This “game” has two Nash equilibria. If we both go home, neither of us regrets it: we can spend time with each other and we've both got our highest utility. If we both stay at work, again, neither of us regrets it: since my girlfriend is at work, I am glad I stayed at work instead of going home, and since I am at work, my girlfriend is glad she stayed at work instead of going home.

Looking at the problem, I believe there is a third equilibrium, a mixed one. Both you and your girlfriend toss a coin, and choose to go home with probability one half, or stay at work with probability one half. This gives you both an expected utility of 2. If you are playing that strategy, then it doesn't matter to your girlfriend whether she stays at work (definite utility of 2) or goes home (50% probability of 1, 50% probability of 3), so she can't do better than tossing a coin.

Incidentally, this is expected from Wilson's oddness theorem (1971) - almost all finite games have an odd number of equilibria.