Units of wave functions in real and reciprocal space

Question

I'm confused about the units of wave functions in reciprocal space and their Fourier transform in real space. On one hand, I believe the Fourier transform of a reciprocal space wave function in 2D is given by

$$ (1) \ \psi(\vec{r}) = \frac{A}{(2 \pi)^2} \int d^2k \ \psi(\vec{k}) e^{i\vec{k}\cdot \vec{r}},$$

$A$ being the area of the system. According to this, the wave functions in real and reciprocal space have the same units, since the $m^2$ from the area cancels with the $\frac{1}{m^2}$ from the reciprocal space integration. On the other hand, according to Parseval's theorem,

$$ (2) \int d^2 r |\psi(\vec{r})|^2 = \frac{A}{(2 \pi)^2} \int d^2k \|\psi(\vec{k})|^2$$

(= 1 for normalised wave functions). For the LHS of (2) to be unitless, the units of $\psi(r)$ must be $\frac{1}{m}$. But if the units of $\psi(k)$ are also $\frac{1}{m}$, as the definition of the Fourier transform (1) implies, the unit of the RHS of 2 would be $\frac{1}{m^2}$, not unitless as it should be.

The context is excitons in a 2D material. The $\vec{r}$ is the relative coordinate of the electron and hole in real space, $\vec{r} = \vec{r}_e - \vec{r}_h $ and $\vec{k}$ is the relative coordinate in reciprocal space: $ \vec{k} = \frac{m_e}{M} \vec{k}_h + \frac{m_h}{M} \vec{k}_e$ (all 2D vectors, $m_e$ and $m_h$ the effective masses of the electron and hole, $M = m_e + m_h$). I learnt that a sum over k turns into an integral in the thermodynamic limit and gets a factor $\frac{V}{(2 \pi)^d}$, see equation (12) in https://cpb-us-w2.wpmucdn.com/u.osu.edu/dist/3/67057/files/2018/09/density_of_states-vjqh7n.pdf

Answer 1

Ok, I think I've found a way to handle it: I want my integrals in $k-$ space to have a prefactor $\frac{A}{(2 \pi)^2}$ from the thermodynamic limit of a sum in $k-$space, but I want to Fourier transform my reciprocal space wave functions into real space and use the normal $\mathcal{L}^2$ scalar product without that factor for the real space wave functions. I also want to use Parseval's theorem to find the relationship between the normalisation of the real space and reciprocal space wave functions. I will adapt the definition of the Fourier transform from Walter Rudin's Functional Analysis, assuming $\psi(k)$ and its Fourier transform $\hat{\psi}(r) \in \mathcal{L}^1 \cap \mathcal{L}^2$. Let's start by adapting the definition of the Fourier transform from $$ (1a) \quad \hat{f}(t) = \int_{\mathbb{R}^2} f(x) e^{-i t \cdot x} dm_2(x), $$ to $$ (1b) \quad \hat{\psi}(r) = \frac{A}{(2 \pi)^2} \int_{\mathbb{R}^2} \psi(k) e^{-i r \cdot k} dm_2(k), $$ $d_{m_n}(x) = (2 \pi)^{-\frac{n}{2}} d^nx$ being the normalised Lebesque measure and $x, t \in \mathbb{R}^2$. The inversion theorem, given as $$ (2a) \quad g(x) = \int_{\mathbb{R}^2} \hat{g}(t) e^{i t \cdot x} dm_2(t),$$ is not fulfilled in our case, but substituting (1b) into the RHS of (2a) gives us \begin{align*} (2b) \quad \int_{\mathbb{R}^2} \hat{\psi}(r) e^{i r \cdot k} dm_2(r) &= \int_{\mathbb{R}^2} \bigg( \frac{A}{(2 \pi)^2} \int_{\mathbb{R}^2} \psi(k') e^{-i r \cdot k'} dm_2(k') \bigg) e^{i r \cdot k} dm_2(r) \\ &= \frac{A}{(2 \pi)^2} \int_{\mathbb{R}^2} \psi(k') dm_2(k')\int_{\mathbb{R}^2} e^{-i r \cdot (k- k')} dm_2(r) \\ & = \frac{A}{(2 \pi)^2} \int_{\mathbb{R}^2} \psi(k') \delta(k-k') d^2k'\\ & = \frac{A}{(2 \pi)^2} \psi(k),\end{align*} (ie. $\psi(k) = \frac{(2 \pi)^2}{A} \int_{\mathbb{R}^2} \hat{\psi}(r) e^{i r \cdot k} dm_2(r))$, having used $\delta(k) = \frac{1}{(2 \pi)^2}\int_{\mathbb{R}^2} e^{ik \cdot r}d^2r.$ With the further condition that f and g are rapidly decreasing functions, the inversion theorem leads to the Parseval formula $$(3a) \quad \int_{\mathbb{R}^2} f\bar{g} = \int_{\mathbb{R}^2} \hat{f}\bar{\hat{g}}$$ (in particular, the $\mathcal{L}^2$ norm of a function and its Fourier transform are the same). In my case, \begin{align*} (3b) \quad \frac{A}{(2 \pi)^2} \int_{\mathbb{R}^2} \psi(k) \bar{\phi}(k) dm_2(k) &= \frac{A}{(2 \pi)^2} \int_{\mathbb{R}^2} \bigg(\frac{(2 \pi)^2}{A} \int_{\mathbb{R}^2} \hat{\psi}(r) e^{i r \cdot k} dm_2(r)\bigg) \bar{\phi}(k)dm_2(k)\\ & = \int_{\mathbb{R}^2} \hat{\psi}(r) \bigg(\int_{\mathbb{R}^2} \bar{\phi}(k) e^{i r \cdot k} dm_2(k) \bigg) dm_2(r) \\ & = \int_{\mathbb{R}^2} \hat{\psi}(r) \frac{(2 \pi)^2}{A} \bar{\hat{\phi}}(r) dm_2(r),\end{align*} ie. $$\frac{A^2}{(2 \pi)^4} \int_{\mathbb{R}^2} \psi(k) \bar{\phi}(k) d^2k = \int_{\mathbb{R}^2} \hat{\psi}(r) \hat{\bar{\phi}}(r) d^2r$$ or $$\frac{A^2}{(2 \pi)^4} \int_{\mathbb{R}^2} |\psi(k) |^2 d^2k = \int_{\mathbb{R}^2} |\hat{\psi}(r) |^2 d^2r.$$ The last equation is the corrected version of (2) from my original question. To answer my question about the units, if I want to define my Fourier transform as in (1) of my original question, the units of the real and reciprocal space wave functions are both $\frac{1}{m}$.

I would appreciate feedback on my answer and on whether it is necessary to keep the factor $\frac{A}{(2 \pi)^2}$ in front of all k-space integrals.