# Tag Info

9

More generally, for an arbitrary integrable function $\eta \in {\cal L}^1(\mathbb{R})$ with unit integral $\int_{\mathbb{R}} \! \mathrm{d}y~\eta(y)=1$, one has the following representation $$\delta(x)~=~\lim_{\varepsilon\to 0^+} \frac{1}{\varepsilon}\eta\left(\frac{x}{\varepsilon}\right) \tag{1}$$ of the Dirac delta distribution $\delta(x)$. In more ...

8

The delta function $\delta(x)$ has unit area, but the function $\delta(2x)$ is "half as wide" and thus has half as much area; thus you can pick up extra factors from 'how fast' you cross the peak of the delta function. The general identity is $$\delta(f(x)) = \sum \frac{\delta(x-x_i)}{\big| df/dx|_{x=x_i} \big|}$$ where the $x_i$ are the roots of $f$. In ...

7

Just in case you are not a native speaker of English, the phrase "golden rule" is an idiom which refers to a rule which is cherished or otherwise held in high regard. Essentially, something made of gold is very special. As for why he may have regarded that rule as being "golden", Farcher answers that question in the comments: The rule was first derived ...

6

As for the derivation of the Fermi golden rule, there is a crystal clear one by myself: http://arxiv.org/abs/1404.4280 http://arxiv.org/abs/1604.06916 There is no hand-waving argument at all. It is completely rigorous in the mathematical sense. Of course, it is based on some assumptions on the continuum spectrum and the couplings. In talking about ...

6

In this answer, we shall not go into measurement-theoretical aspects of quantum mechanics. Here we shall just derive the range where (the derivation of) Fermi's golden rule holds and is trustworthy. I) Let us for simplicity assume that all states are normalizable and live in a Hilbert space, so that we have an absolute notion of probability. (This ...

4

You want to show that a limit of some functions is a particular distribution. Specifically: $$\lim_{\alpha\rightarrow\infty} \frac{\sin^2\alpha x}{\alpha x^2} =\pi\delta(x)$$ So first you can ask what a distribution is. A distribution $F$ is a functional. And specifically it takes a test function $g$ and gives you a scalar. An example is if we start with a ...

4

The key thing to keep in mind is that Fermi's golden rule describes transitions between eigenstates of some unperturbed hamiltonian, and that these eigenstates are assumed to be orthogonal during its derivation. Essentially, this means that your final state will be something of the form $$|\psi_i⟩+a(t)|\psi_f⟩,$$ where the relative phase of the two ...

4

I think you're right being cautious calling the quantity $Q(t)/t$ a "transition rate" and it's a pity that this question has not received a complete answer yet. The point is this: what the derivations of Fermi's Golden Rule (FGR) (e.g. the one given above in Qmechanic's answer) tells me is that there is a certain regime in which the expected value $p$ of ...

4

As mentioned in Semoi's answer, electronic transitions in the x-ray regime have the disadvantage that they will tend to be ionizing transitions, i.e. they will put one electron in the continuum, where it will tend to fly away and not come back. However, in general, that is only true for neutral atoms, but once you remove one or a few electrons, the ...

4

These orbitals have many orbital nodes ($n-1$ when $n$ is the orbital quantum number which can be a few hundred). So the radial wave functions are rapidly oscillating, and contributions to the overlap integral with the dipole operator tend to cancel.

4

I do not understand why one would intuitively expect the transition probability to be linear in $t$, equivalently why we would expect the transition rate to be constant. Sakurai is writing for an audience familiar with Fermi's golden rule, wherein the probability of a quantum system to transition into a particular state (or set of states) with the same ...

3

Consider a real function $\;f(x)\;$ of the real variable $\;x\in\mathbb{R}\;$ for which \begin{align} f(x)\boldsymbol{=}0 \quad & \text{for any} \quad x\boldsymbol{\ne} x_{0} \quad \textbf{and} \tag{01a}\label{01a}\\ \mathcal{I}\boldsymbol{=}\!\!\!\!\int\limits_{\boldsymbol{x_{0}-\varepsilon}}^{\boldsymbol{x_{0}+\varepsilon}}\!\!\!f(x)\mathrm dx\...

3

I shall try here to prove the statement posted in your question, but I have to admit that this is an effort with one particular danger; since I'm not a mathematician I can only rely on others with more knowledge from me, so that, any mistakes in the formality of my proof or any inconsistencies on the mathematics involved may be found and corrected. So, ...

3

It is based on the time-dependent perturbation theory. So, surely it is not an exactly result. As it is a perturbative result, generally it will break down if the perturbation (the coupling in the theory) is too strong. The fermi golden rule is widely used in atomic physics to calculate the life time of an excited state and the result is reasonably good, ...

2

As proposed by Lubos, the delta function you started with $\delta(E_i-E_f)$ forces the final result to be invariant by $E_i \leftrightarrow E_f$.

2

One can see the difference by considering the resolvent. Generic matrix elements of $(E-H)^{-1}$ have poles at the discrete spectrum of $H$, but are square integrable (typically smooth) functions of energy in the continuous spectrum, though with large peaks near resonances (poles in the analytic continuation to the nonphysical sheet). Treating the ...

2

The explanation you are looking for is on the bottom of pg.1 in the notes you linked to: The central result is that for weak perturbations and long times, the transition probability $P_b(t)$ raises linear in time as $P_b(t) = \Gamma_{a \rightarrow b} \times t$ for $b\neq a$. Now the transition probability $P_b(t)$ is by definition the average number of ...

2

The original use of the algorithm known as the Fermi rule is Dirac's calculation of Einstein's absorption coefficient of an ensemble of atoms: P.A.M. Dirac, On the theory of quantum mechanics, Proceedings of the Royal Society of London A: Mathematical, Physical and Engineering Sciences (1926), 112, p. 661-677 http://dx.doi.org/10.1098/rspa.1926.0133 http://...

2

There is no measurement. "Transition" means the possibility of finding the system in a different state than it was prepared in. You say it yourself; Fermi's Golden Rule calculates the probability of transition per unit time Prepare the system in an eigenstate of the unperturbed Hamiltonian $H_0$ $$\Psi(t=0) = a_i \psi_i$$ Now switch on a perturbation $... 2 Your second expression is really the more fundamental one, in the sense that Fermi's Golden Rule (FGR) is just an approximation. You can't really go from FGR to the field theoretic result, but you can see why they're equivalent in a certain limit: As you said, we are using first-order perturbation theory to derive FGR, which means the timescales involved ... 2 This question has been discussed in the comment section, and OP has already found the answer in this wonderful post by QMechanic. In that post, we can clearly see that the magnitude of$t$is not arbitrary: it has to be bounded both below and above, $$\frac{2\pi}{\Delta \omega} \lesssim t \ll \frac{\hbar}{\sup_{f\in F}|V_{fi}|}$$ where the symbols are ... 2 On one hand, Fermi's golden rule states that transition probability rate is $$\frac{dP}{dt}~=~\frac{2\pi}{\hbar} | \langle f | V| i\rangle|^2 \rho_f. \tag{1}$$ It is assumed that the initial state$|i\rangle $is normalized. The final states$|f\rangle $do not have to be normalized. (The latter can be seen by putting the system in a potential box with ... 2 Consider an electron, which is bound to an atom. If it absorbs an X-ray photon, the electron leaves the atom and becomes a free electron. Hence, the inverse process is "not possible" by stimulated emission -- the free electron does not belong to the atom. Furthermore, the X-rays would "immediately" be re-absorbed, because transitions of bounded electrons ... 2 Indeed the matrix element between an outer Rydberg state to a lower lying Rydberg state is much larger than the matrix element between the outer Rydberg state and the ground state. However, the decay rates depend also on the density of electromagnetic modes, which depends on the frequency of the transition. In particular, the decay rate$\Gamma \sim \omega^3$... 2 "Probability density" is nowhere to be found in the WP article you are link-quoting. I'll try to detail @rob 's recommendation that should have sufficed to answer your question, together with the article linked. Set$\hbar=1$here for simplicity, easily reinstatable by elementary dimensional analysis. I sense the confusion might reside in the bra-ket ... 2 If the contribution is from state close to$E_n$, they can assume that the level density doesn't vary much in that within that range in energy. Replace$\rho(E_m)$with its value at$E_n$. The delta comes from$\frac{\sin^2(xT)}{Tx^2} \rightarrow \pi \delta(x), T \rightarrow \infty. 2 Let's dispense with all the vector arrows to simplify the notation. \begin{align} \left \langle k | V | k' \right \rangle =& \int dx' \int dx \langle k \underbrace{| x' \rangle \langle x'|}_\text{identity} V \underbrace{| x \rangle \langle x |}_\text{identity} k' \rangle \, . \end{align} Now the question is what to do with\langle x' | V | x \rangle\$. ...

2

OP's formula is derived from the Poisson kernel representation $$\delta(\varepsilon)~=~\lim_{\Gamma \to 0^+}\frac{1}{\pi}\frac{\Gamma/2}{|\varepsilon+i\Gamma/2|^2}$$ of the Dirac delta distribution.

2

In the comments it became clear that OP is trying to understand the formula $$\lim_{t\to\infty} F\left(t,\frac{E-E_i}{\hbar}\right)~=~\pi t~\delta\left(\frac{E-E_i}{2\hbar}\right)~=~2\pi\hbar t~\delta(E-E_i) , \tag{C-32}$$ taken from Ref. 1, where $$F(t,\omega)~=~\left[\frac{\sin(\omega t/2)}{\omega/2}\right]^2. \tag{C-7}$$ It is clear that eq. (C-32) ...

1

I am not sure if I understand your motivation. Is the goal to get formula for rate of change from given initial density matrix to some specified set of density matrices? Formally, I do not know, but here is an obvious problem with such an idea. The golden rule gives the number of transitions, per unit time, from a single initial state to some state in the ...

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