Lecture 3 Exercise

(1) Please show that the right hand side of (2) is Lorentz invariant.

$$\begin{array}{}\text{(2)}& ⟨\overrightarrow{p}|\overrightarrow{q}⟩=2E_p\times (2\pi {)}^3{\delta }^{(3)}(\overrightarrow{p}-\overrightarrow{q})\end{array}$$

First, we consider the identity for the four-dimensional delta function restricted to the positive-energy mass shell

$$\int {\mathrm{d}}^{4}p\delta (p^2+m^2)\mathrm{\Theta }(p^0)f(p)$$

where $p^2=-(p^0{)}^2+{\overrightarrow{p}}^2$. Since ${\mathrm{d}}^{4}p$ and the constraints $\delta (p^2+m^2)$ and $\mathrm{\Theta }(p^0)$ are Lorentz invariant, any measure derived from this must also be invariant.

We use the property of the delta function

$$\delta (g(x))=\sum _{x_i}\frac{\delta (x-x_i)}{|g^{\prime }(x_i)|}$$

Let $g(p^0)=(p^0{)}^2-({\overrightarrow{p}}^2+m^2)=(p^0{)}^2-E_p^2$. The roots are $p^0=\pm E_p$, but the $\mathrm{\Theta }(p^0)$ picks only $p^0=E_p$.

Then $|g^{\prime }(E_p)|=|2p^0{|}_{p^0=E_p}=2E_p$. Thus

$$\int {\mathrm{d}}^{4}p\delta (p^2+m^2)\mathrm{\Theta }(p^0)=\int {\mathrm{d}}^{3}p\int d{p}^0\frac{\delta (p^0-E_p)}{2E_p}=\int \frac{{\mathrm{d}}^{3}p}{2E_p}$$

Since the integration measure $\frac{{\mathrm{d}}^{3}p}{2E_p}$ is Lorentz invariant, and we have

$$|\overrightarrow{q}⟩=\int {\mathrm{d}}^3\overrightarrow{p}|\overrightarrow{p}⟩⟨\overrightarrow{p}|\overrightarrow{q}⟩$$

So the integration

$$\int {\mathrm{d}}^3\overrightarrow{p}⟨\overrightarrow{p}|\overrightarrow{q}⟩=\int \frac{{\mathrm{d}}^{3}p}{2E_p}2E_p\cdot {\delta }^3(\overrightarrow{p}-\overrightarrow{q})$$

must be a Lorentz invariant. Therefore, we can know that

$$\begin{array}{}\text{(2)}& 2E_p\times (2\pi {)}^3{\delta }^{(3)}(\overrightarrow{p}-\overrightarrow{q})\end{array}$$

is a Lorentz invariant.

(2) Please use the unitarity of the $S$-matrix to prove the optical theorem in the form of (26).

$$\begin{array}{}\text{(26)}& \mathrm{Im}\mathcal{M}(p_1{p}_2\to p_1{p}_2)=\sum _{X\in \text{final states}}{\left(2\pi \right)}^4{\delta }^{(4)}(p_1+p_2-p_X){\left|\mathcal{M}(p_1{p}_2\to X)\right|}^2\end{array}$$

We take the form

$$S=1+i(2\pi {)}^4{\delta }^4(p)M$$

and using unitarity of $S$ matrix, we can know that

$$S{S}^{†}=(1+i(2\pi {)}^4{\delta }^4(p)M)\cdot (1-i(2\pi {)}^4{\delta }^4(p)M^{†})=1+i(2\pi {)}^4{\delta }^4(p)(M-M^{†})+(2\pi {)}^8{\delta }^4(p)M{M}^{†}=1$$

Therefore, we can derive that

$$i(M-M^{†})+(2\pi {)}^4{\delta }^4(p)M{M}^{†}=0$$$$i⟨\psi |M-M^{†}|\psi ⟩+(2\pi {)}^4{\delta }^4(p)⟨\psi |M{M}^{†}|\psi ⟩=0$$

and insert a complete set of states on the right side

$$i⟨\psi |M-M^{†}|\psi ⟩=-\sum _X(2\pi {)}^4{\delta }^4(p)⟨\psi |M|X⟩⟨X|M^{†}|\psi ⟩$$

Notice that $⟨\psi |M^{†}|\psi ⟩=⟨\psi |M^{†}|\psi {⟩}^{\ast }$, thus

$$i⟨\psi |M-M^{†}|\psi ⟩=-2\mathrm{\Im }M(p_1{p}_2\to p_1{p}_2)$$

So we have the optical theorem

$$\mathrm{Im}\mathcal{M}(p_1{p}_2\to p_1{p}_2)=\sum _{X\in \text{final states}}{\left(2\pi \right)}^4{\delta }^{(4)}(p_1+p_2-p_X){\left|\mathcal{M}(p_1{p}_2\to X)\right|}^2$$

(3) Please read through Sec. 1.1-1.3 in Lecture I of [1] (or ask AI directly) and write a short summary of

1. how does causality imply analyticity in classical physics (no quantum mechanics required)

2. what is Kramers-Kronig relation and how is it derived.

[You can certainly use AI but please write the summary yourself.]

1. How causality implies analyticity in classical physics: In classical linear response theory, a system's output $x(t)$ is related to an input signal $f(t)$ via a response function $\chi (\tau )$: $x(t)=\int \chi (t-t^{\prime })f(t^{\prime })d{t}^{\prime }$. The principle of causality states that the effect cannot precede the cause, meaning $\chi (\tau )=0$ for $\tau <0$. When we look at the generalized frequency response $\tilde{\chi }(\omega )={\int }_0^{\mathrm{\infty }}\chi (\tau )e^{i\omega \tau }d\tau$, if we allow $\omega$ to be complex ($\omega ={\omega }_R+i{\omega }_I$), the factor $e^{i\omega \tau }=e^{i{\omega }_R\tau }{e}^{-{\omega }_I\tau }$ acts as a powerful convergence factor for any ${\omega }_I>0$. Because the integral is only over positive $\tau$ (due to causality), $\tilde{\chi }(\omega )$ is guaranteed to be a holomorphic (analytic) function in the upper half-plane of the complex frequency space.

2. The Kramers-Kronig relation and its derivation: The Kramers-Kronig relations relate the real and imaginary parts of a causal response function.

   Derivation: Since $\tilde{\chi }(\omega )$ is analytic in the upper half-plane, we apply Cauchy's Integral Formula for a point $\omega$ on the real axis, using a contour that travels along the real axis and closes via a large semi-circle in the upper half-plane (assuming $\tilde{\chi }$ vanishes at infinity).

   By taking the Cauchy Principal Value ($\mathcal{P}$) of the integral:

   $$\tilde{\chi }(\omega )=\frac{1}{i\pi }\mathcal{P}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{\tilde{\chi }({\omega }^{\prime })}{{\omega }^{\prime }-\omega }d{\omega }^{\prime }$$

   Splitting $\tilde{\chi }$ into real and imaginary parts ($\tilde{\chi }=\text{Re}\chi +i\text{Im}\chi$) yields:

   $$\text{Re}\tilde{\chi }(\omega )=\frac{1}{\pi }\mathcal{P}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{\text{Im}\tilde{\chi }({\omega }^{\prime })}{{\omega }^{\prime }-\omega }d{\omega }^{\prime }\text{and}\text{Im}\tilde{\chi }(\omega )=-\frac{1}{\pi }\mathcal{P}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{\text{Re}\tilde{\chi }({\omega }^{\prime })}{{\omega }^{\prime }-\omega }d{\omega }^{\prime }$$

   This shows that if you know the absorption (imaginary part) of a system across all frequencies, you can calculate its dispersion (real part), and vice versa.

References

[1] S. Mizera, “Physics of the analytic S-matrix,” Phys. Rept. 1047 (2024) 1–92, arXiv:2306.05395 [hep-th].