Compute the Momentum of the Wave Function

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}

$\phi$ 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.
• Thank you, you answer has been very helpful. What would $\phi$ be with this wave equation? How do I go from position space to momentum space? – Tsangares Oct 16 '16 at 6:50
• With a Fourier transform, which you can see by inserting a unity operator expanded in $x$ basis: $$\psi(p) = \langle p|\Psi\rangle = \int d^3x\langle p|x\rangle\langle x|\Psi\rangle = \int d^3 x \exp(ipx)\psi(x)$$ since $$\langle p|x\rangle = \exp(ipx)$$ – Photon Oct 16 '16 at 7:01
• A lot of stuff is starting to make sense. Is $\Psi$ the time dependent wave function? – Tsangares Oct 16 '16 at 21:13
• $\Psi$ is not a wave function, it is a ket, $\psi(x)$ is a wave function. Since in this particular exercise $\psi(x)$ is explicitly given and is time independent, the answer is: No, it is not time dependent. But in principle it could also have been time dependent, the calculation would work the same way. – Photon Oct 16 '16 at 21:18