# Peskin and Schroeder - Derivation of equation (2.45)

1. Oct 20, 2015

### spaghetti3451

I'm having trouble deriving equation (2.45) on page 25. In particular, in the derivation of

$i\frac{\partial}{\partial t}\pi({\bf{x}},t) = -i(-\nabla^{2}+m^{2}) \phi({\bf{x}},t)$,

I need to show that

$\frac{1}{2}\pi({\bf{x}},t) \phi({\bf{x'}},t)(-\nabla^{2}+m^{2}) \phi({\bf{x'}},t) - \frac{1}{2} \phi({\bf{x'}},t)(-\nabla^{2}+m^{2})\phi({\bf{x'}},t)\pi({\bf{x}},t) = -i \delta^{(3)}({\bf{x}}-{\bf{x'}})(-\nabla^{2}+m^{2})\phi({\bf{x'}},t)$.

Now, this is what I've done so far:

$\frac{1}{2}\pi({\bf{x}},t) \phi({\bf{x'}},t)(-\nabla^{2}+m^{2}) \phi({\bf{x'}},t) - \frac{1}{2} \phi({\bf{x'}},t)(-\nabla^{2}+m^{2})\phi({\bf{x'}},t)\pi({\bf{x}},t)$

$=\frac{1}{2}\pi({\bf{x}},t) \phi({\bf{x'}},t)(-\nabla^{2}+m^{2}) \phi({\bf{x'}},t)-\frac{1}{2}\phi({\bf{x'}},t)\pi({\bf{x}},t) (-\nabla^{2}+m^{2}) \phi({\bf{x'}},t) +\frac{1}{2}\phi({\bf{x'}},t)\pi({\bf{x}},t) (-\nabla^{2}+m^{2}) \phi({\bf{x'}},t)- \frac{1}{2} \phi({\bf{x'}},t)(-\nabla^{2}+m^{2})\phi({\bf{x'}},t)\pi({\bf{x}},t)$

$=\frac{1}{2}[\pi({{\bf{x}},t}), \phi({{\bf{x'}},t})](-\nabla^{2}+m^{2}) \phi({\bf{x'}},t) + \frac{1}{2}\phi({\bf{x'}},t)[\pi({\bf{x}},t), (-\nabla^{2}+m^{2}) \phi({\bf{x'}},t)]$

$=\frac{1}{2}(-i)\delta^{(3)}({\bf{x'}}-{\bf{x}})(-\nabla^{2}+m^{2}) \phi({\bf{x'}},t) + ??$.

How do I evaluate the second commutator?

2. Oct 23, 2015

### spaghetti3451

bumpp!

3. Oct 23, 2015

### bapowell

Did you try breaking the second commutator up as $[\pi,m^2\phi] - [\pi,\nabla^2\phi]$ and convincing yourself that both the $m^2$ and $\nabla^2$ can be pulled out of their respective commutators?