Vacuum energy of a free scalar field from path integral

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.

• Note that the "trace identity" you mention here is just the definition of the trace. – AlmostClueless Oct 17 '20 at 13:30
• @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? – hodop smith Oct 17 '20 at 14:01
• 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. :) – AlmostClueless Oct 17 '20 at 14:45
• @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. – Prof. Legolasov Oct 18 '20 at 13:57
• 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. – 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: $$\left< k | k \right> = \left = \int d^4 x \left< k | x \right> \left< x | k \right> = \int d^4 x \; e^{-i k x} e^{i k x} = \int d^4 x = V T.$$

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.

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