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3

Although we can define the momentum as a self-adjoint operator in $L^2[0,1]$ as you proposed, I think it's rather artificial to think about it as having relation to momentum in the case of $L^2(\Bbb{R})$. Realize that the operator $p_1$ with domain $D(p_1)=\{\psi\in\mathcal{H}^1[0,1]\,|\,\psi(1)=\psi(0)\}$, is related to spatial translations via the unitary ...


1

The fact that $D(p_0)$ is not big enough to define a self-adjoint operator does not mean it is not included into the domain of the self-adjoint momentum. The choice of boundary conditions for the function is a specification of the vector, not of the operator. Once you have fixed the self-adjoint extension you are considering (choosing the proper domain, ...


0

I would like to add another point to the ones already mentioned by Kyle Kanos and tpg2114. When you deal with partial differential equations (PDE), you have to take into account more things than with ordinary differential equations. I will try to explain the some of the ideas behind them, but without entering into details. For example in evolution ...


1

Both boundary conditions and initial conditions matter equally when connecting the model to the real world. Consider the flow around a cylinder that you mention. We know that the Reynolds number, $$ {\rm Re}=\frac{u L}{\nu}\tag{1} $$ can characterize laminar or turbulent flows, depending on the values in (1). Below are flows for different Reynolds numbers ...


2

I didn't entirely understand your question, aside from the last bit "what makes a solution of these equations unique?" And you are absolutely right, it is the boundary and initial conditions that entirely determine the solution (which may or may not be unique, that's a very open problem for the Navier-Stokes equations). You say that the situation is ...


0

I am really just motivating a few links. I also hope a few comments may be added to the mix. UPDATE: See Qmechanic's links for extended details relevant to qft. Well I was not sure if I should dare write anything here, but I am sure there are some flaws in my thinking, so someone can correct me. OK. Let's begin with some ...


0

Nakahara's Geometry, Topology and Physics has two chapters covering Fiber Bundles up to Connections on Fiber Bundles with a few applications in Gauge theories. If it is your first time learning Fiber Bundles i would recommend this books, it's rigorous and has a lot of physics-motivated examples.


4

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

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

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, ...


4

You can simply take the semi-derivative of your equation again, which yields $$\begin{align} m\frac{d^2}{dt^2}\underbrace{\frac{d^{\tfrac12}x}{dt^{\tfrac12}}}_{=-\frac mk\frac{d^2x}{dt^2}} &= -k\frac{dx}{dt} \\\Rightarrow m^2\frac{d^4x}{dt^4} &= k^2\frac{dx}{dt} \tag{*} \end{align}$$ and then solve that ODE. But, similarly to squaring an ...


3

I think I good book for that may be C. J. Isham's Modern Differential Geometry for Physicists. I haven't gotten to the chapter of fiber bundles, but what I've read seems to be quite rigorous. And as it is written for physicists, I think it could please your needs.


2

So, what you have here is, as others have mentioned, a fractional order differential equation. Since others have provided graphs, there seems to be little point to adding one. Also, you seem more interested in the qualitative aspect than actual analytical solutions, as you've mentioned. In essence, what you have here is some second order system with some ...


2

An attempt for a more explicit solution. Using the definition of the half-derivative given by JamalS, one can transform the differential equation using the Laplace transform and get $$s^2X(s)-sx(0)-x'(0)=-\gamma^3{\sqrt s}\;X(s)$$ where $\gamma=\sqrt[3]{\frac km}$ is a positive constant. Solving for $X$ gives $$X(s)=\frac{sx(0)+\dot x(0)}{s^2+\gamma^3{\sqrt ...


1

What you can say is that you have a distribution of the results you are going to get (be it a discrete or continuous random variable), and when you calculate the average of a large sample, you are adding the random variables and multiplying by a constant. The addition of random variables translates into a convolution of the probability density functions, ...


3

In general, we can say nothing about finite $n$, but most of the time, we can safely assume some "niceness" of the distributions in question. If, for example, we assume a finite Variance $\sigma^2$ (a quite common feature), we could use Chebyshev's inequality for a rough error estimation of the form $$P(|\bar{X_n} - µ| > \alpha) \leq ...


6

This is the exact reason why we do statistical hypothesis confidence testing. In essence, the confidence interval we get from this test is a quantitative measure of "how far we've converged". For example, consider an experiment to test whether a coin is imbalanced or not. Our null hypothesis is that it is not: in symbols, our hypothesis is ${\rm Pr}(H) = ...


1

Whether or not a PDE allows separation of variables depends not only on the equation, but also the boundary conditions. The following conditions must be satisfied for the method of separation of variables to work: The differential operator should be separable. An example of the one which is not separable is \begin{equation} \frac{\partial^2u}{\partial x^2} ...


2

It is usually very difficult to give a characterization of the domain of self-adjointness of an operator. However, the Harmonic oscillator is a well-known operator. Unluckily, this does not mean there is a completely explicit form of its domain. Anyways, I will give you what in my opinion is the best shot at explicitness: As you may know there are ...


3

First of all, if you focus on proper functions, instead of elements of $L^2$, the domain is much more tough than your candidate ($L$ extended to our domain is again simply essentially self adjoint but not self-adjoint). The self-adjointness domain contains functions which are nowhere differentiable. A trivial example: If you consider the simpler operator: ...


3

1) One way to show that a map constitutes a 1-1 correspondence is by showing that it has an inverse. This is what is done here: elements of $T_eG$ are tangent vectors to $G$ at the unit element, and one-parameter subgroups of $G$ are smooth homomorphisms $\mathbb R\to G$. Note that this is a notion more general than a closed 1-dimensional subgroup, for which ...


1

You found $$x = e^{ik} + e^{-ik} - 2 = 2(\cos(k) - 1) \leq 0 . $$ If the question is whether $g\ge 0$ then the answer is "no" in the sense that it's eigenvalues are non-positive as we have just showed. A bit of intuition We recognize $$\phi(n+1) + \phi(n-1) -2\phi(n)$$ as some kind of second derivative. Consider the first derivative of a function $f$ ...


2

One way to try to solve the equation is transforming it in an ODE. Apply the fractional derivative $D^{1/2}$ again to the equation to find $$D^{1/2}[D^2x(t)]=D^{5/2}x(t)-C_1t^{-3/2}-C_2t^{-5/2}-C_3t^{-7/2},$$ and $$D^{1/2}[D^{1/2}x(t)]=Dx(t)-C_4t^{-3/2}$$ Hence we got $$D^{5/2}x(t)=-\frac{k}{m}Dx(t)+C_1t^{-3/2}+C_2t^{-5/2}+C_3t^{-7/2}$$ But, we also have ...


14

I am no mathematician and am a little afraid that my answer is too simple to be true, but here goes: I use Fourier transforms to define the fractional derivative. $x(\omega)$ is defined such that $$ x(t) = \int_{-\infty}^\infty \, \frac{\text{d}\omega}{2\pi} \text{e}^{i \omega t} \, x(\omega) \, .$$ Then any integer derivatives is $$ ...


20

If $D^n$ denotes the $n$th derivative and $D^{-n}$ the $n$th integral, then we have that, $$D^n f(t) = D^m[D^{-(m-n)}f(t)]$$ providing $m \geq \lceil{n}\rceil$. For our half derivative, we choose $n=1/2$, and $m=2$, in which case we have, $$D^{1/2}f(t) = D^2[D^{-(3/2)}f(t)]$$ There is a general formula for the $n$th integral of a function, one of my ...


3

Consider the case of a free scalar field, governed by the usual Lagrangian, $$\mathcal{L} = \frac{1}{2}\partial_\mu \phi \partial^\mu \phi - \frac{1}{2}m^2 \phi^2$$ The propagator, or equivalently Green's function for the theory is a function which can be though of as a response when we use a delta function as an input in the equations of motion, i.e. ...


0

You can think of a Floquet energy in a similar way to a Bloch state. In the latter case, because space is periodic, the momentum states are repeated at every reciprocal lattice vector, $\textbf{G}$. For a Floquet state, because time is periodic, energy states are repeated every $n\hbar \omega$ where $n$ is an integer and $\hbar \omega$ depends on the time, ...


4

There's a problem of notation here. If you are assuming that $\langle \cdot \rangle$ is the expectation in time, ie: $\langle \cdot \rangle = \int_0^\infty \cdot dt$ then $\frac{d \langle x \rangle}{dt} =0$ by definition. The expectation does not change in time because it's the time-averaged value! However, let's assume that you are using an ensemble ...


1

All these functions (and more) are documented in the Handbook of mathematical functions by Abramowitz and Stegun (which is a staple of many professor libraries)


0

There are no proofs in physics. PERIOD FULL STOP. There are only measurements and models that are fit numerically. If that is not what you are looking for, you will have to stick with mathematics.


1

Both $L^p$-spaces and Sobolev spaces are actually defined via equivalence classes (this is at least one of many equivalent ways). By definition, two functions that do not differ in norm are representants of the same function as already explained. Now, note that the Sobolev norm is just the sum of the $L^p$ norms for weak derivatives (to whichever order you ...


1

However then I say: Why do you make things so complicated? Suppose you want to calculate $\exp(A)$. Why don't you define $$\exp(A)~:=~1+A+1/2 A^2 + \ldots $$ and require convergence with respect to the operator norm. An example: Consider the vectorspace spanned by the monomials $1,x,x^2,\ldots$ and let $A=d/dx$. Then you can perfectly define ...


1

I think expecting "fluid" behaviour in terms of a material that does not support shear, is not useful in the context of the various systems you have listed in the question. Instead, I believe you are intuitively connecting ideas and concepts pertaining to conservation laws. So the idea that in specific systems, conserved charges (in the sense of Noether) ...


2

Incompleteness of a coordinate system is not a canonical definition as that, for instance, of geodesical (in)completeness. It simply means that the domain of the coordinate system does not cover the whole manifold (and perhaps there are several inequivalent extensions of the initial manifold represented by the given domain of the coordinate system). If a ...


2

Typically, the Hilbert spaces one considers in quantum mechanics are $L^2$ spaces. The elements of these spaces are equivalence classes of functions which differ only on a null set of points, i.e. whose distance in terms of the $L^2$ norm is zero, $\|n-\tilde n\|_{L^2}=0$. That is, you are right, but it's the $L^2$ norm that matters, not the Sobolev norm. ...



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