Why would that be necessary? You don't have long distances or excessive common mode voltages between the input circuit and the output circuit.

are you sure about that? according to OP, alarm is something that monitors phone line (that sound pretty long to me). we don't know anything about the alarm and how is it referenced. Raspberry PI has USB etc. if that USB connection is to a desktop computer for example, that would reference it to ground as USB and PC ground are connected to mains earth. sounds to me as we could have a need for isolation. at least i would not want my computer connected to something that is not isolated.

but lets look at requirements too - all we want is to take the signal from one circuit and pass it to another. i repeat, we only need signal, we do not need to introduce galvanic connection between the two (at least I could see no such requirement), so WHY would you want to do that?

and if there is ANY argument against isolation, what would it be?
circuit is simple, cheap and - provides piece of mind. what is wrong with that?

 

to get +3.3V output when 12V is present just swap output of the optocoupler and R3. you could use values such as
R1: 560 Ohm
LED: any common LED
opto: any common optocoupler (PS2501, K817, VOM617A, )
R2: 10k

 

...if there is ANY argument against isolation, what would it be?

It assumes you have enough current at 12V to light the LED. That's probably not a problem for this application.

A small bit of complexity for a noob is about the only other argument I would have. Once you know about isolation and have the parts in your box, it's a good way to go. As I said earlier, it's the elegant solution.

 

you are right of course. that was the assumption but I'm pretty sure it is reasonable. i can't think of a DC output that was not rated for at least 20mA. even logic gates and MCUs can handle that much. on 12V or more, the lowest current rating I've ever seen was 50mA. note to self, never assume...