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10

You may indeed identify the generators in the way you did. However, the Lie algebras and Lie groups are different because – as quickly said by Qmechanic – you must use different reality conditions for the coefficients. A general matrix in the $SU(2)$ group is written as $$ M = \exp[ i( \alpha J_+ + \bar\alpha J_- + \gamma J_0 )] $$ where $\alpha\in ...


7

In QM, a real valued observable $A$ is mathematically represented by a projector valued measure over $\mathbb{R}$, $P^{(A)}$, i.e., if $E$ is a Borel subset of the real line, then $P^{(A)}(E)$ is a projector representing the proposition "the outcome of measuring $A$ falls in $E$". In principle, that's all you need for representing, mathematically, ...


7

The ladder operators do belong to the real Lie algebra $$\quad su(1,1)~\cong~ so(2,1)~\cong~sl(2,\mathbb{R}),$$ but they do not belong to the real Lie algebra $$su(2)~\cong~ so(3).$$ All the above real Lie algebras have complexifications isomorphic to $sl(2,\mathbb{C})$.


3

I) Well, $r=0$ is the boundary of a $d$-dimensional spherical coordinate patch, i.e. an artifact of our choice of coordinate system, but $r=0$ does not correspond to a physical boundary per se, other than what we can deduce from the TISE. The mantra is that a boundary condition (for finite $r$) should only be imposed if it is a consequence of the TISE. See ...


3

There is a mathematical point that can be made, and in my opinion is related to a deeper understanding of what it means to solve a (partial) differential equation. I will try to keep things simple, and consider only linear models. Suppose that you have a space $X$ with some properties, for example it has a topology. We suppose that the state of our system ...


3

In principle, yes, $\varphi$ is a coordinate system, in that it tags every member of $\mathcal S$ with an appropriate $n$-tuple of real numbers. If you know a set of coordinates $(c_1,\ldots,c_n)$, you can use $\varphi^{-1}$ to find the corresponding point in $\mathcal S$. In practice, this will not be useful at all, mostly because the amount of real ...


3

Assuming that your operator has a spectrum consisting of isolated points you can look for all the independent solutions of the eigenvalue equation $$(Q-\lambda I)\xi = 0$$ Let these solutions generate a vector space $V_\lambda$ and then compute the dimension of $V_\lambda$. If it is greater than 1 then the eigenvalue $\lambda$ is degenerate.


2

We are not able, at least for the moment, to define in a rigorous mathematical fashion a meaningful interacting QFT in $3+1$ dimensions that is coherent with the perturbative theory utilized by physicists (in more precise words, that satisfies the Wightman axioms). On the contrary, some rigorous interacting QFTs can be defined in lower (spatial) dimensions. ...


2

The tools and techniques for constructing QFTs are the same whatever the dimension. However you cannot prove conjectures which are false, no matter how powerful your tools are. The issue with 3+1 is that the class of models for which the conjecture "yeah it can be constructed" is in all likelihood true is much more narrow than in 1+1 and 2+1. Basically the ...


2

$\renewcommand{ket}[1]{|#1\rangle}$ As others have pointed out, you can go whole hog and solve the characteristic equation $\text{det}(\hat{Q} - \lambda \hat{I})=0$ and find repeated solutions for $\lambda$. However, there is a simpler, more physically intuitive way to hunt for degeneracy: look for symmetry. When an operator $\hat{Q}$ has a symmetry there ...


2

Comment to the question (v1): No, such decomposition is in general not unique. E.g. the unit element ${\bf 1}_{2\times 2}\in SU(2)$ can be written with parameters $b\in 4\pi\mathbb{Z}$ and $a=0=c$.


1

QFTs in spacetime of dimension $<4$ have their use in real applications - 3D theories to quantum surfaces, and 2D theories to quantum wires. There much of the exceptional behavior of lower-dimensional QFTs can be observed. A famous example is the fractional Hall effect with anyonic (rather than Bose or Fermi) statistics.


1

Comments to the question (v2): To be specific, let us assume that the underlying 2D manifold is the Riemann sphere $S^2\cong \mathbb{C}\cup\{\infty\}$. The group of globally defined conformal transformations is the 6-dimensional group $PSL(2,\mathbb{C})$ of Moebius transformations. Mathematically speaking, one should consider the groupoid of locally ...


1

You are essentially correct. If you start on the real line with the Schrödinger equation $$ \left(-\frac12 \frac{\partial^2}{\partial x^2} + V(x) \right ) \psi(x) = E\,\psi(x),$$ then at every point $x_0$ where $V(x)$ is analytic you are guaranteed that $\psi(x)$ will be analytic in a neighbourhood of that point. However, if $V(x)$ is not analytic then you ...


1

Think of this not as an extremely rigorous way of solving the differential equation, but rather as using your intuition to guess a solution. Often when you are given a differential equation, the solution is not at all obvious, and perhaps the equation isn't even solvable analytically. Instead of giving up, though, sometimes you can identify a parameter ...


1

A good discussion of regularity properties of the Wiener measure is in section II.5 of the book "Functional Integration and Quantum Physics" by Barry Simon. He gives the proofs of most of the relevant theorems except the borderline case $\alpha=\frac{1}{2}$, namely Levy's Theorem showing that $\frac{|x(\tau_1)-x(\tau_2)|}{\sqrt|\tau_1-\tau_2|}$ diverges ...


1

Well, consider a Lax pair for the Harmonic Oscillator; \begin{equation} L = \begin{pmatrix} p & \omega q \\ \omega q & -p \\ \end{pmatrix}, \quad M=\frac{\omega}{2} \begin{pmatrix} 0 & -1 \\ 1& 0 \\ \end{pmatrix} \end{equation} Since the Hamiltonian is $$H(q, p)=\frac{1}{2}(p^{2}+\omega^{2}q^{2})$$ It is eay to check that the Lax ...


1

As Phoenix87 said, you need to solve the asociated eigenvalue problem. $\lambda$ is an eigenvalue if and only if $\hat{Q}|\Psi>=\lambda|\Psi>$. If you have a matrix expression for the operator $\hat{Q}$, then the usual way to solve the problem is writing the above equation as Phoenix87 did, writing everything in the left side: ...


1

For simplicity consider the 1-d case, with $\psi =\sqrt{n} e^{2i\phi}$, then $$i \psi_t =\frac{i}{2} \frac{\dot{n}}{\sqrt{n}} e^{2i\phi} -\sqrt{n} e^{2i\phi} 2\dot{\phi}.$$ Similarly $$ \frac{\partial H}{\partial \psi^*} = \frac{\partial H}{\partial n}\frac{\partial n}{\partial \psi^*} + \frac{\partial H}{\partial \phi}\frac{\partial \phi}{\partial ...


1

Unitarity of the time-evolution operator is exactly the point: Stone's theorem (see e.g. Reed, Simon: Theorems VIII.7, VIII.8) tells us If $A$ is self-adjoint, the spectral theorem holds. This gives us a functional calculus which makes it possible to define $U(t) = e^{itA}$ in the first place. A such defined $U(t)$ is a strongly continuous unitary group. ...



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