My question has been asked two other times: Spinor vacuum energy (misleading title) and Vacuum Energy Calculation using Path Integral. I am not completely satisfied with the answers and it looks like they both have errors in their algebraic steps. Since it has been asked twice, I hope you will look at my VERY DETAILED question which contains and exceeds the clarifications sought in these other questions.

I am using Zee's QFT book and he skipped over far too many steps in section II.5. My questions are about the missing steps. They are in bold below. (I think I will read through Zee's section III and then switch to a non-nutshell book but in the meantime I am working through it.) Let $\varphi$ be a scalar field with ground state $|0\rangle$. We have, by identity, the energy of the vacuum $E_{\text{vac}}$ as

$$ Z=\langle 0|e^{-i\hat H T} |0\rangle=e^{-iE_{\text{vac}}T} $$

and we want to determine exactly what $E_{\text{vac}}$ is. We also let the time $T\to\infty$ so our integrals are over all of spacetime. We write out $Z$ as the generating functional

$$ Z=\int D\varphi e^{ i\int d^4x\frac{1}{2}[(\partial\varphi)^2-m^2\varphi^2 ] } .$$

By a standard Gaussian identity and a magical procedure for "discretizing" infinite dimensional path integrals, and for some "non-essential" stuff $C$, we obtain

$$ Z=C\left( \frac{1}{\det [\partial^2+m^2]} \right) =Ce^{ -\frac{1}{2}\text{Tr}\log(\partial^2+m^2) } .$$

Therefore, setting the exponentials equal, the energy of the vacuum has the form

$$ iE_{\text{vac}}T \varphi= \frac{1}{2}\text{Tr}\log(\partial^2+m^2)\varphi . $$

(Since $C$ has exponential dependence, this gives the additional energy $A$ obtained below.) Now this is where Zee skips some steps. He writes

$$ \text{Tr} \log(\partial^2+m^2)=\int \!d^4x\,\langle x| \log(\partial^2+m^2)|x\rangle . $$

Is this an identity for the trace? I kind of see that by the orthogonality of $|x\rangle$ and $|y\rangle$, we will only pick out the diagonal elements of the operator but he introduces this formula from nowhere. He proceeds to solve the integral inserting the identity twice as

$$ \text{Tr} \log(\partial^2+m^2)= \int\!d^4x\int\!\frac{d^4k}{(2\pi)^4} \int\!\frac{d^4q}{(2\pi)^4} \langle x|k\rangle\langle k| \log(\partial^2+m^2) |q\rangle\langle q| x\rangle. $$

What is $q$? Is it momentum written as a second dummy variable akin to $(k,q)\sim (k_1,k_2)$? As if by magic, Zee uses "we obtain" to write

$$ iE_{\text{vac}}T =\frac{1}{2} VT\int\!\frac{d^4k}{(2\pi)^4} \log(k^2-m^2+i\varepsilon) +A $$

WHAT HAPPENED HERE? (How did he know to insert the identity two times?!?!) I see we get $VT$ from $\int d^4x$, kind of. I see the $i\varepsilon$ appeared magically in the usual way. I don't see what else happened there. Both of the above linked previous questions (Spinor vacuum energy and Vacuum Energy Calculation using Path Integral) try to explain this, but I am not satisfied and I will begin my own computation. Assuming the trace identity, we have

\begin{align} iE_{\text{vac}}T&=\frac{1}{2} \int\!d^4x\int\!\frac{d^4k}{(2\pi)^4} \int\!\frac{d^4q}{(2\pi)^4} \langle x|k\rangle\langle k| \log(\partial^2+m^2) |q\rangle\langle q| x\rangle. \end{align}

Use $\langle x| k\rangle=e^{ikx}$, $\langle q| x\rangle=e^{-iqx}$, and $-i\partial|q\rangle=q|q\rangle$ to obtain

\begin{align} &=\frac{1}{2} \int\!d^4x\int\!\frac{d^4k}{(2\pi)^4} \int\!\frac{d^4q}{(2\pi)^4} e^{ix(k-q)}\log(-q^2+m^2) \langle k |q\rangle . \end{align}

Now I use

$$\delta(k-q)=\int \frac{d^4x}{(2\pi)^4}e^{ix(k-q)}$$

to obtain

\begin{align} &=\frac{1}{2} \int\!\frac{d^4k}{(2\pi)^4} \int \!d^4\!q\,\delta(k-q)\log(-q^2+m^2) \langle k |q\rangle \\ &=\frac{1}{2} \int\!\frac{d^4k}{(2\pi)^4} \log(-k^2+m^2) \langle k |k\rangle \\ &=\frac{1}{2} \int\!\frac{d^4k}{(2\pi)^4} \log(-k^2+m^2) .\\ \end{align}

If I proceed here, I do not get the correct answer. Even if I add $i\varepsilon$ and use an identity for the complex logarithm, there's no way I could get $VT$. The steps are worked out most clearly in Spinor vacuum energy, but I do not like what he has done. For instance, his partial operator should have acted to the right to return $q$ but he has acted to the left to obtain $k$. Seems like he messed up a factor of $(2\pi)^4$ as well. Mostly my question is about why he delayed the creation of the Dirac delta until after the insertion of a third resolution of the identity.

  • $\begingroup$ Note that the "trace identity" you mention here is just the definition of the trace. $\endgroup$ – AlmostClueless Oct 17 '20 at 13:30
  • $\begingroup$ @AlmostClueless I think you must mean "the definition of the trace of an operator with a continuous spectrum," because certainly this is not the usual definition of the trace encountered in matrix or tensor algebra. Where can I find the trace definition for the operator with the continuous spectrum? $\endgroup$ – hodop smith Oct 17 '20 at 14:01
  • $\begingroup$ I have to admit, since i am rather new to QFT and my understanding of formal functional analysis is sadly poor, that i cannot give a rigorous answer to how this trace is defined. To my understanding when dealing with operators which have a continous specrum we need to use trace-class operators, but there are a lot of restrictions to the "traced over" operator which are not satisfied in general. So this "trace definition" is actually a strong abuse of notation. Hopefully someone can elaborate on this topic. :) $\endgroup$ – AlmostClueless Oct 17 '20 at 14:45
  • $\begingroup$ @hodopsmith $\text{tr} \left| \chi \right> \left< \psi \right| = \left< \psi | \chi \right>$ is an identity that holds for the usual traces in matrix or tensor algebra. I'm not sure what you mean. $\endgroup$ – Prof. Legolasov Oct 18 '20 at 13:57
  • $\begingroup$ OP I understand your frustration, but if I were you I would try to avoid jumping to conclusions that textbook authors must have screwed up a calculation if it doesn't come out as yours does. $\endgroup$ – Prof. Legolasov Oct 18 '20 at 14:15

OP's calculation seems to match Zee's calculation; except for the final step. Here OP has made a mistake: $$ \left< k | k \right> = (2 \pi)^4 \delta^{(4)}(0) \neq 1. $$

This is where the factor of $VT$ comes from: $$ \left< k | k \right> = \left<k | 1 | k \right> = \int d^4 x \left< k | x \right> \left< x | k \right> = \int d^4 x \; e^{-i k x} e^{i k x} = \int d^4 x = V T. $$

Below are answers to OP's questions in the bold font.

It is a very well known technique from ordinary quantum mechanics to insert resolutions of identity $$ 1 = \int d^d x \left| x \right> \left< x \right| $$ and $$ 1 = \int \frac{d^d p}{(2\pi)^d} \left| p \right> \left< p \right| $$ in operator equations. Since both are equal to one, they can be inserted anywhere one desires.

Both operators above act on $L_2(\mathbb{R}^d)$. The notation may be a bit confusing to mathematicians, because $\left| x \right>$ itself doesn't belong to $L_2(\mathbb{R}^d)$, but to the distribution space. However, physicists use this bra-ket notation all the time.

The distributional nature of kets is also the reason a singularity equal to the infinite spacetime volume appears in $\left< k | k \right>$. Squares of distributions are always ill defined and care must be taken to make sure the resulting theory makes sense nevertheless.

W.r.t. traces. The identity he uses is: $$ \text{tr} (\left| \psi \right> \left< \chi \right|) = \left< \chi | \psi \right>. $$

This is almost by definition of the trace. Expand both vectors in some orthonormal basis and write the trace explicitly: $$ \text{tr} (\left| \psi \right> \left< \chi \right|) = \left| \psi \right>_a \left< \chi \right|_a = \left< \chi \right|_a \left| \psi \right>_a = \left< \chi | \psi \right>. $$

W.r.t. $k$ and $q$ – they are both just mathematical symbols in the momentum-space resolution of identity. We're allowed to insert as many resolutions as we please, and he chose to insert two.

It is a well-known fact from the theory of Fourier integrals that $$ \left< x | k \right> = e^{i k x}, $$ and so $$ \partial_{\mu} \left| k \right> = i k_{\mu} \left| k \right>. $$

He uses it later to put a differential operator into algebraic form.

  • $\begingroup$ Thanks, you are a cool guy! $\endgroup$ – hodop smith Oct 18 '20 at 16:20

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