# Upper bound for first excited state - variational principle

• lua
In summary, the conversation discussed problem number 5 from a quantum physics course, involving finding an upper bound for the first excited state energy using a trial wave function and the shooting method. The conversation also mentioned the use of Scilab to calculate the ground state energy. The result was an upper bound of 2.423, which was found to be lower than the actual energy of 4.696. The conversation also mentioned the use of a different trial wave function for the ground state energy, which yielded a more accurate result. Finally, the conversation clarified that the first excited state wave function must have a non-zero derivative at the origin.

#### 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")

Last edited:
What is ##\psi(0)## for the first excited state?
Does ##\psi'(0) = 0## for the first excited state?

Last edited:
lua
Oh!
I did't think of that at all. Thank you!

TSny