quantum mechanics - Free particle propagation amplitude (Peskin & Schroeder 2.1)

In section 2.1 on p. 14 of Peskin and Schroeder QFT, they calculate the amplitude of a free non-relativistic particle to propagate from $\mathbf{x_0}$ to $\mathbf{x}$ as $$U(t) = \left<\mathbf{x}|e^{-i\left(\mathbf{p}^2/2m\right)t}|\mathbf{x_0}\right>$$ $$ = \int \frac{d^3p}{\left( 2\pi\right)^3} \left<\mathbf{x}|e^{-i\left(\mathbf{p}^2/2m\right)t}|\mathbf{p}\right>\left<\mathbf{p}|\mathbf{x_0}\right>$$ $$ = \frac{1}{\left( 2\pi\right)^3} \int d^3p \ e^{-i(\mathbf{p}^2/2m)t} e^{i\mathbf{p}\cdot (\mathbf{x}-\mathbf{x_0})}$$ $$ = \ ...$$

using the nonrelativistic Hamiltonian $\mathbf{H} = \mathbf{p}^2/2m$. I think in going from the first to second line, they just inserted the identify operator as a sum over all the momentum projection operators, since the momentum eigenstates form a complete set. Could someone please explain the step from the second line to the third line?

Answer

The transition from the 2nd line to the 3rd simply uses the fact that

$$\langle \mathbf{p}|\mathbf{x} \rangle = e^{-i \mathbf{p}\cdot\mathbf{x}}.$$

The exponent of the Hamiltonian merely picks up the eigenvalue $\mathbf{p}$ from the state $|\mathbf{p}\rangle$, so you can take it out of the inner product, and what you have left is

$$\langle \mathbf{x}|\mathbf{p} \rangle \langle \mathbf{p}|\mathbf{x}_0 \rangle = e^{i \mathbf{p}\cdot(\mathbf{x}-\mathbf{x}_0)},$$

since switching the order of states in the inner product gives you the complex conjugate, that is,

$$\langle \mathbf{x}|\mathbf{p} \rangle = \langle \mathbf{p}|\mathbf{x} \rangle^*.$$

Let me know if this is clear.