Hamiltonian density from Klein-Gordon field

Question

In the solution for Peskin & Shroeder 2.2 where the Hamiltonian density obtained from the Klein-Gordon Lagrangian is given by:
$$
 H = pi^* pi + nabla phi cdot nabla phi^* + m^2 phi^* phi 
 =  pi^* pi + phi ^ * (- nabla ^ 2 + m^2 ) phi
$$
Im confused with how the last equality was obtained. The notes said they used integration by parts and canceling out the surface term. Using the product rule
$$ 
 nabla phi cdot nabla phi^* = frac{1}{2} (nabla ^2(phi phi^*) - phi nabla^2 phi^* - phi^* nabla^2 phi) 
$$
Besides the surface term $nabla^2(phi phi^*)$, I have an extra term which does not seem to evaluate to zero here

t_sanjana · Accepted Answer

So, part (a) of problem 2.2 in P&S asks you to find the expression of the Hamiltonian (which you already found):$$H=int d^3x ,mathcal{H}=int d^3x,(pi^*pi+nablaphi^*cdotnablaphi+m^2phi^*phi).$$To get to the next expression you mentioned, we use the chain rule (Einstein summation convention is assumed)$$nablaphi^*cdotnablaphi=(partial_iphi^*)(partial_iphi)=partial_i(phi^*partial_iphi)-phi^*(nabla^2phi)$$Now, substituting this expression in the Hamiltonian, we find that$$begin{align*}H=&int d^3x,(pi^*pi+(partial_i(phi^*partial_iphi)-phi^*(nabla^2phi))+m^2phi^*phi) =&int d^3x,(pi^*pi-phi^*(nabla^2phi)+m^2phi^*phi)+int d^3x;partial_i(phi^*partial_iphi) =&int d^3x,(pi^*pi-phi^*(nabla^2phi)+m^2phi^*phi)end{align*}$$Note that the last term in step two is a surface term whose integral goes to zero. The integral of the surface term goes to zero because $phi,phi^*rightarrow0$ as $xrightarrowinfty$ which is a necessary condition for integrals like $int_{text{all spacetime}} d^3x,phi^*phi$ (which are clearly present in the Hamiltonian and the Lagrangian) to make sense.
Therefore you have the desired expression for the Hamiltonian density:$$mathcal{H}=pi^*pi+phi^*(-nabla^2+m^2)phi$$
Hope this clarified.

Hamiltonian density from Klein-Gordon field

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