$\rho = \psi_1^* \psi_2$ is the probability density for transition from state 2 to 1. I am having trouble interpreting the probability though $$dP_{1\leftarrow 2}=\psi_1^* \psi_2 d^3 r $$ does this mean the probability in a region of volume $d^3 r$ to go from state 2 to state 1? That does not make sense at all!

EDIT: I agree with Marek's answer. But why is this not the probability density. If we have two states:
$$ -i \hbar \partial_t \psi_{1,2}= -\frac{\hbar^2}{2m} \nabla^2 \psi_{1,2} + V\psi_{1,2}$$

After making standard manipulations (taking conjugate, multiplying by conjugate, subtracting) we would get $$i\hbar \partial_t ( \psi_1^*\psi_2) +\frac{\hbar}{2mi} \nabla\cdot \left( \psi_1^* \nabla \psi_2 - (\nabla \psi_1^* )\psi_2\right)=0$$ This seems consistent with the continuity equation: $$\partial_t \rho +\nabla.\mathbb{j} =0$$ However, we seem to be changing its relation to the probability by making the probability $P=\int \int\rho^*(r,r') \rho(r,r') d^3r d^3r' $ instead of the regular $P=\int \rho(r) d^3r$ For a kind of intuitive "dimensional consistency" can we instead say $P=\int \int \; \sqrt{\rho^* \rho} \; d^3r d^3r'$

Full steps: $\psi_1$ and $\psi_2$ are two states of the same system, i.e both satisfy the SE of the same potential: Taking the SE for $\psi_2$ and multiplying by $\psi_1^*$ : $$ \psi_1^*\times\left( i \hbar \partial_t \psi_2 = -\frac{\hbar^2}{2m} \nabla^2 \psi_2 + V\psi_2\right)$$ Taking the SE for $\psi_1$ complex conjugating the entire equation and multiplying by $\psi_2$ : $$\left( -i \hbar \partial_t \psi_1^*= -\frac{\hbar^2}{2m} \nabla^2 \psi_1^* + V\psi_1^*\right)\times \psi_2$$

($V$ is real, no decay processes etc) On subtracting the second equation from the first

$$i\hbar \partial_t (\psi_1^*\psi_2 )= -\frac{\hbar}{2mi} \left( \psi_1^* \nabla^2 \psi_2 - (\nabla^2 \psi_1^* )\psi_2\right)$$ which is the same as $$ i\hbar \partial_t (\psi_1^*\psi_2 )= -\frac{\hbar}{2mi} \nabla\cdot \left( \psi_1^* \nabla \psi_2 - (\nabla \psi_1^* )\psi_2\right)$$