Google today released an experimental payments app called Hands Free, a service that lets you pay for items in stores without taking your phone out. The iOS and Android app relies on Wi-Fi, Bluetooth, and location services to determine when you're nearby a participating retailer or restaurant. When you get to the cashier, all you have to say is, "I'll pay with Google," and the clerk can confirm your identity with a photo on their end to complete the transaction. The service is only available in the southern end of the San Francisco Bay Area right now, and participating stores include McDonald's, Papa John's, and a handful of local retailers Google doesn't list by name.
Hands Free, which was first announced last May during Google's I/O developer conference, is designed to be a companion to Android Pay, the separate Google-owned payments service. Android Pay now has around 9 million registered members, but the company "wanted to explore what the future of mobile payments could look like," writes Pali Bhat, a senior director on the Hands Free project.
Cashiers have to identify you by checking your initials and the photo you submit to Hands Free. But Google says it's working on implementing an in-store camera system that would automatically confirm your identify by taking a picture and cross-checking it with your Hands Free profile. "All images captured by the Hands Free camera are deleted immediately," Bhat writes.
Eerie photo recognition systems aside, the idea of paying with just your face is not a new one. Both PayPal and Square have tried similar initiatives in the past and both have failed to take off, mostly due to a limited number of participating merchants and the sheer enormity of trying to replace credit cards and cash. Android Pay, which lets you put your phone next to a NFC terminal to pay, is the most mainstream of Google's payment options out there as it doesn't require anything beyond upgrading existing card swiping tech. So it's unclear how far Hands Free can spread, even if it does remove the phone entirely from the equation.