diff --git a/cubic-phase.md b/cubic-phase.md
index 054df8a..e3146b0 100644
--- a/cubic-phase.md
+++ b/cubic-phase.md
@@ -1,5 +1,4 @@
 # Cubic phase gate
-
 The cubic phase gate is defined as
 
 $``P(t) = \exp\left( i t \hat{q}^2 / 2 \hbar \right)``$
@@ -24,4 +23,68 @@ The above equations gives rise to the following decomposition in Hilbert space
 
 $``P(t) = e^{i \varphi}\exp\left( i \theta \hat{a}^\dagger \hat{a} \right) \exp(\tfrac{r}{2}(\hat{a}^{\dagger 2} - \hat{a}^2)) \exp\left( i (\tfrac{\pi}{2} - \theta) \hat{a}^\dagger \hat{a}  \right) ``$
 
-where $``\exp(i \varphi) = \frac{\sqrt[4]{1+\tfrac{t^2}{4}} }{\sqrt{1 - i \tfrac{t}{2}}}``$.
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+where $``\exp(i \varphi) = \frac{\sqrt[4]{1+\tfrac{t^2}{4}} }{\sqrt{1 - i \tfrac{t}{2}}}``$.
+
+We now compute the coherent-state matrix elements of $P(t)$
+
+$$\langle \alpha^* \vert P(t) \vert \beta \rangle = \langle \alpha^* \vert e^{i \varphi}\exp\left( i \theta \hat{a}^\dagger \hat{a} \right) \exp(\tfrac{r}{2}(\hat{a}^{\dagger 2} - \hat{a}^2)) \exp\left( i (\tfrac{\pi}{2} - \theta) \hat{a}^\dagger \hat{a}  \right) \vert \beta \rangle,$$
+
+where we use the following identity to express the squeezing operator as 
+
+$$\exp{\left(\frac{r}{2}(\hat{a}^{\dagger 2} - \hat{a}^{2})\right)} = \exp{\left(\frac{\tanh{r}}{2}\hat{a}^{\dagger 2}\right)}\exp{\left(-[\hat{a}^{\dagger}\hat{a} + 1/2]\ln{\cosh{r}}\right)}\exp{\left(-\frac{\tanh{r}}{2}\hat{a}^{2}\right)},$$
+
+and a second identity to normal-order this expression
+
+$$\exp{(\theta \hat{a}^{\dagger}\hat{a})} = :\exp{([\exp{(\theta)} - 1]\hat{a}^{\dagger}\hat{a})}:,$$
+
+with $:(\hat{a}^{\dagger}\hat{a})^{p}: = \hat{a}^{\dagger p}\hat{a}^{p}$.
+
+Using these two identities, the decomposition of $P(t)$ takes the form
+
+$$P(t) = e^{i\varphi} \exp(i\theta \hat{a}^{\dagger}\hat{a})\exp{\left(\frac{\tanh{r}}{2}\hat{a}^{\dagger 2}\right)}\times:\frac{\exp{([1/\cosh{r}-1]\hat{a}^{\dagger}\hat{a})}}{\sqrt{\cosh{r}}}:\times\exp{\left(-\frac{\tanh{r}}{2}\hat{a}^{2}\right)}\exp{(i(\pi/2-\theta)\hat{a}^{\dagger}\hat{a})}.$$
+
+Evaluating the matrix element $\langle \alpha^* \vert \dots \vert \beta \rangle = e^{i\varphi} \langle \bar{\alpha}^* \vert \dots \vert \bar{\beta} \rangle$ gives
+
+$$e^{i\varphi} \langle \bar{\alpha}^* \vert \exp{\left(\frac{\tanh{r}}{2}\hat{a}^{\dagger 2}\right)}\times:\frac{\exp{([1/\cosh{r}-1]\hat{a}^{\dagger}\hat{a})}}{\sqrt{\cosh{r}}}:\times\exp{\left(-\frac{\tanh{r}}{2}\hat{a}^{2}\right)} \vert \bar{\beta} \rangle$$
+
+where we have introduced the rotated labels $\langle \bar{\alpha}^* \vert \equiv \langle \alpha^* e^{-i\theta} \vert = \langle \alpha^* \vert e^{i\theta \hat{a}^{\dagger}\hat{a}}$ and $\vert \bar{\beta} \rangle \equiv \vert \beta e^{i(\pi/2 - \theta)} \rangle = e^{i(\pi/2 - \theta) \hat{a}^{\dagger}\hat{a}} \vert \beta \rangle$ to simplify the expression. The resulting expression takes the form
+
+$$\sqrt{2}e^{i\varphi} \frac{\exp{[-\frac{1}{2}(\lvert \alpha \rvert^{2} + \lvert \beta \rvert^{2})]}}{\sqrt[4]{4+t^{2}}} \exp{\left(-\frac{1}{2} \begin{bmatrix} \alpha & \beta \end{bmatrix} \begin{bmatrix} f & g \\\\ g & f^* \end{bmatrix} \begin{bmatrix} \alpha \\\\ \beta \end{bmatrix} \right)}$$
+
+where we used the overlap of two coherent states $\langle \bar{\alpha}^* \vert \bar{\beta} \rangle = \exp{(-\frac{1}{2}(\lvert\alpha\rvert^{2} + \lvert\beta\rvert^{2} - 2\bar{\alpha}\bar{\beta}))}$, and the functions $f$ and $g$ are given by
+
+$$\begin{aligned} 
+f &= -\tanh{r}e^{2i\theta}, \\
+g &= -\frac{i}{\cosh{r}},
+\end{aligned}$$
+
+or, in terms of the parameter $t$
+
+$$\begin{aligned} 
+f &= \frac{t(t - 2i)}{4 + t^{2}}, \\
+g &= - \frac{2i}{\sqrt{4 + t^{2}}}.
+\end{aligned}$$
+
+An important result is
+
+$$
+[\hat{a},P(t)] = P\left\lbrace \frac{it}{2}(\hat{a}^{\dagger} + \hat{a}) \right\rbrace.
+$$
+
+We calculate the Fock-state matrix elements of the commutator above
+
+$$
+\langle m|[\hat{a},P(t)]|n \rangle = \sqrt{m+1}\,P_{m+1,n} - \sqrt{n}\,P_{m,n-1} = \frac{it}{2}\left\lbrace \sqrt{n+1}\,P_{m,n+1} + \sqrt{n}\,P_{m,n-1} \right\rbrace,
+$$
+
+with the initial condition $P_{0,0} = \frac{e^{i\varphi}}{\sqrt{\cosh{r}}} = \frac{1}{\sqrt{1 - it/2}}$.
+
+Solving for the term with highest index
+
+$$
+P_{m+1,n} = \frac{1}{\sqrt{m+1}} \left\lbrace \sqrt{n}\left(1 + \frac{it}{2}\right)P_{m,n-1} + \frac{it}{2}\sqrt{n+1}P_{m,n+1} \right\rbrace,
+$$
+
+$$
+P_{m+1,0} = \frac{it}{2\sqrt{m+1}} P_{m,1}.
+$$
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