diff options
Diffstat (limited to 'content/know/concept')
| -rw-r--r-- | content/know/concept/bose-einstein-distribution/index.pdc | 83 | ||||
| -rw-r--r-- | content/know/concept/electric-field/index.pdc | 129 | ||||
| -rw-r--r-- | content/know/concept/fermi-dirac-distribution/index.pdc | 86 | ||||
| -rw-r--r-- | content/know/concept/landau-quantization/index.pdc | 3 | ||||
| -rw-r--r-- | content/know/concept/larmor-precession/index.pdc | 108 | ||||
| -rw-r--r-- | content/know/concept/magnetic-field/index.pdc | 110 |
6 files changed, 518 insertions, 1 deletions
diff --git a/content/know/concept/bose-einstein-distribution/index.pdc b/content/know/concept/bose-einstein-distribution/index.pdc new file mode 100644 index 0000000..2462c68 --- /dev/null +++ b/content/know/concept/bose-einstein-distribution/index.pdc @@ -0,0 +1,83 @@ +--- +title: "Bose-Einstein distribution" +firstLetter: "B" +publishDate: 2021-07-11 +categories: +- Physics +- Statistics +- Quantum mechanics + +date: 2021-07-11T18:22:44+02:00 +draft: false +markup: pandoc +--- + +# Bose-Einstein statistics + +**Bose-Einstein statistics** describe how bosons, +which do not obey the [Pauli exclusion principle](/know/concept/pauli-exclusion-principle/), +will distribute themselves across the available states +in a system at equilibrium. + +Consider a single-particle state $s$, +which can contain any number of bosons. +Since the occupation number $N_s$ is variable, +we turn to the [grand canonical ensemble](/know/concept/grand-canonical-ensemble/), +whose grand partition function $\mathcal{Z_s}$ is as follows, +where $\varepsilon_s$ is the energy per particle, +and $\mu$ is the chemical potential: + +$$\begin{aligned} + \mathcal{Z}_s + = \sum_{N_s = 0}^\infty \Big( \exp\!(- \beta (\varepsilon_s - \mu)) \Big)^{N_s} + = \frac{1}{1 - \exp\!(- \beta (\varepsilon_s - \mu))} +\end{aligned}$$ + +The corresponding [thermodynamic potential](/know/concept/thermodynamic-potential/) +is the Landau potential $\Omega$, given by: + +$$\begin{aligned} + \Omega_s + = - k T \ln{\mathcal{Z_s}} + = k T \ln\!\Big( 1 - \exp\!(- \beta (\varepsilon_s - \mu)) \Big) +\end{aligned}$$ + +The average number of particles $\expval{N_s}$ +is found by taking a derivative of $\Omega$: + +$$\begin{aligned} + \expval{N_s} + = - \pdv{\Omega_s}{\mu} + = k T \pdv{\ln{\mathcal{Z_s}}}{\mu} + = \frac{\exp\!(- \beta (\varepsilon_s - \mu))}{1 - \exp\!(- \beta (\varepsilon_s - \mu))} +\end{aligned}$$ + +By multitplying both the numerator and the denominator by $\exp\!(\beta(\epsilon_s \!-\! \mu))$, +we arrive at the standard form of the **Bose-Einstein distribution** $f_B$: + +$$\begin{aligned} + \boxed{ + \expval{N_s} + = f_B(\varepsilon_s) + = \frac{1}{\exp\!(\beta (\varepsilon_s - \mu)) - 1} + } +\end{aligned}$$ + +This tells the expected occupation number $\expval{N_s}$ of state $s$, +given a temperature $T$ and chemical potential $\mu$. +The corresponding variance $\sigma_s^2$ of $N_s$ is found to be: + +$$\begin{aligned} + \boxed{ + \sigma_s^2 + = k T \pdv{\expval{N_s}}{\mu} + = \expval{N_s} \big(1 + \expval{N_s}\big) + } +\end{aligned}$$ + + + +## References +1. H. Gould, J. Tobochnik, + *Statistical and thermal physics*, 2nd edition, + Princeton. diff --git a/content/know/concept/electric-field/index.pdc b/content/know/concept/electric-field/index.pdc new file mode 100644 index 0000000..ce2c4fc --- /dev/null +++ b/content/know/concept/electric-field/index.pdc @@ -0,0 +1,129 @@ +--- +title: "Electric field" +firstLetter: "E" +publishDate: 2021-07-12 +categories: +- Physics +- Electromagnetism + +date: 2021-07-12T09:46:25+02:00 +draft: false +markup: pandoc +--- + +## Electric field + +The **electric field** $\vb{E}$ is a vector field +that describes electric effects, +and is defined as the field that +correctly predicts the Lorentz force +on a particle with electric charge $q$: + +$$\begin{aligned} + \vb{F} + = q \vb{E} +\end{aligned}$$ + +This definition implies that the direction of $\vb{E}$ +is from positive to negative charges, +since opposite charges attracts and like charges repel. + +If two opposite point charges with magnitude $q$ +are observed from far away, +they can be treated as a single object called a **dipole**, +which has an **electric dipole moment** $\vb{p}$ defined as follows, +where $\vb{d}$ is the vector going from +the negative to the positive charge (opposite direction of $\vb{E}$): + +$$\begin{aligned} + \vb{p} = q \vb{d} +\end{aligned}$$ + +Alternatively, for consistency with [magnetic fields](/know/concept/magnetic-field/), +$\vb{p}$ can be defined from the aligning torque $\vb{\tau}$ +experienced by the dipole when placed in an $\vb{E}$-field. +In other words, $\vb{p}$ satisfies: + +$$\begin{aligned} + \vb{\tau} = \vb{p} \times \vb{E} +\end{aligned}$$ + +Where $\vb{p}$ has units of $\mathrm{C m}$. +The **polarization density** $\vb{P}$ is defined from $\vb{p}$, +and roughly speaking represents the moments per unit volume: + +$$\begin{aligned} + \vb{P} \equiv \dv{\vb{p}}{V} + \:\:\iff\:\: + \vb{p} = \int_V \vb{P} \dd{V} +\end{aligned}$$ + +If $\vb{P}$ has the same magnitude and direction throughout the body, +then this becomes $\vb{p} = \vb{P} V$, where $V$ is the volume. +Therefore, $\vb{P}$ has units of $\mathrm{C / m^2}$. + +A nonzero $\vb{P}$ complicates things, +since it contributes to the field and hence modifies $\vb{E}$. +We thus define +the "free" **displacement field** $\vb{D}$ +from the "bound" field $\vb{P}$ +and the "net" field $\vb{E}$: + +$$\begin{aligned} + \vb{D} \equiv \varepsilon_0 \vb{E} + \vb{P} + \:\:\iff\:\: + \vb{E} = \frac{1}{\varepsilon_0} (\vb{D} - \vb{P}) +\end{aligned}$$ + +Where the **electric permittivity of free space** $\varepsilon_0$ is a known constant. +It is important to point out some inconsistencies here: +$\vb{D}$ and $\vb{P}$ contain a factor of $\varepsilon_0$, +and therefore measure **flux density**, +while $\vb{E}$ does not contain $\varepsilon_0$, +and thus measures **field intensity**. +Note that this convention is the opposite +of the magnetic analogues $\vb{B}$, $\vb{H}$ and $\vb{M}$, +and that $\vb{M}$ has the opposite sign of $\vb{P}$. + +The polarization $\vb{P}$ is a function of $\vb{E}$. +In addition to the inherent polarity +of the material $\vb{P}_0$ (zero in most cases), +there is a possibly nonlinear response +to the applied $\vb{E}$-field: + +$$\begin{aligned} + \vb{P} = + \vb{P}_0 + \varepsilon_0 \chi_e^{(1)} \vb{E} + + \varepsilon_0 \chi_e^{(2)} |\vb{E}| \: \vb{E} + + \varepsilon_0 \chi_e^{(3)} |\vb{E}|^2 \: \vb{E} + ... +\end{aligned}$$ + +Where the $\chi_e^{(n)}$ are the **electric susceptibilities** of the medium. +For simplicity, we often assume that only the $n\!=\!1$ term is nonzero, +which is the linear response to $\vb{E}$. +In that case, we define +the **relative permittivity** $\varepsilon_r \equiv 1 + \chi_e^{(1)}$ +and the **absolute permittivity** $\varepsilon \equiv \varepsilon_r \varepsilon_0$, +so that: + +$$\begin{aligned} + \vb{D} + = \varepsilon_0 \vb{E} + \vb{P} + = \varepsilon_0 \vb{E} + \varepsilon_0 \chi_e^{(1)} \vb{E} + = \varepsilon_0 \varepsilon_r \vb{E} + = \varepsilon \vb{E} +\end{aligned}$$ + +In reality, a material cannot respond instantly to $\vb{E}$, +meaning that $\chi_e^{(1)}$ is a function of time, +and that $\vb{P}$ is the convolution of $\chi_e^{(1)}(t)$ and $\vb{E}(t)$: + +$$\begin{aligned} + \vb{P}(t) + = (\chi_e^{(1)} * \vb{E})(t) + = \int_{-\infty}^\infty \chi_e^{(1)}(t - \tau) \: \vb{E}(\tau) \:d\tau +\end{aligned}$$ + +Note that this definition requires $\chi_e^{(1)}(t) = 0$ for $t < 0$ +in order to ensure causality, +which leads to the [Kramers-Kronig relations](/know/concept/kramers-kronig-relations/). diff --git a/content/know/concept/fermi-dirac-distribution/index.pdc b/content/know/concept/fermi-dirac-distribution/index.pdc new file mode 100644 index 0000000..8820cbb --- /dev/null +++ b/content/know/concept/fermi-dirac-distribution/index.pdc @@ -0,0 +1,86 @@ +--- +title: "Fermi-Dirac distribution" +firstLetter: "F" +publishDate: 2021-07-11 +categories: +- Physics +- Statistics +- Quantum mechanics + +date: 2021-07-11T18:22:37+02:00 +draft: false +markup: pandoc +--- + +# Fermi-Dirac distribution + +**Fermi-Dirac statistics** describe how identical **fermions**, +which obey the [Pauli exclusion principle](/know/concept/pauli-exclusion-principle/), +will distribute themselves across the available states in a system at equilibrium. + +Consider one single-particle state $s$, +which can contain $0$ or $1$ fermions. +Because the occupation number $N_s$ is variable, +we turn to the [grand canonical ensemble](/know/concept/grand-canonical-ensemble/), +whose grand partition function $\mathcal{Z}_s$ is as follows, +where we sum over all microstates of $s$: + +$$\begin{aligned} + \mathcal{Z}_s + = \sum_{N_s = 0}^1 \exp\!(- \beta N_s (\varepsilon_s - \mu)) + = 1 + \exp\!(- \beta (\varepsilon_s - \mu)) +\end{aligned}$$ + +Where $\mu$ is the chemical potential, +and $\varepsilon_s$ is the energy contribution per particle in $s$, +i.e. the total energy of all particles $E_s = \varepsilon_s N_s$. + +The corresponding [thermodynamic potential](/know/concept/thermodynamic-potential/) +is the Landau potential $\Omega_s$, given by: + +$$\begin{aligned} + \Omega_s + = - k T \ln{\mathcal{Z}_s} + = - k T \ln\!\Big( 1 + \exp\!(- \beta (\varepsilon_s - \mu)) \Big) +\end{aligned}$$ + +The average number of particles $\expval{N_s}$ +in state $s$ is then found to be as follows: + +$$\begin{aligned} + \expval{N_s} + = - \pdv{\Omega_s}{\mu} + = k T \pdv{\ln{\mathcal{Z}_s}}{\mu} + = \frac{\exp\!(- \beta (\varepsilon_s - \mu))}{1 + \exp\!(- \beta (\varepsilon_s - \mu))} +\end{aligned}$$ + +By multiplying both the numerator and the denominator by $\exp\!(\beta (\varepsilon_s \!-\! \mu))$, +we arrive at the standard form of +the **Fermi-Dirac distribution** or **Fermi function** $f_F$: + +$$\begin{aligned} + \boxed{ + \expval{N_s} + = f_F(\varepsilon_s) + = \frac{1}{\exp\!(\beta (\varepsilon_s - \mu)) + 1} + } +\end{aligned}$$ + +This tells the expected occupation number $\expval{N_s}$ of state $s$, +given a temperature $T$ and chemical potential $\mu$. +The corresponding variance $\sigma_s^2$ of $N_s$ is found to be: + +$$\begin{aligned} + \boxed{ + \sigma_s^2 + = k T \pdv{\expval{N_s}}{\mu} + = \expval{N_s} \big(1 - \expval{N_s}\big) + } +\end{aligned}$$ + + + +## References +1. H. Gould, J. Tobochnik, + *Statistical and thermal physics*, 2nd edition, + Princeton. diff --git a/content/know/concept/landau-quantization/index.pdc b/content/know/concept/landau-quantization/index.pdc index 4212078..60b1331 100644 --- a/content/know/concept/landau-quantization/index.pdc +++ b/content/know/concept/landau-quantization/index.pdc @@ -13,7 +13,8 @@ markup: pandoc # Landau quantization -When a particle with charge $q$ is moving in a homogeneous magnetic field, +When a particle with charge $q$ is moving in a homogeneous +[magnetic field](/know/concept/magnetic-field/), quantum mechanics decrees that its allowed energies split into degenerate discrete **Landau levels**, a phenomenon known as **Landau quantization**. diff --git a/content/know/concept/larmor-precession/index.pdc b/content/know/concept/larmor-precession/index.pdc new file mode 100644 index 0000000..3affdee --- /dev/null +++ b/content/know/concept/larmor-precession/index.pdc @@ -0,0 +1,108 @@ +--- +title: "Larmor precession" +firstLetter: "L" +publishDate: 2021-07-02 +categories: +- Physics +- Quantum mechanics + +date: 2021-07-02T15:48:41+02:00 +draft: false +markup: pandoc +--- + +# Larmor precession + +Consider a stationary spin-1/2 particle, +placed in a [magnetic field](/know/concept/magnetic-field/) +with magnitude $B$ pointing in the $z$-direction. +In that case, its Hamiltonian $\hat{H}$ is given by: + +$$\begin{aligned} + \hat{H} = - \gamma B \hat{S}_z = - \frac{\hbar}{2} \gamma B \hat{\sigma_z} +\end{aligned}$$ + +Where $\gamma = - q / m$ is the gyromagnetic ratio, +and $\hat{\sigma}_z$ is the Pauli spin matrix for the $z$-direction. +Since $\hat{H}$ is proportional to $\hat{\sigma}_z$, +they share eigenstates $\ket{\downarrow}$ and $\ket{\uparrow}$. +The respective eigenenergies $E_{\downarrow}$ and $E_{\uparrow}$ are as follows: + +$$\begin{aligned} + E_{\downarrow} = \frac{\hbar}{2} \gamma B + \qquad + E_{\uparrow} = - \frac{\hbar}{2} \gamma B +\end{aligned}$$ + +Because $\hat{H}$ is time-independent, +the general time-dependent solution $\ket{\chi(t)}$ is of the following form, +where $a$ and $b$ are constants, +and the exponentials are "twiddle factors": + +$$\begin{aligned} + \ket{\chi(t)} + = a \exp\!(- i E_{\downarrow} t / \hbar) \: \ket{\downarrow} + \:+\: b \exp\!(- i E_{\uparrow} t / \hbar) \: \ket{\uparrow} +\end{aligned}$$ + +For our purposes, we can safely assume that $a$ and $b$ are real, +and then say that there exists an angle $\theta$ +satisfying $a = \sin\!(\theta / 2)$ and $b = \cos\!(\theta / 2)$, such that: + +$$\begin{aligned} + \ket{\chi(t)} = \sin\!(\theta / 2) \exp\!(- i E_{\downarrow} t / \hbar) \: \ket{\downarrow} + \:+\: \cos\!(\theta / 2) \exp\!(- i E_{\uparrow} t / \hbar) \: \ket{\uparrow} +\end{aligned}$$ + +Now, we find the expectation values of the spin operators +$\expval*{\hat{S}_x}$, $\expval*{\hat{S}_y}$, and $\expval*{\hat{S}_z}$. +The first is: + +$$\begin{aligned} + \matrixel{\chi}{\hat{S}_x}{\chi} + &= \frac{\hbar}{2} + \begin{bmatrix} a \exp\!(i E_{\downarrow} t / \hbar) \\ b \exp\!(i E_{\uparrow} t / \hbar) \end{bmatrix}^{\mathrm{T}} + \cdot + \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} + \cdot + \begin{bmatrix} a \exp\!(- i E_{\downarrow} t / \hbar) \\ b \exp\!(- i E_{\uparrow} t / \hbar) \end{bmatrix} + \\ + &= \frac{\hbar}{2} + \begin{bmatrix} a \exp\!(i E_{\downarrow} t / \hbar) \\ b \exp\!(i E_{\uparrow} t / \hbar) \end{bmatrix}^{\mathrm{T}} + \cdot + \begin{bmatrix} b \exp\!(- i E_{\uparrow} t / \hbar) \\ a \exp\!(- i E_{\downarrow} t / \hbar) \end{bmatrix} + \\ + &= \frac{\hbar}{2} \Big( a b \exp\!(i (E_{\downarrow} \!-\! E_{\uparrow}) t / \hbar) + + b a \exp\!(i (E_{\uparrow} \!-\! E_{\downarrow}) t / \hbar) \Big) + \\ + &= \frac{\hbar}{2} \cos\!(\theta/2) \sin\!(\theta/2) \Big( \exp\!(i \gamma B t) + \exp\!(- i \gamma B t) \Big) + \\ + &= \frac{\hbar}{2} \cos\!(\gamma B t) \Big( \cos\!(\theta/2) \sin\!(\theta/2) + \cos\!(\theta/2) \sin\!(\theta/2) \Big) + \\ + &= \frac{\hbar}{2} \sin\!(\theta) \cos\!(\gamma B t) +\end{aligned}$$ + +The other two are calculated in the same way, +with the following results: + +$$\begin{aligned} + \matrixel{\chi}{\hat{S}_y}{\chi} = - \frac{\hbar}{2} \sin\!(\theta) \sin\!(\gamma B t) + \qquad + \matrixel{\chi}{\hat{S}_z}{\chi} = \frac{\hbar}{2} \cos\!(\theta) +\end{aligned}$$ + +The result is that the spin axis is off by $\theta$ from the $z$-direction, +and is rotating (or **precessing**) around the $z$-axis at the **Larmor frequency** $\omega$: + +$$\begin{aligned} + \boxed{ + \omega = \gamma B + } +\end{aligned}$$ + + + +## References +1. D.J. Griffiths, D.F. Schroeter, + *Introduction to quantum mechanics*, 3rd edition, + Cambridge. diff --git a/content/know/concept/magnetic-field/index.pdc b/content/know/concept/magnetic-field/index.pdc new file mode 100644 index 0000000..2ad5fbf --- /dev/null +++ b/content/know/concept/magnetic-field/index.pdc @@ -0,0 +1,110 @@ +--- +title: "Magnetic field" +firstLetter: "M" +publishDate: 2021-07-12 +categories: +- Physics +- Electromagnetism + +date: 2021-07-12T09:46:31+02:00 +draft: false +markup: pandoc +--- + +## Magnetic field + +The **magnetic field** $\vb{B}$ is a vector field +that describes magnetic effects, +and is defined as the field +that correctly predicts the Lorentz force +on a particle with electric charge $q$: + +$$\begin{aligned} + \vb{F} + = q \vb{v} \cross \vb{B} +\end{aligned}$$ + +If an object is placed in a magnetic field $\vb{B}$, +and wants to rotate to align itself with the field, +then its **magnetic dipole moment** $\vb{m}$ +is defined from the aligning torque $\vb{\tau}$: + +$$\begin{aligned} + \vb{\tau} = \vb{m} \times \vb{B} +\end{aligned}$$ + +Where $\vb{m}$ has units of $\mathrm{J / T}$. +From this, the **magnetization** $\vb{M}$ is defined as follows, +and roughly represents the moments per unit volume: + +$$\begin{aligned} + \vb{M} \equiv \dv{\vb{m}}{V} + \:\:\iff\:\: + \vb{m} = \int_V \vb{M} \dd{V} +\end{aligned}$$ + +If $\vb{M}$ has the same magnitude and orientation throughout the body, +then $\vb{m} = \vb{M} V$, where $V$ is the volume. +Therefore, $\vb{M}$ has units of $\mathrm{A / m}$. + +A nonzero $\vb{M}$ complicates things, +since it contributes to the field +and hence modifies $\vb{B}$. +We thus define +the "free" **auxiliary field** $\vb{H}$ +from the "bound" field $\vb{M}$ +and the "net" field $\vb{B}$: + +$$\begin{aligned} + \vb{H} \equiv \frac{1}{\mu_0} \vb{B} - \vb{M} + \:\:\iff\:\: + \vb{B} = \mu_0 (\vb{H} + \vb{M}) +\end{aligned}$$ + +Where the **magnetic permeability of free space** $\mu_0$ is a known constant. +It is important to point out some inconsistencies here: +$\vb{B}$ contains a factor of $\mu_0$, and thus measures **flux density**, +while $\vb{H}$ and $\vb{M}$ do not contain $\mu_0$, +and therefore measure **field intensity**. +Note that this convention is the opposite of the analogous +[electric fields](/know/concept/electric-field/) +$\vb{E}$, $\vb{D}$ and $\vb{P}$. +Also note that $\vb{P}$ has the opposite sign convention of $\vb{M}$. + +Some objects, called **ferromagnets** or **permanent magnets**, +have an inherently nonzero $\vb{M}$. +Others objects, when placed in a $\vb{B}$-field, +may instead gain an induced $\vb{M}$. + +When $\vb{M}$ is induced, +its magnitude is usually proportional +to the applied field strength $\vb{H}$: + +$$\begin{aligned} + \vb{B} + = \mu_0(\vb{H} + \vb{M}) + = \mu_0 (\vb{H} + \chi_m \vb{H}) + = \mu_0 \mu_r \vb{H} + = \mu \vb{H} +\end{aligned}$$ + +Where $\chi_m$ is the **volume magnetic susceptibility**, +and $\mu_r \equiv 1 + \chi_m$ and $\mu \equiv \mu_r \mu_0$ are +the **relative permeability** and **absolute permeability** +of the medium, respectively. +Materials with intrinsic magnetization, i.e. ferromagnets, +do not have a well-defined $\chi_m$. + +If $\chi_m > 0$, the medium is **paramagnetic**, +meaning it strengthens the net field $\vb{B}$. +Otherwise, if $\chi_m < 0$, the medium is **diamagnetic**, +meaning it counteracts the applied field $\vb{H}$. + +For $|\chi_m| \ll 1$, as is often the case, +the magnetization $\vb{M}$ can be approximated by: + +$$\begin{aligned} + \vb{M} + = \chi_m \vb{H} + \approx \chi_m \vb{B} / \mu_0 +\end{aligned}$$ |