S-matrix formula from Lippmann-schwinger eqn

[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.

 

[azntoon]

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.