# Fluctuation spectrum of lipid bilayer membranes

I am interestend in calculating the fluctuation spectrum of a thermally fluctuating 2d membane which is only subject to a surface tension $\sigma$. ($\mathcal{H}=\sigma\int\mathrm{d}A$)

Depending in which form I write my Hamiltonian to begin with my results differ by a factor 2, which should not be the case. Therefore I assume I am messing something up whithout noticing. Below I explain my calculation so far.

I use the Monge presentation and work in the limit of almost flat height profiles $h(x,y)$. By using a Fourier transform I can write the Hamiltonian of the system in the following way: \begin{equation} \mathcal{H} = (2\pi)^2 \frac{\sigma}{2} \int k^2\ \left|\tilde{h}(\mathbf{k})\right|^2 \ \mathrm{d}\mathbf{k} \end{equation} where $\sigma$ is the surface tension of the membrane. Since the integrand is invariant under the mapping $\mathbf{k}\mapsto -\mathbf{k}$ one can also rewrite this Hamiltonian as \begin{equation} \mathcal{H} = (2\pi)^2 \sigma \int_{k_y>0} k^2\ \left|\tilde{h}(\mathbf{k})\right|^2 \ \mathrm{d}\mathbf{k} \end{equation} Where one gets a factor 2 by restricting the integration to the upper half plane of $\mathbf{k}$ - space.

The $(2\pi)^2$ stems from the convention of my Fourier transform which is \begin{align} h(\mathbf{x}) &= \int \tilde{h}(\mathbf{k}) e^{i\mathbf{k}\cdot\mathbf{x}}\mathrm{d}\mathbf{k} \\ \tilde{h}(\mathbf{k}) &= \frac{1}{(2\pi)^2}\int h(\mathbf{x}) e^{-i\mathbf{k}\cdot \mathbf{x}} \mathrm{d}\mathbf{x} \end{align}

I want to calculate $\langle |\tilde{h}(\mathbf{k})|^2\rangle$. Therefore I use $\tilde{h} = a+ib$ And since $h$ is real I know that $\tilde{h}(\mathbf{k})=\tilde{h}(-\mathbf{k})^*$ which leads to $a(\mathbf{k}) = a(-\mathbf{k})$ and $b(\mathbf{k}) = -b(-\mathbf{k})$. Hence I start by calculating $\langle a^2(\mathbf{k})\rangle$. Therefore I establish the partition sum using the functional integral over all height profiles $h$.

\begin{align} Z &= \int \mathcal{D}\tilde{h} \ e^{-\beta \mathcal{H}(\tilde{h}(\mathbf{k})} \\ &= \prod_{k_x}\prod_{k_y}\int_{-\infty}^{\infty}\mathrm{d}a(\mathbf{k})\int_{-\infty}^{\infty}\mathrm{d}b(\mathbf{k})\ \exp\left[-\beta (2\pi)^2 \frac{\sigma}{2} k^2(a^2(\mathbf{k})+b^2(\mathbf{k}))\right] \end{align} Since I evaluate $\langle a^2(\mathbf{k})\rangle$ for a psecific fixed $\mathbf{k}$ the average can be computed by the Gaussian integrals \begin{align} \langle a^2(\mathbf{k})\rangle = \frac{\int_{-\infty}^{\infty} a^2(\mathbf{k}) \exp(-\beta (2\pi)^2\sigma/2 k^2 a^2(\mathbf{k}))\mathrm{d}a(\mathbf{k})}{\int_{-\infty}^{\infty} \exp(-\beta (2\pi)^2\sigma/2 k^2 a^2(\mathbf{k}))\mathrm{d}a(\mathbf{k})} \end{align} And by using the result for the standard exponential integrals this evaluates to \begin{equation} \langle a^2(\mathbf{k})\rangle = \frac{\frac{\sqrt{\pi}}{2(\beta(2\pi)^2 \sigma/2 k^2)^{3/2}}}{\frac{\sqrt{\pi}}{\sqrt{\beta (2\pi)^2\sigma/2 k^2}}} = \frac{1}{(2\pi)^2\beta \sigma k^2} \end{equation}

So far so good. But if I had used my second form of the Hamiltonian where I only integrate of half of $\mathbf{k}$ space I would have obtained a result that differs by a factor of 2 which should not be the case. The partition sum would then be \begin{align} Z &= \int \mathcal{D}\tilde{h} \ e^{-\beta \mathcal{H}(\tilde{h}(\mathbf{k})} \\ &= \prod_{k_x}\prod_{k_y>0}\int_{-\infty}^{\infty}\mathrm{d}a(\mathbf{k})\int_{-\infty}^{\infty}\mathrm{d}b(\mathbf{k})\ \exp\left[-\beta (2\pi)^2 \frac{\sigma}{2} k^2(a^2(\mathbf{k})+b^2(\mathbf{k}))\right] \end{align} But as soon as I calculate $\langle a^2(\mathbf{k})^2\rangle$ I can again cancel many terms and obtain the two integrals

\begin{align} \langle a^2(\mathbf{k})\rangle = \frac{\int_{-\infty}^{\infty} a^2(\mathbf{k}) \exp(-\beta (2\pi)^2\sigma k^2 a^2(\mathbf{k}))\mathrm{d}a(\mathbf{k})}{\int_{-\infty}^{\infty} \exp(-\beta (2\pi)^2\sigma k^2 a^2(\mathbf{k}))\mathrm{d}a(\mathbf{k})} \end{align} which are the same as above, except that $\sigma /2$ is now replace by $\sigma$ only. This would yield the result: \begin{equation} \langle a^2(\mathbf{k})\rangle = \frac{\frac{\sqrt{\pi}}{2(\beta(2\pi)^2 \sigma k^2)^{3/2}}}{\frac{\sqrt{\pi}}{\sqrt{\beta (2\pi)^2\sigma k^2}}} = \frac{1}{2 (2\pi)^2\beta \sigma k^2} \end{equation}

Obviously there must be a mistake, because the result should not depend on in which form I write the same Hamiltonian. Can someone point out what is going wrong in my current approach.

Thanks and best regards khx0