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As per this question, using the Laplace transform is quite difficult for this problem. I would like to post an alternate transformation based on so-called similarity solutions. For diffusion problems where a scalar field is initially uniform and the scalar quantity starts diffusing from one boundary to another boundary very far away (e.g. $u(\infty,t)=0$), ...


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Indeed using a rigged Hilbert Space is almost certainly what you should use and there are tutorials on it on arxiv (I'll add them when/if/I can find them again). However it isn't necessarily going to help you understand what an average physicist is doing on the one hand, and on the other hand there were already many things you were doing that not rigorous ...


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As all the answers are oriented in the theoretical side let me remind everybody of the Balmer series, an experimental observation fitted with a mathematical series, which was first modeled with the Bohr model of the hydrogen atom, and then was the cornerstone of building quantum mechanics, as it came out from the Schrodinger equation. So in this sense a ...


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The answer is no, I am afraid. As you may well know, the self-adjoint Laplace operator $-\Delta$ on $L^2(\mathbb{R})$ has purely absolutely continuous spectrum $\mathbb{R}^+$. Now let $V\in L^{\infty}(\mathbb{R},\mathbb{R}^+)$ be an arbitrary bounded positive function. Then $-\Delta_x +V(x)$, where $V$ acts as a multiplicative operator is self-adjoint and ...


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A helpful yet elementary answer may do the trick, If you are familiar with the Euler-Lagrange equation then it will be straight forward and you can skip ahead a little. If not then you have to accept that there is an equation in physics that generalises classical mechanics called the Euler-Lagrange equation. For a particle moving in one dimension under a ...


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It is difficult to understand conservation from symmetry. But the opposite is much simpler. conservation means the invariance of equation of motion in its form under certain transformation. and the invariance of equation of motions arises as an implication of the underlying symmetry. for example i am taking one equation X2 =1 the solutions are +or-1 which ...


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I am giving you a few differential equations applied in physics. CLASSICAL MECHANICS Newton's second law: Newton's law of cooling Kinematical equations of motion Radioactive decay equation Wave equation QUANTUM MECHANICS This is the home of innumerable differential equations. A few being the Schrodinger's equation,the Klein-Gordon equation, ...


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The whole Newtonian-mechanics blossoms from the famous differential equation: $$m\dfrac{dv}{dt} = F\left(t,x, \frac{dx}{dt}\right)$$ Then waves are one of the great examples: the wave equation is a PDE : $$\frac{\partial ^2 \phi}{\partial t^2} = c^2\frac{\partial ^2 \phi}{\partial x^2} $$. Then the famous Maxwell's equations & the great Schroedinger's ...


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Air resistance or air drag is basically proportional to square of the velocity of the moving object ,that is, F=bv^2 since air has turbulent motion where b is the air drag constant.Drag constant depends on the dimensions of the object. So air drag is independent of length or volume of object, rather it depends on the surface area of the object. The ...


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First, we will compute $\text{div}\,F$. The partial derivatives are given by$${{\partial F}\over{\partial x_i}} = q {\partial\over{\partial x_i}}\left({{x_i}\over{r^3}}\right) = q\left({1\over{r^3}} - {{3r^2 {{x_i}\over{r}} x_i}\over{r^6}}\right) = 0.$$Thus, $\text{div}\,F = 0$ away from the origin. Consider now a ball $B$ of radius $r$ centered at the ...


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Ref.1 is already in eq. (3.1) considering a functional integral over the scalar field $\phi:M\to \mathbb{R}$. Here $M$ is spacetime. For a rigorous treatment of functional integrals, Ref. 1 points in the beginning of Section 3 to its Ref. [3.2]. In this answer we will just take an intuitive heuristic approach, and try to construct the functional integral as ...


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tl;dr: The canonical commutation relations require unbounded operators. Your truncation procedure seems ill-defined because it destroys the Hilbert space structure: Option one: You truncate to the span of finitely many vectors. The canonical commutation relations are not representable on finite-dimensional Hilbert spaces, this space is useless (because ...


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I am not sure if you mean $x$ one or two dimensional variable. I don't have the article at hand but I guess we probably mean $D=1$?? The argument goes as follows: you have to use Bochner's theorem, which says that $W(x-y):=W_2(x,y)$ is positive distribution if and only if the Fourier transformation yields is a positive measure $\tilde W(p) d^Dp$. In this ...


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Yes, they can. It's not great form, but they can. For a simple example, consider the hamiltonian for a 1D particle in a potential $V(x)$, $$ \hat H=\frac1{2m}\hat{ p}^2+V(\hat{x}). $$ This holds equally well if the potential is a delta function at $x_0$, $V(x)=\kappa \delta(x-x_0)$, in which case $$ \hat H=\frac1{2m}\hat{ p}^2+\kappa\delta(\hat{x}-x_0). $$ ...


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Once you admit the tunnelling the "topology" of the problem changes to "particle in a circle with an infinite barrier". In fact, to avoid arguments with energy when crossing the barrier, it can be described as "particle in a circle with an infinite barrier in one point". So physically it can be argued to be a different problem.


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Technically speaking, manifolds are by definition topological spaces, which resemble locally an inner-product space. Since there are vector spaces (with a dot product) of infinite dimension, then there shall be infinately-dimensional manifolds as well. The infinity of the dimension is not a problem for the tensors as well - each multi-linear function over a ...


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For an analytic function $f$, $f(A)$ can be defined as the Maclaurin series, as described in the OP. However, when $A$ is an element of a C*-algebra rather than just a Banach algebra, one can also consider continuous functions over the spectrum of $A$, which can still be defined as limits of certain polynomials of $A$ (as a consequence of the ...


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The term to look for is Coulomb wave. These wavefunctions are well explained in the corresponding Wikipedia article. Depending on your mathematical background, you should be ready for a bit of a formula jolt, as these wavefunctions rely very intimately on the confluent hypergeometric function. If you want the short of it, then I can tell you that the ...


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I think the point is that if you took the limits to infinity without doing anything else, you'd be implicitly redefining the matrix element in order to make the equations consistent. So he just calls the redefined matrix element $T_{ni}$ instead of $V_{ni}$. Later he solves for $T_{ni}$ in terms of $V_{nj}$ (see the section "Solving for the T matrix").



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