Upper bound for first excited state - variational principle

[lua]

Homework Statement

Find the upper bound of the first excited energy eigenstate in the quartic potential using variational principle.

Relevant Equations

Time-independent Schrodinger equation:
$$
-\frac{\hbar}{2m}\frac{\partial^2 \psi}{\partial x^2}+\alpha x^{4} \psi=E \psi
$$
Dimensionless time-independent Schrodinger equation
$$
-\frac{1}{2}\frac{\partial^2 \psi}{\partial u^2}+(u^{4} - e) \psi=0
$$
Variational principle:
$$
E_{1}\leqslant \int_{-\infty }^{\infty }\psi^{*}(x)H\psi(x)dx
$$

I'm solving problem number 5 from https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/resources/mit8_05f13_ps2/.

(a) Here I got:
$$
\beta = \frac{\hbar^{\frac{1}{3}}}{(\alpha m)^\frac{1}{6}}
$$
and:
$$
E = \left ( \frac{\alpha \hbar^4}{m^2} \right )^\frac{1}{3}e
$$
(b) Using Scilab I found that for the ground state energy:
$$
e_{0}=0.667986
$$
(c) A candidate wavefunction for the variational principle is:
$$
\psi _{1}=Axe^{-\frac{ax^2}{2}}
$$
(It is chosen to be orthogonal to $\psi _{0}$.)
Here I got that upper bound for the first excited state is:
$$
h_{1}=\frac{3}{4}\left ( 10^{\frac{1}{3}} + \frac{5}{100^{\frac{1}{3}}} \right )=2.423
$$
(d) But here, by using shooting method, I got that:
$$
e_{1}=4.696
$$
or, the upper bound is less than the actual energy of the first excited state.
I'm not sure what went wrong.

I have to mention that I checked this for the ground state energy. By using the trial wave function:
$$
\psi _{0}=Be^{-\frac{ax^2}{2}}
$$
I got the upper bound for the ground state energy:
$$
h_{0}=\frac{1}{4}\left ( 6^{\frac{1}{3}} + \frac{3}{36^{\frac{1}{3}}} \right )=0.68142
$$
and exact ground state energy obtained by using the shooting method is:
$$
e_{1}=0.667986
$$
which is OK: exact value of the ground state energy is little bellow the upper bound.
The Scilab source follows:
```
//Solve a second order differential equation
//Y’’= 2(x^4-e)y, y(0)=1 and y’(0)=0
//e0=0.6679863 - dimensionless gruound state energy
funcprot(0)

function dy=f(x, y)
dy(1)=y(2);
dy(2)=2*(x^4-0.6679863)*y(1);
endfunction

x0=0;
xmax=10;
x=x0:0.1:xmax;
y0=1;
dy0=0;

y=ode([y0;dy0],x0,x,f);

clf;

plot(x,y(1,:),"r")
plot(x,y(2,:),"g")

xtitle("$\frac{d^2 y}{dx^2} =2*(x^4-0.6679863)*y $","x","f(x,y)");

legend("x","dy/dx")

 

[TSny]

What is $\psi(0)$ for the first excited state?
Does $\psi'(0) = 0$ for the first excited state?

 

[lua]

Oh!
I did't think of that at all. Thank you!