φHφdx=-(h2/872m)/(k-x2)d2(kxdx2dx =h2/(4m2m)/(x2-kx)dx h2/(242m) ∫φ*dx=∫x2(-x)2dx=F/30 E6=5h2(4mPm)≥h2/(8m/)=E1
f* H f dx = -(h2 /8 2m) (lx-x 2 ) d2 (lx-x 2 )/dx2 dx = h2 /(4 2m) (x2 - lx) dx = h2 l 3 /(24 2m) f*f dx = x 2 (l-x)2 dx = l 5 /30 Ef = 5h2 /(4 2 l 2m) h 2 /(8ml 2 ) = E1
Variational method (1)Construct a wave function p(C1,C2,00,C (2 )Calculate the energy of o E6≡E6(C1C2。cn) ()Choose c,(i-1, 2, .ee, m) so that eo is minimum
(1) Construct a wave function f(c1 ,c2 , •••,cm ) (2) Calculate the energy of f: Ef Ef (c1 ,c2 , •••,cm ) (3) Choose {cj *} (i=1,2,•••,m) so that Ef is minimum Variational Method
Example: one-dimensional harmonic oscillator Potential V(x)=(1/2)kx2=(1/2)mo2x2=2T'mvx2 Trial wave function for the ground state p(x)=exp(-cx2) So*HOdx=-( h2/8T2m)Sexp(cx2)d?lexp(-cx2 )dx2 dx +212mv2 x2 exp(-2cx2)dx (h2/4m2m)(πc8)12+m2my2(π8c)l2 ∫φ*ψdx=∫exp(2cx)dx=(/2)l2c12 E6=W=(h2872m)c+(r2)my2/c
Example: one-dimensional harmonic oscillator Potential: V(x) = (1/2) kx2 = (1/2) m2x 2 = 2 2m 2x 2 Trial wave function for the ground state: f(x) = exp(-cx2 ) f* H f dx = -(h2 /8 2m) exp(-cx2 ) d2 [exp(-cx2 )]/dx2 dx + 2 2m 2 x 2 exp(-2cx2 ) dx = (h2 /4 2m) (c/8)1/2 + 2m 2 (/8c3 ) 1/2 f*f dx = exp(-2cx2 ) dx = (/2)1/2 c -1/2 Ef = W = (h2 /8 2m)c + ( 2 /2)m 2 /c
To minimize w 0=dW/dc=h2/8T2m-(T2/2)mv2c-2 c=2 vm/h W=(1/2)hv
To minimize W, 0 = dW/dc = h2 /8 2m - ( 2 /2)m 2c -2 c = 2 2m/h W= (1/2) h
Extension of variation method EEE V For a wave function o which is orthogonal to the ground state wave function v,i.e dτdv1=0 E=dP/Jdrp≥E2 the first excited state energy
. . . E3 y3 E2 y2 E1 y1 Extension of Variation Method For a wave function f which is orthogonal to the ground state wave function y1 , i.e. dt f * y1 = 0 Ef = dt f *Hf / dt f *f > E2 the first excited state energy