# Show for $r \in \mathbb R$, $e^{(a_1 ^\dagger a_2^\dagger - a_1a_2)r}|0,0 \rangle = \frac{1}{\cosh r} \sum_{j=0}^\infty (\tanh r)^j |j,j \rangle$

Let $$a_1, a_2$$ be annihilation operators for the first and second component in the product state $$|m,n \rangle$$ using Fock basis.

Following "Quantum continuous variables, A primer of Theoretical Methods" by A.Serafini, page 93, I want to show for $$r \in \mathbb R$$,

$$e^{(a_1 ^\dagger a_2^\dagger - a_1a_2)r}|0,0 \rangle = \frac{1}{\cosh r} \sum_{j=0}^\infty (\tanh r)^j |j,j \rangle\tag{5.23}$$ holds.

I am given the hint to differentiate both sides with respect to $$r$$ and check if they are equal, but I am having trouble with that.

Any help is appreciated.

$$\frac{d}{dr}\bigg(\frac{1}{\cosh r} \sum_{j=0} ^{\infty} (\tanh r)^j |j,j \rangle \bigg) \\=\sum_{j=0}^\infty \big(\frac{1}{\cosh^3 r}j(\tanh r)^{j-1}-\frac{1}{\cosh r} (\tanh r)^{j+1} \big) |j,j \rangle =\\ \frac{1}{\cosh r} \sum_{j=0} ^{\infty} \big( j(\tanh r)^{j-1} -(j+1) (\tanh r)^{j+1} \big) |j,j \rangle \\=\frac{1}{\cosh r} \sum_{j=0} ^{\infty} (\tanh r)^j \big(a_1 ^\dagger a_2^\dagger - a_1a_2 \big) |j,j \rangle \\=\big(a_1 ^\dagger a_2^\dagger - a_1a_2 \big) \big(\frac{1}{\cosh r} \sum_{j=0} ^{\infty} (\tanh r)^j |j,j \rangle \big)$$
Thus, together with the fact that putting $$r=0$$ in the right hand side yields $$|0,0 \rangle$$, $$\frac{1}{\cosh r} \sum_{j=0} ^{\infty} (\tanh r)^j |j,j \rangle=e^{(a_1 ^\dagger a_2^\dagger - a_1a_2)r}|0,0 \rangle$$