# S-matrix formula from Lippmann-schwinger eqn

• muppet
In summary: So in summary, the expression for the S-matrix can be derived using the matrix elements $T^{\pm}_{\beta\alpha}$, but it requires the use of a distribution rather than a regular function.
muppet
Hi all,
In chapter 3.2 of Weinberg's QFT text he asserts that one can derive the expression
$$S_{\beta\alpha}=\delta(\beta-\alpha)-2\pi i\delta(E_\alpha -E_\beta)T_{\beta\alpha}$$
for the S-matrix in terms of the matrix elements $$T^{\pm}_{\beta \alpha}=(\Phi_\beta , V\Psi^{\pm}_\alpha)$$, where I'm using Weinberg's "linear algebraic" notation for the inner product of state vectors, V is a scattering potential, $$\Phi_{\alpha}$$ is an eigenstate of a free hamiltonian $$H_0$$ with eigenvalue $$E_\alpha$$, and the $$\Psi^{\pm}$$ are in/out states, from the Lippmann -Schwinger equations
$$\Psi^{\pm}_\alpha=\Phi_{\alpha}+(E_\alpha-H_0 \pm i\epsilon )^{-1} V\Psi^{\pm}_\alpha$$.

However, he never actually gets around to it. Here's my attempt, which seems to get me close but contains either a mistake or a pathologically strange representation of the delta function.

Consider the matrix element $$(\Psi_{\beta}^-,V\Psi_\alpha^+)$$. We apply the Lippman-Schwinger equations to both the 'in' and 'out' states and equate the results.
Note that I'll use $$\Psi^{\pm}=\int d\gamma(\Phi_\gamma,\Psi^{\pm})\Phi_\gamma$$ a few times.
Starting with the in state:
$$(\Psi_{\beta}^-,V\Psi_\alpha^{\plus})=(\Psi_{\beta}^-,V\Phi_{\alpha}+V(E_\alpha-H_0 + i\epsilon )^{-1} V\Psi^{+}_\alpha)$$
$$=(\Psi_{\beta}^-,V\Phi_{\alpha})+(\Psi_{\beta}^-,V(E_\alpha-H_0 + i\epsilon )^{-1} V\Psi^{+}_\alpha)$$
$$=(T^{-}_{\alpha \beta})^*+\int d \gamma (\Psi_{\beta}^-,V(E_\alpha-H_0 + i\epsilon )^{-1} T^+_{\gamma\alpha}\Phi_\gamma)$$
$$=T^{-}_{ \beta \alpha}+\int d \gamma (E_\alpha-E_\gamma + i\epsilon )^{-1} T^+_{\gamma\alpha} (T^-_{\gamma \beta})^*=T^{-}_{ \beta \alpha}+\int d \gamma (E_\alpha-E_\gamma + i\epsilon )^{-1} T^-_{\beta \gamma}T^+_{\gamma\alpha}$$

Now the out state:
$$(\Psi_{\beta}^-,V\Psi_\alpha^{+})=(\Phi_{\beta}+(E_\blpha-H_0 - i\epsilon )^{-1} V\Psi^-_\beta,V\Psi_\alpha^{+})$$
$$=(\Phi_{\beta},V\Psi_\alpha^{+})+(\Psi^-_\beta,V(E_\beta-H_0 + i\epsilon )^{-1} V\Psi_\alpha^{+})$$
$$=T^{+}_{\beta \alpha}+ \int d \gamma (\Psi_{\beta}^-,V(E_\alpha-H_0 + i\epsilon )^{-1} T^+_{\gamma\alpha}\Phi_\gamma)$$
$$=T^{+}_{ \beta \alpha}+\int d \gamma (E_\beta-E_\gamma + i\epsilon )^{-1} T^-_{\beta \gamma}T^+_{\gamma\alpha}$$

Now we can get a relationship between $$T^{+}_{ \beta \alpha}$$ and $$T^{-}_{ \beta \alpha}$$ by expanding the defining matrix element for in states using the completeness relation for the out states:
$$T^{+}_{ \beta \alpha}=(\Phi_\beta,V\Psi_\alpha^+)=\int d\gamma (\Phi_\beta,V(\Psi^-_\gamma,\Psi_\alpha^+)\Psi^-_\gamma)=\int d\gamma S_{\gamma \alpha}T^-_{\beta\gamma}$$

So equating the two expansions of $$(\Psi_{\beta}^-,V\Psi_\alpha^{\plus})$$ and rearranging for $$T^+_{\beta\alpha}$$ leads to
$$T^+_{\beta\alpha}=\int d\gamma S_{\gamma \alpha}T^-_{\beta\gamma}$$
$$=T^{-}_{ \beta \alpha}+\int d \gamma (E_\alpha-E_\gamma + i\epsilon )^{-1} T^-_{\beta \gamma}T^+_{\gamma\alpha} - (E_\beta-E_\gamma + i\epsilon )^{-1} T^-_{\beta \gamma}T^+_{\gamma\alpha}$$
$$=\int d \gamma T^-_{\beta \gamma}(\delta(\alpha-\gamma)+\{(E_\alpha-E_\gamma + i\epsilon )^{-1} - (E_\beta-E_\gamma + i\epsilon )^{-1} \}T^+_{\gamma\alpha} )$$

Now this argument is only correct if
$$\{(E_\alpha-E_\gamma + i\epsilon )^{-1} - (E_\beta-E_\gamma + i\epsilon )^{-1} \} = -2\pi i\delta(E_\alpha -E_\beta)$$
inside an integral, which really doesn't appear to be true.

Any help would be greatly appreciated, I've spent most of today deriving this and then trying to correct my error. Thanks in advance.

Last edited:
A:I think the problem is that you are not using the right expression for the delta function.$\delta(E_\alpha-E_\beta)$ is a Dirac delta function, which means that it is equal to $0$ everywhere expect for $E_\alpha=E_\beta$, where it is equal to $+\infty$. This means that it does not appear as part of an integral.The expression $\frac{1}{2\pi i}\bigg(\frac{1}{E_\alpha-E_\beta+i\epsilon}-\frac{1}{E_\alpha-E_\beta-i\epsilon}\bigg)$ is a distribution, and it is equal to $\delta(E_\alpha-E_\beta)$ when it appears as part of an integral.That is why Weinberg's expression for the S-matrix contains this term.

## What is the S-matrix formula from Lippmann-schwinger eqn?

The S-matrix formula from Lippmann-schwinger eqn is a mathematical expression that describes the scattering of particles in quantum mechanics. It relates the initial and final states of a system to the potential interaction between particles.

## How is the S-matrix formula derived?

The S-matrix formula is derived from the Lippmann-schwinger equation, which is a integral equation used to solve for the wave function of a system. By solving this equation and applying certain mathematical techniques, the S-matrix formula can be obtained.

## What is the significance of the S-matrix formula?

The S-matrix formula is an important tool in studying the behavior of particles in quantum mechanics. It allows for the prediction and calculation of scattering amplitudes, which are essential in understanding the properties and interactions of particles.

## Can the S-matrix formula be applied to all types of interactions?

Yes, the S-matrix formula can be applied to both strong and weak interactions in quantum physics. It is a general formula that can be used to study the behavior of particles in a wide range of systems.

## Is the S-matrix formula experimentally verified?

Yes, the predictions made by the S-matrix formula have been experimentally verified in various experiments, confirming its validity in describing particle interactions. It has also been successfully used in predicting the behavior of particles in high-energy collisions, such as those in particle accelerators.

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