I'm just learning the Brakets formalisms for QM and I'm having struggles solving a simple problem.
For an harmonic oscillator, particularly Griffiths' Introduction to Quantum Mechanics P3.34:
I want to measure the expected value of momentum $p$ as:
$$\langle p \rangle = \langle \Psi | p | \Psi \rangle$$
considering the wavefunction
$$\Psi(x,t) = \sum\limits_{n=0}^{1} c_n \,\psi_n \,e^{-iE_nt/\hbar}$$
so, my first thought was to insert $\Psi$ in $\langle p \rangle$ as:
$$\langle c_0 \,\psi_0 \,e^{-iE_0t/\hbar} + c_1 \,\psi_1 \,e^{-iE_1t/\hbar} \,|\, p \,|\, c_0 \,\psi_0 \,e^{-iE_0t/\hbar} + c_1 \,\psi_1 \,e^{-iE_1t/\hbar} \rangle$$
but I recognize this is too much "brute force" and clearly shows to me that I don't understand well how to calculate with bras and kets operations (and also what's the benefit of this).
Following my lecturer I understood these being eigenvalues and eigenvectors of $\psi$ respectively so I think I can treat the operation as an inner product (?) pulling the coefficients outside the operation respecting the order when $c_i^* c_j$ products appear.
Anyways, honestly, I don't see the obvious: how should I proceed in a practical way? Why does the outcome has the form of a product? Something like
$$(c_0^* \langle \psi_0|p|\; {e^{-iE_0 t/\hbar}}^* + c_1^* {e^{-iE_1 t/\hbar}}^* \; \langle \psi_1|p|)(c_0 |\psi_0\rangle e^{-iE_0 t/\hbar} + c_1 |\psi_1 \rangle e^{-iE_1 t/\hbar})$$
D.J Griffiths himself states that:
I'm aware my reasoning is not correct and I don't want to bother anyone with the question. I'm just a little confused about it and wanting to understand more.
EDIT: Following what JEB
and Cosmas Zachos
are suggesting:
since $\Psi$ can be represented as
$$|\Psi \rangle = \frac{1}{\sqrt{2}} [|0\rangle + e^{i\phi}|1\rangle] \equiv \frac{1}{\sqrt{2}} \begin{pmatrix} \psi_0 \\ \psi_1 e^{i\phi} \end{pmatrix}$$
and the momentum expected value is $\langle \Psi | \hat{p} | \Psi \rangle$ one can write
$$\langle \Psi | = (|\Psi\rangle)^{\dagger} = \frac{1}{\sqrt{2}}[\langle 0|+e^{-i\phi}\langle 1 |]$$
then
$$\langle \Psi | \hat{p} | \Psi \rangle = \frac{1}{2} [\langle0| +e^{-i\phi}\langle 1 | p | 0 \rangle + e^{i\phi} |1\rangle]$$
being $\hat{p} = i\sqrt{\frac{\hbar m \omega}{2}}(\hat{a_+}-\hat{a_{-}})$ so
$$\langle \Psi | \hat{p} | \Psi \rangle = 1/2 \, i\sqrt{\frac{\hbar m \omega}{2}}[\langle 0 | + e^{-i\phi} \langle 1 | \Big| \hat{a_+} |0\rangle + \hat{a_+} e^{i\phi} |1\rangle - \hat{a_{-}}|0\rangle - \hat{a_{-}}e^{i\phi} |1\rangle]$$
then distribuiting the bras to the resulting kets by the right:
$$ = 1/2\, i\sqrt{\frac{\hbar m \omega}{2}} ( \langle 0 |(\hat{a_+} |0\rangle + \hat{a_+} e^{i\phi} |1\rangle - \hat{a_{-}}|0\rangle - \hat{a_{-}}e^{i\phi} |1\rangle) + e^{-i\phi} \langle 1| (\hat{a_+} |0\rangle + \hat{a_+} e^{i\phi} |1\rangle - \hat{a_{-}}|0\rangle - \hat{a_{-}}e^{i\phi} |1\rangle) )$$
Now all the raising and lowering operators act on the kets next to them, following
$$\hat{a} |n\rangle = \sqrt{n} |n-1\rangle$$ $$\hat{a}^{\dagger} |n \rangle = \sqrt{n+1} |n+1 \rangle$$
and I get inner products of the states $\psi_0$, $\psi_1$ and $\psi_2$ ponderated by $\sqrt{n}$ and $\sqrt{n+1}$.
This results in:
$$\langle p \rangle = \frac{1}{2} \sqrt{\frac{m\omega \hbar}{2}}i [\langle 0 | 1 \rangle + e^{i\phi}\langle 0| 2\rangle - e^{i\phi}\langle 0| 0\rangle +e^{-i\phi} \langle 1|1 \rangle +\sqrt{2} \langle 1|2 \rangle - \langle 1|0 \rangle]$$
What should I do next?
Being the states represented by an orthonormal basis, the inner product $\psi_n^*\psi_{n'}$ is 0 if $n \neq n'$? i.e.,
$$\langle p \rangle = \frac{1}{2} \sqrt{\frac{m\omega \hbar}{2}}i [\langle 0 | 1 \rangle + e^{i\phi}\langle 0| 2\rangle - e^{i\phi}\langle 0| 0\rangle +e^{-i\phi} \langle 1|1 \rangle +\sqrt{2} \langle 1|2 \rangle - \langle 1|0 \rangle] = \frac{1}{2} \sqrt{\frac{m\omega \hbar}{2}}i [0 + 0 - e^{i\phi}\langle 0| 0\rangle +e^{-i\phi} \langle 1|1 \rangle +0 - 0] $$