# Schrodinger equation

1. Dec 24, 2009

### oddiseas

1. The problem statement, all variables and given/known data

I have just started quantum mechanics bacuase i want to prepare for my class starting in march. I must say so far i find it very confusing: I could use some help with this problem and in addition some explanation on the logic of what it all means.

i get that this theory combines classical mechanics with the wave nature of particles and that the eigenfunctions correspond to standing waves at certain energy levels. But if a particle is localised, what information are we actually getting from its wave function? and how is the wave produced?

Anyway the question i am working on deals with an infinite square well.

Q1)
At t=0 a quantum mechanical system is described by the eigenfunction
$$\Psi$$=iA(L-x^2)

-L<=X<=L

a) Clearrly stating your reasons indicate whether or not $$\Psi$$ is an eigenfunction of the kinetic energy operation;

b) assuming the system is in a well defined eigenstate with total energy E, find the wave function $$\Psi$$

2. Relevant equations

3. The attempt at a solution

a)$$\Psi_x_x$$=-2iA. This does not match the kinetic operator from schrodingers equation. Anyhow i dont understand what the operator means, so if anyone does i could use some help.

b)Solving for the initial conditions, and then finding A by evaluating the integral of|$$\Psi^2$$| i get:
$$\Psi_n(x)$$= $$\frac{1}{2\sqrt{L}}$$sin(npix/L)

Then i tried using the fourier series to find the coefficients
but i get something that looks wrong,

i get one value for n=0, a different one for n= even and another for n=odd.

Anyway i dont have the answer to this question and i am getting confused so it would be great if i could see a worked solution.

Last edited: Dec 24, 2009
2. Dec 24, 2009

### diazona

The idea of quantum mechanics (as you may know) is that you replace the plain old variables of classical mechanics with operators. With position and momentum, you have to pick one of them to remain a variable,
$$x \to x$$
and then the other one becomes a derivative,
$$p \to i\hbar\frac{\partial}{\partial x}$$
Then you can build up all sorts of other operators using familiar formulas from classical mechanics. Like $H = p^2/2m$ (the Hamiltonian). In quantum mechanics, that becomes
$$H = -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}$$
When you want to figure out whether a wavefunction is an eigenfunction of that operator, you check to see whether
$$H\Psi = E\Psi$$
for some value $E$ - that is, when you apply the operator (in this case, a second derivative and then multiplying by a constant) to the function, do you get back a multiple of the same function? If so, it's an eigenfunction of that operator. If not, it's not. That's all that's involved in part (a); just check the result you got from applying the operator to $\Psi$ and see if it's a multiple of $\Psi$.

The significance of the eigenfunctions is that they're the only wavefunctions for which the value of the observable corresponding to that operator is well defined. For example, considering the operator $H$, only eigenfunctions of $H$ have a definite value of energy. Non-eigenfunctions are a mix of different states with different values of energy. So in part (b), it's telling you to assume that $\Psi$ (different $\Psi$ from part (a)) is an eigenstate, or eigenfunction, of the Hamiltonian operator. That means that you know it has to satisfy the equation
$$H\Psi = E\Psi$$
for some value $E$. (They didn't give you a numeric value for E, but consider it a "known quantity;" you can use it in your solution.) When you substitute in the actual definition of $H$, you get a differential equation, which you can solve to get one unique solution. (If you didn't "know" the value of $E$, then yes, you'd have to find a generic solution, like you did. But if you consider $E$ known, and remember that you're talking about one eigenstate with one particular value of the energy, you can find a specific solution to the equation.)

3. Dec 24, 2009

### oddiseas

Thanks, thats helpful, but i am still having trouble finding the solution.If it satisfies the differential equation, then the general solution is Acosmx +Bsinmx, where m=$$\sqrt{2mK}$$/h and the solution becomes:

$$\sum_{n=0}^\infty\ D*sin(npix/L)\*exp(iEt/h)$$

Where the value D is determined by the initila eigen function:

If someone can solve this question i would like to see what the solution is.

4. Dec 24, 2009

### diazona

Stop right there. Assuming that by $K$ you mean $E$, you've got it. Now figure out what conditions A and B need to satisfy so that they are consistent with the required boundary conditions.