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I'm following this book on QFT called "Quantum Field Theory of Point Particles and Strings" by Brian Hatfield. After the end of the scalar field theory section on Exercise 3.6, it asks us to express the number operator

$$N=\int \frac{d^3k}{(2\pi)^3}\frac{1}{2\omega_k}a^\dagger(\vec{k}) a(\vec{k})$$

In terms of $\varphi(x,t)$ and $\pi(x,t)$.

My attempt

My first thought was to express the creation and annihilation operators in terms of the field and conjugate momentum

$$a(\vec{k})=\int d^3x \, e^{ik\cdot x}\,(\omega_k\varphi(x)+i\pi(x))$$ $$a^\dagger(\vec{k})=\int d^3x \, e^{-ik\cdot x}\,(\omega_k\varphi(x)-i\pi(x))$$ So substituting these definitions into the number operator we get the following $$N=\int \frac{d^3k}{(2\pi)^3}\frac{1}{2\omega_k}\int d^3x \int d^3x' \, e^{i(x-x')\cdot k}\,(\omega_k\varphi(x')-i\pi(x'))(\omega_k\varphi(x)+i\pi(x))$$ So, I rearrange the expression in the following way $$N=\int d^3x \int d^3x' \int \frac{d^3k}{(2\pi)^3}\frac{1}{2\omega_k} \, e^{i(x-x')\cdot k}\,(\omega_k\varphi(x')-i\pi(x'))(\omega_k\varphi(x)+i\pi(x))$$

However, I do not know how to do the $k$ integration over the exponential term to get the delta function, because we also have to integrate over the energy terms which may give a different result.

$$\int \frac{d^3 k}{(2\pi)^3} e^{i(\vec{x}-\vec{x}')\cdot \vec{k}} =\delta(\vec{x}-\vec{x}') $$ $$\int \frac{d^3 k}{(2\pi)^3} \frac{1}{2\omega_k} e^{i(\vec{x}-\vec{x}')\cdot \vec{k}} = \int \frac{d^3 k}{(2\pi)^3} \frac{1}{2\sqrt{|\vec{k}|^2 +m^2}} e^{i(\vec{x}-\vec{x}')\cdot \vec{k}} = ?$$

Also, I knew from inspection that if you expanded it out, there aren't any terms that you can combine into a commutator, so the commutation relations would not be useful.

I was thinking the number operator may take the form

$$N = \int d^3 x \varphi(x)\pi(x)$$

As in the complex case (judging by the charge operator), it takes

$$N_A - N_B = \int d^3 x( \varphi(x)\pi_\varphi(x) - \varphi^*(x)\pi_{\varphi^*}(x))$$

How would you find an operator which is originally in terms of creation and annihilation operators to an operator that is dependent on fields?

Note: This question is about how do you actually reverse the process to find an operator in terms of the field operators where using this computation as an example and may be useful to other users.

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  • $\begingroup$ Thst is a non-local expression of the fields at fixed time. I am busy now, later I will write the expression explicitly. $\endgroup$ Aug 17, 2022 at 17:17
  • $\begingroup$ Up to coefficients, $N = \frac{1}{2}\int d^3x \:\phi(x) (\sqrt{-\Delta + m^2} \phi)(x) + \frac{1}{2}\int d^3x\: \pi(x) (\sqrt{-\Delta + m^2}^{-1} \phi)(x)$ $\endgroup$ Aug 17, 2022 at 17:21
  • $\begingroup$ It is just a re-phrasing of your first expression of $N$ in terms of three integrals, there is no hope to find a better expression. In the limit of large mass it takes a more easy aspect. $\endgroup$ Aug 17, 2022 at 17:25
  • $\begingroup$ @ValterMoretti How would you compare this result to $j^0$. Also what process did you use to derive the expression? $\endgroup$ Aug 17, 2022 at 17:27
  • $\begingroup$ In that case there is another local addend (which cancels here) and $N_a$, $N_b$ expand as the same non local itegral plus/minus the local term. Taking the difference only two times the local term survives. That is $j^0$ up to a factor $i$. $\endgroup$ Aug 17, 2022 at 17:44

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I stress that your expression for the total charge is wrong. The total charge arising from the symmetry $U(1)$ by the Noether therem reads $$Q=i \int d^3x \left(\varphi^*(x)\pi(x) - \pi^*(x) \varphi(x)\right)\:.$$ Notice that the two fields are mixed up and the factor $i$ takes place.

Let us start from your expression for $N$. I perform all computations at $t=0$, it would be easy to insert this variable, nothing would change, only all fiedls would be computed at that time

$\varphi(x)\to \varphi(x,t)$, $\varphi(x')\to \varphi(x',t)$,

$\pi(x)\to \pi(x,t)$, $\pi(x')\to \pi(x',t)$

You found $$N=\int d^3x \int d^3x' \int \frac{d^3k}{(2\pi)^3}\frac{1}{2\omega_k} \, e^{i(x-x')\cdot k}\,(\omega_k\varphi(x')-i\pi(x'))(\omega_k\varphi(x)+i\pi(x)).$$ It can be expanded as $$N=\frac{1}{2}\int d^3x \varphi(x) \int \frac{d^3k}{(2\pi)^{3/2}}e^{ix\cdot k} \omega_k \, \int \frac{d^3x'}{(2\pi)^{3/2}} e^{-ix'\cdot k}\,\varphi(x') $$ $$ + \frac{1}{2}\int d^3x \pi(x) \int \frac{d^3k}{(2\pi)^{3/2}}e^{ix\cdot k}\frac{1}{\omega_k} \int \frac{d^3x'}{(2\pi)^{3/2}}\, e^{-i-x'\cdot k}\,\pi(x')$$ $$+\frac{i}{2} \int d^3x \left(\varphi(x)\pi(x) - \pi(x) \varphi(x)\right)\:.$$ In the last term I used $$ \int \frac{d^3k}{(2\pi)^{3}} \, e^{i(x-x')\cdot k}= \delta(x-x')\:.$$ The last term vanishes. The remaining ones can be re-written as follows. $$N=\frac{1}{2}\int d^3x \;\varphi(x) \left(\sqrt{-\Delta +m^2}\varphi\right)(x) + \frac{1}{2}\int d^3x \:\pi(x)\left(\left(\frac{1}{\sqrt{-\Delta + m^2}}\right)\pi\right)(x)\tag{1}$$ where I used the fact that, in Fourier representation, $\sqrt{-\Delta +m^2}$ is just $\omega_k$.

What happens if we consider complex fields? The result is similar and one finds $$N_a=\frac{1}{2}\int d^3x \;\varphi^*(x) \left(\sqrt{-\Delta +m^2}\varphi\right)(x) + \frac{1}{2}\int d^3x \:\pi^*(x)\left(\left(\frac{1}{\sqrt{-\Delta + m^2}}\right)\pi\right)(x)$$ $$+\frac{i}{2} \int d^3x \left(\varphi^*(x)\pi(x) - \pi^*(x) \varphi(x)\right)\:.$$ and (using also the fact that $\sqrt{-\Delta +m^2}$ and $1/\sqrt{-\Delta +m^2}$ are selfadjoint) $$N_b=\frac{1}{2}\int d^3x \;\varphi^*(x) \left(\sqrt{-\Delta +m^2}\varphi\right)(x) + \frac{1}{2}\int d^3x \:\pi^*(x)\left(\left(\frac{1}{\sqrt{-\Delta + m^2}}\right)\pi\right)(x)$$ $$-\frac{i}{2} \int d^3x \left(\varphi^*(x)\pi(x) - \pi^*(x) \varphi(x)\right)\:.$$ Therefore, the total charge is the integral of a local object ($j^0$) $$Q= N_a-N_b= i \int d^3x \left(\varphi^*(x)\pi(x) - \pi^*(x) \varphi(x)\right)$$

ADDENDUM. I would like to complete the solution of the proposed exercise with the formulae I introduced. Let us start from the equations arising from the Heisenberg evolution of the field $\varphi$ and taking the Klerin Gordon equation into account, $$i[H, \varphi(x,t)]= \pi(x,t)\:\quad i[H, \pi(x,t)] = \Delta \varphi(x,t) - m^2\varphi(x,t)\:.$$

Using these equarions in (1) we find $$i[H,N] = \frac{1}{2}\int d^3x \;i[H,\varphi(x)] \left(\sqrt{-\Delta +m^2}\varphi\right)(x) + \frac{1}{2}\int d^3x \;\varphi(x) \left(\sqrt{-\Delta +m^2}i[H, \varphi]\right)(x)$$ $$+ \frac{1}{2}\int d^3x \:i[H,\pi(x)]\left(\left(\frac{1}{\sqrt{-\Delta + m^2}}\right)\pi\right)(x) + \frac{1}{2}\int d^3x \:\pi(x)\left(\left(\frac{1}{\sqrt{-\Delta + m^2}}\right)i[H,\pi]\right)(x)$$ $$ = \frac{1}{2}\int d^3x \;\pi(x) \left(\sqrt{-\Delta +m^2}\varphi\right)(x) + \frac{1}{2}\int d^3x \;\varphi(x) \left(\sqrt{-\Delta +m^2} \pi\right)(x)$$ $$- \frac{1}{2}\int d^3x \:((-\Delta +m^2)\varphi)(x)\left(\left(\frac{1}{\sqrt{-\Delta + m^2}}\right)\pi\right)(x) - \frac{1}{2}\int d^3x \:\pi(x)\left(\left(\frac{1}{\sqrt{-\Delta + m^2}}\right)(-\Delta+m^2)\varphi\right)(x)$$ $$= \frac{1}{2}\int d^3x \;\pi(x) \left(\sqrt{-\Delta +m^2}\varphi\right)(x) + \frac{1}{2}\int d^3x \;\varphi(x) \left(\sqrt{-\Delta +m^2} \pi\right)(x)$$ $$- \frac{1}{2}\int d^3x \;\varphi(x) \left(\sqrt{-\Delta +m^2} \pi\right)(x)- \frac{1}{2}\int d^3x \;\pi(x) \left(\sqrt{-\Delta +m^2}\varphi\right)(x) =0\:.$$ Above I used the fac that $\sqrt{-\Delta+m^2}$ is selfadjoint and thus I can move it from one entry to the other one of the scalar product given by the variuos integrals appearing above.

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  • $\begingroup$ So in the complex case when you find the difference, they cancel out to give the expression for the charge, however, in the real case, there is no other number operator to find the difference. That's quite hard to see, are you sure this is what the author intended? Or is there an easier explanation? $\endgroup$ Aug 17, 2022 at 22:31
  • $\begingroup$ No, I am by no means sure! But I do not see any other way. However the number operator is non local in QFT. $\endgroup$ Aug 18, 2022 at 5:27
  • $\begingroup$ @Joshua Pasa I completed the answer to the exercise in the textbook... $\endgroup$ Aug 18, 2022 at 14:23
  • $\begingroup$ Hi, I'm using the convention where the Fourier transform is using $1/(2\pi)^3$ not $1/(2\pi)^{3/2}$, so would that give equivalent results but with the additional factor of $(2\pi)^3$? $\endgroup$ Aug 24, 2022 at 16:14
  • $\begingroup$ Instead of assuming it is self-adjoint I used the chain rule to rewrite some of the terms and found that all terms after that contain a total derivative so the integral vanishes. $\endgroup$ Aug 24, 2022 at 20:20

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