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I'm trying to compute the following path integral \begin{equation} Z = \int\mathcal{D}\phi\exp\left(-\int_{\mathbb{R}^d_+}\frac{d^dx}{2}\phi(-\partial_\mu^2 + m^2)\phi \right) \propto \frac{1}{\sqrt{\det(-\partial_\mu^2 + m^2)}} = \exp\left( -\frac{\text{tr}\log(-\partial_\mu^2 + m^2)}{2} \right) \ , \end{equation} near a boundary placed at $x_\perp = 0$, where $\mathbb{R}^d_+ = \{x_\parallel \in \mathbb{R}^{d - 1}, x_\perp \geq 0\}$. At the boundary I consider Dirichlet b.c.'s \begin{equation} \phi(x_\parallel, x_\perp = 0) = 0 \ . \end{equation} The trace is by definition given by \begin{equation} \text{tr}\log(-\partial_\mu^2 + m^2) = \int_{\mathbb{R}^d_+}d^dx \langle x_\parallel, x_\perp| \log(-\partial_\mu^2 + m^2) | x_\parallel, x_\perp\rangle \ , \quad | x_\parallel, x_\perp\rangle = | x_\parallel\rangle \otimes |x_\perp\rangle \ . \end{equation} How do I proceed from here?

Without the boundary, I can Fourier transform the Hilbert state \begin{equation} |x\rangle = \int_{\mathbb{R}^d}\frac{d^dk}{(2\pi)^d}e^{-ikx}|k\rangle \ , \end{equation} and then the derivatives only act on the exponential \begin{equation} \text{tr}\log(-\partial_\mu^2 + m^2) = \int_{\mathbb{R}^d}d^dx\int_{\mathbb{R}^d}\frac{d^dk'}{(2\pi)^d}\int_{\mathbb{R}^d}\frac{d^dk}{(2\pi)^d} \log(k_\mu^2 + m^2)e^{i(k' - k)x} \langle k'| k\rangle \ . \end{equation} Here we could factor out the logarithm from the states since its argument is now just a number (eigenvalue), and not a differential operator. We normalize the states s.t. \begin{equation} \int_{\mathbb{R}^d}\frac{d^dk'}{(2\pi)^d} f(k') \langle k'| k\rangle = f(k) \quad\Rightarrow\quad \text{tr}\log(-\partial_\mu^2 + m^2) = \int_{\mathbb{R}^d}d^dx\int_{\mathbb{R}^d}\frac{d^dk}{(2\pi)^d} \log(k_\mu^2 + m^2) \ . \end{equation} This is of course a divergent integral and therefore needs to be renormalized. However, it's the result of the path integral.

Now in my case I can use the same method for the parallel Hilbert state $|x_\parallel\rangle$. However, I'm confused about $|x_\perp\rangle$. Since translation invariance is broken in the direction of $x_\perp$, I'm unsure whether it even has a well-defined Fourier transform.

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  • $\begingroup$ Let me write my attempt at finding this trace. I don't know whether this is entirely correct though. Firstly, since $|x_\perp\rangle$ is a Hilbert state, it has to satisfy the Dirichlet b.c. Using method of images we expect it to be on the form \begin{equation} |x_\perp\rangle = \int_{\mathbb{R}}\frac{d^dk_\perp}{2\pi}(e^{-ik_\perp x_\perp} - e^{+ik_\perp x_\perp})|k_\perp\rangle \ . \end{equation} Here I'm unsure though whether we should integrate over the entire $\mathbb{R}$ or not. $\endgroup$
    – A.Dunder
    Jul 12 at 7:26
  • $\begingroup$ However, by making the variable change $k_\perp \rightarrow - k_\perp$ on the second term we find \begin{equation} |x_\perp\rangle = \int_{\mathbb{R}}\frac{dk_\perp}{2\pi}e^{-ikx}|k_-\rangle \ , \quad |k_-\rangle \equiv |k_\perp\rangle - |-k_\perp\rangle \ , \end{equation} which is on the form of a Fourer transform. The state $|k_-\rangle$ describes the momenta being absorbed/emitted by the boundary. I think this looks quite nice, although I'm not sure whether it's correct at all. $\endgroup$
    – A.Dunder
    Jul 12 at 7:29

1 Answer 1

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Taking a step back, the reason why you usually do a Fourier transform is because this new basis diagonalises $-i\frac{d}{dx}$ so in particular $-\frac{d^2}{dx^2}$, but it is the last operator that is relevant to calculate your trace. This is how you can generalise to different boundary conditions: diagonalise $-\frac{d^2}{dx^2}$ on $\mathbb R_+^*$ with Dirichlet boundary conditions. Physically, it's solving a quantum, free particle on a half-line. The relevant basis is thus $\psi_k = \sin(kx)$ with $k>0$ and the "orthonormality" conditions: $$ \langle \psi_k|\psi_l\rangle = \int_0^\infty dx \sin(kx)\sin(lx)\\ = \frac{2\pi}{4}(\delta(k-l)-\delta(k+l))\\ = \frac{\pi}{2}\delta(k-l) $$

This is another way of seeing @A.Dundler's method of images, only it lacks the proper normalisation.

You can now calculate:

$$ \text{tr} \log(\partial^2_\mu+m^2) = \int_{\mathbb R^{d-1}}\frac{d^{d-1}k_{||}}{(2\pi)^{d-1}}\int_0^\infty\frac{dk_\perp}{\pi/2}\log(k_{||}^2+k_\perp^2+m^2)(2\pi)^{d-1}\frac{\pi}{2}\\ =m^d\frac{2\pi^{\frac{d-1}{2}}}{\Gamma\left(\frac{d-1}{2}\right)}\int_0^\infty dk_{||}k_{||}^{d-2}\int_0^\infty dk_\perp\log(k_{||}^2+k_\perp^2+1) $$

I don't know how calculate generally the final integral, but it only makes sense for $d<0$ and you'll need to do the usual analytic continuation to extract information for the physically relevant dimensions.

Hope this helps.

Edit (answer to the comments)

I took the basis $|k_\perp\rangle$ such that $\langle x_\perp|k_\perp\rangle = \sin(k_\perp x_\perp)$, ie $|k_\perp\rangle = \int_{-\infty}^{+\infty}dx_\perp\sin(k_\perp x_\perp) |x_\perp\rangle $. You just took basis with a different norm, but it would be easier to check the normalisation if you express it as a function of $|x_\perp\rangle$ rather than the opposite, I have a hard time following your calculation.

For the calculation of the trace, you did the same mistake I did, you forgot the extra multiply by the weight in front of the Dirac delta, which cancels the denominators in of the $dk$. This is why the normalisation of the basis does not matter in the end, and we both get the same result. Check out the new formula. The net effect is to simply restrict the integration domain to the half-space.

Indeed, switching to vN BC doesn't change the trace. In this case, a basis would be $|k_\perp$ for $k_\perp\geq 0$ (this time include $0$) with $k_\perp\geq0$, with $\langle x_\perp|k_\perp\rangle = \cos(k_\perp x_\perp)$. This isn't too crazy as the detail of the boundary condition shouldn't matter much, the bulk effect being more important.

Edit 2 (answer to the comments) It's all equivalent. It's just that to first define the new basis, it's best to do it explicitly from the old one, rather than implicitly. It makes the calculation of the normalisations easier. Once you finished the checkup, the implicit formula is useful for applying the operators, but you can also do it from the explicit expression.

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  • $\begingroup$ Thanks for the help. I didn't realize that for the normal direction I could draw wisdom from QM on a half-line. Just to clarify: I can use method of images to find \begin{equation} |x_\perp\rangle = \int_0^\infty \frac{dk_\perp}{2\pi} \frac{e^{-ik_\perp x_\perp} - e^{+ik_\perp x_\perp}}{\sqrt{2}} |k_\perp\rangle \ , \end{equation} which is normalized s.t. \begin{equation} \langle x_\perp'|x_\perp\rangle = \delta(x_\perp' - x_\perp) \ . \end{equation} $\endgroup$
    – A.Dunder
    Jul 12 at 12:09
  • $\begingroup$ The orthogonality follows from: which can be seen using \begin{equation} \frac{dk_\perp}{2\pi} f(k_\perp') \langle k_\perp'|k_\perp\rangle = f(k_\perp) \, \\ \delta(x_\perp' + x_\perp) \ , \quad \text{since $x_\perp', x_\perp > 0$} \ . \end{equation} This gives me: \begin{equation} \text{trace}\log(-\partial_\mu^2 + m^2) = \int_{\mathbb{R}^d_+}\int_{\mathbb{R}^{d-1}}\frac{d^{d-1}k_\parallel}{(2\pi)^{d-1}}\int_0^\infty\frac{dk_\perp}{2\pi}\log(k_\parallel^2 + k_\perp^2 + m^2) \, \end{equation} where the vanishes due to $\delta(k_\perp' + k_\perp) = 0$ similar to $x_\perp$. $\endgroup$
    – A.Dunder
    Jul 12 at 12:16
  • $\begingroup$ Note that in my first attempt I integrated over the entire $k_\perp \in \mathbb{R}$, while now I only integrate over $k_\perp > 0$. Is this correct? What about momentum conservation in the normal direction? I expect it not to be conserved. Furthermore, where does the b.c. enter in the final integral? E.g. if I would consider Neumann b.c. I would shift $e^{-ik_\perp x_\perp} - e^{+ik_\perp x_\perp}$ in $|x_\perp\rangle$ to $e^{-ik_\perp x_\perp} + e^{+ik_\perp x_\perp}$. However, it seems like the image-term ($e^{+ik_\perp x_\perp}$) doesn't contribute to the final result. $\endgroup$
    – A.Dunder
    Jul 12 at 12:19
  • $\begingroup$ $\textbf{Comment for your edit: }$ What I'm confused about now is how I can choose to work in a momentum basis from the start. Take the homogeneous case without a boundary. Then how do I find $\log(-\partial_\mu^2+m^2)|k>$? What I did in my original post was to write $\log(-\partial_\mu^2+m^2)|x>=\int_{\mathbb{R}^d}\frac{d^k}{(2\pi)^d}\log(-\partial_\mu^2+m^2)e^{-ikx}|k>$, where the $-\partial_\mu^2$ could now act on the exponential. This is also why I wanted to write $|x_\perp>$ in terms of $|k_\perp>$ in the prescence of a boundary. $\endgroup$
    – A.Dunder
    Jul 13 at 14:59
  • $\begingroup$ $\textbf{Comment for your edit 2: }$ Thanks I think I understand now. I'm still a bit puzzled as to why the b.c. doesn't affect the general result that much. I would have found it odd if it were a PDE. One last question: do we integrate over $k_\perp \geq 0$ due to momentum conservation? $\endgroup$
    – A.Dunder
    Jul 14 at 1:09

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