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]
>
>

```

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