# Initial condition for Schrödinger equation

1. Dec 3, 2013

### kengrimwood

(If the equation below do not appear correctly, you can read all of the question in the attached file.)

Solving the time dependent 1D Schrödinger equation, one can show that in all points (x,t),

$$i\bar{h}\frac{\partial}{\partial t}\Psi(x,t)=-\frac{\bar{h}^2}{2m}\frac{\partial^2}{\partial x^2}\Psi(x,t)+V(x)\Psi(x,t)=E\Psi(x,t)$$

for a certain value of the energy E. This means that the wave function is necessarily an eigenfunction of the hamiltonian operator.

On the other hand, we may specify an initial condition for this problem. My question is: what if the initial condition DOES NOT verify the above equations. Namely, if $$-\frac{\bar{h}^2}{2m}\frac{\partial^2}{\partial x^2}\Psi(x,t)+V(x)\Psi(x,t)$$ is not proportional to $$\Psi(x,t)$$.

Does it mean that not all initial conditions will give a solution or am I wrong somewhere? Thank you for your help!

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2. Dec 3, 2013

### Jazzdude

Your assertion that all possible wavefunctions are eigenfunctions of the Hamiltonian is wrong. You mixed the stationary and the non-stationary Schroedinger equations. The stationary one is an eigenvector equation and gives you a basis of the solution space together with the eigenfrequencies. Any solution to the time dependent S.E. can be decomposed into that eigenbasis and each component then evolves with stationary magnitude and given angular frequency for the phase.

Cheers,

Jazz

3. Dec 3, 2013

### Staff: Mentor

The equations did not display correctly originally because you had a blank space inside each [tex ] and [/tex ] tag, as shown here. I removed those spaces for you. (It's rather difficult to show those tags as ordinary text because the forum software insists on interpreting them as actual tags and then hiding them!)

4. Dec 3, 2013

### universal_101

There can be many solutions to the T.I.S.E(including non-normalizable), but normally only those are of any interest which are proportional to $$\Psi(x,t)$$(eigenfunction), since only these solutions give you the $$E$$(eigenvalue) as constant, which in turn are stationary states with definite energy when we apply the initial conditions.

5. Dec 3, 2013

### amitbashyal

Plainly speaking, if I am getting your problem right, if you cannot solve for the initial wave function, then you cannot determine its solution after time 't'. Its because the time dependent solution is because we consider that after finding the initial solution, we can find the wave evolved over time through the initial one. Its too hard to right the mathematics from mobile, so i just conveyed my opinion on words.