# 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? – Klein Four Jun 17 '16 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. – Sean Pohorence Jun 17 '16 at 22:24