Sorry, your browser cannot access this site
This page requires browser support (enable) JavaScript
Learn more >

click here to save the pdf version of this article

Mathematical Physics

Based on Qiao Gu’s and Liang’s.

Complex Function

Basic Definitions

For a complex function $f(z)=f(x+iy)=u(x,y)+iv(x,y)$, and an interior point $z=x+iy$ in its domain. It’s derivative at $z$ if and only if $u$ and $v$ are all differentiable at $(x,y)$ and the function satisfy Cauchy-Riemann condition at that point.

C-R condition:
$$\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}$$
$$
\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}
$$

Also, we have the polar coordinate version C-R condition:

$$
\frac{\partial u}{\partial \rho}=\frac{1}{\rho}\frac{\partial v}{\partial \phi}
$$

$$
\frac{1}{\rho}\frac{\partial u}{\partial \phi}=-\frac{\partial v}{\partial \rho}
$$

If function $f(z)$ is differentiable at any point in a neighborhood of $z_0$, then we call it a holomorphic function at $z_0$. (equivalent to analytic because of some mathematical reason)

Tips

$\mathtt{Q}$: If you know the function is holomorphic, but you only have the $f(x,y)$ form, how to quickly gain the compact $f(z)$ form?

$\mathtt{A}$: There’s a clever trick for form-reshaping: let all $y$ in $f(x,y)$ be zero and let $x$ in $f(x,y)$ be $z$, then you get the answer! You probably wonder why, and the reason behind is the uniqueness of a power series. For an analytic function, its power series’ form is unique, so that when $z$ is real ($z=x+i0$), the expression of power series shall remain.

Harmonic functions

The real and imaginary parts of a holomorphic function are all harmonic, which can be easily verified by operating C-R condition.

Taylor Series

Know that the Taylor series is unique, and wherever the function is holomorphic, it can be expanded. Thus the convergent radius must be the distance between the nearest singularity and the central point for expansion. (If not, the small neighborhoods at the fringe could combine and expanded the convergent circle into a larger one, contradiction)

Laurent Series

Laurent theorem’s statement: If function $f(z)$ is holomorphic in a disk ${R_1 < |z-z_0| < R_2}$, then it could be expressed in the form
$$
f(z)=\sum_{n=-\infty}^{\infty} C_n (z-z_0)^n
$$
$$
C_n:=\frac{1}{2\pi i}\oint_{C_R} \frac{f(\xi)}{(\xi-z_0)^{n+1}} d\xi
$$
The theorem is complicated as you can see, but the theorem itself is not the key point. The key advantage is the uniqueness of Laurent series, which implies that no matter how you get the expanded series’ expression, it’s unique and correct. Thus you can apply the formula of geometric power series to get the result, etc. Besides, the Laurent theorem is almost the Residue theorem.

Integrals

To simplify the framework, you only need to remember the Residue Theorem, because Cauchy Formula and Cauchy Theorem are all special case of Residue Theorem.
$$
\oint_l f(z)dz=2\pi i \sum_{j=1}^{n} \text{Res}f(b_j)
$$

where $b_j$ are isolated singularities, and the function must be analytic inside of $l$ except at these singularities.

Note that Residue Theorem works as well for essential singularities! The only obstacle is how to calculate the Residue there though.

Jordan’s Lemma

If $g(z)$ is continuous outside a sufficient large circle in $z$-plane and $\lim_{z\to\infty} g(z)=0$, then for $\alpha>0$ we have:
$$
\lim_{R\to\infty}\int_{\Gamma_R} g(z)e^{i\alpha z}dz=0
$$
where $\Gamma_R$ is an arc with radius $R$.

Useful Variable Changes

When the integral interval is $0\to 2\pi$, let $z=e^{ix}$

When the integral interval is $-\infty\to\infty$ then let $z=x$ and apply Jordan’s Lemma. ($zf(z)\rightrightarrows 0$)

When the integral interval is $0\to\infty$, first convert it to $-\infty\to\infty$.

For $\int_0^\infty F(x)\cos mx dx$ or so, first consider $\text{Re}\int_0^{\infty} F(z) e^{imz}dz$. ($F(z)\rightrightarrows 0$).

If you don’t convert trigonometrical functions (boundless) into exponential functions, the Residue Theorem is not allowed to use! Because the validness is guaranteed by Jordan’s Lemma which require exponential form.

Fourier Theory

Fourier Integral

$$
f(x)=\int_0^\infty [A(\omega)\cos\omega x+B(\omega) \sin\omega x]d\omega
$$ where
$$
A(\omega)=\frac{1}{\pi}\int_{-\infty}^\infty f(x)\cos\omega x dx
$$
$$
B(\omega)=\frac{1}{\pi}\int_{-\infty}^\infty f(x)\sin\omega x dx
$$

Q: why is the interval like this?
A: $f(x)=\int_0^\infty\cdots$,comes from the Fourier series $\sum\limits_0^\infty$.
$A(\omega),B(\omega)=\int_{-\infty}^{\infty}$ comes from $\lim\limits_{L\rightarrow\infty}\int_{-L}^L$.

An Example

Dirichlet integral :
considering
the function below and its Fourier integral
$$
f(x)=\begin{cases}
0,&(|x|\leq1)\\
1,&(|x|>1)
\end{cases}
$$ we get $$
\int_0^\infty\frac{\sin\omega}{\omega}d\omega=\frac{\pi}{2}
$$

This could be derived later by FT more generally.

Fourier Transfom

$$
F(\omega)=\int_{-\infty}^\infty f(x) e^{-i\omega x}dx
$$
$$
f(x)=\frac{1}{2\pi}\int_{-\infty}^\infty F(\omega) e^{i\omega x}d\omega
$$

Property

  • Linear

  • $\frac{d}{dx}f(x)\rightarrow i\omega F(\omega)$

  • $xf(x)\rightarrow i\frac{d}{d\omega}F(\omega)$

  • $\int f(x)dx\rightarrow\frac{F(\omega)}{i\omega}$

  • $f(x+\zeta)\rightarrow e^{i\omega\zeta}F(x)$

  • convolution

The methods to verify properties involving integral and derivative are similar to Laplacian Transform, the detail of it will be shown there.

Two Examples

when f is even
$f(x)*\cos\omega x=F(\omega)\cos\omega x$
$f(x)*\sin\omega x=F(\omega)\sin\omega x$

$\delta$ Function

We view $\delta$ function as a generalized function.

In the form $\int f(x) dy$, what exactly does $dy$ mean? A mathematical perspective is the measure $\sigma$ of a set. Riemann coincidentally defined his integral using the simplest measure namely $x_{i+1}-x_i$ , i.e. the measure of interval $[x_i,x_{i+1})$ in partition. When $\delta (x)$ is entailed, $\int f(x)d\sigma$ shall be viewed as a Lebesgue Integral where only $f(0)$ gained a measure of 1, any other value weighed 0, such that the integral $=\Sigma f(x_i) \sigma_i=f(0)\times 1=f(0)$.

Also, we may define it ($W$) using inner product: $\langle L(W),\phi\rangle=\langle W,L^*(\phi)\rangle$, where $L$ represents any linear differential operator, and $L^*$ its adjoint, $\phi$ is a test function of great property.

Besides, Riesz Representation Theorem guarantees the equality of an inner product and a linear functional, thus Linear Functional is also a good choice to describe generalized functions.

The mathematical interpretation shall stop and physical introduction begins.


The convolution of $\delta(x)$ and $f(x)$ $$
\delta(x-a)*f(x)=f(x-a)
$$ satisfy the eigen-equation of coordinates:
$$ x\delta(x-x_0)=x_0\delta(x-x_0) $$ where $x$ is the operator of coordinates.

It’s obvious that delta function is orthogonal to each other and the completeness is given by:
$$f(x)=\int_{-\infty}^{\infty}f(\zeta)\delta(\zeta-x)d\zeta$$ where $f(\zeta)$ are the parameters.

its FT & Applications

$$
\mathscr{F}[\delta(x)]=1
$$ $$
\delta(x)=\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{i\omega x}d\omega
$$ or write it as $$ \delta(p-p’)=\frac{1}{2\pi\hbar}\int_{-\infty}^{\infty}e^{\frac{i}{\hbar}(p-p’)x}dx$$ which implies the orthogonality of momentum

the normalized eigen-vector of momentum is:

$$
\psi_p(x)=\frac{1}{\sqrt{2\pi\hbar}}e^{\frac{i}{\hbar}px}
$$

as for its completeness, considering the Fourier transform of $f(x)$:

$$
f(x)
=\frac{1}{2\pi}\int_{-\infty}^{\infty}F(\omega)e^{i\omega x}d\omega
=\frac{1}{2\pi\hbar}\int_{-\infty}^{\infty}F(\omega)e^{(i/\hbar)(\hbar\omega) x}d(\hbar\omega)
$$
$$
=\frac{1}{2\pi\hbar}\int_{-\infty}^{\infty}F(\omega)e^{(i/\hbar)(\hbar\omega) x}d(\hbar\omega)
=\frac{1}{2\pi\hbar}\int_{-\infty}^{\infty}\tilde{F}(p)e^{(i/\hbar)p x}dp
$$
$$
=\frac{1}{\sqrt{2\pi\hbar}}\int_{-\infty}^{\infty}\tilde{F}(p)\psi_p(x)dp
$$

where
$$
\tilde{F}(p)=F\left(\frac{p}{\hbar}\right)=\int_{-\infty}^{\infty}f(x)e^{-i\frac{p}{\hbar}}dx
$$

Delta function in concrete forms

common ones:

$$ V(x)=\lim\limits_{\beta\rightarrow\infty}\frac{\sin\beta x}{\pi x} $$

$$
G(x)=\lim\limits_{\beta\rightarrow 0}\frac{1}{\sqrt{\pi \beta}}\exp({-\frac{x^2}{\beta}})
$$

$$ L(x)=\lim\limits_{\beta\rightarrow 0}\frac{1}{\pi}\frac{\beta}{x^2+\beta^2}
$$

Dirichlet Kernel

$$
D_m(x)=\frac{1}{2\pi}\frac{\sin(m+\frac{1}{2}x)}{\sin\frac{1}{2}x}=\frac{1}{2\pi}(1+2\sum\limits_{k=1}^m\cos kx)
$$

and

$$
\lim\limits_{m\rightarrow\infty}D_m(x)=\delta(x)
$$

  • The fatal loophole in Qiao Gu’s proof of Dirichlet theorem is that he apply the property of selection.

Delta function here corresponds to Good-Kernel in math, which requires three essential aspects to satisfy, while Dirichlet Kernel violates one of them. So Dirichlet Kernel cannot be used to approach a function, i.e. cannot apply the property of selection.

Example of $\delta$ function

Though $\sin kx , \cos kx$ aren’t absolutely integrable, but we can discuss their Fourier Transform in physics.

$$
F_{\sin}(\omega)=\int_{-\infty}^{\infty}\sin kx e^{-i\omega x}dx
=\frac{1}{2i}\int_{-\infty}^{\infty}(e^{ikx}-e^{-ikx})e^{-i\omega x}dx
=i\pi[\delta(\omega+k)-\delta(\omega-k)]
$$
$$
F_{\cos}(\omega)=\int_{-\infty}^{\infty}\cos kx e^{-i\omega x}dx
=\frac{1}{2}\int_{-\infty}^{\infty}(e^{ikx}+e^{-ikx})e^{-i\omega x}dx
=\pi[\delta(\omega+k)+\delta(\omega-k)]
$$

important transform pairs

$$
\mathscr{F}[\frac{1}{1+x^2}]\rightleftharpoons e^{-|x|}
$$

Gauss function: $g(x)=\exp({-\frac{ax^2}{2}})$

$$
\begin{align*}
G(\omega)&=\int_{-\infty}^{\infty}\exp({-\frac{ax^2}{2}}) e^{-i\omega x}dx
\newline

\newline
&=\dots(\text{integration by parts})
\newline

\newline
&=\frac{ia}{\omega}\mathscr{F}[x\exp(-\frac{ax^2}{2})]
\newline

\newline
&=\frac{ia}{\omega}i\frac{d}{d\omega}G(\omega)
\end{align*}
$$

solve an ode to get $G(\omega)=\sqrt{\frac{2\pi}{a}}\exp(-\frac{\omega^2}{2a})$
How to calculate the FT of a step function?

$$
u(x)=\begin{cases}
0,&(x<0)\\
1,&(x\geq0)
\end{cases}
$$

view it as the $\lim\limits_{\beta\rightarrow 0^+}f(x)$
where

$$
f(x)=\begin{cases}
0,&(x<0)\\
e^{-\beta x},&(x\geq0)
\end{cases}
$$

hence the FT of $u(t)$ is

$$
U(\omega)=\lim\limits_{\beta\rightarrow 0^+}F(\omega)
=\lim\limits_{\beta\rightarrow 0^+}\frac{1}{\beta +i \omega}
=\pi\delta(\omega)+\frac{1}{i\omega}
$$

The final step see $L(x)$ there.

Apply Residue Theorem

The above examples show some Physical methods to deal with special functions. While for a wide range of integrations we are able to calculate them using Residue Theorem, a much more handy way than Qiao Gu’s Fourier method in his book.

Usual Steps (for trigonometric and exponential):

1.Adjust the interval to $(-\infty,\infty)$

2.Show proper form like $\int_{-\infty}^{\infty} e^{imx}f(x) dx$

3.If m<0, $\int_{-\infty}^{\infty} e^{-i|m|x}f(x)dx=\int_{-\infty}^{\infty} e^{i|m|y}f(-y)dy$

4.Examine whether $f(z)$ converges uniformly to 0 at $\infty$.

5.Jordan’s Lemma.

For instance (P62):

$$
\begin{align*}
&\quad\int_0^\infty \frac{\cos\omega t}{\beta^2+\omega^2}d\omega
=\frac{1}{2}\int_{-\infty}^\infty \frac{\cos\omega t}{\beta^2+\omega^2}d\omega
\newline

\newline
=&\frac{1}{2}\int_{-\infty}^\infty \frac{\cos\omega t+i\sin \omega t }{\beta^2+\omega^2}d\omega
\newline

\newline
=&\frac{1}{2}\int_{-\infty}^\infty \frac{e^{i\omega t}}{\beta^2+\omega^2}d\omega ~~(\text{when}~t>0)
\newline

\newline
=&\pi i{\text{Res}[\frac{e^{i\omega t}}{\beta^2+\omega^2},i\beta]} =\frac{2\beta}{\pi}e^{-\beta t}
\end{align*}
$$

for $t<0$, replace $-\omega$ with $p$:

$$
=\frac{1}{2}\int_{-\infty}^\infty \frac{e^{-ip t}}{\beta^2+p^2}dp
=\frac{2\beta}{\pi}e^{\beta t}
$$

We already apply the Jordan’s Lemma. Be careful when you meet trigonometric functions, because they’re not bounded in $\mathbb{C}$, that’s why Jordan’s Lemma is an essential. In the above example the even and odd function helps us to achive the form of Jordan’s Lemma. When the function you met is neither even nor odd, please write like this:$\int f(x)\cos x =\text{Re}(\int f(x) e^{ix})=\cdots$

Note again that Jordan’s Lemma could only be utilized when $m$ in $\int e^{imx}dx$ is larger than 0 (which explained the absolute value in the end), and only the singularities above the x-axis are needed to be considered.

Another example (P38 in Qiao’s book):

$$
\begin{align*}
&\quad\int_{-\infty}^{\infty}\frac{\sin a\omega}{\omega}e^{i\omega x}d\omega
=\int_{-\infty}^{\infty}\frac{e^{i(a+x)\omega}-e^{i(x-a)\omega}}{2i\omega}d\omega
\newline

\newline
=&\frac{1}{2i}\left(\int_{-\infty}^{\infty}\frac{e^{i(a+x)\omega}}{\omega}d\omega-\int_{-\infty}^{\infty}\frac{e^{i(x-a)\omega}}{\omega}d\omega\right)
\newline

\newline
=&\frac{\pi i}{2i}\left(
\begin{cases}
\text{Res}[\frac{e^{i(x+a)\omega}}{\omega},0],&(x>-a)\\
\text{Res}[-\frac{e^{-i(x+a)\omega}}{\omega},0],&(x<-a)
\end{cases}
-\begin{cases}
\text{Res}[\frac{e^{i(x-a)\omega}}{\omega},0],&(x>a)\\
\text{Res}[-\frac{e^{-i(x-a)\omega}}{\omega},0],&(x<a)
\end{cases}\right)
\newline

\newline
=&\begin{cases}
0,& |x|>a\\
\pi,& |x|<a
\end{cases}
\end{align*}
$$

Note that the singularity is 0, which stuck in the x-axis, so $2\pi i \times \text{Res}$ is not allowed (only works for singularities above the x-axis). Instead, the multipliers should be replaced by $\pi i$.
A drawback of Residue Theorem is that we’re not able to calculate the value of such an integral at $x=\pm a$, which could be solved by Dirichlet’s Theorem.

Again, the two cases above entail polar points (a kind of singularities) only. For Essential singularities, the Residue Theorem works as well. The only extra obstacle is how to calculate the Residue of an essential singularity, and you’d better calculate its Laurent series stiffly and examine the $C_{-1}$ term.

When the singularity isn’t isolated, Residue Theorem is no longer legal to use.

Laplacian Transform

The integral of Laplacian transform converges absolutely and uniformly, such that shifting the order of operators is legal.

LT’s Properties

Let’s appreciate their symmetry:

$$
\frac{d}{ds}\mathscr{L}[f(t)]=\mathscr{L}[-tf(t)]
$$
$$
\mathscr{L}[\frac{d}{dt}f(t)]=s\mathscr{L}[f(t)]-f(0^+)
$$
$$
\int_s^\infty\mathscr{L}[f(t)]du=\mathscr{L}[\frac{f(t)}{t}]
$$
$$
\mathscr{L}[\int_0^tf(\tau)d\tau]=\frac{1}{s}\mathscr{L}[f(t)]
$$

How to verify these properties smoothly?
(I mean treat them individually, do not quote others results)

  • $\frac{d}{ds}\mathscr{L}[f(t)]=\mathscr{L}[-tf(t)]$
    proof
    $$
    \frac{d}{ds}\mathscr{L}[f(t)]
    =\frac{d}{ds}\int_0^\infty f(t)e^{-st}dt
    $$
    $$
    =\int_0^\infty -tf(t)e^{-st}dt
    $$
    $$
    =\mathscr{L}[-tf(t)]
    $$

  • $\mathscr{L}[\frac{d}{dt}f(t)]=s\mathscr{L}[f(t)]-f(0^+)$
    proof

$$
\mathscr{L}[\frac{d}{dt}f(t)]
=\int_0^\infty e^{-st}df(t)
$$
$$
=\left[f(t)e^{-st}\right]_0^\infty-\int_0^\infty -sf(t)e^{-st}dt
$$
$$
=s\mathscr{L}[f(t)]-f(0^+)
$$

  • $\int_s^\infty\mathscr{L}[f(t)]du=\mathscr{L}[\frac{f(t)}{t}]$
    proof

$$
\int_s^\infty\mathscr{L}[f(t)]du=\int_s^\infty\left(\int_0^\infty f(t)e^{-ut}dt\right)du
$$
$$
=\int_0^\infty\left(\int_s^\infty f(t)e^{-ut}du\right)dt
$$
$$
=\int_0^\infty \frac{1}{t}f(t)e^{-st}dt
=\mathscr{L}[\frac{f(t)}{t}]
$$

  • $\mathscr{L}[\int_0^tf(\tau)d\tau]=\frac{1}{s}\mathscr{L}[f(t)]$
    proof
    $$
    \mathscr{L}[\int_0^tf(\tau)d\tau]=\int_0^\infty e^{-st}\left(\int_0^tf(\tau)d\tau \right )dt
    $$
    $$
    =-\frac{1}{s}\int_0^\infty (\int_0^tf(\tau)d\tau) d(e^{-st})
    $$
    $$
    =\frac{1}{s}\int_0^\infty e^{-st}d\left(\int_0^tf(\tau)d\tau\right)
    $$
    $$
    =\frac{1}{s}\int_0^\infty e^{-st}f(t)dt
    =\frac{1}{s}\mathscr{L}[f(t)]
    $$

Conclusion:
If the integral and derivative sign is in $\mathscr{L}[\quad]$,the method is Integration by Parts;
If the integral and derivative sign is out of $\mathscr{L}[\quad]$,the method is Changing the Order of integral operators.

Translations:(easy to derive)

$$
e^{-st_0}\mathscr{L}[f(t)]=\mathscr{L}[f(t-t_0)]
$$
$$
\mathscr{L}[e^{s_0 t}f(t)]=\mathscr{L}[f(t)] (s-s_0)
$$

Warning: Laplacian Transform focus on the right side of y-axis, i.e. $f(x)=f(x)u(x)$. Functions like $f(t-1)$may cause misunderstandings. A better way to avoid such consequence is to apply Convolution Property instead of Translation in practice.

Convolution:

$$
\mathscr{L}[f(t)]\times\mathscr{L}[g(t)]=\mathscr{L}[f*g(t)]
$$

E.g.
$$
\mathscr{L^{-1}}{e^{-st_0}\mathscr{L}[f(t)]}
=\mathscr{L^{-1}}[e^{-st_0}]*f(t)
$$
$$
=\delta(t-t_0)*f(t)=\int_0^t \delta(t-t_0-\tau)f(\tau)d\tau
$$
$$
=\begin{cases}
f(t-t_0) &(t-t_0>0) \\
0 &(t-t_0<0)
\end{cases}
$$

If we simply copied the conclusion of Translation, we happily got $f(t-t_0)$, whose $(0,t_0)$ part is actually wrong.

评论