# Question on Step in Lancaster's "Quantum Field Theory for the Gifted Amateur"

I'm having trouble understanding a single step in Lancaster's book. In Chapter 16, the propagator is derived and proved to be the Green's function of the Schrodinger equation. The derivation is pretty straightforward, but I don't understand this one step (Equation 16.27):

$\left(\hat{H}_x - i \frac{\partial}{\partial t_x}\right)G^{+}(x,t_x,y,t_y) = -i\delta(t_x - t_y)\sum_n \phi_n(x) \phi_n(y)^* e^{-i E_n (t_x - t_y)} = -i\delta(t_x - t_y)\delta(x - y),$

where $\hat{H}_x$ is the Hamitlonian (only acting on $x$), $G^{+}$ is the retarded propagator, and the $\phi_n$ are the eigenfunctions of the Hamiltonian with eigenvalues $E_n$. I don't see how the sum is turned into a delta function in the last step. Is there some identity I'm missing? Any help is appreciated.

If the eigenfunctions $\phi_n(x)$ are an orthonormal basis of the Hilbert space, then the sum $$\sum_n\phi_n(x)\phi_n(y)^*$$ is the integral kernel for the identity operator. That is, when we multiply this by any function $f(y)$ and integrate over $y$ we get $$\int \sum_n \phi_n(x)\phi_n(y)^*f(y)dy = \sum_n\phi_n(x)\int \phi_n(y)^*f(y)dy = \sum_n c_n\phi_n(x) = f(x)$$ where $c_n = \langle \phi_n \left|\, f\right\rangle$. This is exactly the property of the delta function $\delta(x-y)$ that we expect: $$\int \delta(x-y)f(y)dy = f(x).$$
• And we can just set $t_x = t_y$ inside the exponential because the delta function makes it zero otherwise? Jun 17, 2016 at 22:19
• The exponential part just adds in the time dependence of the eigenstates: we could write $e^{-iE_n t_x}\phi_n(x)\left(e^{-iE_n t_y}\phi_n(y)\right)^*$ instead of the expression you wrote above. Then, you would want to use the fact that on orthonormal basis will evolve into another orthonormal basis under the Hamiltonian evolution. Jun 17, 2016 at 22:24