Calculating the Hamiltonian of complex scalar field

Question

I am working on the Peskin's QFT problems, and i'm finding difficulties in calculating the Hamiltonian of complex scalar field. The Hamiltonian read: $$H = \int d^3x (\pi^*\pi+\nabla\phi^*\nabla\phi+m^2\phi^*\phi)$$ Then I substitute the expression of $\phi$ and $\phi^*$ into it: $$\phi(x)=\int\frac{d^3p}{(2\pi)^3}\frac 1{\sqrt{2E_p}}(a_pe^{-ipx}+b_p^+e^{ipx})$$ $$\phi^*(x)=\int\frac{d^3p}{(2\pi)^3}\frac 1{\sqrt{2E_p}}(b_pe^{-ipx}+a_p^+e^{ipx})$$ Integrate over $x$ and get the delta function, thereby cancel one of the integral variable q, I get: $$H=\int\frac{d^3p}{(2\pi)^3}[(-\frac{E_p}2+p^2+\frac{m^2}{2E_p})(b_{-p}a_pe^{-i2E_pt}+a_p^+b_{-p}e^{i2E_pt})+(\frac{E_p}2+p^2+\frac{m^2}{2E_p})(b_{p}b_{p}^++a_p^+a_p)]$$ I don't know how to continue. In some answer book I find the next step change the integral variable back to $x$, and get $\int d^3x\frac{E_p^2+p^2+m^2}{2E_p}(b_{p}b_{p}^++a_p^+a_p)$ but I don't know how and why.

Answer 1

I'm using $1\over2\omega_p$ convention which is basically equivalent to the convention that you have used.

First of all, we have to define each variable we are going to use.

$$\phi(x,t)=\int \frac{d^3p}{(2\pi)^32\omega_p}(a(\vec{p})e^{-ip\cdot x}+b^\dagger(\vec{p})e^{ip\cdot x})$$

$$\bar{\phi}(x,t)=\int \frac{d^3p}{(2\pi)^32\omega_p}(b(\vec{p})e^{-ip\cdot x}+a^\dagger(\vec{p})e^{ip\cdot x})$$

$$\pi_\phi(x,t)=\int \frac{d^3p}{(2\pi)^32\omega_p}(-i\omega_p)(b(\vec{p})e^{-ip\cdot x}-a^\dagger(\vec{p})e^{ip\cdot x})$$

$$\pi_\bar{\phi}(x,t)=\int \frac{d^3p}{(2\pi)^32\omega_p}(-i\omega_p)(a(\vec{p})e^{-ip\cdot x}-b^\dagger(\vec{p})e^{ip\cdot x})$$

$$\nabla\phi(x,t)=\int \frac{d^3p}{(2\pi)^32\omega_p}(i\vec{p})(a(\vec{p})e^{-ip\cdot x}-b^\dagger(\vec{p})e^{ip\cdot x})$$

$$\nabla\bar{\phi}(x,t)=\int \frac{d^3p}{(2\pi)^32\omega_p}(i\vec{p})(b(\vec{p})e^{-ip\cdot x}-a^\dagger(\vec{p})e^{ip\cdot x})$$

Now that we have defined our variables lets substitute this into the equation for the hamiltonian $$H = \int d^3x (\pi^*\pi+\nabla\phi^*\nabla\phi+m^2\phi^*\phi)$$

$$H=\int d^3x \frac{d^3p}{(2\pi)^32\omega_p} \frac{d^3p'}{(2\pi)^32\omega_{p'}}[(-i\omega_{p}\cdot-i\omega_{p'})(b(\vec{p})e^{-ip\cdot x}-a^\dagger(\vec{p})e^{ip\cdot x})(a(\vec{p}')e^{-ip'\cdot x}-b^\dagger(\vec{p}')e^{ip'\cdot x})+(i\vec{p}\cdot i\vec{p}')(a(\vec{p})e^{-ip\cdot x}-b^\dagger(\vec{p})e^{ip\cdot x})(b(\vec{p}')e^{-ip'\cdot x}-a^\dagger(\vec{p}')e^{ip'\cdot x})+m^2(b(\vec{p}')e^{-ip\cdot x}+a^\dagger(\vec{p})e^{ip\cdot x})(b(\vec{p})e^{-ip'\cdot x}+a^\dagger(\vec{p}')e^{ip'\cdot x})]$$

Now we can expand and simplify the coefficients

$$H=\int d^3x \frac{d^3p}{(2\pi)^32\omega_p} \frac{d^3p'}{(2\pi)^32\omega_{p'}}[(-\omega_{p}\omega_{p'})(b(\vec{p})a(\vec{p}')e^{-i(p+p')\cdot x}-b(\vec{p})b^\dagger(\vec{p}')e^{-i(p-p')\cdot x}-a^\dagger(\vec{p})a(\vec{p}')e^{i(p-p')\cdot x}+a^\dagger(\vec{p})b^\dagger(\vec{p}')e^{i(p+p')\cdot x})-(p\cdot p')(a(\vec{p})b(\vec{p}')e^{-i(p+p')\cdot x}-a(\vec{p})a^\dagger(\vec{p}')e^{-i(p-p')\cdot x}-b^\dagger(\vec{b})b(\vec{p}')e^{i(p-p')\cdot x}+b^\dagger(\vec{p})a^\dagger(\vec{p}')e^{i(p+p')\cdot x})+m^2(a(\vec{p})b(\vec{p}')e^{-i(p+p')\cdot x}+a(\vec{p})a^\dagger(\vec{p}')e^{-i(p-p')\cdot x}+b^\dagger(\vec{b})b(\vec{p}')e^{i(p-p')\cdot x}+b^\dagger(\vec{p}')a^\dagger(\vec{p}')e^{i(p+p')\cdot x})]$$

Now, by using the identities $$\int d^3x e^{-i(p-p')\cdot x}=\int d^3x e^{-i(\omega_p-\omega_{p'})t} e^{-i(\vec{p}-\vec{p}')\cdot \vec {x}}=e^{-i(\omega_p-\omega_{p'})t}(2\pi)^3 \delta^3(\vec{p}-\vec{p}')=(2\pi)^3 \delta^3(\vec{p}-\vec{p}')$$

$$\int d^3x e^{-i(p+p')\cdot x}=\int d^3x e^{-i(\omega_p+\omega_{p'})t} e^{-i(\vec{p}-\vec{p}')\cdot \vec {x}}=e^{-i(\omega_p+\omega_{p'})t}(2\pi)^3 \delta^3(\vec{p}+\vec{p}')=e^{-2i\omega_pt}(2\pi)^3 \delta^3(\vec{p}+\vec{p}') $$

$$\int d^3x e^{i(p-p')\cdot x}=\int d^3x e^{i(\omega_p-\omega_{p'})t} e^{i(\vec{p}-\vec{p}')\cdot \vec {x}}=e^{i(\omega_p-\omega_{p'})t}(2\pi)^3 \delta^3(\vec{p}-\vec{p}')=(2\pi)^3 \delta^3(\vec{p}-\vec{p}')$$

$$\int d^3x e^{i(p+p')\cdot x}=\int d^3x e^{i(\omega_p+\omega_{p'})t} e^{i(\vec{p}-\vec{p}')\cdot \vec {x}}=e^{i(\omega_p+\omega_{p'})t}(2\pi)^3 \delta^3(\vec{p}+\vec{p}')=e^{2i\omega_pt}(2\pi)^3 \delta^3(\vec{p}+\vec{p}') $$

We can now do the x integral and we get

$$H=\int \frac{d^3p}{(2\pi)^32\omega_p} \frac{d^3p'}{(2\pi)^32\omega_{p}}[(-\omega_{p}\omega_{p'})(b(\vec{p})a(\vec{p}') e^{-2i\omega_pt} (2\pi)^3 \delta^3 (\vec{p}+\vec{p}')- b(\vec{p})b^\dagger(\vec{p}')(2\pi)^3 \delta^3(\vec{p}-\vec{p}')-a^\dagger(\vec{p})a(\vec{p}')(2\pi)^3 \delta^3(\vec{p}-\vec{p}')+a^\dagger(\vec{p})b^\dagger(\vec{p}')e^{2i\omega_pt} (2\pi)^3 \delta^3(\vec{p}+\vec{p}'))-(p\cdot p')(a(\vec{p})b(\vec{p}')e^{-2i\omega_pt}(2\pi)^3 \delta^3(\vec{p}-\vec{p}')-a(\vec{p})a^\dagger(\vec{p}')(2\pi)^3 \delta^3(\vec{p}-\vec{p}')-b^\dagger(\vec{b})b(\vec{p}')(2\pi)^3 \delta^3(\vec{p}-\vec{p}')+b^\dagger(\vec{p})a^\dagger(\vec{p}')e^{2i\omega_pt}(2\pi)^3 \delta^3(\vec{p}+\vec{p}'))+m^2(a(\vec{p})b(\vec{p}')e^{-2i\omega_pt}(2\pi)^3 \delta^3 (\vec{p}+\vec{p}')+a(\vec{p})a^\dagger(\vec{p}')(2\pi)^3 \delta^3(\vec{p}-\vec{p}')+b^\dagger(\vec{b})b(\vec{p}')(2\pi)^3 \delta^3(\vec{p}-\vec{p}')+b^\dagger(\vec{p}')a^\dagger(\vec{p}')e^{2i\omega_pt}(2\pi)^3 \delta^3(\vec{p}+\vec{p}'))]$$

We can now factor out a $(2\pi)^3$ and use the delta functions to do the $p'$ integral

Remember $\omega_{-p}=\sqrt{(-p)^2+(m)^2}=\omega_{p}$

$$H=\int \frac{d^3p}{(2\pi)^32\omega_p} \frac{1}{(2\pi)^32\omega_{p}} (2\pi)^3 [(-\omega_{p}^2)(b(\vec{p})a(-\vec{p}) e^{-2i\omega_pt} - b(\vec{p})b^\dagger(\vec{p}) -a^\dagger(\vec{p})a(\vec{p})+a^\dagger(\vec{p})b^\dagger(-\vec{p})e^{2i\omega_pt})-(p\cdot p)(-a(\vec{p})b(-\vec{p})e^{-2i\omega_pt}-a(\vec{p})a^\dagger(\vec{p}) -b^\dagger(\vec{b})b(\vec{p})-b^\dagger(\vec{p})a^\dagger(-\vec{p})e^{2i\omega_pt})+m^2(a(\vec{p})b(-\vec{p})e^{-2i\omega_pt} +a(\vec{p})a^\dagger(\vec{p})+b^\dagger(\vec{b})b(\vec{p})+b^\dagger(\vec{p}')a^\dagger(-\vec{p})e^{2i\omega_pt})]$$

Multiply the through by $-1$

$$H=\int \frac{d^3p}{(2\pi)^32\omega_p} \frac{1}{(2\pi)^32\omega_{p}} (2\pi)^3 [(-\omega_{p}^2)(b(\vec{p})a(-\vec{p}) e^{-2i\omega_pt} - b(\vec{p})b^\dagger(\vec{p}) -a^\dagger(\vec{p})a(\vec{p})+a^\dagger(\vec{p})b^\dagger(-\vec{p})e^{2i\omega_pt})+(p\cdot p)(a(\vec{p})b(-\vec{p})e^{-2i\omega_pt}+a(\vec{p})a^\dagger(\vec{p}) +b^\dagger(\vec{b})b(\vec{p})+b^\dagger(\vec{p})a^\dagger(-\vec{p})e^{2i\omega_pt})+m^2(a(\vec{p})b(-\vec{p})e^{-2i\omega_pt} +a(\vec{p})a^\dagger(\vec{p})+b^\dagger(\vec{b})b(\vec{p})+b^\dagger(\vec{p}')a^\dagger(-\vec{p})e^{2i\omega_pt})]$$

We see that $p \cdot p$ and $m^2$ have the same coefficient so we can factor

$$H=\int \frac{d^3p}{(2\pi)^32\omega_p} \frac{1}{(2\pi)^32\omega_{p}} (2\pi)^3 [(-\omega_{p}^2)(b(\vec{p})a(-\vec{p}) e^{-2i\omega_pt} - b(\vec{p})b^\dagger(\vec{p}) -a^\dagger(\vec{p})a(\vec{p})+a^\dagger(\vec{p})b^\dagger(-\vec{p})e^{2i\omega_pt})+(p\cdot p+m^2)(a(\vec{p})b(-\vec{p})e^{-2i\omega_pt}+a(\vec{p})a^\dagger(\vec{p}) +b^\dagger(\vec{b})b(\vec{p})+b^\dagger(\vec{p})a^\dagger(-\vec{p})e^{2i\omega_pt})]$$

We can simplify $(p\cdot p+m^2)$ into $\omega_p^2$ $$H=\int \frac{d^3p}{(2\pi)^32\omega_p} \frac{1}{(2\pi)^32\omega_{p}} (2\pi)^3 [(-\omega_{p}^2)(b(\vec{p})a(-\vec{p}) e^{-2i\omega_pt} - b(\vec{p})b^\dagger(\vec{p}) -a^\dagger(\vec{p})a(\vec{p})+a^\dagger(\vec{p})b^\dagger(-\vec{p})e^{2i\omega_pt})+\omega_p^2(a(\vec{p})b(-\vec{p})e^{-2i\omega_pt}+a(\vec{p})a^\dagger(\vec{p}) +b^\dagger(\vec{b})b(\vec{p})+b^\dagger(\vec{p})a^\dagger(-\vec{p})e^{2i\omega_pt})]$$

Expand the $-1$ in the first part of the integral and factor out the $\omega_p^2$ $$H=\int \frac{d^3p}{(2\pi)^32\omega_p} \frac{1}{(2\pi)^32\omega_{p}} (2\pi)^3 \omega_{p}^2 [a^\dagger(\vec{p})a(\vec{p})+ b(\vec{p})b^\dagger(\vec{p}) - b(\vec{p})a(-\vec{p}) e^{-2i\omega_pt} - a^\dagger(\vec{p})b^\dagger(-\vec{p})e^{2i\omega_pt}+a(\vec{p})b(-\vec{p})e^{-2i\omega_pt}+a(\vec{p})a^\dagger(\vec{p}) +b^\dagger(\vec{b})b(\vec{p})+b^\dagger(\vec{p})a^\dagger(-\vec{p})e^{2i\omega_pt}]$$

$$H=\int \frac{d^3p}{(2\pi)^32\omega_p} \frac{1}{(2\pi)^32\omega_{p}} (2\pi)^3 \omega_{p}^2 [a^\dagger(\vec{p})a(\vec{p})+a(\vec{p})a^\dagger(\vec{p}) + b(\vec{p})b^\dagger(\vec{p}) +b^\dagger(\vec{b})b(\vec{p}) + b^\dagger(\vec{p})a^\dagger(-\vec{p})e^{2i\omega_pt} - a^\dagger(\vec{p})b^\dagger(-\vec{p})e^{2i\omega_pt}+a(\vec{p})b(-\vec{p})e^{-2i\omega_pt}- b(\vec{p})a(-\vec{p}) e^{-2i\omega_pt} ]$$

$$H=\int \frac{d^3p}{(2\pi)^32\omega_p} \frac{1}{(2\pi)^32\omega_{p}} (2\pi)^3 \omega_{p}^2 [a^\dagger(\vec{p})a(\vec{p})+a(\vec{p})a^\dagger(\vec{p}) + b(\vec{p})b^\dagger(\vec{p}) +b^\dagger(\vec{b})b(\vec{p}) + a^\dagger(-\vec{p})b^\dagger(\vec{p})e^{2i\omega_pt} - a^\dagger(\vec{p})b^\dagger(-\vec{p})e^{2i\omega_pt}+a(\vec{p})b(-\vec{p})e^{-2i\omega_pt}- a(-\vec{p})b(\vec{p}) e^{-2i\omega_pt} ]$$

We can change variables for $a(-\vec{p})b(\vec{p}) e^{-2i\omega_pt}$ and $a^\dagger(\vec{p})b^\dagger(-\vec{p})e^{2i\omega_pt}$ to $\vec{p} \rightarrow -\vec{p}$

$$H =\int \frac{d^3p}{(2\pi)^32\omega_p} \frac{1}{(2\pi)^32\omega_{p}} (2\pi)^3 \omega_{p}^2 [a^\dagger(\vec{p})a(\vec{p})+a(\vec{p})a^\dagger(\vec{p}) + b(\vec{p})b^\dagger(\vec{p}) +b^\dagger(\vec{b})b(\vec{p}) + a^\dagger(-\vec{p})b^\dagger(\vec{p})e^{2i\omega_pt} - a^\dagger(-\vec{p})b^\dagger(\vec{p})e^{2i\omega_pt}+a(\vec{p})b(-\vec{p})e^{-2i\omega_pt}- a(\vec{p})b(-\vec{p}) e^{-2i\omega_pt} ]$$

$$H =\int \frac{d^3p}{(2\pi)^32\omega_p} \frac{1}{(2\pi)^32\omega_{p}} (2\pi)^3 \omega_{p}^2 [a^\dagger(\vec{p})a(\vec{p})+a(\vec{p})a^\dagger(\vec{p}) + b(\vec{p})b^\dagger(\vec{p}) +b^\dagger(\vec{b})b(\vec{p})]$$

$$H =\int \frac{d^3p}{(2\pi)^32\omega_p} \frac{1}{2} \omega_{p} [a^\dagger(\vec{p})a(\vec{p})+a(\vec{p})a^\dagger(\vec{p}) + b(\vec{p})b^\dagger(\vec{p}) +b^\dagger(\vec{b})b(\vec{p})]$$

Remember that $a^\dagger(\vec{p})a(\vec{p})+[a(\vec{p}),a^\dagger(\vec{p})]=a(\vec{p})a^\dagger(\vec{p})$ and $b^\dagger(\vec{p})b(\vec{p})+[b(\vec{p}),b^\dagger(\vec{p})]=b(\vec{p})b^\dagger(\vec{p})$ so we can substitute in

$$H =\int \frac{d^3p}{(2\pi)^32\omega_p} \frac{1}{2} \omega_{p} [a^\dagger(\vec{p})a(\vec{p})+a^\dagger(\vec{p})a(\vec{p})+[a(\vec{p}),a^\dagger(\vec{p})] + b^\dagger(\vec{p})b(\vec{p})+[b(\vec{p}),b^\dagger(\vec{p})] +b^\dagger(\vec{b})b(\vec{p})]$$

$$H =\int \frac{d^3p}{(2\pi)^32\omega_p} \frac{1}{2} \omega_{p} [2a^\dagger(\vec{p})a(\vec{p})+2b^\dagger(\vec{p})b(\vec{p})+[a(\vec{p}),a^\dagger(\vec{p})] +[b(\vec{p}),b^\dagger(\vec{p})]$$

$$H =\int \frac{d^3p}{(2\pi)^32\omega_p} \frac{1}{2} \omega_{p} [2a^\dagger(\vec{p})a(\vec{p})+2b^\dagger(\vec{p})b(\vec{p})] + \int \frac{d^3p}{(2\pi)^32\omega_p} \frac{1}{2}([a(\vec{p}),a^\dagger(\vec{p})] + [b(\vec{p}),b^\dagger(\vec{p})])$$

$$H =\int \frac{d^3p}{(2\pi)^32\omega_p} \omega_{p} [a^\dagger(\vec{p})a(\vec{p})+b^\dagger(\vec{p})b(\vec{p})] + \int \frac{d^3p}{(2\pi)^32\omega_p} \frac{1}{2}([a(\vec{p}),a^\dagger(\vec{p})] + [b(\vec{p}),b^\dagger(\vec{p})])$$

We know $[b(\vec{p}),b^\dagger(\vec{p})]=[a(\vec{p}),a^\dagger(\vec{p})]=(2\pi)^32\omega_p\delta^3(0)$

$$H =\int \frac{d^3p}{(2\pi)^32\omega_p} \omega_{p} [a^\dagger(\vec{p})a(\vec{p})+b^\dagger(\vec{p})b(\vec{p})] + \int \frac{d^3p}{(2\pi)^32\omega_p} \frac{1}{2}((2\pi)^32\omega_p\delta^3(0) + (2\pi)^32\omega_p\delta^3(0))$$ $$H =\int \frac{d^3p}{(2\pi)^32\omega_p} \omega_{p} [a^\dagger(\vec{p})a(\vec{p})+b^\dagger(\vec{p})b(\vec{p})] + \delta^3(0) \int \frac{d^3p}{(2\pi)^32\omega_p} \frac{1}{2}((2\pi)^32\omega_p + (2\pi)^32\omega_p)$$ $$C = \delta^3(0) \int \frac{d^3p}{(2\pi)^32\omega_p} \frac{1}{2}((2\pi)^32\omega_p + (2\pi)^32\omega_p)$$ $$H =\int \frac{d^3p}{(2\pi)^32\omega_p} \omega_{p} [a^\dagger(\vec{p})a(\vec{p})+b^\dagger(\vec{p})b(\vec{p})] + C$$

We only can measure differences in energy, so we can remove the infinite constant through renormalization. This, finally, leaves us with $$H =\int \frac{d^3p}{(2\pi)^32\omega_p} \omega_{p} [a^\dagger(\vec{p})a(\vec{p})+b^\dagger(\vec{p})b(\vec{p})]$$ This shows that the Energy of the field depends on the number of particles and anti-particles in the fock-space.

Answer 2

The other answer can't be correct because it doesn't have units of energy $\omega_k$. Making an expansion using

$$ \phi(x) = \int \frac{d^3 k}{(2 \pi)^3 \sqrt{2 \omega_k}} \ \big(a_{\vec{k}} e^{-ik\cdot x} + b^{\dagger}_{\vec{k}} e^{ik\cdot x} \big )$$

I found (after neglecting the infinite constant)

$$ H = \int \frac{d^3k}{(2\pi)^3} \omega_k (a_{\vec k}^\dagger a_{ \vec k} + b_{ \vec k}^\dagger b_{ \vec k}) $$