Re: Schroedinger EQ

*To*: mathgroup at smc.vnet.net*Subject*: [mg122055] Re: Schroedinger EQ*From*: "Kevin J. McCann" <Kevin.McCann at umbc.edu>*Date*: Tue, 11 Oct 2011 04:23:21 -0400 (EDT)*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com*References*: <j6q61b$p1g$1@smc.vnet.net> <j6uafl$f2c$1@smc.vnet.net>

There are a few problems. First, I would load the Notation package and then Symbolize a_so, V_so, etc. or else just use aso, Vso, ... Next, in one of your equations you have simply R instead of a subscripted one, I assume you left this off. In your function eq[V_,x_,x0_] you have ... == -e = y[x]. I am not sure what you mean by this, but the syntax is not good. Also the "e" you use should be either "E" or esc-ee-esc, i.e. 2.71828 I didn't get any further, because there are others I think. Kevin On 10/10/2011 4:30 AM, raj kumar wrote: > sorry, A =82 > > > On Oct 9, 2:50 am, raj kumar<rajesh779... at gmail.com> wrote: >> dear esteemed experts, >> >> i wonder if somebody can help me >> >> i have been trying to find a a certain value of a parameter V that >> will "match" the logarithmic derivative (of solutions to the time >> independent Schroedinger eq ) at both sides of a particular point >> called the matching point. But cannot seem to find the correct V value >> that will make >> bc1[V_] = bc2[V_].. ...mathematica keeps giving an error message .See >> below for the code. >> >> any help will be most appreciated >> >> a=0.63; >> A8; >> j=9/2; >> L=4; >> mu=(931.5 (208 1.))/(208+1.); >> Z=82; >> Subscript[a, so]=0.5; >> R=1.25 A^(1/3); >> Subscript[V, so]=7; >> Subscript[R, c]=1.25 A^(1/3); >> Subscript[R, so]=1.1 A^(1/3); >> V1[x_,V_]:=-(V/(E^((x-R)/a)+1)); V2[x_]:=-((2 ((j+1) j-L (L+1)-3/4= > ) >> Subscript[V, so] E^((x-Subscript[R, so])/Subscript[a, so]))/ >> (Subscript[a, so] (E^((x-Subscript[R, so])/Subscript[a, so])+1)^2)); >> pott[x_,V_]=V1[x,V]+V2[x]; >> >> emin=-55; >> emax= -5; >> xmax=10; >> xmin=0.1; >> xmatch=4.5; >> e=3.94; >> >> eq[V_, x_, x0_] = {-( >> \!\(\*SuperscriptBox["y", "\[Prime]\[Prime]", >> MultilineFunction->None]\)[x]/( >> 2 mu)) + (pott[x, V] + L (L + 1)/(2 mu (x^2))) y[x] == -e= > y[x], >> y[x0] == 0, >> \!\(\*SuperscriptBox["y", "\[Prime]", >> MultilineFunction->None]\)[x0] == 1/10^6}; >> y1[V_, x_] := y[x] /. NDSolve[eq[V, x, xmin], y, {x, xmin, xmatch}]; >> bc1[V_] := \!\( >> \*SubscriptBox[\(\[PartialD]\), \(x\)]\(y1[x]\)\)/y1[x] /. x -> >> xmatch; >> y2[V_, x_] := y[x] /. NDSolve[eq[V, x, xmax], y, {x, xmax, xmatch}]; >> bc2[V_] := \!\( >> \*SubscriptBox[\(\[PartialD]\), \(x\)]\(y2[x]\)\)/y2[x] /. x -> >> xmatch; >> bc[V_?NumericQ] := bc1[V] - bc2[V]; >> Vvalue = V /. >> If[emax == emin, V, >> FindRoot[bc[V], {V, emin, emax}, AccuracyGoal -> 10, >> WorkingPrecision -> 20]]; >> Print["the value of V is =" , Vvalue] > >