# How was this one probability amplitude derived by Mattuck?

I'm reading A Guide to Feynman Diagrams in the Many-Body Problem by Richard D. Mattuck (2nd edition). You can look at the relevant pages here.

On page 45, he presents a formula for $D_t c_p(t)$. However, he writes no formula for $c_p(t)$ so I'm at a loss for how he derived that equation.

I just pretended to know how he got that formula and moved on to page 46 and presents (3.27). In this case, I'm not even sure what he did. I'd appreciate a more in detail walk-through of how he turned the wave function into a laplacian then a quadratic term in the integral evaluation.

Here's my shameful attempt at reconciling what's going on in (3.26). $c_p$ is the probability amplitude of a particle being in state $\phi_p$ at time $t_0$. This is almost the same definition as the Green function (propagator), (3.10), divided by the delta function. Taking the derivative of that you get something close to (3.26). Perturbing $\epsilon$ and then taking a derivative makes the $\epsilon_a-\epsilon_b$ term believable. However, I don't even heuristically know how we get the summation or the matrix potentials though.

As for (3.27) I'm even more clueless. Conversion from the potential to the integral left me scratching my head. In addition, the manipulations inside are confusing. I don't really know what to make of (3.27).

Considering this book is about the most intuitive intro to the subject that exists, my chances don't seem to be boding well...any help is appreciated.

• Comment to the question (v2): It would be good if OP (or somebody else?) could try to make the question formulation self-contained, so one doesn't have to open the book to understand the question. – Qmechanic Jul 30 '15 at 18:26

## 1 Answer

What follows is a derivation of the expression directly using Schrodinger's equation.

Assume we have a state, at general time $t$ (setting $t_0 = 0$ for convenience): $$\psi(t) = \sum_l c_l(t) \phi_l e^{-i\epsilon_l t}$$ where the $\phi_l$ are (time-independent) eigenfunctions of the Hamiltonian $H_0$, with eigenvalues $\epsilon_l$. We apply a perturbation $V$ to the Hamiltonian $H_0$ to get the new Hamiltonian: $H = H_0 + V$. We are interested in how $\psi$ evolves with time under this Hamiltonian, or equivalently, how the coefficients $c_i$ change with time. Now, according to Schrodinger's equation: $$i\hbar \frac{\partial}{\partial t} \psi = H\psi = H_0\psi + V\psi$$ Expanding out $\psi$, $$i\hbar \sum_l \left[\dot{c_l} \phi_l e^{-i\epsilon_l t} + \epsilon_l c_l \phi_l e^{-i\epsilon_l t}\right] = \sum_l \left(c_l H_0 \phi_l + c_lV\phi_l\right)e^{-i\epsilon_l t}$$ Now, since the $\phi_l$ are eigenfunctions of $H_0$, $H_0\phi_l = \epsilon_l\phi_l$ The second term on the left and the first on the right are therefore equal, and we are left with: $$\hbar\sum_l \dot{c_l} \phi_l e^{-i\epsilon_l t} = -i\sum_l c_l V \phi_l e^{-i\epsilon_l t}$$

Now, we multiply both sides of the equation with $\phi^\ast_p$, integrate and make use of the orthonormality of eigenfunctions: $$\int \phi^\ast_p \phi_l d\tau = \delta_{pl}$$ where $\delta_{pl}$ is the Kronecker delta, so that: $$\hbar\dot{c_p} = -i\sum_l \left[\int \phi^\ast_p V \phi_l d\tau \right] c_l e^{i(\epsilon_p -\epsilon_l) t}$$ We recognize the integral as the matrix element of $V$ in this representation, leading to the desired expression: $$\hbar \dot{c_p}(t) = -i\sum_l V_{pl} c_l e^{i(\epsilon_p -\epsilon_l) t}$$

• Thanks, that makes a lot of sense. I also figured out that second part. I didn't know $p^2$ could actually refer to the operator. – Zach466920 Aug 3 '15 at 15:44