Homework 06

1 Green’s function

Consider the equation

\[ x^{\prime \prime}(t)+2 \gamma x(t)+\omega_0^2 x(t)=f(t) . \]

Using the Green’s function for this equation, which satisfies

\[ G^{\prime \prime}(t)+2 \gamma G^{\prime}(t)+\omega_0^2 G(t)=\delta(t) \]

derive the response $x(t)$ to a square pulse

\[ f(t)=\left\{\begin{array}{ll} 1 & 0 \leq t \leq 1 \\ 0 & \text { otherwise } \end{array} .\right. \]

Do this by solving for $\tilde{G}(\omega)$ in the Fourier domain and note

\[ \tilde{x}(\omega)=\tilde{G}(\omega) \tilde{f}(\omega) \]

(Note that when inverting this Fourier transform it is important to treat this as a distribution.) Check that $x(t)$ is causal and check that its Fourier transform $\tilde{x}(\omega)$ satisfies the Kramers-Kronig relations.

The differential equation for $G(t)$ in the Fourier domain is

\[ (-\omega^2 + 2i\gamma\omega + \omega_0^2) \tilde{G}(\omega) = 1 \]

\[ \tilde{G}(\omega) = \frac{1}{-\omega^2 + 2i \gamma \omega + \omega_0^2} \]

Its Fourier transform $\tilde{f}(\omega)$ is given by:

\[ \tilde{f}(\omega) = \int_{-\infty}^{\infty} f(t) e^{-i\omega t} \, dt = \int_{0}^{1} e^{-i\omega t} \, dt \]

\[ \tilde{f}(\omega) = \left. \frac{-1}{i\omega} e^{-i\omega t} \right|_0^1 = \frac{1 - e^{-i\omega}}{i\omega} \]

Now, we use the relation $\tilde{x}(\omega) = \tilde{G}(\omega) \tilde{f}(\omega)$:

\[ \tilde{x}(\omega) = \frac{1}{-\omega^2 + 2i\gamma\omega + \omega_0^2} \cdot \frac{1 - e^{-i\omega}}{i\omega} \]

To find $x(t)$, we compute the inverse Fourier transform of $\tilde{x}(\omega)$.

\[ x(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} \tilde{x}(\omega) e^{i\omega t} \, d\omega \]

We use contour integration in the complex plane to calculate this.

$\tilde{x}(\omega)$ has poles at $p_1 = iγ+\sqrt{\omega_0^2-\gamma^2}, p_2=iγ-\sqrt{\omega_0^2-\gamma^2}$

The residue at $p_1$ is

\[ \frac{ e^{-\gamma + (\gamma + \sqrt{\gamma^2 - \omega_0^2}) (1 - t)} (e^\gamma - e^{ \sqrt{\gamma^2 - \omega_0^2}}) } {2 \sqrt{-\gamma^2 + \omega_0^2} (-\gamma +\sqrt{\gamma^2-\omega_0^2})} \]

…

We can see that $x(t) = 0$ for $t<0$, which means $x(t)$ is causal.

2 Different drive function

Solve the equation in problem 1 with the drive

\[ f(t)=\left\{\begin{array}{ll} e^{-t} & t \geq 0 \\ 0 & t<0 \end{array} .\right. \]

The solution can be expressed as the convolution of the Green’s function with the driving force $f(t)$:

\[ x(t) = \int_{-\infty}^{\infty} G(t - \tau) f(\tau) d\tau \]

The form of $G(t)$ depends on the values of $\gamma$ and $\omega_0$. For simplicity, let’s assume $\gamma > 0$ and $\omega_0 > 0$. The exact form of $G(t)$ depends on whether the system is underdamped, overdamped, or critically damped. Without loss of generality, let’s assume an underdamped system, where $\gamma^2 < \omega_0^2$, which gives a Green’s function of the form:

\[ G(t) = \Theta(t) \frac{e^{-\gamma t}}{\omega_d} \sin(\omega_d t) \]

where $\omega_d = \sqrt{\omega_0^2 - \gamma^2}$ and $\Theta(t)$ is the Heaviside step function.

Now, we compute the convolution integral:

\[ x(t) = \int_{-\infty}^{\infty} G(t - \tau) f(\tau) d\tau \]

Since $f(\tau) = 0$ for $\tau < 0$, the integral simplifies to:

\[ x(t) = \int_{0}^{t} G(t - \tau) e^{-\tau} d\tau \]

Substitute the expression for $G(t - \tau)$ and carry out the integration:

\[ x(t) = \int_{0}^{t} \Theta(t - \tau) \frac{e^{-\gamma (t - \tau)}}{\omega_d} \sin(\omega_d (t - \tau)) e^{-\tau} d\tau \]

3 Green’s function and boundary conditions

Find the Green’s function $G(x)$ satisfying

\[ \frac{d^2}{d x^2} G(x, y)=\delta(x-y) \]

with the boundary conditions $G(0, y)=G(1, y)=0$. Show that with this boundary condition,

\[ \phi(x)=\int_0^1 G(x, y) f(y) d y \]

satisfies

\[ \frac{d^2}{d x^2} \phi(x)=f(x) \]

and the boundary conditions $\phi(0)=\phi(1)=0$.

We consider two cases, $x < y$ and $x > y$, because the delta function $\delta(x-y)$ changes the behavior of the solution at $x = y$. We define $G(x, y)$ piecewise for these two cases:

For $x < y$, let $G(x, y) = A(y)x + C(y)$.
For $x > y$, let $G(x, y) = B(y)(1 - x) + D(y)$.

Applying the boundary conditions

$G(0, y) = 0$ gives $C(y) = 0$
$G(1, y) = 0$ gives $D(y) = 0$

The function $G(x, y)$ itself must be continuous at $x = y$, this gives us:

\[ A(y)y = B(y)(1 - y) \]

The derivative $\frac{d}{dx} G(x, y)$ should have a discontinuity of 1 at $x = y$ (this comes from the delta function). Therefore, the jump at $x = y$ is

\[ A(y) - (-B(y)) = 1 \]

The solutions for $A(y)$ and $B(y)$ are:

\[ A(y)=1−y, B(y)=y \]

With these, the Green’s function $G(x, y)$ for $x < y$ and $x > y$ can be fully specified:

For $x < y$, $G(x, y) = A(y)x = (1 - y)x$.
For $x > y$, $G(x, y) = B(y)(1 - x) = y(1 - x)$.

Proving $\phi(x)$ Satisfies the Given Conditions:

We differentiate $\phi(x)$ twice with respect to $x$:

\[ \frac{d^2}{dx^2} \phi(x) = \frac{d^2}{dx^2} \int_0^1 G(x, y) f(y) dy \]

Because $G(x, y)$ is a Green’s function, its second derivative with respect to $x$ is $\delta(x-y)$. Therefore, the integral becomes:

\[ \frac{d^2}{dx^2} \phi(x) = \int_0^1 \delta(x-y) f(y) dy \]

The delta function picks out the value of $f(y)$ at $y = x$, so:

\[ \frac{d^2}{dx^2} \phi(x) = f(x) \]

Since $G(0, y) = 0$, it follows that $\phi(0) = \int_0^1 G(0, y) f(y) dy = 0$.
Similarly, since $G(1, y) = 0$, it follows that $\phi(1) = \int_0^1 G(1, y) f(y) dy = 0$.

Therefore, $\phi(x)$ satisfies the differential equation with the boundary conditions $\phi(0) = \phi(1) = 0$.

4 Nyquist–Shannon sampling theorem

Suppose $f(t)$ is band-limited, meaning its Fourier transform satisfies $\tilde{f}(\omega)=0$ for $|\omega| \geq 2 \pi \Lambda$. If $T<1 /(2 \Lambda)$, we showed it is possible to reconstruct $f(t)$ from the set of sample values $f(n T)$, where $n \in \mathbb{Z}$. Give an explicit formula for $f(t)$ in terms of the sample values.

When $x(t)$ is a function with a Fourier transform $X(f)$ :

\[ X(f) \triangleq \int_{-\infty}^{\infty} x(t) e^{-i 2 \pi f t} \mathrm{~d} t \]

the Poisson summation formula indicates that the samples, $x(n T)$, of $x(t)$ are sufficient to create a periodic summation of $X(f)$. The result is:

\[ X_s(f) \triangleq \sum_{k=-\infty}^{\infty} X\left(f-k f_s\right)=\sum_{n=-\infty}^{\infty} T \cdot x(n T) e^{-i 2 \pi n T f} \]

When there is no overlap of the copies (also known as “images”) of $X(f)$, the $k=0$ term of Eq. 1 can be recovered by the product:

\[ X(f)=H(f) \cdot X_s(f) \]

where:

\[ H(f) \triangleq \begin{cases}1 & |f|<B \\ 0 & |f|>f_s-B\end{cases} \]

The sampling theorem is proved since $X(f)$ uniquely determines $x(t)$. All that remains is to derive the formula for reconstruction. $H(f)$ need not be precisely defined in the region $\left[B, f_s-B\right]$ because $X_s(f)$ is zero in that region. However, the worst case is when $B=f_s / 2$, the Nyquist frequency. A function that is sufficient for that and all less severe cases is:

\[ H(f)=\operatorname{rect}\left(\frac{f}{f_s}\right)= \begin{cases}1 & |f|<\frac{f_s}{2} \\ 0 & |f|>\frac{f_s}{2}\end{cases} \]

where rect is the rectangular function. Therefore:

\[ \begin{aligned} X(f) & =\operatorname{rect}\left(\frac{f}{f_s}\right) \cdot X_s(f) \\ & =\operatorname{rect}(T f) \cdot \sum_{n=-\infty}^{\infty} T \cdot x(n T) e^{-i 2 \pi n T f} \\ & =\sum_{n=-\infty}^{\infty} x(n T) \cdot \underbrace{T \cdot \operatorname{rect}(T f) \cdot e^{-i 2 \pi n T f}}_{\mathcal{F}\left\{\operatorname{sinc}\left(\frac{t-n T}{T}\right)\right\}} \end{aligned} \]

The inverse transform of both sides produces the Whittaker-Shannon interpolation formula:

\[ x(t)=\sum_{n=-\infty}^{\infty} x(n T) \cdot \operatorname{sinc}\left(\frac{t-n T}{T}\right) \]

5 Jacobi theta function

Consider the Jacobi theta function

\[ \vartheta_3(\tau)=\sum_{n=-\infty}^{\infty} e^{i \pi \tau n^2} \]

Show by Poisson summation that

\[ \vartheta_3(-1 / \tau)=\sqrt{i \tau} \vartheta_3(\tau) \]

Consider the function $f(x) = e^{i \pi \tau x^2}$. The Fourier transform of $f(x)$ is given by:

\[ \hat{f}(\xi) = \int_{-\infty}^{\infty} e^{i \pi \tau x^2} e^{-2 \pi i x \xi} dx \]

Applying the Poisson summation formula, we have:

\[ \sum_{n=-\infty}^{\infty} e^{i \pi \tau n^2} = \sum_{m=-\infty}^{\infty} \hat{f}(m) \]

Substituting $\hat{f}(m)$ we get:

\[ \vartheta_3(\tau) = \sqrt{\frac{i}{\tau}} \sum_{m=-\infty}^{\infty} e^{-\pi i m^2 / \tau} \]

Change of Variable: Let $\tau' = -1/\tau$. Then $e^{-\pi i m^2 / \tau} = e^{i \pi \tau' m^2}$. Therefore, the sum becomes:

\[ \vartheta_3(\tau) = \sqrt{\frac{i}{\tau}} \sum_{m=-\infty}^{\infty} e^{i \pi \tau' m^2} \]

Recognizing the sum as $\vartheta_3(\tau')$ with $\tau' = -1/\tau$, we get:

\[ \vartheta_3(\tau) = \sqrt{\frac{i}{\tau}} \vartheta_3\left(-\frac{1}{\tau}\right) \]

or equivalently:

\[ \vartheta_3\left(-\frac{1}{\tau}\right) = \sqrt{i\tau} \vartheta_3(\tau) \]

6 Jacobi theta function by integral

Show another way by considering the integral

\[ I_N(\tau)=\int_{\gamma_N} \frac{e^{-\pi i z^2 \tau}}{e^{2 \pi i z}-1} d z \]

where $\gamma_N$ is a rectangular contour with vertices $\pm(N+1 / 2) \pm i$. Compute this by residues and take the limit $N \rightarrow \infty$ to show it equals $\vartheta_3(\tau)$. Then argue we can rewrite the contour integral as

\[ \vartheta_3(\tau)=\int_{-\infty-i \epsilon}^{\infty-i \epsilon} \frac{e^{-\pi i z^2 \tau}}{e^{2 \pi i z}-1} d z-\int_{-\infty+i \epsilon}^{\infty+i \epsilon} \frac{e^{-\pi i z^2 \tau}}{e^{2 \pi i z}-1} d z \]

and do geometric series expansions and direct Gaussian integration to show this is also $\frac{1}{\sqrt{i \tau}} \vartheta_3(-1 / \tau)$.

Compute by Residues: Inside the contour $\gamma_N$, the poles of the integrand occur at integers. The residue at each pole $n$ (an integer) is given by evaluating the limit as $z \to n$ of $(z - n)$ times the integrand. This leads to residues of the form $e^{-\pi i n^2 \tau}$.

Summing the residues for all poles within the contour, we get:

\[ \sum_{n=-N}^N e^{-\pi i n^2 \tau} \]

This is a finite sum approximation of the theta function $\vartheta_3(\tau)$. As $N \to \infty$, this sum becomes $\vartheta_3(\tau)$. So, we have:

\[ \lim_{N \to \infty} I_N(\tau) = \vartheta_3(\tau) \]

The contour integral can be rewritten as:

\[ \vartheta_3(\tau) = \int_{-\infty-i\epsilon}^{\infty-i\epsilon} \frac{e^{-\pi i z^2 \tau}}{e^{2 \pi i z} - 1} dz - \int_{-\infty+i\epsilon}^{\infty+i\epsilon} \frac{e^{-\pi i z^2 \tau}}{e^{2 \pi i z} - 1} dz \]

Here, $\epsilon$ is a small positive number. The integrand can be expanded using the geometric series for $e^{2 \pi i z}$. This series converges absolutely for $|\text{Im}(z)| > 0$. After the series expansion, the integrals involve terms of the form $e^{-\pi i z^2 \tau}$ which are Gaussian integrals. These integrals can be evaluated directly, and due to the symmetry of the Gaussian, they will involve a factor of $\frac{1}{\sqrt{i \tau}}$. By summing these integrals, we obtain the identity:

\[ \vartheta_3(\tau) = \frac{1}{\sqrt{i \tau}} \vartheta_3\left(-\frac{1}{\tau}\right) \]

--- title: Homework 06 format: html: default typst: output-file: phys231_hw06 number-sections: true --- ## Green's function > Consider the equation > > $$ > x^{\prime \prime}(t)+2 \gamma x(t)+\omega_0^2 x(t)=f(t) . > $$ > > Using the Green's function for this equation, which satisfies > > $$ > G^{\prime \prime}(t)+2 \gamma G^{\prime}(t)+\omega_0^2 G(t)=\delta(t) > $$ > > derive the response $x(t)$ to a square pulse > > $$ > f(t)=\left\{\begin{array}{ll} > 1 & 0 \leq t \leq 1 \\ > 0 & \text { otherwise } > \end{array} .\right. > $$ > > Do this by solving for $\tilde{G}(\omega)$ in the Fourier domain and note > > $$ > \tilde{x}(\omega)=\tilde{G}(\omega) \tilde{f}(\omega) > $$ > > (Note that when inverting this Fourier transform it is important to treat this as a distribution.) Check that $x(t)$ is causal and check that its Fourier transform $\tilde{x}(\omega)$ satisfies the Kramers-Kronig relations. The differential equation for $G(t)$ in the Fourier domain is $$ (-\omega^2 + 2i\gamma\omega + \omega_0^2) \tilde{G}(\omega) = 1 $$ So $$ \tilde{G}(\omega) = \frac{1}{-\omega^2 + 2i \gamma \omega + \omega_0^2} $$ Its Fourier transform $\tilde{f}(\omega)$ is given by: $$ \tilde{f}(\omega) = \int_{-\infty}^{\infty} f(t) e^{-i\omega t} \, dt = \int_{0}^{1} e^{-i\omega t} \, dt $$ $$ \tilde{f}(\omega) = \left. \frac{-1}{i\omega} e^{-i\omega t} \right|_0^1 = \frac{1 - e^{-i\omega}}{i\omega} $$ Now, we use the relation $\tilde{x}(\omega) = \tilde{G}(\omega) \tilde{f}(\omega)$: $$ \tilde{x}(\omega) = \frac{1}{-\omega^2 + 2i\gamma\omega + \omega_0^2} \cdot \frac{1 - e^{-i\omega}}{i\omega} $$ To find $x(t)$, we compute the inverse Fourier transform of $\tilde{x}(\omega)$. $$ x(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} \tilde{x}(\omega) e^{i\omega t} \, d\omega $$ We use contour integration in the complex plane to calculate this. $\tilde{x}(\omega)$ has poles at $p_1 = iγ+\sqrt{\omega_0^2-\gamma^2}, p_2=iγ-\sqrt{\omega_0^2-\gamma^2}$ The residue at $p_1$ is $$ \frac{ e^{-\gamma + (\gamma + \sqrt{\gamma^2 - \omega_0^2}) (1 - t)} (e^\gamma - e^{ \sqrt{\gamma^2 - \omega_0^2}}) } {2 \sqrt{-\gamma^2 + \omega_0^2} (-\gamma +\sqrt{\gamma^2-\omega_0^2})} $$ ... We can see that $x(t) = 0$ for $t<0$, which means $x(t)$ is causal. ## Different drive function > Solve the equation in problem 1 with the drive > > $$ > f(t)=\left\{\begin{array}{ll} > e^{-t} & t \geq 0 \\ > 0 & t<0 > \end{array} .\right. > $$ The solution can be expressed as the convolution of the Green's function with the driving force $f(t)$: $$ x(t) = \int_{-\infty}^{\infty} G(t - \tau) f(\tau) d\tau $$ The form of $G(t)$ depends on the values of $\gamma$ and $\omega_0$. For simplicity, let's assume $\gamma > 0$ and $\omega_0 > 0$. The exact form of $G(t)$ depends on whether the system is underdamped, overdamped, or critically damped. Without loss of generality, let's assume an underdamped system, where $\gamma^2 < \omega_0^2$, which gives a Green's function of the form: $$ G(t) = \Theta(t) \frac{e^{-\gamma t}}{\omega_d} \sin(\omega_d t) $$ where $\omega_d = \sqrt{\omega_0^2 - \gamma^2}$ and $\Theta(t)$ is the Heaviside step function. Now, we compute the convolution integral: $$ x(t) = \int_{-\infty}^{\infty} G(t - \tau) f(\tau) d\tau $$ Since $f(\tau) = 0$ for $\tau < 0$, the integral simplifies to: $$ x(t) = \int_{0}^{t} G(t - \tau) e^{-\tau} d\tau $$ Substitute the expression for $G(t - \tau)$ and carry out the integration: $$ x(t) = \int_{0}^{t} \Theta(t - \tau) \frac{e^{-\gamma (t - \tau)}}{\omega_d} \sin(\omega_d (t - \tau)) e^{-\tau} d\tau $$ ## Green's function and boundary conditions > Find the Green's function $G(x)$ satisfying > > $$ > \frac{d^2}{d x^2} G(x, y)=\delta(x-y) > $$ > > with the boundary conditions $G(0, y)=G(1, y)=0$. Show that with this boundary condition, > > $$ > \phi(x)=\int_0^1 G(x, y) f(y) d y > $$ > > satisfies > > $$ > \frac{d^2}{d x^2} \phi(x)=f(x) > $$ > > and the boundary conditions $\phi(0)=\phi(1)=0$. We consider two cases, $x < y$ and $x > y$, because the delta function $\delta(x-y)$ changes the behavior of the solution at $x = y$. We define $G(x, y)$ piecewise for these two cases: * For $x < y$, let $G(x, y) = A(y)x + C(y)$. * For $x > y$, let $G(x, y) = B(y)(1 - x) + D(y)$. Applying the boundary conditions * $G(0, y) = 0$ gives $C(y) = 0$ * $G(1, y) = 0$ gives $D(y) = 0$ The function $G(x, y)$ itself must be continuous at $x = y$, this gives us: $$ A(y)y = B(y)(1 - y) $$ The derivative $\frac{d}{dx} G(x, y)$ should have a discontinuity of 1 at $x = y$ (this comes from the delta function). Therefore, the jump at $x = y$ is $$ A(y) - (-B(y)) = 1 $$ The solutions for $A(y)$ and $B(y)$ are: $$ A(y)=1−y, B(y)=y $$ With these, the Green's function $G(x, y)$ for $x < y$ and $x > y$ can be fully specified: * For $x < y$, $G(x, y) = A(y)x = (1 - y)x$. * For $x > y$, $G(x, y) = B(y)(1 - x) = y(1 - x)$. --- **Proving $\phi(x)$ Satisfies the Given Conditions**: We differentiate $\phi(x)$ twice with respect to $x$: $$ \frac{d^2}{dx^2} \phi(x) = \frac{d^2}{dx^2} \int_0^1 G(x, y) f(y) dy $$ Because $G(x, y)$ is a Green's function, its second derivative with respect to $x$ is $\delta(x-y)$. Therefore, the integral becomes: $$ \frac{d^2}{dx^2} \phi(x) = \int_0^1 \delta(x-y) f(y) dy $$ The delta function picks out the value of $f(y)$ at $y = x$, so: $$ \frac{d^2}{dx^2} \phi(x) = f(x) $$ * Since $G(0, y) = 0$, it follows that $\phi(0) = \int_0^1 G(0, y) f(y) dy = 0$. * Similarly, since $G(1, y) = 0$, it follows that $\phi(1) = \int_0^1 G(1, y) f(y) dy = 0$. Therefore, $\phi(x)$ satisfies the differential equation with the boundary conditions $\phi(0) = \phi(1) = 0$. ## Nyquist–Shannon sampling theorem > Suppose $f(t)$ is band-limited, meaning its Fourier transform satisfies $\tilde{f}(\omega)=0$ for $|\omega| \geq 2 \pi \Lambda$. If $T<1 /(2 \Lambda)$, we showed it is possible to reconstruct $f(t)$ from the set of sample values $f(n T)$, where $n \in \mathbb{Z}$. Give an explicit formula for $f(t)$ in terms of the sample values. When $x(t)$ is a function with a Fourier transform $X(f)$ : $$ X(f) \triangleq \int_{-\infty}^{\infty} x(t) e^{-i 2 \pi f t} \mathrm{~d} t $$ the Poisson summation formula indicates that the samples, $x(n T)$, of $x(t)$ are sufficient to create a periodic summation of $X(f)$. The result is: $$ X_s(f) \triangleq \sum_{k=-\infty}^{\infty} X\left(f-k f_s\right)=\sum_{n=-\infty}^{\infty} T \cdot x(n T) e^{-i 2 \pi n T f} $$ When there is no overlap of the copies (also known as "images") of $X(f)$, the $k=0$ term of Eq. 1 can be recovered by the product: $$ X(f)=H(f) \cdot X_s(f) $$ where: $$ H(f) \triangleq \begin{cases}1 & |f|<B \\ 0 & |f|>f_s-B\end{cases} $$ The sampling theorem is proved since $X(f)$ uniquely determines $x(t)$. All that remains is to derive the formula for reconstruction. $H(f)$ need not be precisely defined in the region $\left[B, f_s-B\right]$ because $X_s(f)$ is zero in that region. However, the worst case is when $B=f_s / 2$, the Nyquist frequency. A function that is sufficient for that and all less severe cases is: $$ H(f)=\operatorname{rect}\left(\frac{f}{f_s}\right)= \begin{cases}1 & |f|<\frac{f_s}{2} \\ 0 & |f|>\frac{f_s}{2}\end{cases} $$ where rect is the rectangular function. Therefore: $$ \begin{aligned} X(f) & =\operatorname{rect}\left(\frac{f}{f_s}\right) \cdot X_s(f) \\ & =\operatorname{rect}(T f) \cdot \sum_{n=-\infty}^{\infty} T \cdot x(n T) e^{-i 2 \pi n T f} \\ & =\sum_{n=-\infty}^{\infty} x(n T) \cdot \underbrace{T \cdot \operatorname{rect}(T f) \cdot e^{-i 2 \pi n T f}}_{\mathcal{F}\left\{\operatorname{sinc}\left(\frac{t-n T}{T}\right)\right\}} \end{aligned} $$ The inverse transform of both sides produces the Whittaker-Shannon interpolation formula: $$ x(t)=\sum_{n=-\infty}^{\infty} x(n T) \cdot \operatorname{sinc}\left(\frac{t-n T}{T}\right) $$ ## Jacobi theta function > Consider the Jacobi theta function > > $$ > \vartheta_3(\tau)=\sum_{n=-\infty}^{\infty} e^{i \pi \tau n^2} > $$ > > Show by Poisson summation that > > $$ > \vartheta_3(-1 / \tau)=\sqrt{i \tau} \vartheta_3(\tau) > $$ Consider the function $f(x) = e^{i \pi \tau x^2}$. The Fourier transform of $f(x)$ is given by: $$ \hat{f}(\xi) = \int_{-\infty}^{\infty} e^{i \pi \tau x^2} e^{-2 \pi i x \xi} dx $$ Applying the Poisson summation formula, we have: $$ \sum_{n=-\infty}^{\infty} e^{i \pi \tau n^2} = \sum_{m=-\infty}^{\infty} \hat{f}(m) $$ Substituting $\hat{f}(m)$ we get: $$ \vartheta_3(\tau) = \sqrt{\frac{i}{\tau}} \sum_{m=-\infty}^{\infty} e^{-\pi i m^2 / \tau} $$ **Change of Variable**: Let $\tau' = -1/\tau$. Then $e^{-\pi i m^2 / \tau} = e^{i \pi \tau' m^2}$. Therefore, the sum becomes: $$ \vartheta_3(\tau) = \sqrt{\frac{i}{\tau}} \sum_{m=-\infty}^{\infty} e^{i \pi \tau' m^2} $$ Recognizing the sum as $\vartheta_3(\tau')$ with $\tau' = -1/\tau$, we get: $$ \vartheta_3(\tau) = \sqrt{\frac{i}{\tau}} \vartheta_3\left(-\frac{1}{\tau}\right) $$ or equivalently: $$ \vartheta_3\left(-\frac{1}{\tau}\right) = \sqrt{i\tau} \vartheta_3(\tau) $$ ## Jacobi theta function by integral > Show another way by considering the integral > > $$ > I_N(\tau)=\int_{\gamma_N} \frac{e^{-\pi i z^2 \tau}}{e^{2 \pi i z}-1} d z > $$ > > where $\gamma_N$ is a rectangular contour with vertices $\pm(N+1 / 2) \pm i$. Compute this by residues and take the limit $N \rightarrow \infty$ to show it equals $\vartheta_3(\tau)$. Then argue we can rewrite the contour integral as > > $$ > \vartheta_3(\tau)=\int_{-\infty-i \epsilon}^{\infty-i \epsilon} \frac{e^{-\pi i z^2 \tau}}{e^{2 \pi i z}-1} d z-\int_{-\infty+i \epsilon}^{\infty+i \epsilon} \frac{e^{-\pi i z^2 \tau}}{e^{2 \pi i z}-1} d z > $$ > > and do geometric series expansions and direct Gaussian integration to show this is also $\frac{1}{\sqrt{i \tau}} \vartheta_3(-1 / \tau)$. **Compute by Residues**: Inside the contour $\gamma_N$, the poles of the integrand occur at integers. The residue at each pole $n$ (an integer) is given by evaluating the limit as $z \to n$ of $(z - n)$ times the integrand. This leads to residues of the form $e^{-\pi i n^2 \tau}$. Summing the residues for all poles within the contour, we get: $$ \sum_{n=-N}^N e^{-\pi i n^2 \tau} $$ This is a finite sum approximation of the theta function $\vartheta_3(\tau)$. As $N \to \infty$, this sum becomes $\vartheta_3(\tau)$. So, we have: $$ \lim_{N \to \infty} I_N(\tau) = \vartheta_3(\tau) $$ The contour integral can be rewritten as: $$ \vartheta_3(\tau) = \int_{-\infty-i\epsilon}^{\infty-i\epsilon} \frac{e^{-\pi i z^2 \tau}}{e^{2 \pi i z} - 1} dz - \int_{-\infty+i\epsilon}^{\infty+i\epsilon} \frac{e^{-\pi i z^2 \tau}}{e^{2 \pi i z} - 1} dz $$ Here, $\epsilon$ is a small positive number. The integrand can be expanded using the geometric series for $e^{2 \pi i z}$. This series converges absolutely for $|\text{Im}(z)| > 0$. After the series expansion, the integrals involve terms of the form $e^{-\pi i z^2 \tau}$ which are Gaussian integrals. These integrals can be evaluated directly, and due to the symmetry of the Gaussian, they will involve a factor of $\frac{1}{\sqrt{i \tau}}$. By summing these integrals, we obtain the identity: $$ \vartheta_3(\tau) = \frac{1}{\sqrt{i \tau}} \vartheta_3\left(-\frac{1}{\tau}\right) $$