0
$\begingroup$

I'm trying to use mathematica to normalize this trial wave function which requires finding the following integral and solving for A so that the total propability equals 1. 'a' is the constant width of the potential well and alpha is the variational parameter. I'm having a lot of dificulty with mathematica accepting non-numeric bounds such as a and while still managing to get numerical results but it also can't be done purely symbolically without encountering a ton of erf functions.

enter image description here

8*A^2*Integrate[((1 - x^2*y^2)/a^4/E^(\[Alpha]*((x^2 + y^2)/a^2)))^2, {y, 0, a}, {x, y, a}] + 8*A^2*Integrate[((1 - x^2)/a^4/E^(\[Alpha]*((x^2 + y^2)/a^2)))^2, {y, 0, a}, {x, a, Infinity}]

f1[(x_)?NumericQ, (y_)?NumericQ] := NIntegrate[((1 - x^2*y^2)/a^4 E^(\[Alpha]*((x^2 + y^2)/a^2)))^2, {x, y, a}]

NIntegrate[f1[y], {y, 0, a}]

NIntegrate::nlim: y = a is not a valid limit of integration.


N[8*A^2*Integrate[((1 - x^2*y^2)/a^4/
   E^(\[Alpha]*((x^2 + y^2)/a^2)))^2, {y, 0, a}, 
 {x, y, a}] + 8*A^2*Integrate[
 ((1 - x^2)/a^4/E^(\[Alpha]*((x^2 + y^2)/a^2)))^
  2, {y, 0, a}, {x, a, Infinity}]]

((1/(a^6*\[Alpha]^5))*0.001953125*A^2*
    (8.*a^4*\[Alpha]*(-32.*\[Alpha]^2 + 
            a^4*(3. + 4.*\[Alpha])^2) - 10.026513098524001*
         2.718281828459045^(2.*\[Alpha])*a^4*Sqrt[\[Alpha]]*
         (-32.*\[Alpha]^2 + 3.*a^4*(3. + 4.*\[Alpha]))*
         Erf[1.4142135623730951*Sqrt[\[Alpha]]] + 
       3.141592653589793*2.718281828459045^
           (4.*\[Alpha])*(9.*a^8 - 32.*a^4*\[Alpha]^2 + 
            256.*\[Alpha]^4)*Erf[1.4142135623730951*
               Sqrt[\[Alpha]]]^2))/2.718281828459045^
   (4.*\[Alpha]) + ((1/(a^6*\[Alpha]^3))*0.0625*A^2*
    Erf[1.4142135623730951*Sqrt[\[Alpha]]]*
    (5.0132565492620005*a^2*Sqrt[\[Alpha]]*
         (-8.*\[Alpha] + a^2*(3. + 4.*\[Alpha])) + 
       3.141592653589793*2.718281828459045^
           (2.*\[Alpha])*(3.*a^4 - 8.*a^2*\[Alpha] + 
     16.*\[Alpha]^2)*
         Erfc[1.4142135623730951*Sqrt[\[Alpha]]]))/
 2.718281828459045^(2.*\[Alpha])
$\endgroup$
1
  • $\begingroup$ I mean, you have to make a choice: either use numeric bounds, or accept the fact that A will be in terms of error functions (and what's wrong with error functions?). $\endgroup$ – march Dec 11 '19 at 5:45
1
$\begingroup$

Analytical integration yields good results when applying assumptions.

int = 8*A^2*
        Integrate[((1 - x^2*y^2)/a^4/E^(\[Alpha]*((x^2 + y^2)/a^2)))^2, {y,
  0, a}, {x, y, a}, Assumptions -> a > 0 && \[Alpha] > 0] + 
     8*A^2*Integrate[((1 - x^2)/a^4/
   E^(\[Alpha]*((x^2 + y^2)/a^2)))^2, {y, 0, a}, {x, a, Infinity},
 Assumptions -> a > 0 && \[Alpha] > 0]

Result is a long expression. Solve for A:

AA[a_, \[Alpha]_] = A /. Solve[1 == int, A]

Get A for a given a = 1:

AA[1, \[Alpha]] // FullSimplify[# \[Alpha] > 0] &

Since A^2 is used, you have two solutions:

Plot3D[Evaluate@AA[a, \[Alpha]], {a, 0, 5}, {\[Alpha], 0, 4}, 
    PlotStyle -> {Blue, Red}]
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.