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Introductory Quantum Mechanics (Fitzpatrick)
https://phys.libretexts.org/Bookshelves/Quantum_Mechanics/Introductory_Quantum_Mechanics_(Fitzpatrick)
$\newcommand {\ltapp} {\stackrel {_{\normalsize<}}{_{\normalsize \sim}}}$ $\newcommand {\gtapp} {\stackrel {_{\normalsize>}}{_{\normalsize \sim}}}$ $\newcommand {\btau}{\mbox{\boldmath$\tau$}}$ ... $\newcommand {\ltapp} {\stackrel {_{\normalsize<}}{_{\normalsize \sim}}}$ $\newcommand {\gtapp} {\stackrel {_{\normalsize>}}{_{\normalsize \sim}}}$ $\newcommand {\btau}{\mbox{\boldmath$\tau$}}$ $\newcommand {\bmu}{\mbox{\boldmath$\mu$}}$ $\newcommand {\bsigma}{\mbox{\boldmath$\sigma$}}$ $\newcommand {\bOmega}{\mbox{\boldmath$\Omega$}}$ $\newcommand {\bomega}{\mbox{\boldmath$\omega$}}$ $\newcommand {\bepsilon}{\mbox{\boldmath$\epsilon$}}$
6: Three-Dimensional Quantum Mechanics
https://phys.libretexts.org/Bookshelves/Quantum_Mechanics/Introductory_Quantum_Mechanics_(Fitzpatrick)/06%3A_Three-Dimensional_Quantum_Mechanics
$\newcommand {\ltapp} {\stackrel {_{\normalsize<}}{_{\normalsize \sim}}}$ $\newcommand {\gtapp} {\stackrel {_{\normalsize>}}{_{\normalsize \sim}}}$ $\newcommand {\btau}{\mbox{\boldmath$\tau$}}$ ... $\newcommand {\ltapp} {\stackrel {_{\normalsize<}}{_{\normalsize \sim}}}$ $\newcommand {\gtapp} {\stackrel {_{\normalsize>}}{_{\normalsize \sim}}}$ $\newcommand {\btau}{\mbox{\boldmath$\tau$}}$ $\newcommand {\bmu}{\mbox{\boldmath$\mu$}}$ $\newcommand {\bsigma}{\mbox{\boldmath$\sigma$}}$ $\newcommand {\bOmega}{\mbox{\boldmath$\Omega$}}$ $\newcommand {\bomega}{\mbox{\boldmath$\omega$}}$ $\newcommand {\bepsilon}{\mbox{\boldmath$\epsilon$}}$
1.2: Combining Probabilities
https://phys.libretexts.org/Bookshelves/Quantum_Mechanics/Introductory_Quantum_Mechanics_(Fitzpatrick)/01%3A_Probability_Theory/1.02%3A_Combining_Probabilities
Furthermore, the number of pairs of systems in the ensemble ${\mit\Sigma}\otimes {\mit\Sigma}$ that exhibit the outcome $X$ in the first system and the outcome $Y$ in the second system is simply...Furthermore, the number of pairs of systems in the ensemble ${\mit\Sigma}\otimes {\mit\Sigma}$ that exhibit the outcome $X$ in the first system and the outcome $Y$ in the second system is simply the product of the number of systems that exhibit the outcome $X$ and the number of systems that exhibit the outcome $Y$ in the original ensemble, so that ${\mit\Omega}(X\otimes Y) = {\mit\Omega}(X) \,{\mit\Omega}(Y).$ It follows from the basic definition of probability that
2.2: Plane-Waves
https://phys.libretexts.org/Bookshelves/Quantum_Mechanics/Introductory_Quantum_Mechanics_(Fitzpatrick)/02%3A_Wave-Particle_Duality/2.02%3A_Plane-Waves
It follows, by comparison with Equation (2.2.4), that the wave maxima consist of a series of parallel planes, normal to the wavevector, that are equally spaced a distance $\lambda$ apart, and that p...It follows, by comparison with Equation (2.2.4), that the wave maxima consist of a series of parallel planes, normal to the wavevector, that are equally spaced a distance $\lambda$ apart, and that propagate in the ${\bf k}$ -direction at the velocity $v$ .
14.2: Born Approximation
https://phys.libretexts.org/Bookshelves/Quantum_Mechanics/Introductory_Quantum_Mechanics_(Fitzpatrick)/14%3A_Scattering_Theory/14.02%3A_Born_Approximation
The Born approximation yields $f({\bf k},{\bf k}') \simeq \frac{m}{2\pi\,\hbar^{\,2}} \int {\rm e}^{\,{\rm i}\,({\bf k}-{\bf k'})\cdot{\bf r}'}\,V({\bf r'})\,d^{\,3}{\bf r}'.$ Thus, \(f({\bf k},{\bf...The Born approximation yields $f({\bf k},{\bf k}') \simeq \frac{m}{2\pi\,\hbar^{\,2}} \int {\rm e}^{\,{\rm i}\,({\bf k}-{\bf k'})\cdot{\bf r}'}\,V({\bf r'})\,d^{\,3}{\bf r}'.$ Thus, $f({\bf k},{\bf k}')$ becomes proportional to the Fourier transform of the scattering potential $V({\bf r})$ with respect to the wavevector ${\bf q} = {\bf k}-{\bf k}'$ .
8.4: Rydberg Formula
https://phys.libretexts.org/Bookshelves/Quantum_Mechanics/Introductory_Quantum_Mechanics_(Fitzpatrick)/08%3A_Central_Potentials/8.04%3A_Rydberg_Formula
According to Equation ([e9.55]), the energy of the electron will change by ${\mit\Delta} E = E_0\left(\frac{1}{n_f^{\,2}}-\frac{1}{n_i^{\,2}}\right).$ If ${\mit\Delta} E$ is negative then we would...According to Equation ([e9.55]), the energy of the electron will change by ${\mit\Delta} E = E_0\left(\frac{1}{n_f^{\,2}}-\frac{1}{n_i^{\,2}}\right).$ If ${\mit\Delta} E$ is negative then we would expect the electron to emit a photon of frequency $\nu=- {\mit\Delta}E/h$ . [See Equation ([ee3.15]).] Likewise, if ${\mit\Delta} E$ is positive then the electron must absorb a photon of energy $\nu={\mit\Delta}E/h$ .
7.1: Angular Momenum Operators
https://phys.libretexts.org/Bookshelves/Quantum_Mechanics/Introductory_Quantum_Mechanics_(Fitzpatrick)/07%3A_Orbital_Angular_Momentum/7.01%3A_Angular_Momenum_Operators
In classical mechanics, the vector angular momentum, L, of a particle of position vector ${\bf r}$ and linear momentum ${\bf p}$ is defined...
2.6: Quantum Theory of Light
https://phys.libretexts.org/Bookshelves/Quantum_Mechanics/Introductory_Quantum_Mechanics_(Fitzpatrick)/02%3A_Wave-Particle_Duality/2.06%3A_Quantum_Theory_of_Light
Special relativity also gives the following relationship between the energy $E$ and the momentum $p$ of a massless particle , $p = \frac{E}{c}.$ Note that the previous relation is consistent wit...Special relativity also gives the following relationship between the energy $E$ and the momentum $p$ of a massless particle , $p = \frac{E}{c}.$ Note that the previous relation is consistent with Equation (2.4.12), because if light is made up of a stream of photons, for which $E/p=c$ , then the momentum density of light must be the energy density divided by $c$ .
12.2: Two-State System
https://phys.libretexts.org/Bookshelves/Quantum_Mechanics/Introductory_Quantum_Mechanics_(Fitzpatrick)/12%3A_Time-Dependent_Perturbation_Theory/12.02%3A_Two-State_System
Consider a system in which the time-independent Hamiltonian possesses two eigenstates, denoted $\begin{aligned} H_0\,\psi_1 &= E_1\,\psi_1,\\[0.5ex] H_0\,\psi_2&=E_2\,\psi_2.\end{aligned}$ Suppose, ...Consider a system in which the time-independent Hamiltonian possesses two eigenstates, denoted $\begin{aligned} H_0\,\psi_1 &= E_1\,\psi_1,\\[0.5ex] H_0\,\psi_2&=E_2\,\psi_2.\end{aligned}$ Suppose, for the sake of simplicity, that the diagonal elements of the interaction Hamiltonian, $H_1$ , are zero: that is, $\langle 1|H_1|1\rangle = \langle 2|H_1|2\rangle = 0.$ The off-diagonal elements are assumed to oscillate sinusoidally at some frequency $\omega$ : that is, \[\langle 1|H_1|2\rangle…
10.2: Angular Momentum in Hydrogen Atom
https://phys.libretexts.org/Bookshelves/Quantum_Mechanics/Introductory_Quantum_Mechanics_(Fitzpatrick)/10%3A_Addition_of_Angular_Momentum/10.02%3A_Angular_Momentum_in_Hydrogen_Atom
$\label{e11.31} J^{\,2}\,\psi^{(2)}_{l,1/2;j,m+1/2}= j\,(j+1)\,\hbar^{\,2}\,\psi^{(2)}_{l,1/2;j,m+1/2},$ where [see Equation ([e11.12])] \[\label{e11.32} J^{\,2} = L^2+S^{\,2} +2\,L_z\,S_z+ L_+\,S_-... $\label{e11.31} J^{\,2}\,\psi^{(2)}_{l,1/2;j,m+1/2}= j\,(j+1)\,\hbar^{\,2}\,\psi^{(2)}_{l,1/2;j,m+1/2},$ where [see Equation ([e11.12])] $\label{e11.32} J^{\,2} = L^2+S^{\,2} +2\,L_z\,S_z+ L_+\,S_-+L_-\,S_+.$ Moreover, according to Equations ([e11.28]) and ([e11.29]), we can write $\label{e11.33} \psi^{(2)}_{l,1/2;j,m+1/2} = \alpha\,Y_{l,m}\,\chi_+ + \beta\, Y_{l,m+1}\,\chi_-.$ Recall, from Equations ([eraise]) and ([elow]), that \[\begin{aligned} \label{e11.34} L_+\,Y_{l,m} &= [l\,(l+1)-…
2.4: Classical Light-Waves
https://phys.libretexts.org/Bookshelves/Quantum_Mechanics/Introductory_Quantum_Mechanics_(Fitzpatrick)/02%3A_Wave-Particle_Duality/2.04%3A_Classical_Light-Waves
Suppose that the wave is polarized such that this electric field oscillates in the $y$ -direction. (According to standard electromagnetic theory, the magnetic field oscillates in the $z$ -direction,...Suppose that the wave is polarized such that this electric field oscillates in the $y$ -direction. (According to standard electromagnetic theory, the magnetic field oscillates in the $z$ -direction, in phase with the electric field, with an amplitude which is that of the electric field divided by the velocity of light in vacuum. ) Now, the electric field can be conveniently represented in terms of a complex wavefunction:

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