2
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ Perhaps it would help if you posted your expression for $\pi(x)$ as well. $\endgroup$
    – J. Murray
    Dec 20, 2019 at 4:23
  • $\begingroup$ I think it may be a typo, the integral is over momentum space. I've posted a full derivation below, I have a slightly different form since I used a different convention $\endgroup$ Apr 10, 2020 at 2:59
  • $\begingroup$ Let me know if that helps. $\endgroup$ May 28, 2020 at 16:03

2 Answers 2

8
$\begingroup$

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.

$\endgroup$
5
  • $\begingroup$ I can't belive you worte it being so long!! Why don't you try to simplify the post by talking a little bit more and just leaving the formulae for the crucial steps? $\endgroup$
    – vin92
    May 29, 2020 at 18:30
  • $\begingroup$ @vin92 Which bits aren't clear? $\endgroup$ May 29, 2020 at 21:37
  • $\begingroup$ Everything is clear and very well written I just thought that it can be a little bit overwhelming to read. Anyway nice answer! $\endgroup$
    – vin92
    May 29, 2020 at 23:45
  • $\begingroup$ @vin92 Thanks! It took so long, and I don't even think the person who asked the question is active anymore, so it might of been a bit of a waste. Hopefully someone else might find it useful one day. $\endgroup$ May 29, 2020 at 23:49
  • $\begingroup$ That's the spirit! At least you get my vote! C: $\endgroup$
    – vin92
    May 29, 2020 at 23:50
0
$\begingroup$

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)

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

$\endgroup$
2
  • $\begingroup$ If you act the creation operator on the vacuum state and apply the hamiltonian operator it gives the correct result (with my answer). There are different conventions for the Lorentz invariant integration measure which I stated. The results are equivalent. $\endgroup$ Jan 3, 2022 at 8:30
  • $\begingroup$ Also, it does have units of energy, the creation and annihilation operators ensure that, the units of your answer also has energy, but the creation and annihilation operator have different units to my one $\endgroup$ Jan 8, 2022 at 8:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.