Lie algebraically, the eigenvalue of the spherical function \begin{align*} \phi_{\lambda}(g)=\int_{K} e^{(i \lambda+\rho)(A(k g))} \mathrm{d} k \quad (g \in G,\,\lambda\in\mathfrak{a}^*) \end{align*} on a real semisimple group $G=NAK$, where $A(g) \in \mathfrak{a}$ and $u(g) \in K$ denote the unique elements such that $g \in N \exp A(g) u(g)$, $\rho$ denotes the half-sum of restricted positive roots of $\mathfrak{g}$ and $\langle\cdot,\cdot\rangle$ denotes the inner product induced on $\mathfrak{a}^{*}$ by the Killing form $B$, is given by \begin{align*} \Delta \phi_{\lambda}=-(\langle\rho,\rho\rangle+\langle\lambda,\lambda\rangle)\, \phi_{\lambda}. \end{align*} In trying to calculate this value for $G=\mathrm{SL}(2,\mathbb{R})$, i.e., $\mathfrak{g}=\mathfrak{sl}(2,\mathbb{R})$ from this formula, I am getting $\langle\rho,\rho\rangle+\langle\lambda,\lambda\rangle=\frac{1}{8}(1+\lambda^2)\,(\lambda\in\mathbb{R})$, whereas this eigenvalue for the radial Laplacian on $\mathbb{H}=\mathrm{SL}(2,\mathbb{R})/\mathrm{SO}(2,\mathbb{R})$ is well known to be \begin{align*} \Delta \phi_{\frac{1}{2}+i\lambda}=-\Big(\frac{1}{4}+\lambda^2\Big)\, \phi_{\frac{1}{2}+i\lambda}\quad(\lambda\in\mathbb{R}). \end{align*} So, where am I going wrong? my calculations are as follows:
For $G=\mathrm{SL}(2,\mathbb{R})$, i.e., $\mathfrak{g}=\mathfrak{sl}(2,\mathbb{R})$, I am taking the standard basis \begin{align*} H=\begin{pmatrix} 1 & 0\\ 0 & -1\end{pmatrix},\quad X=\begin{pmatrix} 0 & 1\\ 0 & 0\end{pmatrix}, \quad Y=\begin{pmatrix} 0 & 0\\ 1 & 0\end{pmatrix}, \end{align*} which satisfy the relations \begin{align*} [X,Y]=H,\quad [H,X]=2X,\quad [H,Y]=-2Y. \end{align*} Therefore, in the adjoint representation, we have \begin{align*} \mathrm{ad} (H)=\begin{pmatrix} 0 & 0 & 0\\ 0 & 2 & 0\\ 0 & 0 & -2 \end{pmatrix},\quad \mathrm{ad} (X)=\begin{pmatrix} 0 & 0 & 1\\ -2 & 0 & 0\\ 0 & 0 & 0 \end{pmatrix},\quad \mathrm{ad} (Y)=\begin{pmatrix} 0 & -1 & 0\\ 0 & 0 & 0\\ 2 & 0 & 0 \end{pmatrix}. \end{align*} The matrix of the Killing form $B$ is then given by \begin{align*} \begin{pmatrix} 8 & 0 & 0\\ 0 & 0 & 4\\ 0 & 4 & 0 \end{pmatrix}. \end{align*} Therefore, writing \begin{align*} Q_1=\begin{pmatrix}a_1 & b_1\\ c_1 & -a_1\end{pmatrix}\in \mathfrak{sl}(2,\mathbb{R}) \quad \text{and}\quad Q_2=\begin{pmatrix}a_2 & b_2\\ c_2 & -a_2\end{pmatrix}\in \mathfrak{sl}(2,\mathbb{R}) \end{align*} in the above standard basis $\{H,X,Y\}$, we calculate the Killing form on $\mathfrak{sl}(2,\mathbb{R})$ as \begin{align*} B(Q_1,Q_2)&=\begin{pmatrix} a_1 & b_1 & c_1\end{pmatrix}\begin{pmatrix} 8 & 0 & 0\\ 0 & 0 & 4\\ 0 & 4 & 0 \end{pmatrix} \begin{pmatrix} a_2 \\ b_2 \\ c_2\end{pmatrix},\\ &=4(2a_1 a_2+b_1 c_2+b_2 c_1),\\ &=4\mathrm{tr} (Q_1 Q_2). \end{align*} Then taking $e\in\mathfrak{a}^*$ given by the assignment $(\begin{smallmatrix}r & 0\\ 0 & -r\end{smallmatrix})\mapsto r$ as a basis of $\mathfrak{a}^*$, we calculate the roots of $\mathfrak{g}$ to be $2e$ and $-2e$ as \begin{align*} \bigg[\begin{pmatrix}r & 0\\ 0 & -r\end{pmatrix},\begin{pmatrix} 0 & 1\\ 0 & 0\end{pmatrix}\bigg]&=2r \begin{pmatrix} 0 & 1\\ 0 & 0\end{pmatrix}= 2e \bigg(\begin{pmatrix}r & 0\\ 0 & -r\end{pmatrix}\bigg)\begin{pmatrix} 0 & 1\\ 0 & 0\end{pmatrix},\\ \bigg[\begin{pmatrix}r & 0\\ 0 & -r\end{pmatrix},\begin{pmatrix} 0 & 0\\ 1 & 0\end{pmatrix}\bigg]&=-2r\begin{pmatrix} 0 & 0\\ 1 & 0\end{pmatrix} = -2e \bigg(\begin{pmatrix}r & 0\\ 0 & -r\end{pmatrix}\bigg)\begin{pmatrix} 0 & 0\\ 1 & 0\end{pmatrix}. \end{align*} Of these we take $2e$ to be the positive root. So the half-sum of positive roots is given by $\rho=e$.
Next, since \begin{align*} B \bigg(\begin{pmatrix} \frac{1}{8} & 0 \\ 0 & -\frac{1}{8}\end{pmatrix},\begin{pmatrix}r & 0\\ 0 & -r\end{pmatrix}\bigg)=e \bigg(\begin{pmatrix}r & 0\\ 0 & -r\end{pmatrix}\bigg), \end{align*} the Killing form $B(X,Y)=4\mathrm{tr}(XY)$ assigns $e\in\mathfrak{a}^*$ to $H_e= \big(\begin{smallmatrix} 1/8 & 0 \\ 0 & -1/8\end{smallmatrix}\big)\in \mathfrak{a}$. So the induced scalar product $\langle \cdot , \cdot \rangle$ on $\mathfrak{a}^*$ is given by \begin{align*} \langle e,e \rangle=B \bigg(\begin{pmatrix} \frac{1}{8} & 0 \\ 0 & -\frac{1}{8}\end{pmatrix},\begin{pmatrix} \frac{1}{8} & 0 \\ 0 & -\frac{1}{8}\end{pmatrix}\bigg)=\frac{1}{8}\,, \end{align*} thereby giving $\langle\rho,\rho\rangle+\langle\lambda,\lambda\rangle=\frac{1}{8}(1+\lambda^2)$, (with the abuse of notation $\lambda=\lambda e\,(\lambda\in\mathbb{R})$.
So where am I going wrong or what is the connection between these two formulas?