Little Quantum Universe

We can convert the differential equation $-\psi''(x) - \sqrt{x}\psi(x) \; = \; \psi(x) \hspace{2cm} \psi(0) = 1\,,\,\psi'(0) = 0$ to the Volterra-type integral equation $\psi(x) \; = \; 1 - \int_0^x (x-u)(\sqrt{u} + 1)\psi(u)\,du$ Consider the linear operator $X \,:\, \mathcal{C}[0,1] \to \mathcal{C}[0,1]$ defined by $(Xf)(x) \; = \; -\int_0^x (x-u)(\sqrt{u} + 1)f(u)\,du$ We can show that if $|f(x)| \le Kx^a$ for all $x$ , then $|(Xf)(x)| \le \tfrac{2K}{(a+1)(a+2)}x^{a+2}$ , which implies that $\Vert X^nf \Vert_\infty \; \le \; \tfrac{2^n}{(2n)!}\Vert f\Vert_\infty \hspace{2cm} f \in \mathcal{C}[0,1]\,,\, n \in \mathbb{N}$ Thus we have a linear operator $Y \,:\, \mathcal{C}[0,1] \to \mathcal{C}[0,1]$ defined by $Yf \; = \; \sum_{n=0}^\infty X^nf$ with $\Vert Yf\Vert_\infty \,\le \, \cosh(\sqrt{2}) \Vert f\Vert_2$ . This operator has the property that $Yf \;=\; f + XYf \hspace{2cm} f \in \mathcal{C}[0,1]$ Moreover, if $Y_N(f) \; = \; \sum_{n=0}^N X^nf$ then $\Vert Y_Nf - Yf\Vert_\infty \; \le \; \left(\cosh(\sqrt{2}) -\sum_{n=0}^N \frac{2^n}{(2n)!}\right)\Vert f \Vert_\infty$ We want the function $\psi = Y1$ . Then $\Vert \psi - Y_31\Vert_\infty \, \le \, 4 \times 10^{-4}$ , which means that $Y_31$ is a good approximation for $\psi$ . Now $(Y_31)(x) \; = \; \begin{array}{l} 1 - \frac12x^2 - \frac{4}{15}x^{\frac{5}{2}} + \frac{1}{24}x^4 + \frac{46}{945}x^{\frac{9}{2}} + \frac{1}{75}x^5 - \frac{1}{720}x^6 \\ {} - \frac{683}{270270}x^{\frac{13}{2}} - \frac{293}{198450}x^7 - \frac{4}{14625}x^{\frac{15}{2}} \end{array}$ and we can use this in place of $\psi$ to approximate the probability as $\boxed{0.697917}$ , which is accurate to $4$ decimal places. Using $Y_41$ would give an even better approximation.

The problem can easily be solved numerically with these two lines in Mathematica:

$s(\text{t\$\_\$})\text{:=}\text{NDSolveValue}\left[\left\{\psi ''(x)+\left(\sqrt{x}+1\right) \psi (x)=0,\psi (0)=1,\psi '(0)=0\right\},\{\psi (t)\},\{x,0,1\}\right]$

$P=\frac{\text{NIntegrate}\left[s(t)^2,\left\{t,0,\frac{1}{2}\right\}\right]}{\text{NIntegrate}[s(t)^2,\{t,0,1\}]}$

which gives us $P=0.697901$ .