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Diffstat (limited to 'source')
| -rw-r--r-- | source/know/concept/diffie-hellman-key-exchange/index.md | 26 | ||||
| -rw-r--r-- | source/know/concept/fabry-perot-cavity/index.md | 14 | ||||
| -rw-r--r-- | source/know/concept/ficks-laws/index.md | 21 | ||||
| -rw-r--r-- | source/know/concept/gram-schmidt-method/index.md | 3 | ||||
| -rw-r--r-- | source/know/concept/legendre-transform/index.md | 82 | ||||
| -rw-r--r-- | source/know/concept/matsubara-sum/index.md | 137 | ||||
| -rw-r--r-- | source/know/concept/sokhotski-plemelj-theorem/index.md | 47 |
7 files changed, 194 insertions, 136 deletions
diff --git a/source/know/concept/diffie-hellman-key-exchange/index.md b/source/know/concept/diffie-hellman-key-exchange/index.md index 4735209..3525881 100644 --- a/source/know/concept/diffie-hellman-key-exchange/index.md +++ b/source/know/concept/diffie-hellman-key-exchange/index.md @@ -7,18 +7,19 @@ categories: layout: "concept" --- -In cryptography, the **Diffie-Hellman key exchange** is a method -for two parties to securely agree on an encryption key, -when they can only communicate over an insecure channel. +In cryptography, the **Diffie-Hellman key exchange** is a method for two parties, +who can only communicate over an insecure channel, +to securely agree on an encryption key. The fundamental assumption of the Diffie-Hellman scheme, upon which its security rests, is that the following function $$f(n)$$ is a **trapdoor function**, which means that calculating $$f$$ is easy, -but its inverse $$f^{-1}$$ is extremely hard to find: +but its inverse $$f^{-1}$$ is extremely hard: $$\begin{aligned} - f(n) = g^n \bmod p + f(n) + \equiv g^n \bmod p \end{aligned}$$ Where $$n$$ is a natural number, and $$p$$ is a prime. @@ -39,9 +40,11 @@ Alice and Bob each choose a secret number from $$\{0, ..., p \!-\! 2\}$$, respec and then privately calculate $$A$$ and $$B$$ as follows: $$\begin{aligned} - A = g^a \bmod p - \qquad \quad - B = g^b \bmod p + A + \equiv g^a \bmod p + \qquad \qquad + B + \equiv g^b \bmod p \end{aligned}$$ Finally, they transmit these numbers $$A$$ and $$B$$ @@ -50,13 +53,16 @@ and then each side calculates $$k$$, which is the desired secret key: $$\begin{aligned} \boxed{ - k = A^b \bmod p = B^a \bmod p = g^{ab} \bmod p + k + \equiv A^b \bmod p + = B^a \bmod p + = g^{ab} \bmod p } \end{aligned}$$ The point is that $$k$$ includes both $$a$$ *and* $$b$$, but each side only needs to know *either* $$a$$ *or* $$b$$. -And, due to the trapdoor assumption, +Thanks to the trapdoor assumption, the eavesdropper knows $$A$$ and $$B$$, but cannot recover $$a$$ or $$b$$. diff --git a/source/know/concept/fabry-perot-cavity/index.md b/source/know/concept/fabry-perot-cavity/index.md index c88e62d..d5ea0ea 100644 --- a/source/know/concept/fabry-perot-cavity/index.md +++ b/source/know/concept/fabry-perot-cavity/index.md @@ -95,10 +95,10 @@ $$\begin{aligned} \end{bmatrix} \end{aligned}$$ -We want non-trivial solutions, where we -cannot simply satisfy the system by setting $$A_1$$, $$A_2$$, $$A_3$$ and -$$A_4$$; this constraint will give us an equation for $$k_m$$. Therefore, we -demand that the system matrix is singular, i.e. its determinant is zero: +We do not want to simply satisfy this equation +by setting $$A_1$$, $$A_2$$, $$A_3$$ and $$A_4$$, +so we demand that the system matrix is not invertible, +i.e. its determinant is zero: $$\begin{aligned} 0 = @@ -180,7 +180,7 @@ $$\begin{aligned} \end{bmatrix} \end{aligned}$$ -Again, we demand that the determinant is zero, in order to get non-trivial solutions: +Again, we demand that the determinant is zero in order to get non-trivial solutions: $$\begin{aligned} 0 @@ -225,8 +225,8 @@ $$\begin{aligned} Note that we have not demanded continuity of the electric field. This is because the mirrors are infinitely thin "magic" planes; -had we instead used the full mirror structure, -then we would have demanded continuity, as you maybe expected. +had we instead included the full microscopic mirror structure, +then we would have demanded continuity as before. diff --git a/source/know/concept/ficks-laws/index.md b/source/know/concept/ficks-laws/index.md index b205af9..8d5da7d 100644 --- a/source/know/concept/ficks-laws/index.md +++ b/source/know/concept/ficks-laws/index.md @@ -14,6 +14,7 @@ A diffusion process that obeys Fick's laws is called **Fickian**, as opposed to **non-Fickian** or **anomalous diffusion**. + ## Fick's first law **Fick's first law** states that diffusing matter @@ -21,13 +22,14 @@ moves from regions of high concentration to regions of lower concentration, at a rate proportional to the difference in concentration. Let $$\vec{J}$$ be the **diffusion flux** (with unit $$\mathrm{m}^{-2} \mathrm{s}^{-1}$$), -whose magnitude and direction describe the "flow" of diffusing matter. +whose magnitude and direction describes the "flow" of diffusing matter. Formally, Fick's first law predicts that the flux is proportional to the gradient of the concentration $$C$$ (with unit $$\mathrm{m}^{-3}$$): $$\begin{aligned} \boxed{ - \vec{J} = - D \: \nabla C + \vec{J} + = - D \: \nabla C } \end{aligned}$$ @@ -43,6 +45,7 @@ but they say more about those systems than about diffusion in general. + ## Fick's second law To derive **Fick's second law**, we demand that matter is conserved, @@ -65,8 +68,7 @@ $$\begin{aligned} = - \int_V \nabla \cdot \vec{J} \dd{V} \end{aligned}$$ -For comparison, we differentiate the definition of $$M$$, -and exploit that the integral ignores $$t$$: +For comparison, we can also just differentiate the definition of $$M$$ directly: $$\begin{aligned} \dv{M}{t} @@ -74,7 +76,8 @@ $$\begin{aligned} = \int_V \pdv{C}{t} \dd{V} \end{aligned}$$ -Both $$\idv{M}{t}$$ are equal, so stripping the integrals leads to this **continuity equation**: +Above, we have two valid expressions for $$\idv{M}{t}$$, +which must be equal, so stripping the integrals leads to this **continuity equation**: $$\begin{aligned} \pdv{C}{t} @@ -101,6 +104,7 @@ $$\begin{aligned} \end{aligned}$$ + ## Fundamental solution Fick's second law has exact solutions for many situations, @@ -110,11 +114,12 @@ where the initial concentration $$C(x, 0)$$ is a [Dirac delta function](/know/concept/dirac-delta-function/): $$\begin{aligned} - C(x, 0) = \delta(x - x_0) + C(x, 0) + = \delta(x - x_0) \end{aligned}$$ -According to Fick's second law, -the concentration's time evolution of $$C$$ turns out to be: +By solving Fick's second law with this initial condition, +$$C$$'s time evolution turns out to be: $$\begin{aligned} H(x - x_0, t) diff --git a/source/know/concept/gram-schmidt-method/index.md b/source/know/concept/gram-schmidt-method/index.md index a62522e..1f5a0bd 100644 --- a/source/know/concept/gram-schmidt-method/index.md +++ b/source/know/concept/gram-schmidt-method/index.md @@ -58,8 +58,7 @@ into an orthonormal set $$\ket{n_1}, \ket{n_2}, ...$$ as follows to all previous $$\ket{n_m}$$ with $$m < k$$. This is known as the **modified Gram-Schmidt method**. -3. Normalize the resulting ortho*gonal* vector $$\ket{g_j}$$ to make it - ortho*normal*: +3. Normalize the resulting ortho*gonal* vector $$\ket{g_j}$$ to make it ortho*normal*: $$\begin{aligned} \ket{n_j} diff --git a/source/know/concept/legendre-transform/index.md b/source/know/concept/legendre-transform/index.md index 51c003e..c4fdeb4 100644 --- a/source/know/concept/legendre-transform/index.md +++ b/source/know/concept/legendre-transform/index.md @@ -9,76 +9,90 @@ layout: "concept" --- The **Legendre transform** of a function $$f(x)$$ is a new function $$L(f')$$, -which depends only on the derivative $$f'(x)$$ of $$f(x)$$, and from which -the original function $$f(x)$$ can be reconstructed. The point is, -analogously to other transforms (e.g. [Fourier](/know/concept/fourier-transform/)), -that $$L(f')$$ contains the same information as $$f(x)$$, just in a different form. +which depends only on the derivative $$f'(x)$$ of $$f(x)$$, +and from which the original function $$f(x)$$ can be reconstructed. +The point is that $$L(f')$$ contains the same information as $$f(x)$$, +just in a different form, +analogously to e.g. the [Fourier transform](/know/concept/fourier-transform/). -Let us choose an arbitrary point $$x_0 \in [a, b]$$ in the domain of -$$f(x)$$. Consider a line $$y(x)$$ tangent to $$f(x)$$ at $$x = x_0$$, which has -a slope $$f'(x_0)$$ and intersects the $$y$$-axis at $$-C$$: +Let us choose an arbitrary point $$x_0 \in [a, b]$$ in the domain of $$f(x)$$. +Consider a line $$y(x)$$ tangent to $$f(x)$$ at $$x = x_0$$, +which must have slope $$f'(x_0)$$, and intersects the $$y$$-axis at $$-C$$: $$\begin{aligned} - y(x) = f'(x_0) (x - x_0) + f(x_0) = f'(x_0) x - C + y(x) + = f'(x_0) (x - x_0) + f(x_0) + = f'(x_0) \: x - C \end{aligned}$$ -The Legendre transform $$L(f')$$ is defined such that $$L(f'(x_0)) = C$$ -(or sometimes $$-C$$) for all $$x_0 \in [a, b]$$, -where $$C$$ corresponds to the tangent line at $$x = x_0$$. This yields: +Where $$C \equiv f'(x_0) \: x_0 - f(x_0)$$. +We now define the Legendre transform $$L(f')$$ such that +for all $$x_0 \in [a, b]$$ we have $$L(f'(x_0)) = C$$ +(some authors use $$-C$$ instead). +Renaming $$x_0$$ to $$x$$: $$\begin{aligned} - L(f'(x)) = f'(x) \: x - f(x) + L(f'(x)) + = f'(x) \: x - f(x) \end{aligned}$$ -We want this function to depend only on the derivative $$f'$$, but -currently $$x$$ still appears here as a variable. We fix that problem in -the easiest possible way: by assuming that $$f'(x)$$ is invertible for all -$$x \in [a, b]$$. If $$x(f')$$ is the inverse of $$f'(x)$$, then $$L(f')$$ is -given by: +We want this function to depend only on the derivative $$f'$$, +but currently $$x$$ still appears here as a variable. +We fix this problem in the easiest possible way: +by assuming that $$f'(x)$$ is invertible for all $$x \in [a, b]$$. +If $$x(f')$$ is the inverse of $$f'(x)$$, then $$L(f')$$ is given by: $$\begin{aligned} \boxed{ - L(f') = f' \: x(f') - f(x(f')) + L(f') + = f' \: x(f') - f(x(f')) } \end{aligned}$$ The only requirement for the existence of the Legendre transform is thus -the invertibility of $$f'(x)$$ in the target interval $$[a,b]$$, which can -only be true if $$f(x)$$ is either convex or concave, i.e. its derivative -$$f'(x)$$ is monotonic. +the invertibility of $$f'(x)$$ in the target interval $$[a,b]$$, +which can only be true if $$f(x)$$ is either convex or concave, +meaning its derivative $$f'(x)$$ is monotonic. -Crucially, the derivative of $$L(f')$$ with respect to $$f'$$ is simply -$$x(f')$$. In other words, the roles of $$f'$$ and $$x$$ are switched by the -transformation: the coordinate becomes the derivative and vice versa. -This is demonstrated here: +The derivative of $$L(f')$$ with respect to $$f'$$ is simply $$x(f')$$. +In other words, the roles of $$f'$$ and $$x$$ are switched by the transformation: +the coordinate becomes the derivative and vice versa: $$\begin{aligned} \boxed{ - \dv{L}{f'} = \dv{x}{f'} \: f' + x(f') - \dv{f}{x} \dv{x}{f'} = x(f') + \dv{L}{f'} + = f' \dv{x}{f'} + x(f') - f' \dv{x}{f'} + = x(f') } \end{aligned}$$ -Furthermore, Legendre transformation is an *involution*, meaning it is -its own inverse. Let $$g(L')$$ be the Legendre transform of $$L(f')$$: +Furthermore, Legendre transformation is an *involution*, +meaning it is its own inverse. +To show this, let $$g(L')$$ be the Legendre transform of $$L(f')$$: $$\begin{aligned} - g(L') = L' \: f'(L') - L(f'(L')) - = x(f') \: f' - f' \: x(f') + f(x(f')) = f(x) + g(L') + = L' \: f'(L') - L(f'(L')) + = x(f') \: f' - f' \: x(f') + f(x(f')) + = f(x) \end{aligned}$$ -Moreover, the inverse of a (forward) transform always exists, because -the Legendre transform of a convex function is itself convex. Convexity -of $$f(x)$$ means that $$f''(x) > 0$$ for all $$x \in [a, b]$$, which yields -the following proof: +Moreover, a Legendre transform is always invertible, +because the transform of a convex function is itself convex. +Convexity of $$f(x)$$ means that $$f''(x) > 0$$ for all $$x \in [a, b]$$, +so a proof is: $$\begin{aligned} L''(f') + = \dv{}{f'} \Big( \dv{L}{f'} \Big) = \dv{x(f')}{f'} = \dv{x}{f'(x)} = \frac{1}{f''(x)} > 0 \end{aligned}$$ +And an analogous proof exists for concave functions where $$f''(x) < 0$$. + Legendre transformation is important in physics, since it connects [Lagrangian](/know/concept/lagrangian-mechanics/) and [Hamiltonian](/know/concept/hamiltonian-mechanics/) mechanics to each other. diff --git a/source/know/concept/matsubara-sum/index.md b/source/know/concept/matsubara-sum/index.md index aef8379..0e04455 100644 --- a/source/know/concept/matsubara-sum/index.md +++ b/source/know/concept/matsubara-sum/index.md @@ -14,60 +14,88 @@ which notably appears as the inverse [Matsubara Green's function](/know/concept/matsubara-greens-function/): $$\begin{aligned} - S_{B,F} - = \frac{1}{\hbar \beta} \sum_{i \omega_n} g(i \omega_n) \: e^{i \omega_n \tau} + \boxed{ + S_{B,F} + \equiv \frac{1}{\hbar \beta} \sum_{n = -\infty}^\infty g(i \omega_n) \: e^{i \omega_n \tau} + } \end{aligned}$$ -Where $$i \omega_n$$ are the Matsubara frequencies -for bosons ($$B$$) or fermions ($$F$$), -and $$g(z)$$ is a function on the complex plane -that is [holomorphic](/know/concept/holomorphic-function/) +$$g(z)$$ is a *meromorphic* function on the complex frequency plane, +i.e. it is [holomorphic](/know/concept/holomorphic-function/) except for a known set of simple poles, -and $$\tau$$ is a real parameter -(e.g. the [imaginary time](/know/concept/imaginary-time/)) -satisfying $$-\hbar \beta < \tau < \hbar \beta$$. +and $$\tau \in [-\hbar \beta, \hbar \beta]$$ is a real parameter. +The Matsubara frequencies $$i \omega_n$$ are defined as follows +for bosons (subscript $$B$$) or fermions (subscript $$F$$): + +$$\begin{aligned} + \omega_n \equiv + \begin{cases} + \displaystyle\frac{2 n \pi}{\hbar \beta} + & \mathrm{bosons} + \\ + \displaystyle\frac{(2 n + 1) \pi}{\hbar \beta} + & \mathrm{fermions} + \end{cases} +\end{aligned}$$ -Now, consider the following integral -over a (for now) unspecified counter-clockwise contour $$C$$, -with a (for now) unspecified weighting function $$h(z)$$: +How do we evaluate Matsubara sums? +Given a counter-clockwise closed contour $$C$$, +recall that the [residue theorem](/know/concept/residue-theorem/) +turns an integral over $$C$$ into a sum of the residues +of all the integrand's simple poles $$p_g$$ that are enclosed by $$C$$: $$\begin{aligned} - \oint_C \frac{g(z) h(z)}{2 \pi i} e^{z \tau} \dd{z} - = \sum_{z_p} e^{z_p \tau} \: \underset{z \to z_p}{\mathrm{Res}}\big( g(z) h(z) \big) + \oint_C \frac{g(z) \: e^{z \tau}}{i 2 \pi} \dd{z} + = \sum_{p_g} \underset{z \to p_g}{\mathrm{Res}}\Big\{ g(z) \: e^{z \tau} \Big\} + = \sum_{p_g} \underset{z \to p_g}{\mathrm{Res}}\Big\{ g(z) \Big\} \: e^{p_g \tau} \end{aligned}$$ -Where we have applied the [residue theorem](/know/concept/residue-theorem/) -to get a sum over all simple poles $$z_p$$ -of either $$g$$ or $$h$$ (but not both) enclosed by $$C$$. -Clearly, we could make this look like a Matsubara sum, -if we choose $$h$$ such that it has poles at $$i \omega_n$$. +Now, the trick is to manipulate this relation +until a Matsubara sum appears on the right. -Therefore, we choose the weighting function $$h(z)$$ as follows, -where $$n_B(z)$$ is the [Bose-Einstein distribution](/know/concept/bose-einstein-distribution/), -and $$n_F(z)$$ is the [Fermi-Dirac distribution](/know/concept/fermi-dirac-distribution/): +Let us introduce a (for now) unspecified weight function $$h(z)$$, +which crucially does not share any simple poles with $$g(z)$$, +so $$\{p_g\} \cap \{p_h\} = \emptyset$$. +This constraint allows us to split the sum: + +$$\begin{aligned} + \oint_C \frac{g(z) \: h(z) \: e^{z \tau}}{i 2 \pi} \dd{z} + &= \sum_{p_g} \underset{z \to p_g}{\mathrm{Res}}\Big\{ g(z) \: h(z) \: e^{z \tau} \Big\} + + \sum_{p_h} \underset{z \to p_h}{\mathrm{Res}}\Big\{ g(z) \: h(z) \: e^{z \tau} \Big\} + \\ + &= \sum_{p_g} \underset{z \to p_g}{\mathrm{Res}}\Big\{ g(z) \Big\} \: h(p_g) \: e^{p_g \tau} + + \sum_{p_h} g(p_h) \: \underset{z \to p_h}{\mathrm{Res}}\Big\{ h(z) \Big\} \: e^{p_h \tau} +\end{aligned}$$ + +Here, we could make the rightmost term look like a Matsubara sum +if we choose $$h$$ such that it has poles at $$i \omega_n$$. +We make the following choice, +where $$n_B(z)$$ is the [Bose-Einstein distribution](/know/concept/bose-einstein-distribution/) for bosons, +and $$n_F(z)$$ is the [Fermi-Dirac distribution](/know/concept/fermi-dirac-distribution/) for fermions: $$\begin{aligned} h(z) - = + \equiv \begin{cases} n_{B,F}(z) & \mathrm{if}\; \tau \ge 0 \\ -n_{B,F}(-z) & \mathrm{if}\; \tau \le 0 \end{cases} - \qquad \qquad - n_{B,F}(z) - = \frac{1}{e^{\hbar \beta z} \mp 1} \end{aligned}$$ -The distinction between the signs of $$\tau$$ is needed -to ensure that the integrand $$h(z) e^{z \tau}$$ decays for $$|z| \to \infty$$, -both for $$\Real(z) > 0$$ and $$\Real(z) < 0$$. -This choice of $$h$$ indeed has poles at the respective +The distinction between the signs of $$\tau$$ is necessary +to ensure that $$h(z) \: e^{z \tau} \to 0$$ for all $$z$$ when $$|z| \to \infty$$ +(take a moment to convince yourself of this). +The sign flip for $$\tau \le 0$$ is also needed, +as negating the argument negates the residues +$$\mathrm{Res}\{ n_{B,F}(-i \omega_n) \} = -\mathrm{Res}\{ n_{B,F}(i \omega_n) \}$$. + +Indeed, this choice of $$h$$ has poles at the respective Matsubara frequencies $$i \omega_n$$ of bosons and fermions, -and the residues are: +and the residues are given by: $$\begin{aligned} - \underset{z \to i \omega_n}{\mathrm{Res}}\!\big( n_B(z) \big) + \underset{z \to i \omega_n}{\mathrm{Res}}\!\Big\{ n_B(z) \Big\} &= \lim_{z \to i \omega_n}\!\bigg( \frac{z - i \omega_n}{e^{\hbar \beta z} - 1} \bigg) = \lim_{\eta \to 0}\!\bigg( \frac{i \omega_n + \eta - i \omega_n}{e^{i \hbar \beta \omega_n} e^{\hbar \beta \eta} - 1} \bigg) \\ @@ -75,7 +103,7 @@ $$\begin{aligned} = \lim_{\eta \to 0}\!\bigg( \frac{\eta}{1 + \hbar \beta \eta - 1} \bigg) = \frac{1}{\hbar \beta} \\ - \underset{z \to i \omega_n}{\mathrm{Res}}\!\big( n_F(z) \big) + \underset{z \to i \omega_n}{\mathrm{Res}}\!\Big\{ n_F(z) \Big\} &= \lim_{z \to i \omega_n}\!\bigg( \frac{z - i \omega_n}{e^{\hbar \beta z} + 1} \bigg) = \lim_{\eta \to 0}\!\bigg( \frac{i \omega_n + \eta - i \omega_n}{e^{i \hbar \beta \omega_n} e^{\hbar \beta \eta} + 1} \bigg) \\ @@ -84,37 +112,36 @@ $$\begin{aligned} = - \frac{1}{\hbar \beta} \end{aligned}$$ -In the definition of $$h$$, the sign flip for $$\tau \le 0$$ -is introduced because negating the argument also negates the residues, -i.e. $$\mathrm{Res}\big( n_F(-z) \big) = -\mathrm{Res}\big( n_F(z) \big)$$. -With this $$h$$, our contour integral can be rewritten as follows: +With this, our contour integral can now be rewritten as follows: $$\begin{aligned} - \oint_C \frac{g(z) h(z)}{2 \pi i} e^{z \tau} \dd{z} - &= \sum_{z_p} e^{z_p \tau} n_{B,F}(z_p) \: \underset{z \to z_p}{\mathrm{Res}}\big( g(z) \big) - + \sum_{i \omega_n} e^{i \omega_n \tau} g(i \omega_n) \: \underset{z \to i \omega_n}{\mathrm{Res}}\!\big( n_{B,F}(z) \big) + \oint_C \frac{g(z) \: h(z) \: e^{z \tau}}{i 2 \pi} \dd{z} + &= \sum_{p_g} \underset{z \to p_g}{\mathrm{Res}}\Big\{ g(z) \Big\} \: n_{B,F}(p_g) \: e^{p_g \tau} + + \sum_{i \omega_n} g(i \omega_n) \underset{z \to i \omega_n}{\mathrm{Res}}\!\Big\{ n_{B,F}(z) \Big\} \: e^{i \omega_n \tau} \\ - &= \sum_{z_p} e^{z_p \tau} n_{B,F}(z_p) \: \underset{z \to z_p}{\mathrm{Res}}\big( g(z) \big) - \pm \frac{1}{\hbar \beta} \sum_{i \omega_n} g(i \omega_n) \: e^{i \omega_n \tau} + &= \sum_{p_g} \underset{z \to p_g}{\mathrm{Res}}\Big\{ g(z) \Big\} \: n_{B,F}(p_g) \: e^{p_g \tau} + \pm \frac{1}{\hbar \beta} \sum_{n = -\infty}^\infty g(i \omega_n) \: e^{i \omega_n \tau} \end{aligned}$$ -Where $$+$$ is for bosons, and $$-$$ for fermions. -Here, we recognize the last term as the Matsubara sum $$S_{F,B}$$, -for which we isolate, yielding: +Where the top sign ($$+$$) is for bosons, +and the bottom sign ($$-$$) is for fermions. +Here, we recognize the last term as the Matsubara sum $$S_{F,B}$$. +Isolating for that yields: $$\begin{aligned} S_{B,F} - = \mp \sum_{z_p} e^{z_p \tau} n_{B,F}(z_p) \: \underset{z \to z_p}{\mathrm{Res}}\big( g(z) \big) - \pm \oint_C \frac{g(z) h(z)}{2 \pi i} e^{z \tau} \dd{z} + = \mp \sum_{p_g} \underset{z \to p_g}{\mathrm{Res}}\Big\{ g(z) \Big\} \: n_{B,F}(p_g) \: e^{p_g \tau} + \pm \oint_C \frac{g(z) \: h(z) \: e^{z \tau}}{i 2 \pi} \dd{z} \end{aligned}$$ -Now we must choose $$C$$. Assuming $$g(z)$$ does not interfere, -we know that $$h(z) e^{z \tau}$$ decays to zero -for $$|z| \to \infty$$, so a useful choice would be a circle of radius $$R$$. -If we then let $$R \to \infty$$, the contour encloses -the whole complex plane, including all of the integrand's poles. -However, thanks to the integrand's decay, -the resulting contour integral must vanish: +Now we must choose $$C$$. +Earlier, we took care that $$h(z) \: e^{z \tau} \to 0$$ for $$|z| \to \infty$$, +so a good choice would be a circle of radius $$R$$. +If $$R \to \infty$$, then $$C$$ encloses the whole complex plane, +including all of the integrand's poles. +However, because the integrand decays for $$|z| \to \infty$$, +we conclude that the contour integral must vanish +(also for other choices of $$C$$): $$\begin{aligned} C @@ -131,7 +158,7 @@ for bosonic and fermionic Matsubara sums $$S_{B,F}$$: $$\begin{aligned} \boxed{ S_{B,F} - = \mp \sum_{z_p} e^{z_p \tau} n_{B,F}(z_p) \: \underset{ {z \to z_p}}{\mathrm{Res}}\big(g(z)\big) + = \mp \sum_{p_g} \underset{ {z \to p_g}}{\mathrm{Res}}\Big\{ g(z) \Big\} \: n_{B,F}(p_g) \: e^{p_g \tau} } \end{aligned}$$ diff --git a/source/know/concept/sokhotski-plemelj-theorem/index.md b/source/know/concept/sokhotski-plemelj-theorem/index.md index 66e89bc..445b029 100644 --- a/source/know/concept/sokhotski-plemelj-theorem/index.md +++ b/source/know/concept/sokhotski-plemelj-theorem/index.md @@ -9,8 +9,8 @@ categories: layout: "concept" --- -The goal is to evaluate integrals of the following form, where $$a < 0 < b$$, -and $$f(x)$$ is assumed to be continuous in the integration interval $$[a, b]$$: +The goal is to evaluate integrals of the following form, +where $$f(x)$$ is assumed to be continuous in the integration interval $$[a, b]$$: $$\begin{aligned} \lim_{\eta \to 0^+} \int_a^b \frac{f(x)}{x + i \eta} \dd{x} @@ -22,12 +22,14 @@ into its real and imaginary parts (limit hidden): $$\begin{aligned} \int_a^b \frac{f(x)}{x + i \eta} \dd{x} &= \int_a^b \frac{x - i \eta}{x^2 + \eta^2} f(x) \dd{x} - = \int_a^b \bigg( \frac{x}{x^2 + \eta^2} - i \frac{\eta}{x^2 + \eta^2} \bigg) f(x) \dd{x} + \\ + &= \int_a^b \frac{x}{x^2 + \eta^2} f(x) \dd{x} - i \int_a^b \frac{\eta}{x^2 + \eta^2} f(x) \dd{x} \end{aligned}$$ -To evaluate the real part, -we notice that for $$\eta \to 0^+$$ the integrand diverges for $$x \to 0$$, -and thus split the integral as follows: +In the real part, notice that the integrand diverges +for $$x \to 0$$ when $$\eta \to 0^+$$; +more specifically, there is a singularity at zero. +We therefore split the integral as follows: $$\begin{aligned} \lim_{\eta \to 0^+} \int_a^b \frac{x f(x)}{x^2 + \eta^2} \dd{x} @@ -56,7 +58,7 @@ $$\begin{aligned} The expression $$m / \pi (1 + m^2 x^2)$$ is a so-called *nascent delta function*, meaning that in the limit $$m \to +\infty$$ it converges to -the [Dirac delta function](/know/concept/dirac-delta-function/): +the [Dirac delta function](/know/concept/dirac-delta-function/) $$\delta(x)$$: $$\begin{aligned} \lim_{\eta \to 0^+} \int_a^b \frac{\eta \: f(x)}{x^2 + \eta^2} \dd{x} @@ -66,7 +68,8 @@ $$\begin{aligned} By combining the real and imaginary parts, we thus arrive at the (real version of the) -so-called **Sokhotski-Plemelj theorem** of complex analysis: +**Sokhotski-Plemelj theorem** of complex analysis, +which is important in quantum mechanics: $$\begin{aligned} \boxed{ @@ -82,29 +85,33 @@ This awkwardly leaves $$\mathcal{P}$$ behind: $$\begin{aligned} \frac{1}{x + i \eta} - = \mathcal{P} \Big( \frac{1}{x} \Big) - i \pi \delta(x) + = \mathcal{P} \frac{1}{x} - i \pi \delta(x) \end{aligned}$$ -The full, complex version of the Sokhotski-Plemelj theorem -evaluates integrals of the following form -over a contour $$C$$ in the complex plane: +That was the real version of the theorem, +which is a special case of a general result in complex analysis. +Consider the following function: $$\begin{aligned} - \phi(z) = \frac{1}{2 \pi i} \oint_C \frac{f(\zeta)}{\zeta - z} \dd{\zeta} + \phi(z) = \oint_C \frac{f(\zeta)}{\zeta - z} \dd{\zeta} \end{aligned}$$ Where $$f(z)$$ must be [holomorphic](/know/concept/holomorphic-function/). -The Sokhotski-Plemelj theorem then states: +For all $$z$$ not on $$C$$, this $$\phi(z)$$ exists, +but not for $$z \in C$$, since the integral diverges then. +However, in the limit when approaching $$C$$, we can still obtain a value for $$\phi$$, +with a caveat: the value depends on the direction we approach $$C$$ from! +The full Sokhotski-Plemelj theorem then states, for all $$z$$ on the closed contour $$C$$: $$\begin{aligned} \boxed{ - \lim_{w \to z} \phi(w) - = \frac{1}{2 \pi i} \mathcal{P} \oint_C \frac{f(\zeta)}{\zeta - z} \dd{\zeta} \pm \frac{f(z)}{2} + \lim_{y \to z} \phi(y) + = \mathcal{P} \oint_C \frac{f(\zeta)}{\zeta - z} \dd{\zeta} \pm \: i \pi f(z) } \end{aligned}$$ -Where the sign is positive if $$z$$ is inside $$C$$, and negative if it is outside. -The real version follows by letting $$C$$ follow the whole real axis, -making $$C$$ an infinitely large semicircle, -so that the integrand vanishes away from the real axis, +Where $$\pm$$ is $$+$$ if $$C$$ is approached from the inside, and $$-$$ if from outside. +The above real version follows by making $$C$$ an infinitely large semicircle +with its flat side on the real line: +the integrand vanishes away from the real axis, because $$1 / (\zeta \!-\! z) \to 0$$ for $$|\zeta| \to \infty$$. |