题12.19 求出试探波函数-|||-_(0)(x)=A(e)^-dfrac (1{2)}(x)^2(x)^2-|||-其中,A是归一化常数,式中的参数α之值可以给出非简谐振子Hamilton 量-|||-=-dfrac ({h)^2}(2mu )dfrac ({d)^2}(d{x)^2}+lambda (x)^4 λ是常数-|||-的基态能量的最好近似.下列积分可能有用:-|||-.(int )_(-infty )^+infty (e)^-a(x^2)dx=sqrt (dfrac {pi )(alpha )} . (int )_(-infty )^+infty (x)^2(e)^-a(x^2)dx=dfrac (1)(2)sqrt (dfrac {pi )({a)^3}} . (int )_(-infty )^+infty (x)^4(e)^-a(x^2)dx=dfrac (3)(4)sqrt (dfrac {pi )({a)^5}}