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12

Creating anti-protons is straightforward in principle because any high energy collision produces a shower of protons, antiprotons and various types of pions. The pions decay in a few nanoseconds, so you just have to wait for the pions to decay then separate the antiprotons from the protons. At Fermilab a 120GeV proton beam was collided with a nickel target ...


10

Be aware that a "mathematical derivation" of a physical principle is, in general, not possible. Mathematics does not concern the real world, we always need empirical input to decide which mathematical frameworks correspond to the real world. However, the Schrödinger equation can be seen arising naturally from classical mechanics through the process of ...


7

The factor of $1/2\pi$ is an artifact of the normalization convention being used for the momentum eigenstates. To begin to see how this is so, let us note that the choice of normalization of a Dirac-orthogonal continuous basis completely determines the form of the resolution of the identity. Writing an arbitrary state $|\psi\rangle$ in a given ...


6

The momentum operator $P$ in the infinite well can be defined as a self-adjoint operator by infinitely many ways with respect to the boundary conditions by: $$P_\theta=-i\hbar\frac{d}{dx}\\ \mathcal{D}(P_\theta)=\left\{\psi\in \mathcal{H}^1[0,a]:\psi(a)=e^{i\theta}\psi(0)\right\},$$ where $\mathcal{H}^1[0,a]$ is the Sobolev space, on the interval $[0,a]$. In ...


4

As you say, pictures in QM are merely different ways to compute the exact same thing - the expectation value of observables $\mathcal{O}$. Now, in all pictures, we can derive Ehrenfest's theorem $$ \frac{\mathrm{d}}{\mathrm{d}t}\langle \mathcal{O} \rangle = \frac{1}{\mathrm{i}\hbar}\langle [\mathcal{O},H]\rangle + \langle\frac{\partial}{\partial ...


4

(1) We could very well write $0|\psi\rangle=0$ but we must keep in mind that the first $0$ is a scalar, the one that belongs to the field over which the vector space (where $|\psi\rangle$ lives) is defined, while the second $0$ is the zero vector of that v.s. (we always use the same symbol to denote both, provided we always are careful when writing these ...


3

It is an entangled state, as you concluded yourself it cannot be written as a direct tensor product. To recap again: $$ \left( \begin{array}{c} \alpha_0\\ \alpha_1\\ \end{array} \right) \otimes \left( \begin{array}{c} \beta_0\\ \beta_1\\ \end{array} \right) = \left( \begin{array}{c} \alpha_0 \beta_0\\ \alpha_0 \beta_1\\ \alpha_1 \beta_0\\ \alpha_1 \beta_1\\ ...


3

I'm not a fan of explicit calculations, so let us answer a bit more general question: Given two two-dimensional Hilbert spaces $\mathcal{H}_1$ and $\mathcal{H}_2$ with basis $\lvert \uparrow \rangle_i$, $\lvert \downarrow \rangle_i$, when is a state in $\mathcal{H}_1 \otimes \mathcal{H}_2$ entangled? Observe that $\lvert \psi \rangle_i = u_i\lvert \uparrow ...


3

First, the classical and semiclassical adjectives are not quite synonyma. "Semiclassical" means a treatment of a quantum system whose part is described classically, and another part quantum mechanically. Fields may be classical, particle positions' inside the fields quantum mechanical; metric field may be classical and other matter fields are quantum ...


3

Yes, operators in quantum mechanics can be understood basically as infinite matrices, $|r\rangle$ as basis vectors and $\psi(r)\equiv \langle\psi|r\rangle \sim \psi_r \sim "\psi_i"$ as components of the state vector numbered by a continuous index $r$. $\langle r|F|r'\rangle$ are indeed just matrix components of the operator $F$. Generally $\langle ...


3

Yes, the device is called a Faraday mirror and it consists of a normal mirror following a Faraday rotator. The latter is a magneto-optical device that rotates the state of polarization of light passing through it in a non-reciprocal manner. The most well-known application of Faraday rotators is to provide optical isolation. The Faraday effect is wavelength ...


3

Unfortunately, I am not so sure what ⟨r|F|r′⟩ is ... in order to evaluate this expectation value it's not an expectation, it is a matrix element; think of it as the components of the operator $F$ on the position basis. If the operator is 'diagonal' on the position basis then $\langle r|F|r' \rangle$ is zero except when $r = r'$. Thus, for example, ...


3

In a certain sense, what you said that every operator might be represented in position representation as a integral operator may be true for many operators only if you allow distributions to be used, and even though that's not always the case. You are kind of confusing things when you say about the dual of distributions. What is a distribution is the ...


3

Short answer: complexifications facilitate representation theory. In physics, we typically want to find representations of a Lie algebra $\mathfrak g$, and often times determining the representations of its complexification $\mathfrak g_\mathbb C$ is easier. Moreover, we have the following theorem (see ref 1. Proposition 4.6) which tells us that ...


3

He's doing a linear approximation. Suppose $\Delta x$ is very small. Then $\langle x - \Delta x | \alpha \rangle$ is almost equal to $\langle x | \alpha \rangle$, but not quite, because $\Delta x$ isn't zero. So we do a first order approximation: Let's write $\langle x | \alpha \rangle$ as $f(x)$. Then $f(x - \Delta x) \approx f(x) - \Delta x ...


2

Technically the electron and proton are both orbiting the barycenter of the system, both in classical and quantum mechanics, just as in gravitational systems. You find the same dynamics for the system if you assume the proton and electron are moving independently about the barycenter, or if you convert to a one-body problem of a single "particle" with the ...


2

Fundamental laws of physics cannot be derived (turtles all the way down and all that). However, they can be motivated in various ways. Direct experimental evidence aside, you can argue by analogy - in case of the Schrödinger equation, comparisons to Hamiltonian mechanics and the Hamilton-Jacobi equation, fluid dynamics, Brownian motion and optics have been ...


2

The most elementary formulation of Quantum Mechanics (the one usually formulated in Hilbert spaces) can be constructed starting form a lattice of all the elementary propositions which can be tested on a given quantum system obtaining, as the outcome, YES or NOT. This can be similarly done for classical mechanics and, in that case the elementary ...


2

This definition of bound and scattering states is not quite correct, although it holds for many potentials. There are counterexamples to this fact that have roots in a paper by von Neumann and Wigner. One is the spherical potential $$V(r)=32\sin r \frac{\sin r-(1+r)\cos r}{\left(2+2r-\sin 2r\right)^2}$$ It is not hard to check that $V(r)$ is a bounded ...


2

The wave mechanics dispersion relation you cite is for EM waves propagating in free space. In other media, the dispersion relation is not necessarily linear (it can be quadratic or have some more complex dependence). So in this context, there's nothing special about quantum mechanics. More generally, the dispersion relation tells us about the phase speed ...


2

Can someone explain what I'm doing wrong You're not doing anything wrong but you haven't taken the next logical step. The potential is an even function so there (may) be both an even and an odd solution (proving this is also a problem in this chapter of the textbook). For an even solution, and using your coefficients, we require $$A = D$$ and $$B = ...


2

A superposition is a perfectly valid state in quantum mechanics. It simply doesn't have a correspondence in classical physics, which is the result of quantum mechanics (and not the other way around!). "In other words, the aforementioned particle doesn't have a definite state until it is "looked at" (measured)." Yes, the particle (or better "the ...


2

If you assume separability of the wave function, i.e., $\psi(\mathbf x)=u(x)v(y)w(z)$, you can solve the individual components separately: \begin{align} -\frac{\hbar^2}{2\mu}\frac{d^2u(x)}{dx^2}+V_1(x)u(x)&=E_1u(x)\\ -\frac{\hbar^2}{2\mu}\frac{d^2v(y)}{dy^2}+V_2(y)v(y)&=E_2v(y)\tag{1}\\ ...


2

From a mathematical perspective, to develop Lie algebra representation theory most efficiently, we need the field $\mathbb{F}$ of the Lie algebra to be algebraically closed. See e.g. Ref. 1, where this assumption is used already in the beginning of Chapter II. The situation for Lie algebras is similar to when we in linear algebra try to diagonalize, say, a ...


1

There are results that are mathematically rigorous concerning the semiclassical limit of quantum theories. It is in fact an ongoing and interesting theme of research in mathematical physics. However you need to be rather well versed in analysis to understand the results. The bibliography is quite huge, but I would like to mention the following (some quite ...


1

Any (analytical) function $f(x)$ can be written as a Taylor expansion, $$ f(x) = \sum_{k=0}^\infty a_k x^k $$ As you see, in this expansion, only powers of $x$ and some constants appear. If we want to build a function of $\mathbf{S}$ which gives us a vector, the only way we can do it using only $\mathbf{S}$ is taking its vector product with himself, ...


1

In quantum physics, a particle can be in a superposition of two states until it is measured. In other words, the aforementioned particle doesn't have a definite state until it is "looked at" (measured). A superposition state is a state. So a particle in such a state does have a state before being measured. Since certain properties (i.e., an ...


1

I also don't quite understand what you mean by "imaginary $\psi$", but let me give you some general, more mathematically correct view. In general, what we measure is the spectrum of a self-adjoint operator. For the energy, this operator is of course the Hamiltonian. Now, since it is self-adjoint, its spectrum will lie on the real line. To have stability, we ...


1

The integral form is not an alternative way of writing the completeness relation. The integral form is the definition of the completeness relation ONLY when a continuous variable is involved. What you have included in your question, equation (2) only shows how to compute the completeness relation when the number of states are infinite i.e the spatial ...


1

The question you ask is a tough one and everyone has his own opinion about the answer (look at the comments of your question). In particular, one has to adopt a more or less clear philosophical position about what science ought to tell us in order to reply. That being said, the decoherence programme tries to address some of the questions you wonder about. ...



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