Quantum Mechanics: Momentum operator questions I'm asked to determine $\hat{P}|\Psi_0\rangle$, $\langle{\hat{P}}\rangle$, and $\langle\hat{P}^2\rangle$ for 
$$\Psi_0(u) = \psi_0 + 2\psi_1$$
I understand how to make the matrix for $P$ in regards to the original $\psi_0$ and $\psi_1$ functions, but I'm clueless for the $\Psi $ function that contains the two.
 A: You say that you know the results of
$$
\hat P \psi_0\\
\hat P \psi_1
$$
in which case simply write
$$
\Psi = \psi_0 + 2\psi_1\\
\hat P \Psi = \hat P \psi_0 + 2 \hat P \psi_1
$$
though you might need to check your normalisation.
PS. I'm not sure that I understood your question. You write that $\hat P$ is a matrix but $\psi$, $\Psi$ are functions; you seem to be mixing representations.
A: I will give a try at the answer. First I will say what can be said without knowing what $\psi_0$ and $\psi_1$ are. 
First I will normalize $\Psi_0$. $\Psi_0$ becomes $\frac{1}{\sqrt{5}}(\psi_0 + 2\psi_1)$. As innisfree says, we have 
$$\hat{P}\Psi_0 =\frac{1}{\sqrt{5}}(\hat{P}\psi_0 + 2\hat{P}\psi_1)  $$.
To get $\langle{\hat{P}}\rangle$, we just calculate 
\begin{equation}
\begin{aligned}
&\langle{\Psi_0|\hat{P}}|\Psi_0\rangle\\
=&\langle{\frac{1}{\sqrt{5}}(\psi_0 + 2\psi_1)|\hat{P}}|\frac{1}{\sqrt{5}}(\psi_0 + 2\psi_1)\rangle\\
=&\frac{1}{5}(\langle{\psi_0 |\hat{P}}|\psi_0\rangle +\langle{2\psi_1 |\hat{P}}|\psi_0\rangle +\langle{\psi_0|\hat{P}}|2\psi_1\rangle +\langle{2\psi_1 |\hat{P}}|2\psi_1\rangle)\\
=& \frac{1}{5}(\langle{\psi_0 |\hat{P}}|\psi_0\rangle +4\langle{\psi_1 |\hat{P}}|\psi_1\rangle+2 \mathrm{Re}(\langle{2\psi_1 |\hat{P}}|\psi_0\rangle))\\
\end{aligned}
\end{equation}
Similarly,
\begin{equation}
\begin{aligned}
&\langle{\Psi_0|\hat{P}^2}|\Psi_0\rangle\\
=& \frac{1}{5}(\langle\psi_0 |\hat{P}^2|\psi_0\rangle +4\langle\psi_1 |\hat{P}^2|\psi_1\rangle+2 \mathrm{Re}(\langle 2\psi_1 |\hat{P}^2|\psi_0\rangle))\\
\end{aligned}
\end{equation}
Now judging by your comment "$\psi_0$ and $\psi_1$ are wave functions similar to ${\frac{1}{i\pi} \exp(-u^2 /2)}$ or such". I will assume that the $\psi_i$ are eigenstates of the harmonic oscillator hamiltonian. For simplicity, let's take $m=\omega = \hbar = 1$.
Then we have a lowering operator $a=\frac{1}{\sqrt{2}}(\hat{X} + i \hat{P})$ and a raising operator $a^\dagger$. These operators act on the eigenstates $|n\rangle$ by $a |n\rangle = \sqrt{n} | n-1 \rangle$ and $a^\dagger |n\rangle = \sqrt{n+1} | n+1 \rangle$.
Then $\hat{P} |n \rangle = \frac{1}{i \sqrt{2}}(a-a^\dagger) = \frac{1}{i \sqrt{2}}(\sqrt{n} | n-1 \rangle - \sqrt{n+1} | n+1 \rangle)$.
Plugging this into our previous expressions, we find 
\begin{equation}
\begin{aligned}
&\hat{P}\Psi_0 \\
=&\frac{1}{\sqrt{5}}(\hat{P}\psi_0 + 2\hat{P}\psi_1)\\
=&\frac{1}{i\sqrt{10}}(-\psi_1 + 2\psi_0-2\sqrt{3}\psi_3).
\end{aligned}
\end{equation}
Now to find $\langle{\Psi_0|\hat{P}}|\Psi_0\rangle$, we can simply take the inner product between $\langle \Psi_0|$ and $\hat{P}|\Psi_0\rangle$. We find that the answer is $\frac{1}{i \sqrt{10}}(2-2)=0$.
Now to find $\langle{\Psi_0|\hat{P}^2}|\Psi_0\rangle$, we just need to take the inner product of $\hat{P}|\Psi_0\rangle$ with itself. We get $\frac{1}{10}(1+4+12)=\frac{17}{10}$.
