Tag Info

Hot answers tagged

8

For time-independent Hamiltonians, $U(t) = \mathrm{e}^{\mathrm{i}Ht}$ follows from Stone's theorem on one-parameter unitary groups, since the Schrödinger equation $$ H\psi = \mathrm{i}\partial_t \psi$$ is just the statement that $H$ is the infinitesimal generator of a one-parameter group parameterized by the time $t$. For time-dependent Hamiltonians $H(t) = ...


5

This rather detailed answer will potentially chart out the issues and assumptions that are needed for a straightforward idea. Understanding how it is done might be more useful than actually doing it. Assumption 0) The Hamiltonian is time independent. To be honest, I never find arguments about taking derivatives of operators very appealing. We will have a ...


3

A very simple proof proceeds along the following lines. $$\Psi(t_1+\Delta t)=\Psi(t_1)+\frac{\partial \Psi}{\partial t} \Delta t $$ where we omitted higher order $\Delta t$ terms (will disappear when we let $\Delta t \to 0$ later). Applying this formula recursively, we get $$\Psi(t_1+2\Delta t)=\Psi(t_1)+2\frac{\partial \Psi (t_1)}{\partial t} \Delta ...


3

An intuitive answer to motivate the Stone theorem that ACuriousMind's answer cites. Take a quantum system. For now, let it be a finite dimensional one (the Stone theorem is needed to make the reasoning work in a separable, infinite dimensional Hilbert space). Let it evolve for time $t_1$. The evolved state is some linear operator $U(t_1)$ imparted on the ...


3

First of all you need the vectors $|O_k\rangle$ to form an orthonormal system. Then to prove what you need you start with polynomials and verify that, e.g. $O^2$ has the desired form etc... For a continuous function $f$, assuming that $O$ is a bounded operator you can then invoke the Stone-Weierstrass theorem.


3

The proof is given in great generality by the beautiful and powerful spectral theorem; in its functional calculus form. This theorem clarifies the meaning of your first expression, and specifies the functions for which you can write the second (that however are many). In addition, it applies to any self-adjoint operator, bounded or unbounded. Sadly, the ...


3

Question: Does there exist a nontrival non-Legendre transformation T such that the function defined by F(q,p,t)=T[L(q,q˙,t)] contains the full dynamics of the system? Answer: any function that produces the equations of motion under some sort of rules that you state is an allowed function to describe the dynamics. In particular any function that you can ...


3

TL;DR: It is the wedge product $\wedge$ and the exterior derivative/differential $d$ (which squares to zero $d^2=0$) that give rise to Grassmann-odd elements and supersymmetry. More concretely, Ref. 1 first writes down a (non-relativistic) SUSY algebra ${\cal A}$ $$\tag{10} \{Q_1,Q_1\}_+~=~2H~=~\{Q_2,Q_2\}_+, \qquad \{Q_1,Q_2\}_+~=~0, $$ spanned by two ...


2

Some time ago I have asked a similar question; although I have accepted one of the answers, It did not satisfy my main interest concerning the physics of the case when the diagonal $\Delta$ is not removed. Now I have some more information that I can share with you. Most of the authors give two reasons for the removal of the diagonal: If we include the ...


2

Hints: The square root function has a Taylor expansion around $a>0$ $$\tag{1} \sqrt{a+b}~=~\sum_{n=0}^{\infty} \begin{pmatrix}\frac{1}{2} \cr n\end{pmatrix}a^{\frac{1}{2}-n}b^n, \qquad |b| ~<~a. $$ One may show that a possible non-commutative generalization reads $$ \sqrt{A+B}~=~\sqrt{A}+\sum_{n=1}^{\infty} \begin{pmatrix}\frac{1}{2} \cr ...


2

So, it looks like you're trying to answer a question like this: you've got a plate which is held at $T=0$ at $x = 0, x = L$ for all $y$, with $T= T_0(x)$ at $y=0$. What is $T(x, y)$ in this space $y \ge 0, 0\le x\le L?$ Then the answer is that on this strip we can solve Laplace's equation $\nabla^2T = 0$ with the functions $T(x, y) = A \exp(-\alpha y) ...


2

First of all, a small historical note. As far as I know, in constructive mathematics you can construct uncountable sets, e.g. the reals, since the usual way of introducing them (by Dedekind cuts) is constructible, i.e. involves a countable process. Of course it is not done in a finite number of steps method, but indeed constructible. Most of the people that ...


2

Let $\mathcal{M}$ be our spacetime. Then, a gauge theory is given by a connection form $A$ on a principal bundle over it (that locally projects onto the spacetime in a way compatible with gauge transformations), which is the gauge potential. Maxwell's equations1 (in vacuum) are the equations of motion for the gauge field for the Yang-Mills action coupled to ...


1

Hints to the question (v2): First of all, one should realize the abuse of notation in eq. (6.69) of Ref. 1 where $\phi$ is used in two meanings: both as an external parameter and as internal integration/dummy variable. It is more properly written as $$ \widehat {Z}[\phi]~=~\frac{{e}^{iS[\phi]}}{{\cal N}}, \qquad {\cal N}~:=~\int\!{\cal D}\phi~e^{iS[\phi]} ...


1

If the eigenvectors of $\hat H$ don't depend explicitly on $t$ (the eigenvalues can) then we can easily make sense of $\hat H~U = i \partial_t U$ by working in the eigenbasis of $\hat H$ where $H$ is diagonal, because if $\hat H = \operatorname{diag}(\lambda_1, \lambda_2, \dots)$ then $\exp(-i~\hat H~t) = \operatorname{diag}(e^{-i\lambda_1t}, ...


1

As said in the commentaries, the first one comes from the Dirac formalism. Simply put, it deals with quantum states as vectors $\lvert a \rangle$ whose components contain the projections of the system into different eigenstates. For a system, the set of state eigenvectors $\lvert a_i \rangle$ must be linearly independent $\langle a_i \lvert a_j ...


1

An elementary way to proceed is as follows. Let's put an explicit factor of $\epsilon$ in $B$. The problem is then to solve the following equation for the matrix $X$: $$A + \epsilon B = X^{2}$$ We want to do this perturbatively, so we assume that $X$ can be represented as: $$X = \sum_{n=0}^{\infty}\epsilon^{n}X^{(n)}$$ We can then write: $$ ...



Only top voted, non community-wiki answers of a minimum length are eligible