Free particle propagation amplitude (Peskin & Schroeder 2.1)

Question

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 1

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.