Second quantise the one-body kinetic energy operator


 * 1) Identify the one-body kinetic energy operator in terms of the mode energy \(\frac{\left(\hbar\mathbf{k}\right)^2}{2m}\hat{n}_{\mathbf{k}}\): \(\hat{T} = \int \frac{d\mathbf{k}}{\left(2\pi\right)^3}\frac{\left(\hbar\mathbf{k}\right)^2}{2m}\hat{n}_{\mathbf{k}}\)
 * 2) Take the Fourier transform of \(\hat{T}\): \(\hat{T} = \int \int \int dx dy \frac{d\mathbf{k}}{\left(2\pi\right)^3}\frac{\left(\hbar\mathbf{k}\right)^2}{2m}e^{-i\mathbf{k}(\mathbf{x} - \mathbf{y})}\hat{a}_{\mathbf{x}}^{\dagger}\hat{a}_{\mathbf{y}}\)
 * 3) Substitute in \(\partial_x \partial_y e^{-i\mathbf{k}(\mathbf{x} - \mathbf{y})} = \mathbf{k}^2 e^{-i\mathbf{k}(\mathbf{x} - \mathbf{y})}\) \(\hat{T} = \int \int \int dx dy \frac{d\mathbf{k}}{\left(2\pi\right)^3}\frac{\hbar^2}{2m}\partial_x \partial_y e^{-i\mathbf{k}(\mathbf{x} - \mathbf{y})}\hat{a}_{\mathbf{x}}^{\dagger}\hat{a}_{\mathbf{y}}\)
 * 4) Integrate by parts over \(k\): \(\hat{T} = \int \int dx dy \frac{\hbar^2}{2m}\partial_x\hat{a}_{\mathbf{x}}^{\dagger} \partial_y \hat{a}_{\mathbf{y}}\delta(\mathbf{x}-\mathbf{y})\)
 * 5) Integrate over \(y\): \(\hat{T} = \int dx \frac{\hbar^2}{2m}\partial_x\hat{a}_{\mathbf{x}}^{\dagger} \partial_x \hat{a}_{\mathbf{x}}\)
 * 6) Integrate by parts to simplify: \(\hat{T} = -\int dx \frac{\hbar^2}{2m}\hat{a}_{\mathbf{x}}^{\dagger}\partial_x^2 \hat{a}_{\mathbf{x}}\)
 * 7) Substitute \(\hat{p} = i\hbar\partial_x\): \(\hat{T} = \int dx \frac{1}{2m}\hat{a}_{\mathbf{x}}^{\dagger}\hat{p}^2 \hat{a}_{\mathbf{x}}\)