How to Calculate Wavefunction for Arbitrary Time?

[jimmycricket]

For a particle, given the normalised eigenfunctions of the Hamiltonian, the associated energy eigenfunctions and the wavefunction describing the state of the particle at time t=0 how does one calculate the wavefunction for arbitrary t? I know you could solve the time dependent Schroedinger equation but is there not an easier way than that?

 

[Orodruin]

Each eigenstate of the Hamiltonian evolves independently and the evolution is trivially given by the time-dependent SE (if you know the eigenvalue, it becomes a very simple equation for the coefficient).

 

[jtbell]

Expand the initial wave function as a linear combination of the energy eigenfunctions $\psi_k(x)$, that is, find the coefficients $c_k$ in $$\Psi(x,0) = \sum_{k=0}^\infty c_k \psi_k(x)$$ The initial wave function then evolves as $$\Psi(x,t) = \sum_{k=0}^\infty c_k \psi_k(x) e^{-iE_k t / \hbar}$$

 

[jimmycricket]

Thanks for the quick replies. I'm seriously lacking an understanding of this. Can you tell me what the coefficient $c_k$ corresponds to?

 

[jimmycricket]

is it correct that $c_k=\left\langle \psi_n |\psi \right\rangle$

 

[Orodruin]

Only if n = k.

 

[jimmycricket]

silly typo there. thanks

 

[Nugatory]

Each eigenstate of the Hamiltonian evolves independently and the evolution is trivially given by the time-dependent SE (if you know the eigenvalue, it becomes a very simple equation for the coefficient).

Provided that the potential is not also a function of time, right? Of course that will be the case for a very large number of very important problems.

 

[Orodruin]

Provided that the potential is not also a function of time, right? Of course that will be the case for a very large number of very important problems.

You are right of course, but based on the OP I would say we are dealing with a level where this is the case. It is also true in the adiabatic limit (where the Et in jt's post becomes an integral of E over time).

 

[jimmycricket]

So say If I'm given for example $\psi_n=\sqrt{\frac{2}{L}}sin(\frac{n\pi x}{L})$ what will $\left\langle \psi_n |\psi \right\rangle$ be?
Is it the same as $\left\langle \psi_n |\psi_n \right\rangle$
$$=\frac{2}{L}\int sin^2(\frac{n\pi x}{L})dx$$

Im not sure if this is turning into a homework help style question or not so please say if I should repost this elsewhere.

 

[jimmycricket]

oops think that last post is wrong as $\left\langle \psi_n |\psi_n \right\rangle=1$ since the wavefunction is normalized

 

[DrClaude]

$$
c_n = \langle \psi_n | \Psi \rangle = \int_{-\infty}^{\infty} \psi_n^*(x) \Psi(x) dx
$$