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. $\endgroup$ Commented Oct 17, 2020 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? $\endgroup$ Commented Oct 17, 2020 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. :) $\endgroup$ Commented Oct 17, 2020 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. $\endgroup$ Commented Oct 18, 2020 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. $\endgroup$ Commented Oct 18, 2020 at 14:15

1 Answer 1


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.

  • 1
    $\begingroup$ Thanks, you are a cool guy! $\endgroup$ Commented Oct 18, 2020 at 16:20

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.