Optical theorem Peskin and Schroeder

Question

I'm trying to understand the optical theorem of Peskin and Schroeder

$$\tag{7.50} \text{Im} M(k_1,k_2\rightarrow k_1,k_2)=2E_{cm}p_{cm}\sigma_{tot}(k_1,k_2\rightarrow\text{anything})$$

which Peskin and Schroeder says follows from

$$\tag{7.49} -i[M(a\rightarrow b)-M^\ast(b\rightarrow a)]=\sum_f\int d\Pi_f M^\ast(b\rightarrow f)M(a\rightarrow f)$$

and $$\tag{4.79} d\sigma=\frac{1}{2E_A2E_B|v_A-v_B|}\times\prod_f\frac{d^3 p_f}{(2\pi)^3}\frac{1}{2E_f}\times |M(p_A,p_B\rightarrow\{p_f\})|^2(2\pi)^4\delta^{(4)}(p_A+p_B-\sum p_f)$$

and $$\tag{4.80} \int d\Pi_n=\prod_f\frac{d^3 p_f}{(2\pi)^3}\frac{1}{2E_f}\times(2\pi)^4\delta^{(4)}(P-\sum p_f)$$

But how did we integrate (4.79) to get (7.50)?

Answer 1

From your comments it looks like you're mainly confused about the $E_1 E_2 | v_1 - v_2 |$ prefactor, so I'll try to be very explicit about that part. Start from Eq. (7.49) and take the initial and final states to be the same (that is, $b = a$). Then we get

$$ 2 {\rm \:Im \:} \mathcal{M}(a\rightarrow a) = \sum_f \int d \Pi_f | \mathcal{M}(a \rightarrow f) |^2 . $$

Now the formula for the cross section (Peskin's Eq. (4.79)) relates the right hand side of the above equation to a cross section: $$ d\sigma(a \rightarrow f) = \frac{1}{4 E_1 E_2 |v_1 - v_2|} d \Pi_f |\mathcal{M}(a \rightarrow f) |^2 . $$ Combining these two equations we get $$ {\rm \:Im \:} \mathcal{M}(a\rightarrow a) = 2 E_1 E_2 |v_1 - v_2| \times \sum_f \sigma(a \rightarrow f) = 2 E_1 E_2 |v_1 - v_2| \times \sigma(a \rightarrow {\rm anything}) . $$

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Now it is just a matter of working out some kinematics. Working in the CM frame, we can express the momenta of the two incoming particles as $p_1^\mu = (E_1, \vec{p})$ and $p_2^\mu = (E_2, -\vec{p})$. Their velocities are $v_1 = |\vec{p}|/E_1$ and $v_2 = -|\vec{p}|/E_2$ (note the relative sign because they point in opposite directions).

Then we can calculate $$ E_1 E_2 |v_1 - v_2| = E_1 E_2 \left( \frac{|\vec{p}|}{E_1} + \frac{|\vec{p}|}{E_2} \right) \\ = |\vec{p}| (E_1 + E_2) \\ = p_{cm} E_{cm} . $$ Note the definitions which Peskin gives immediately below Eq. (7.50): $p_{cm} = |\vec{p}|$ is the magnitude of either momentum in the CM frame, while $E_{cm} = E_1 + E_2$ is the total energy in the CM frame.

Anyway, plugging this equation into the previous one gives the desired result, $$ {\rm \:Im \:} \mathcal{M}(a\rightarrow a) = 2 E_1 E_2 |v_1 - v_2| \times \sum_f \sigma(a \rightarrow f) = 2 E_{cm} p_{cm} \times \sigma(a \rightarrow {\rm anything}) . $$

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By the way, you can write the cross section formula in a manifestly Lorentz-invariant way (see e.g. Griffith's elementary particles text): $$ d\sigma(a \rightarrow f) = \frac{1}{4 \sqrt{(p_1 \cdot p_2)^2 - m_1^2 m_2^2}} d \Pi_f |\mathcal{M}(a \rightarrow f) |^2 . $$ You can also calculate the prefactor using this expression. Instead of $E_1 E_2 |v_1 - v_2|$, we have (still working in the CM frame with $p_1^\mu = (E_1, \vec{p})$ and $p_2^\mu = (E_2, -\vec{p})$): $$ \left[ (p_1 \cdot p_2)^2 - m_1^2 m_2^2 \right]^{1/2} = \left[ (E_1 E_2 + |\vec{p}|^2 )^2 - (E_1^2 - |\vec{p}|^2)(E_2^2 - |\vec{p}|^2) \right]^{1/2} \\ = \left[ |\vec{p}|^2 (E_1 + E_2)^2 \right]^{1/2} \\ = E_{cm} p_{cm} . $$ As expected we get the same result in the end.

Answer 2

I am also reading the book, I may not provide an answer, as far as I can understand: $-i[M(k_1,k_2\rightarrow k_1,k_2)-M^*(k_1,k_2\rightarrow k_1,k_2)]=2 Im (M(k_1,k_2\rightarrow k_1,k_2))$

In eqn (4.79), if we consider $(E_A,\vec{p_A})=(E_1,\vec{p_{cm}})$ and $(E_B,\vec{p_B})=(E_2,-\vec{p_{cm}})$ as two incoming particles, and if we are considering two particles of same mass, ie $m_1=m_2$, we have $E_1=E_2$ and the particles are traveling in z-direction only, we have $2E_A=E_A+E_B=E_1+E_2=E_{cm}$ then in eqn (4.79)

$2E_12E_2|v_1-v_2|=(E_1+E_2)(2)|\vec{p}_{cm}-(-\vec{p}_{cm})|=E_{cm}(4)p_{cm}$

If we integrate over every $p_f$ in 4.79 and sum over them, and define

$\sigma_{tot}(k_1,k_2\rightarrow anything)$

$ =\frac{1}{4E_{cm}p_{cm}} \Sigma_n(\Pi_i^n\int \frac{d^3q_i}{(2\pi)^3}\frac{1}{2E_i})|M(k_1k_2\rightarrow {q_i})|^2 \times (2\pi)^4\delta^{(4)}(k_1+k_2-\Sigma q_i)$

... work from (4.79)

$\Sigma_n(\Pi_i^n\int \frac{d^3q_i}{(2\pi)^3}\frac{1}{2E_i})|M(k_1k_2\rightarrow {q_i})|^2 \times (2\pi)^4\delta^{(4)}(k_1+k_2-\Sigma q_i)$

is just the RHS of 7.49, so

$2 Im (M(k_1,k_2\rightarrow k_1,k_2))$

$=\Sigma_n(\Pi_i^n\int \frac{d^3q_i}{(2\pi)^3}\frac{1}{2E_i})|M(k_1k_2\rightarrow {q_i})|^2 \times (2\pi)^4\delta^{(4)}(k_1+k_2-\Sigma q_i)$

$=4E_{cm}p_{cm}\sigma_{tot}(k_1,k_2\rightarrow anything)$