Inverse Problems of Single Molecule Localization Microscopy

Abstract

Single molecule localization microscopy is a recently developed superresolution imaging technique to visualize structural properties of single cells. The basic principle consists in chemically attaching fluorescent dyes to the molecules, which after excitation with a strong laser may emit light. To achieve superresolution, signals of individual fluorophores are separated in time. In this paper we follow the physical and chemical literature and derive mathematical models describing the propagation of light emitted from dyes in single molecule localization microscopy experiments via Maxwell’s equations. This forms the basis of formulating inverse problems related to single molecule localization microscopy. We also show that the current status of reconstruction methods is a simplification of more general inverse problems for Maxwell’s equations as discussed here.

Appendices

Appendix 1: Derivation of Particular Fourier Transforms

Before reading through the appendix it might be useful to recall the notation of Table 2. The derivation in Lemma 3 uses the residual theorem.

Table 2 Some abbreviation used in Lemma 3 and its proof, as well as in Appendix 1 and 2

Theorem 2 (Residual Theorem)

Let the function $z \in \mathbb {C} \to \tilde {f}(z):=f(z) \mathrm e^{\mathrm i a z}$ with a > 0 satisfy the following properties:

• 1. f is analytic with at most finitely many poles p _i , i = 1, …, m, which do not lie on the real axis.

• 2. There exists M, R > 0 such that for every $z \in \mathbb {C}$ satisfying ℑ(z) ≥ 0 and $\left | z\right | \geq R$

  $$\displaystyle \begin{aligned} \left| f(z)\right| \leq \frac{M}{\left| z\right| }. \end{aligned} $$

  (108)

Then

$$\displaystyle \begin{aligned} \int_{\mathbb{R}} \tilde{f}(x) dx = 2 \pi \mathrm i \sum_{i=1}^m \mathit{\text{Res}}(\tilde{f};p_i). \end{aligned}$$

Using this lemma we are able to prove the following result used in Lemma 3:

Lemma 10

Let the assumptions and notation of Lemma 3 hold (in particular this means that $r_3, r_3^{\boldsymbol {\Psi }} \in \mathbb {R}$ satisfy $r_3-r_3^{\boldsymbol {\Psi }} > 0$ ), then

$$\displaystyle \begin{aligned} \begin{array}{rcl} {} & \frac{1}{\sqrt{2\pi}} \mathcal{F}_3^{-1} \left[k_3 \to \left(\boldsymbol{\Psi} + \frac{(\boldsymbol{\Psi} \times \mathbf{k}) \times \mathbf{k}}{\left| \mathbf{ k}\right|{}^2 - \kappa_\varepsilon^2} \right)\mathrm e^{-\mathrm i k_3 r_3^{\boldsymbol{\Psi}}}\right](r_3)\\ & = \Psi_3 {\mathbf{e}}_3\tilde \delta(r_3 - r_3^{\boldsymbol{\Psi}}) + \frac{\mathrm i \mathrm e^{\mathrm i q_{\varepsilon} (r_3-r_3^{\boldsymbol{\Psi}})}}{ 2q_{\varepsilon} } (\boldsymbol{\Psi} \times \mathbf{ k}_{q_{\varepsilon}}) \times {\mathbf{k}}_{q_{\varepsilon}}. \end{array} \end{aligned} $$

(109)

First, we note that

$$\displaystyle \begin{aligned} \boldsymbol{\Psi} + \frac{(\boldsymbol{\Psi} \times \mathbf{k}) \times \mathbf{k}}{k_3^2 - q_{\varepsilon}^2} = \left(\boldsymbol{\Psi} - \Psi_3 {\mathbf{e}}_3 + \frac{(\boldsymbol{\Psi} \times \mathbf{k}) \times \mathbf{k}}{ k_3^2 - q_{\varepsilon}^2} \right) + \Psi_3 {\mathbf{e}}_3. \end{aligned}$$

Now, we calculate the Fourier-transform of each of the two terms on the right-hand side. The second term can be calculated from Eq. (9) and is given by

$$\displaystyle \begin{aligned} \mathcal F^{-1}_3[k_3 \mapsto \mathrm e^{-\mathrm i k_3 r_3^{\boldsymbol{\Psi}}}](r_3) &= \sqrt{2\pi} \tilde \delta(r_3 - r_3^{\boldsymbol{\Psi}}) {}. \end{aligned} $$

(110)

For the calculation of the first term, let

$$\displaystyle \begin{aligned} z \in \mathbb{C} \to f(z):= \frac{1}{z^2 - q_{\varepsilon}^2} \left( {\boldsymbol{\Psi}} \times \begin{pmatrix}k_1\\k_2\\z \end{pmatrix} \right) \times \begin{pmatrix}k_1\\k_2\\z \end{pmatrix} + {\boldsymbol{\Psi}} - \Psi_3 {\mathbf{e}}_3, \end{aligned} $$

(111)

and we use the residual theorem:

• Clearly f is analytic with potential poles at k ₃ = ±q _ε.

• To verify Eq. (108) we use the elementary calculation rules for ∇_r×, summarized in Eq. (45) and get

  $$\displaystyle \begin{aligned} ~ & \frac{(\boldsymbol{\Psi} \times \mathbf{k}) \times \mathbf{k}}{ k_3^2 - q_{\varepsilon}^2} =-\frac{1}{ k_3^2 - q_{\varepsilon}^2} \\ &\left(\!\!k_3^2 (\boldsymbol{\Psi} - \Psi_3 {\mathbf{e}}_3) - k_3 \begin{pmatrix} \Psi_3 k_1 \\ \Psi_3 k_2 \\ \Psi_1 k_1 {+} \Psi_2 k_2\end{pmatrix} {+} (k_1^2+k_2^2) \boldsymbol{\Psi} {-} \begin{pmatrix} (\!\Psi_1 k_1 + \Psi_2 k_2)k_1 \\ (\Psi_1 k_1 + \Psi_2 k_2)k_2 \\0 \end{pmatrix} \!\!\right) \\ = &-\frac{1}{k_3^2 - q_{\varepsilon}^2} \left(\!k_3^2 \left( \boldsymbol{\Psi} - \Psi_3 {\mathbf{e}}_3\right) - k_3 \begin{pmatrix} \Psi_3 k_1 \\ \Psi_3 k_2 \\ \Psi_1 k_1 + \Psi_2 k_2\end{pmatrix} \right.\\ &\left. + (k_1^2+k_2^2) \left( \boldsymbol{\Psi} - \Psi_3 {\mathbf{e}}_3\right) + \begin{pmatrix} -\Psi_1 k_1^2 - \Psi_2 k_2k_1 \\ -\Psi_2 k_2^2 - \Psi_1 k_1k_2 \\(k_1^2+k_2^2) \Psi_3 \end{pmatrix} \!\! \right) \\ = &-\frac{1}{ k_3^2 - q_{\varepsilon}^2} \left((k_3^2-q_{\varepsilon}^2) \left( \boldsymbol{\Psi} - \Psi_3 {\mathbf{e}}_3\right) - k_3 \begin{pmatrix} \Psi_3 k_1 \\ \Psi_3 k_2 \\ \Psi_1 k_1 + \Psi_2 k_2\end{pmatrix} + \kappa^2 \left( \boldsymbol{\Psi} - \Psi_3 {\mathbf{e}}_3\right)\right.\\ &\left. + \begin{pmatrix} -\Psi_1 k_1^2 - \Psi_2 k_2k_1 \\ -\Psi_2 k_2^2 - \Psi_1 k_1k_2 \\(k_1^2+k_2^2) \Psi_3 \end{pmatrix} \right) = - \boldsymbol{\Psi} + \Psi_3 {\mathbf{e}}_3 + \frac{k_3}{k_3^2 - q_{\varepsilon}^2} \mathbf{a} - \frac{1}{k_3^2 - q_{\varepsilon}^2} \mathbf{b} \end{aligned} $$

  where

  $$\displaystyle \begin{aligned} \mathbf{a} = \begin{pmatrix} \Psi_3 k_1 \\ \Psi_3 k_2 \\ \Psi_1 k_1 + \Psi_2 k_2\end{pmatrix} \text{ and } \mathbf{b} = \begin{pmatrix} (\kappa^2 - k_1^2) \Psi_1 -\Psi_2 k_1 k_2 \\ (\kappa^2 - k_2^2) \Psi_2 - \Psi_1 k_1 k_2 \\ (k_1^2 + k_2^2)\Psi_3 \end{pmatrix}. \end{aligned}$$

  Using that

  $$\displaystyle \begin{aligned} \frac{k_3}{k_3^2-q_{\varepsilon}^2} = \frac{1}{2} \left(\! \frac{1}{k_3-q_{\varepsilon}} + \frac{1}{k_3+q_{\varepsilon}} \!\right) \text{ and } \frac{1}{k_3^2-q_{\varepsilon}^2} = \frac{1}{2q_{\varepsilon}} \left( \frac{1}{k_3-q_{\varepsilon}} - \frac{1}{k_3+q_{\varepsilon}} \right) \end{aligned}$$

  we get

  $$\displaystyle \begin{aligned} \frac{(\boldsymbol{\Psi} \times \mathbf{k}) \times \mathbf{k}}{ k_3^2 - q_{\varepsilon}^2} + \boldsymbol{\Psi} - \Psi_3 {\mathbf{e}}_3 = \frac{1}{k_3-q_{\varepsilon}} \left( \frac{1}{2} \mathbf{a} - \frac{1}{2q_{\varepsilon}} \mathbf{b} \right) + \frac{1}{k_3+q_{\varepsilon}} \left( \frac{1}{2}\mathbf{a} + \frac{1}{2q_{\varepsilon}} \mathbf{b} \right). \end{aligned} $$

  (112)

  This shows Eq. (108).

Therefore we can apply the residual Theorem 2 for f defined in (111). Then, since q _ε is complex with positive imaginary part, f has only one pole in the upper half plane, and we get

$$\displaystyle \begin{aligned} \int_{\mathbb{R}} f(k_3) \mathrm e^{\mathrm i k_3 (r_3 - r_3^{\boldsymbol{\Psi}})} d k_3 &= 2 \pi \mathrm i \text{Res} \left(f(z) \mathrm e^{\mathrm i z (r_3 - r_3^{\boldsymbol{\Psi}})}; z=q_{\varepsilon}\right) \\ &= 2 \pi \mathrm i \left( \frac{1}{2q_{\varepsilon}} \left( {\boldsymbol{\Psi}} \times \begin{pmatrix}k_1\\k_2\\q_{\varepsilon} \end{pmatrix} \right) \times \begin{pmatrix}k_1\\k_2\\q_{\varepsilon} \end{pmatrix} \mathrm e^{\mathrm i q_{\varepsilon} (r_3 - r_3^{\boldsymbol{\Psi}})} \right). \end{aligned}$$

This implies

$$\displaystyle \begin{aligned} &\frac{1}{\sqrt{2\pi}} \mathcal{F}_3^{-1} \left[k_3 \to \left(\boldsymbol{\Psi} + \frac{(\boldsymbol{\Psi} \times \mathbf{k}) \times \mathbf{k}}{\left| \mathbf{ k}\right|{}^2 - \kappa_\varepsilon^2} \right)\mathrm e^{-\mathrm i k_3 r_3^{\boldsymbol{\Psi}}}\right](r_3)\\ & = \Psi_3 {\mathbf{e}}_3\tilde \delta(r_3 - r_3^{\boldsymbol{\Psi}}) + \frac{\mathrm i \mathrm e^{\mathrm i q_{\varepsilon} (r_3-r_3^{\boldsymbol{\Psi}})}}{ 2q_{\varepsilon} } (\boldsymbol{\Psi} \times \mathbf{ k}_{q_{\varepsilon}}) \times {\mathbf{k}}_{q_{\varepsilon}}. \end{aligned}$$

Appendix 2: Derivation of the Far Field Approximation for the Electric Field

Lemma 11

Let ζ be the function defined in Eq.(61). That is, for all ${\mathbf {k}}_{12} \in \mathbb {R}^2$ ζ(k ₁₂) = k ₁₂ ⋅v + q(k ₁₂), with $q = \sqrt {\kappa ^2 - k_1^2 - k_2^2}$ ; note the formal definition of q in Eq.(42) is via the limit ε → 0. Then the gradient of ζ is given by

$$\displaystyle \begin{aligned} \nabla \zeta ({\mathbf{k}}_{12}) = \mathbf{v}- \frac{{\mathbf{k}}_{12}}{\sqrt{\kappa^2 - \left| \mathbf{ k}_{12}\right|{}^2}}, \end{aligned} $$

(113)

which vanishes for

$$\displaystyle \begin{aligned} \hat{\mathbf{k}}_{12} := \frac{\kappa}{\sqrt{1+\left| \mathbf{v}\right|{}^2}}\mathbf{v}. \end{aligned} $$

(114)

Consequently r = r ₃ e ₃+r ₃ ( v1 ) ^T as in Eq.(59) satisfies

$$\displaystyle \begin{aligned} \frac{\mathbf{r}}{\left| \mathbf{r}\right| } = \frac{1}{\sqrt{1+\left| \mathbf{v}\right|{}^2}} \begin{pmatrix} \mathbf{v}\\ 1 \end{pmatrix}. \end{aligned} $$

(115)

Moreover, with q as defined in Eq.(42), we get the following identities

$$\displaystyle \begin{aligned} & \zeta(\hat{\mathbf{k}}) = \hat{\mathbf{k}}_{12} \cdot \mathbf{v} + q =\kappa \sqrt{1+\left| \mathbf{v}\right|{}^2} \quad \mathit{\text{ and }}\\ & q = q(\hat{\mathbf{k}}) = \sqrt{ \kappa^2 - \hat{k}_1^2-\hat{k}_2^2} = \frac{\kappa}{\sqrt{1+\left| \mathbf{ v}\right|{}^2}}. \end{aligned} $$

(116)

The Hessian of ζ is given by

$$\displaystyle \begin{aligned} H(\zeta)({\mathbf{k}}_{12}) = \begin{pmatrix} - \frac{1}{\sqrt{\kappa^2 - \left| {\mathbf{k}}_{12}\right|{}^2}} - \frac{k_1^2}{(\kappa^2 - \left| {\mathbf{k}}_{12}\right|{}^2)^{3/2}} & - \frac{k_1 k_2}{(\kappa^2 - \left| {\mathbf{k}}_{12}\right|{}^2)^{3/2}} \\- \frac{k_1 k_2}{(\kappa^2 - \left| {\mathbf{k}}_{12}\right|{}^2)^{3/2}} & - \frac{1}{\sqrt{\kappa^2 - \left| {\mathbf{k}}_{12}\right|{}^2}} - \frac{k_2^2}{(\kappa^2 - \left| {\mathbf{k}}_{12}\right|{}^2)^{3/2}} \end{pmatrix}, \end{aligned}$$

which evaluated at $\hat {\mathbf {k}}$ gives

$$\displaystyle \begin{aligned} &H(\zeta)(\hat{\mathbf{k}}) = \\ &\begin{pmatrix} - \frac{c}{\omega}\sqrt{1+\left| \mathbf{v}\right|{}^2} - \frac{ c^3}{\omega^3 }(1+\left| \mathbf{v}\right|{}^2)^{3/2} \cdot \frac{v_1^2 \omega^2}{c^2 (1+ \left| \mathbf{v}\right|{}^2)} &- \frac{ c^3}{\omega^3 }(1+\left| \mathbf{v}\right|{}^2)^{3/2} \cdot \frac{v_1v_2 \omega^2}{c^2 (1+ \left| \mathbf{v}\right|{}^2)} \\ - \frac{ c^3}{\omega^3 }(1+\left| \mathbf{v}\right|{}^2)^{3/2} \cdot \frac{v_1v_2 \omega^2}{c^2 (1+ \left| \mathbf{ v}\right|{}^2)} & - \frac{c}{\omega}\sqrt{1+\left| \mathbf{v}\right|{}^2} - \frac{ c^3}{\omega^3 }(1+\left| \mathbf{v}\right|{}^2)^{3/2} \cdot \frac{v_2^2 \omega^2}{c^2 (1+ \left| \mathbf{v}\right|{}^2)} \end{pmatrix}, \\ & = - \frac{c}{\omega} \sqrt{1+\left| \mathbf{v}\right|{}^2} \begin{pmatrix} 1 + v_1^2 & v_1 v_2 \\v_1 v_2 & 1+v_2^2 \end{pmatrix}, \end{aligned} $$

and the determinant satisfies

$$\displaystyle \begin{aligned} = \frac{c^2(1+\left| \mathbf{v}\right|{}^2)^2}{\omega^2} > 0. \end{aligned}$$

Appendix 3: Bessel Identities

For $x \in \mathbb {R}$ and φ ₀ ∈ [0, 2π) the Bessel-identities hold:

$$\displaystyle \begin{aligned} 2 \pi (-1)^m \mathrm i^m J_m(x) \cos (m \varphi_0) &= \int_{0}^{2 \pi} \mathrm e^{-\mathrm i x \cos (\varphi - \varphi_0)} \cos (m \varphi) d \varphi, \\ 2 \pi (-1)^m \mathrm i^m J_m(x) \sin (m \varphi_0) &= \int_{0}^{2 \pi} \mathrm e^{-\mathrm i x \cos (\varphi - \varphi_0)} \sin (m \varphi) d \varphi, \end{aligned} $$

(117)

where J _m is the Bessel function of the first kind of order m.