Interesting case. From the wikipedia article, an average white dwarf has a mass of ~0.6 Msun and a radius of ~0.015 Rsun. If we want it to have the same effective temperature as the Sun, and Earth to end up with the same insolation, then the size a of the orbit is determined by the scaling relation
Originally Posted by Xibalba
Rsun^2 / (1 AU)^2 ~ (0.015 Rsun)^2 / a^2 ===> a ~ 0.015 AU ~ 2*10^9 m
which is only five times the distance to the Moon. Tidal forces depend on mass and the inverse cube of distance, which gives us in terms of the lunar tidals:
Ftidal ~ ((1.2*10^30 kg)/(7*10^22 kg)) / 5^3 Ftidal|lunar ~ 1.5*10^5 Ftidal|lunar
Instead of a tidal bulge on the order of a metre, it would be on the order of 100 km. Meaning that until tidal lock is achieved, the only permanent bodies of water would be lakes with sufficiently steep sides. Anything with shallow sides, including the oceans, would lose its water to two big blobs which would do their best to remain stationary with respect to the new sun while the planet rotates along underneath them. I guess life better stays in those sealed bunkers for the duration. The ordinary scaling law for time to tidal lock is something like
Tlock ~ (10^10 years) / ((Trot in days) (Ftidal/Ftidal|lunar)^2) ~ (1 year) / (Trot in days)
where Trot is the original (unlocked) day-length with respect to the tide-inducing body. Unless Earth spins a lot faster upon capture than it does now, the result is on the order of 1 year. If, on the other hand, Earth spins a lot slower, which seems like the more likely scenario to me, then the day-length would be mainly determined by the year-length, which works out on the order of 1 day (as in 24 hours), funnily enough.
Mind you, I'm not sure if this situation might not be too extraordinary for the scaling relation to apply in that form. It assumes that the combination of atmospheric and ocean tides and planetary deformation is sufficient to actually dissipate rotational energy into heat at the maximal rate, which seems questionable. Assuming that it does, the power output would be vast:
Ptidal ~ Erot / Tlock
Ptidal ~ (1/2 I w^2) / ((1 year) / (Trot in days))
Ptidal ~ (1/2 (2/5 M R^2) ((2 pi)/Trot)^2) / ((1 year) * (1 day) / Trot)
Ptidal ~ (7 * (6*10^24 kg) * (6*10^6 m)^2) / ((3*10^7 s) * (10^5 s) * Trot)
Ptidal ~ (5*10^26 kg m^2 / s^2) / Trot
Ptidal ~ Lsun / (Trot in seconds)
I'm not sure where all these numerical coincidences come from, but that aside, it does give one quite a good sense of scale: The Earth has about one ten-thousandth the surface of the Sun, and Trot should be on the order of one hundred thousand, so for the Earth to dissipate that amount of power, it would have to output only one order of magnitude less power per unit surface than the Sun does. Since black-body power output scales with T^4, that means more than half the temperature of the sun. And as rock begins to melt by as little as 1,000 K, this would directly liquify the upper portions of the planet. Assuming the interior has been frozen solid in the interim, one ends up with a sort of inverted planet whose surface is hotter than its core. In the long run, what is of more practical import is that this might just be hot enough to boil away (away as in all the way into space) all of our water, even in as little as that one year we're talking about.
As I said, the planet might well not actually have the capacity to dissipate power at that rate, so things might not get quite that bad. The price for that, though, would be a longer time to lock, so the not-quite-that-bad conditions would last a lot longer. Conclusion: Unless the planet is already as close as no matter to tidal lock when it assumes a habitable-zone orbit, which seems highly unlikely, the process of locking it would pretty much make it uninhabitable... Catch-22. The only upside I see is that this might indeed just about do what you were hoping for, in the long run, and kick-start the planet's dormant geological activity again.