1. The problem statement, all variables and given/known data The Schrodinger equation (in atomic units) of an electron moving in one dimension under the influence of the potential -delta(x) [dirac delta function] is: (-1/2.d2/dx2-delta(x)).psi=E.psi use the variation method with the trial function psi'=Ne-a.x2 to show that -pi-1 is an upper bound to the exact ground state energy (which is -0.5). You will need the integral: int(-inf..inf)dx. x2me-a.x2=(2m)!pi1/2/22mm!.am+1/2 2. Relevant equations a) 1D Shrodinger eq: (-1/2.d2/dx2-delta(x)).phi=E.psi b) trial wave function: psi'=Ne-a.x2 c) Useful integral: int(-inf..inf)dx. x2me-a.x2=(2m)!pi1/2/22mm!.am+1/2 d) variation method: E0<= <psi'|H|psi'>/|<psi'|psi'>|2 3. The attempt at a solution You basically need to plug in the wave function psi' (equation b) into the variation method (equation d) using the hamiltonian in equation a). I do this but do not arrive at -1/pi. Instead I get the answer: a-(2a)1/2/pi1/2 Can anyone find the error in my logic/arithmetic below? First determine the normalization constant N in psi' by computing: <psi'|psi'> = 1 and solving for N. This gives: N2int(-inf..inf)dxe-2ax2 using equation c) for this integral with m = 0 I obtain: N2pi1/2/(2a)1/2=1 therefore N2=(2a)1/2/pi1/2 The hamiltonian has two terms the kinetic energy (Ke) and the potential energy (V) term. <psi'|H|psi'> = <psi'|Ke|psi'> + <psi'|V|psi'> where Ke = -1/2d2/dx2 and V = -delta(x) First compute the Ke term in the integral <psi'|H|psi'> Ke integral = <psi'|-1/2d2/dx2|psi'> The first derivative of psi' is: d/dx psi' = N(-2ax.e-ax2) Therefore the second derivative is: d2/dx2 psi' = N( (2ax)2 -2a) e-ax2 using this the Ke integral becomes: =-1/2 N2int(-inf..inf).dx ( (2ax)2 -2a) e-2ax2 the first term in the brackets corresponds to m = 1 in the useful integral (equation c) and the second term in the brackets corresponds to m = 0 in the useful integral (equation c) So the Ke integral becomes (after cancelling terms and rearranging): <psi'|Ke|psi'> = (1/2)N2((2a)1/2pi1/2) The potential energy integral is the following: <psi'|V|psi'> = <psi'|-delta(x)|psi'> = -N2 ea.0 =-N2 so the whole term < psi'|H|psi'> is: (1/2)N2((2a)1/2pi1/2) -N2 now all that remains is to plug in N2 which we computed at the beginning: this reduces the above to: a-(2a)1/2/pi1/2 this is not -1/pi given in the question..... so can anyone tell me what algebra/logic error have I committed? Thanks in advance.