- #1

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- 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")

(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: