Is the Hamilton Density Diagonalizable?

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Homework Statement

I'm working with a complex scalar field with the lagrange density L= \partial_{\mu} \phi^{\ast} \partial^{\mu} \phi - m^2 \phi^{\ast} \phi And I've shown that's its hamilton density H= \int d^3 x ( \pi^{\ast} \pi + \nabla \phi^{\ast} \cdot \nabla \phi + m^2 \phi^{\ast} \phi )

Now the annihilation and creation operateurs are introduced. And I have to show that hamilton density H is diagonalizable by writing
\phi(x) = d^3 p \left( \frac{1}{(2 \pi)^3} \frac{1}{\sqrt{2E_p}} (a_{\vec{p}} e ^{-i p \cdot x} + b_{\vec{p}}^{\dagger} e^{ip \cdot x}) \right)

I also have to show that the theory contains two particles with mass m. How do I do this? I don't really know how to get started? Neither with how to show it is diagonalizable or how to find the masses:(

 

[gabbagabbahey]

Homework Statement

I'm working with a complex scalar field with the lagrange density L= \partial_{\mu} \phi^{\ast} \partial^{\mu} \phi - m^2 \phi^{\ast} \phi And I've shown that's its hamilton density H= \int d^3 x ( \pi^{\ast} \pi + \nabla \phi^{\ast} \cdot \nabla \phi + m^2 \phi^{\ast} \phi )

Now the annihilation and creation operateurs are introduced. And I have to show that hamilton density H is diagonalizable by writing
\phi(x) = \int d^3 p \left( \frac{1}{(2 \pi)^3} \frac{1}{\sqrt{2E_p}} (a_{\vec{p}} e ^{-i p \cdot x} + b_{\vec{p}}^{\dagger} e^{ip \cdot x}) \right)

Use this to calculate an expression for \phi^*(x) (Most physicists use {}^{\dagger} instead of {}^{*} for Hermitian conjugation ), \mathbf{\nabla}\phi(x), (\mathbf{\nabla}\phi(x))^* and then H. Carry out the integration over x[/itex], and you should get something like <br /> <br /> H=\int\int d^3pd^3p&amp;#039; f(\textbf{p},\textbf{p}&amp;#039;) \delta(\textbf{p}-\textbf{p}&amp;#039;)<br /> <br /> The presence of the delta function means that all non-diagonal elements vanish (since it is zero when \textbf{p}\neq\textbf{p}&amp;#039; ).