You do not need to drop down the voltage. You actually need to limit the current. The LED will drop its rated voltage, if the current is limited by some means.
Use a simple 1k resistor in series with your IR-LED. The current throgh it would be approximately
(9 - 1.5)/1k = 7.5 mA
Visible LED ususally works well with currents in the order of 10 mA. I don't know about IR-LEDs, they should have same characteristics. If necessary reduce this resistance.
It's the current you have to worry about, anyway, not the voltage. 9 volt
power supply, the LED takes 1.5 volts, that leaves 7.5 volts for the
resistor.
Pick the resistor which will give the correct current at 7.5 volts.
Normally IR-LED takes current from 20 mA to 500 mA, checkup the current requirement and find out the value and power rating of resistor to meet your requirement. If you use 270 ohm, current flow limited to (7.5/270) 27.7 mA. It would require 1/4 (P=7.5 * 27.7 mA) Watt.
Without knowing the fwd current of the IR-LED, it would at best be hard to tell you what resistance to use. Also, I am surprised you can not use ohm's law to find the answeer for yourself, shame on the ones who responded and answered a very simple question any reasonable good technician should be able to figure out in under a minute (without a calculator). For yourself, go pickup a basic electronics book and learn how to use ohm's law and NO, I will not give you the answer, to be an engineer or technician you will have to put in some effort on your own. Would you give your kids the answers to their school work?