rayleigh scattering is in the main elastic scattering from small particles whose size is less than that of the wavelength of the photon. The scattering can occur of atoms or molecules and ...


1

The meaning of eigenvalue and eigenvector(or eigenstate if you want)depends on what operator and what observables you are considering. If the operator is now a hamiltonian, the eigenvalue you get will be the energy of the system, and the eigenvector tell you its "state" So for the SHO system,the eigenvalue of the hamiltonian is (n+1/2)hf=Energy and n ...


1

$\newcommand{\real}{\mathbb R}\newcommand{\field}{\mathbb F}\newcommand{\cx}{\mathbb C}\newcommand{\ip}[2]{\left< #1,#2\right>}$We need to dive into mathematics of vector spaces and inner products in order to understand what a vector means and what is it mean to take a scalar product of two vectors. There is a long post ahead so bear with me even ...



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