3
$\begingroup$

I am having the following problem understanding the Born approximation in the case of the Lippmann-Schwinger equation. This exercise is for something which is entitled "computational physics lab course" and the following equations are all taken from the lab manual.

Following the derivation in Chapter 3 of Weinberg's The Quantum Theory of Fields in a similar fashion one obtains the Lippmann-Schwinger-Equation: $$ |\Psi_\mathbf{k}^\pm\rangle = |\mathbf{k}\rangle + \lim_{\epsilon\rightarrow 0^+} [E_k\pm i\epsilon - H_0 ]^{-1} V |\Psi_\mathbf{k}^+\rangle.$$

From this one can obtain the tranfer matrix $T$ by defining $$T|\mathbf{k}\rangle = V | \Psi_\mathbf{k}^+\rangle.$$ Then following holds: $$T|\mathbf{k}\rangle> = V|\mathbf{k}\rangle + V \lim_{\epsilon\rightarrow 0^+} [E+i\epsilon-H_0]^{-1} T|\mathbf{k}\rangle.$$ From which in turn an expression analogous to eq. 3.5.1 in Weinberg can be obtained:

$$\langle\mathbf{l}| T |\mathbf{k}\rangle = T(l,k)=\hat{V}(\mathbf{l}-\mathbf{k}) + \lim_{\epsilon\rightarrow 0^+} \frac{1}{(2\pi)^3}\int_{\mathbb{R}^3} \hat{V}(\mathbf{l} -\mathbf{p})\langle \mathbf{p}|[E+i\epsilon-H_0]^{-1} T|\mathbf{k}\rangle d^3p,$$

where $\hat{V}$ is the Fourier transformation of the potential. After a partial-wave decomposition one obtains:

$$\hat{V}(\mathbf{k}'-\mathbf{k}) = 4\pi\sum_{l=0}^{\infty} (2l+1)\hat{V}_l(k',k) P_l(\cos\theta).$$

After further algebra we end up with the following expression for $T$:

$$T(k',k) = \hat{V}(k',k)+\frac{2}{\pi} \int_0^{\infty} \frac{2m/\hbar^2}{k^2-q^2}\left[ q^2\hat{V}(k',q)T(q,k) - k^2 \hat{V}(k',k)T(k,k)\right]dq - 2i\frac{mk}{\hbar^2} \hat{V}(k',k)T(k,k).$$

Now a real kernel is defined by:

$$K(k',k)=\hat{V}(k',k)+\frac{2}{\pi} \int_0^{\infty} \frac{2m/\hbar^2}{k^2-q^2}\left[ q^2\hat{V}(k',q)T(q,k) - k^2 \hat{V}(k',k)T(k,k)\right]dq.$$

My problem lies with the following:

Given a potential

$$V(r) = V_0 \frac{1}{\mu r} \cdot (e^{-2\mu r} - e^{-\mu r})$$

(where actually $r=||\mathbf{r}||_2$) and using the following expression:

$$\hat{V}(k',k) = \sum_i \frac{V_i}{\mu_i}\frac{1}{k'k}Q_l\left( \frac{k^2+k'^2+\mu_i^2}{2kk'}\right),$$

where the sum is in this case over just two components. I have to solve the equation for $K$ for $l=0,1$ in Born approximation.

Unfortunately, the only expressions for a Born approximation I have are for a state $\Psi_\mathbf{k}^\pm(r)$(Wikipedia says essentially the same) and not for a matrix equation.

Now it took me some time and the equation 3.5.3 from Weinberg

$$T_{\beta\alpha}^+ = V_{\beta\alpha}+\int\frac{V_{\beta\gamma}V_{\gamma\alpha} }{E_\alpha - E_\gamma +i\epsilon}+\dots$$

to get to the following conclusion:

The Born approximation in physics is the same as one step of a Picard iteration in mathematics.

Where the latter one is an iterative solution of a Cauchy problem $y(0)=y_0, \dot{y}= f(t,y)$ in the form:

$$y^{(0)}(t) = y_0, y^{(n+1)}(t) = y_0 + \int_0^t f(t', y^{(n)}(t'))dt'.$$

So in this case I would obtain for $K$:

$$K(k',k)=\hat{V}(k',k)+\frac{2}{\pi} \int_0^{\infty} \frac{2m/\hbar^2}{k^2-q^2}\left[ q^2\hat{V}(k',q)\hat{V}(q,k) - k^2 \hat{V}(k',k)\hat{V}(k,k)\right]dq.$$

Here my question then:

Is this reasoning right or am I missing something fundamental?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.