## Saturday, January 5, 2013

### The Uncertainty Principle and Energy Non-Conservation and Why Your Textbook is Wrong

I read this all the time, in physics books and articles and on the internet: Apparent violation of conservation of energy is possible at the quantum scale for very short periods of time due to the Heisenberg uncertainty relation:
∆E∆t ≥h/2π
In that equation, ∆E is the "uncertainty" in the energy and ∆t the "uncertainty" in the time, meaning the accuracy to which we are able to measure these values.  The h in the equation is Planck's constant (which I didn't write as h-bar because I didn't want to encode LaTeX for one equation).  The two are inversely proportional, so as one goes up, the other must go down, so for short times, you can get enough "free" energy to send a particle through an energy barrier.

This is wrong.  Wrong, wrong, wrong.

If you've ever taken a science class you may have heard the uncertainty principle stated in simple terms, or even seen that equation or its cousin
∆x∆p ≥ h/2π
put up on a board.  It's one of those weird and crazy things about quantum mechanics that's just weird and crazy but makes you sound smart to say at parties.

But if you've ever taken second-semester undergrad quantum, then you have actually derived the uncertainty principle from the commutation relations, and if you recall the process of so doing, then you will remember that the greater-than sign "≥" is extremely, absolutely, necessary.  It is not an equal sign "=" or an order-of sign "~", and definitely not a less-than sign "≤".

The product of uncertainties does not have to bound within phase space of total area h-bar.  It must exceed total area h-bar.

So let's take the canonical argument.  Due to the uncertainty principle, you can have apparent violations of conservation of energy provided that the time is short.  Doing some division, the amount of "free-energy" a particle can borrow is like
∆E ~ h/2π∆t.
Here you can see, the argument goes, that if ∆t gets really big (for long times), then ∆E gets really small (for small energy), but if ∆t is small, then ∆E is big.

Except that, as I said, Heisenberg's uncertainty principle is not a "on the order of" relation, it is an inequality.  The actual algebra gives you

∆E ≥ h/2π∆t.

What happens when ∆t is really large?  Then ∆E must be equal to some really small number, or it must be larger than some really small number... and that's most numbers.

Say ∆t = 1,000,000,000 seconds.  Then h/2π∆t is on the order of 10^{-43} Joules.  The amount of energy you can borrow must be LARGER than that number.

So howabout, ∆E = 1,000,000,000,000 Joules?  Is that larger than 10^{-43}?

Yes.  It is.  Much larger.  Heisenberg is happy.

Is 1,000,000,000 seconds a short time?  No, not even close.  It's something like two millennia.

According to this argument, one of the particles in St. Ignatius's body, just before his martyrdom, might have "borrowed" an atomic-bomb's worth of energy and only have to worry about paying it back right about now.  And that's only one particle.

And it only gets more true (i.e. "worse") as the energy and time get larger and larger and larger.  If ∆E=∞, ∆t=∞, then ∞•∞=∞>>hbar.  Heisenberg is still happy.

If this argument is valid, if particles are allowed to "borrow" energy for short periods of time in accordance with the uncertainty principle, then every particle everywhere in the universe could "borrow" as much energy as it wanted, forever, whenever it wanted.

I'm not saying that quantum mechanics does not allow for interactions to occur despite an apparent energy barrier, nor that the energy barrier that can be crossed is not inversely proportional to the time of the interaction.  I'm saying that this argument does not explain why this is possible.  If this is what happens, then there must be a better (read: valid) argument for it, somewhere.

If you know this argument, then please don't hesitate to point it out to me.  I would really like this issue laid to rest in my mind.

And I'd also like for people to stop using this bad explanation.