# Box potential problem

1. Nov 5, 2007

### quantum_prince

Taking a particle m with box potential (one dimensional) where V(x) = 0 when mod(x) <=a and V(x) = infinity when mod(x) > a and where wave function phi(x) = A (phi1(x) + ph2(x)) where phi1(x) and phi2(x) are normalized wave functions of the ground state and first excited state

We need to Assume that the wave function (x, t) at time t = 0 is given by $$\psi (x,0) = \phi (x)$$ need to find $$\psi (x, t) and \psi (x, t)^2$$ e need to express the latter in terms of sine or cosine functions, eliminating the
exponentials with the help of Euler’s formula.

Abbreviation to be used is:

$$w = \pi ^2 h/8 m a^2$$

How do I proceed in this regard.

I know that

ih dow W /dow t = -h^2/2m dow^2/dow x^2 + V(x) psi

Since V(x)=0 that term disappears so we have

$$ih dow \psi / dow t = -h^2 /2m dow^2/dow x^2$$

Euler formula.

$$e^i[\theta] = \cos\theta + i \sin\theta$$

Regards QP

2. Nov 5, 2007

### Galileo

Do you know how phi1 and phi2 evolve in time? (Hint: They're stationary states, i.e. eigenstates of the Hamiltonian), Once you know that, you can easily write down the evolution of any linear combination of phi1 and phi2 by using the linearity of Hthe Hamiltonian.

3. Nov 7, 2007

### quantum_prince

Hi Galileo,

This is how I would proceed.I have attempted to find a solution to

the problem.Please guide me if am wrong as you did before.

General solution to Schroedinger equation:

psi(x,t) = phi(x) exp(-i(n+1/2)wt)

Now using phi(x) given in our problem

pxi(x,t) = A(ph1(x) + ph2(x)) exp(-i(n+1/2)wt)

since w = pi^2h/8ma^2

phi(x,t) = A(phi1(x) + phi2(x)) exp(-i(n+1/2)pi^2/8ma^2*t)

Z = (n+1/2)pi^2/8ma^2*t

phi(x,t) = A (ph1(x) + phi2(x)) (cos Z - i sinZ)

Regards.

QP.