# Momentum Eigenfunction

1. Sep 19, 2015

### Safinaz

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

$\psi_1$ and $\psi_2$ are momentum eigenfunctions corresponding to
different momentum eigenvalues $p_1 \not= p_2$. Is $\psi_1$ + $\psi_2$ also momentum eigenfunction ?

2. Relevant equations

Yes
No
It Depends ?

3. The attempt at a solution

I think yes, because

$$\frac{h}{i} \frac{d}{dx} \psi_1 = p_1 \psi_1,$$
$$\frac{h}{i} \frac{d}{dx} \psi_2 = p_2 \psi_2,$$
Then
$$\frac{h}{i} \frac{d}{dx} (\psi_1+\psi_2) = p_1+p_2 (\psi_1+\psi_2),$$

is valid also

2. Sep 19, 2015

### blue_leaf77

Think again how you should calculate
$$\frac{h}{i} \frac{d}{dx} (\psi_1+\psi_2)$$
from
$$\frac{h}{i} \frac{d}{dx} \psi_1 = p_1 \psi_1,$$
$$\frac{h}{i} \frac{d}{dx} \psi_2 = p_2 \psi_2,$$

3. Sep 19, 2015

### Safinaz

Ya ..

$$\frac{h}{i} \frac{d}{dx} (\psi_1+\psi_2) = (p_1 \psi_1+ p_2 \psi_2),$$
so $(\psi_1+\psi_2)$ is not an Eigenfunction .

4. Sep 19, 2015

### blue_leaf77

No, it's not.