Something about all this stuff I am not getting. Given the following wave function
$$\psi = Ne^{\frac{-x^2}{\Delta^2}}e^{ik_0 x}$$
Where $N$ is the normalization constant, $N= \left(\frac{1}{\Delta\sqrt{\pi}}\right)^{1/2}$. I need to find the average momentum, $\langle p\rangle$, two ways.
$$
\langle p\rangle = \int_{-\infty}^{\infty}~\mathrm dx\psi^*\left(-i\hbar\frac{\mathrm d}{\mathrm dx}\right)\psi=\int_{-\infty}^{\infty}\frac{~\mathrm dp}{2\pi}|\phi|^2p
$$
Question: What is this second integral? I do not know what $\phi$ is. Any help for the context of this would be greatly appreciated. I do not know where to start with that integral.
Solving the first integral: For the first integral, which is using the momentum operator. I evaluated the derivative of the wave function first.
$$\frac{\mathrm d}{\mathrm dx}e^{-\frac{x^2}{\Delta^2}+ik_0x}=\left(\frac{-2x}{\Delta^2}+ik_0\right)e^{-\frac{x^2}{\Delta^2}+ik_0x}$$
Taking note that $\psi^*\psi$ removes the complex part and multiplies a 2 to the real part of the exponent. I should get something like this:
$$\langle p\rangle=(-i\hbar)N^2\int_{-\infty}^{\infty}~\mathrm dx \left(\frac{-2x}{\Delta^2}+ik_0\right)e^{-\frac{2x^2}{\Delta^2}}$$
Which I can split into two integrals. One of the integrals is odd and evaluates to $0$. The other remains.
$$=(-i\hbar)N^2\int_{-\infty}^{\infty}~\mathrm dx(ik_0)e^{-\frac{2x^2}{\Delta^2}}$$
Which is a the Gaussian integral and the solution to this is
$$\begin{align}
&=N^2k_0\sqrt{\frac{\pi\Delta}{2}}=\frac{k_0}{\Delta\sqrt{\pi}}\sqrt{\frac{\pi\Delta}{2}}\\
&=\sqrt{\frac{k_0^2}{2\Delta}}=\langle p\rangle\end{align}
$$
Best Answer
$\phi$ is the wave function in momentum space: If $\psi(x)=\langle x|\Psi\rangle$, then $\phi(p)=\langle p|\Psi\rangle$. This is why the momentum operator can be evaluated directly: $$\hat p|\phi\rangle = p|\phi\rangle$$ and the integration is going over the momenta.
The calculation of the first integral looks fine to me upon a quick look, though I didn't check all the details.