# Quantum mechanics, wavefunction problem

Gold Member

## Homework Statement

Consider the wavefunction $\Psi (x,t)=c_1 \psi _1 (x)e^{-\frac{iE_1t}{\hbar}}+c_2 \psi _2 (x)e^{-\frac{iE_2t}{\hbar}}$ where $\psi _1 (x)$ and $\psi _2 (x)$ are normalized and orthogonal. Knowing $\Psi (x,0)$, find the values of $c_1$ and $c_2$.

## Homework Equations

$C^2 \int _{-\infty}^{\infty} |\Psi (x,t)|^2 dx=1$. I also know that the product of psi 1 by psi 2 is worth 0 (they are orthogonal) so this simplifies the expression to integrate.
But I'm still left with the integration of both lower case psi functions that I don't know how to handle.

## The Attempt at a Solution

I'm thinking on how to use the fact that I know Psi at t=0 but so far I'm out of ideas.

ehild
Homework Helper
$\psi _i (x)$-s are orthonormal, what does it mean on the integrals of $\psi _i ^* (x) \psi _j (x)$?

ehild

Gold Member
$\psi _i (x)$-s are orthonormal, what does it mean on the integrals of $\psi _i ^* (x) \psi _j (x)$?

ehild

Sorry for being almost 1 month late. That they are worth 1?

Edit: Is what I've done right?:
$\int _{-\infty}^{\infty} \Psi _1 (x) \Psi (x,0)dx=c_1 \int _{-\infty}^{\infty} \Psi _1 ^2 (x)dx=c_1$.
In the same fashion I get $c_2=\int _{-\infty}^{\infty} \Psi _2 (x) \Psi (x,0)dx$.
But I didn't use any complex conjugate... I must be missing something.

Last edited:
vela
Staff Emeritus
Homework Helper
That's almost correct. It should be
$$c_i = \int \psi_i^*(x)\Psi(x,0)\,dx$$When you calculate the inner product, you use the complex conjugate of the first function in the integrand.

Gold Member
That's almost correct. It should be
$$c_i = \int \psi_i^*(x)\Psi(x,0)\,dx$$When you calculate the inner product, you use the complex conjugate of the first function in the integrand.

Let's see if I understand:
$\int _{- \infty}^{\infty} \Psi _1 ^* (x) \Psi (x,0)dx=c_1 \int _{- \infty}^{\infty} \Psi _1 ^* (x) \Psi _1 (x)dx=c_1 \int _{-\infty} ^{\infty} |\Psi _1 (x)|^2 dx=c_1$.
Am I right if I say that since $\Psi _1(x)$ and $\Psi _2 (x)$ are orthogonal, so is $\Psi _1^* (x)$ with $\Psi _2 (x)$? Because I'm using this fact!

vela
Staff Emeritus
Homework Helper
Yup. When you say the two are orthogonal, you mean that
$$\int \psi_1^*(x)\psi_2(x)\,dx = 0.$$It does not mean that
$$\int \psi_1(x)\psi_2(x)\,dx = 0.$$The complex conjugation is a vital part of taking the inner product.

Gold Member
Yup. When you say the two are orthogonal, you mean that
$$\int \psi_1^*(x)\psi_2(x)\,dx = 0.$$It does not mean that
$$\int \psi_1(x)\psi_2(x)\,dx = 0.$$The complex conjugation is a vital part of taking the inner product.
Oh, I understand. Wow, thank so you much. My understanding was so poor before I asked help for this problem... now things are getting clearer.