## Sunday, May 26, 2013

### The Uncertainty Principle and Energy Non-Conservation, part 2

Quantum mechanics is typically interpreted to mean that the conservation of energy can be violated as long as the time scales involved are short.  An old professor of mine used to summarize it as "there is such a thing as a free lunch, if you can eat it fast enough."

Here's how the argument goes.  From quantum mechanics, we get the uncertainty relation
$$\Delta E \Delta t \geq \hbar,$$
where $\Delta E$ is the uncertainty -- or statistical spread -- of the energy, and $\Delta t$ is the uncertainty of the time.

Following this, physicists reinterpret the uncertainty $\Delta E$.  Rather than representing a quantification of our lack of knowledge about the energy of a system, this is interpreted as being, somehow, the amount of "free" energy that a system can borrow in violation of the First Law of Thermodynamics.  So if we have mean energy $E$ and uncertainty $\Delta E$, it means we "actually have" energy $E$, and then Nature gracefully lends us $\Delta E$ to overcome some energy barrier, which we quickly repay in time $\Delta t$.

However, that puts us at
$$\Delta E \geq \hbar/\Delta t$$
which puts no limitation how much energy we can borrow.  Or, rather, it puts a lower bound; we must borrow at least $\hbar/\Delta t$ worth of energy.  Or, we could borrow even more!   If this is true, then we have infinite energy forever!

The oil companies will go bankrupt!

To get around this, physicists then re-reinterpret the uncertainty relation, changing the $\geq$ sign in to a sort of $\simeq$ sign.  Exactly what $\simeq$ means isn't supposed to be exact, but it implies that the size of the numbers in $\hbar/\Delta t$ are about the same as the size of the numbers in $\Delta E$.  So if $\hbar/\Delta t$ is something like $5.7\times10^{-22}$, then $\Delta E$ might be something like $9.3\times10^{-21}$ (I just made those numbers up, they aren't real numbers).

Therefore, the amount of energy we can borrow, $\Delta E$, is inversely proportional to the time scale $\Delta t$ of the problem.

Frankly, I think the whole thing is pretty fishy, but it's the standard interpretation of this.  I recently went through one of my old, beloved textbooks by R. Shankar on quantum mechanics.  (Let me just say, as an aside, that this book is a fantastic text to use if you wish to prepare undergrads for graduate-level quantum mechanics and I'm highly partial towards it.)  Anyway, I looked up his argument on the time-energy uncertainty relation, and I thought he argument was a particularly clear illustration of how this is wrong and how to fix it.  Here I quote the final section in his chapter on the uncertainty relations:
There exists an uncertainty relation
$\Delta E \Delta t \geq \hbar/2$
which does not follow from Eq (9.2.12) since time $t$ is not a dynamical variable but a parameter.  The content of this equation is quite different from the others involving just dynamical variables.  The rough meaning of this inequality is that the energy of a system that has been in existence for only a finite time $\Delta t$ has a spread (or uncertainty) of at least $\Delta E$, where $\Delta E$ and $\Delta t$ are related by (9.5.1).  To see how this comes about, recall that eigenstates of energy have a time dependence $e^{-iEt/\hbar}$, i.e., a definite energy is associated with a definite frequency, $\omega=E/\hbar$.  Now, only a wave train that that is infinitely long long in time (that is to say, a system that has been in existence only for an infinite time) has a well-defined frequency.  Thus a system that has been in existence only for a finite time, even if its time dependence goes as $e^{-iEt/\hbar}$ during this period, is not associated with a pure frequency $\omega=E/\hbar$ or definite energy $E$.
Consider the following example.  At time $t=0$, we turn on light of frequency $\omega$ on an ensemble of hydrogen atoms all in their ground state.  Since the light is supposed to consist of photons of energy $\hbar\omega$, we expect transitions to take place only to a level (if it exists) $\hbar\omega$ above the ground state.  It will however be seen that initially the atoms make transitions to several levels not obeying this constraint.  However, as $t$ increases, the deviation $\Delta E$ from the expected final-state energy will decrease according to $\Delta E \simeq \hbar/t$.  Only as $t\rightarrow \infty$ do we have a rigid law of conservation of energy in the classical sense.  We interpret this result by saying that the light source is not associated with a definite frequency (i.e. does not emit photons of definite energy) if it has been in operation only for a finite time, even if the dial is set at a definite frequency $\omega$ during this time.  [The output of the source is not just $e^{-i\omega t}$ but rather $\theta(t)e^{-i\omega t}$, whose transform is not a delta function peaked at $\omega$.]  Similarly when the excited atoms get deexcited and drop to the ground state, they do not emit photons of a definite energy $E=E_e-E_g$ (the subscripts $e$ and $g$ stand or "excited" and "ground") but rather with a spread $\Delta E \simeq \hbar/\Delta t$, $\Delta t$ being the duration for which there were in the excited state. [The time dependence of the atomic wave function is not $e^{-i\omega t}$ but $\theta(t)\theta(T-t)e^{-i\omega t}$ assuming it abruptly got excited to this state at $t=0$ and abruptly got deexcited $t=T$.]  We shall return to this point when we discuss the interaction of atoms with radiation in a later chapter.
Another way to describe this uncertainty relation is to say that violations in the classical energy conservation law by $\Delta E$ are possible over times $\Delta t \sim \hbar/\Delta E$.  The following example should clarify the meaning of this statement.
Imagine two ice skaters each equipped with several snowballs, and skating toward each other on trajectories that are parallel but separated by some perpendicular distance.  When skater $A$ reaches some point $x_1$let him throw a snowball toward $B$.  He ($A$) will then recoil away from $B$ and start moving along a new straight line.  Let $B$ now catch the snowball.  He too will recoil as a result, as shown in the figure.  If this whole process were seen by someone who could not see the snow balls, he would conclude that there is a repulsive force between $A$ and $B$.  If $A$ (or $B$) can throw the ball at most 10 ft, the observer would conclude that the <i>range of the force</i> is 10 ft, meaning $A$ and $B$ will not affect each other if the perpendicular distance between them exceeds 10 ft.
This is roughly how elementary particles interact with each other: if they throw photons at each other the force is called the electromagnetic force and the ability to throw and catch photons is called electric charge.  If the projectiles are pions the force is called the nuclear force.  We would like to estimate the range of the nuclear force using the uncertainty principle.  Now, unlike the two skaters endowed with snowballs, the protons and neutrons (i.e. nucleons) in the nucleus do not have a ready supply of pions, which have a mass $\mu$ and energy $\mu c^2$.  A nucleon can, however, produce a pion from nowhere (violating the classical law of energy conservation by $\simeq \mu c^2$) provided it is caught by the other nucleon within a time $\Delta t$ such that $\Delta t \simeq \hbar/\Delta E \simeq \hbar/\mu c^2$.  Even if the pion travels toward the receiver at the speed of light, it can only cover a distance $r=c\Delta t=\hbar/\mu c$, which is called the <i>Compton wavelength</i> of the pion and is a measure of the range of nuclear force.  The value of $r$ is approximately 1 Fermi=$10^{-13}$ cm.
So let me try to fix this.  We have a photon gun (aka "light bulb") that has a dial where we can set the desired energy of outgoing photons.  We turn it on.  As Shankar rightly notes, the time profile of the outgoing beam of light is not a simple wave like $e^{-i\omega t}$ but rather involves a step function, $\theta(t)e^{-i\omega t}$, representing the fact that it gets switched on right at $t=0$.  This means the frequency profile is not a simple delta function, but rather has some spread; there are $\omega' \neq \omega$ terms involved.

Let me show this in pictures, for people not familiar with the words I'm using.

Here is a picture representing a planewave, here the real part of $e^{-i\omega t}=\cos\omega t$, for the case $\omega=5$.  The $x$-axis is time, and the $y$ axis is the value of $\cos\omega t$.  Notice that it extends to beyond $t=0$ to negative times; in fact, it extends back to $t=-\infty$.  Because of this, there is only one frequency, $\omega=5$.  If I represent this graphically, it would jus be what is called a delta function, $\delta(\omega-5)$, which is basically just a line at $\omega=5$.  There is only one frequency in the beam, and so all of the beam is at $\omega=5$.
 $\delta(\omega-5)$, $x$-axis is $\omega$. Note that it's just a vertical line at 5, and zero elsewhere

However, we do not have just a planewave, but a planewave modulated by a step function.  We assume the that before $t=0$, there is no beam at all, and right at $t=0$ and thereafter, there is a planewave.

 planewave with step function
If we include, then, the step function part, this adds as extra term to our frequency profile.  We still get a delta function, $\delta(\omega-5)$, but we also now have an imaginary term $\frac{i}{\omega-5}.$  All told, the frequency profile is
$$f(\omega) = \delta(\omega-5) + \frac{iA}{\omega-5},$$
meaning the time profile can be re-written
$$f(t) = e^{-i5t} + iA\int_{-\infty}^{\infty}\frac{e^{i\omega't}}{\omega'-5}d\omega'.$$
All of this to say, that for short times, we have contributions from all possible frequencies $\omega$ going in to our photon wave functions.  There is not only one frequency in our wave function.
 The imaginary part of frequency distribution, $x$-axis is $\omega$ Unlike the $\delta$-function, is non-zero for frequencies other than $\omega=5$.

So when we then measure that some of the atoms are excited beyond $\hbar\omega$ (and when we also observe that some are excited BELOW $\hbar\omega$, or not excited) does it make more sense to conclude that the conservation of energy has been violated, or that our lack of precise knowledge of the energy of each photon in the beam has been revealed?

Remember, the term $\Delta E$ is supposed to be a quantitative measurement of our lack of knowledge of the energy of a photon; that's what "uncertainty" means.  We could think of this statistically, that in the beam there are many photons, some with more and some with less energy, but with a standard deviation $\Delta E$; the photons with more excite atoms more, those with less, less.  Or we could think of this individually, that each particular photon exists in a linear superposition of energy eigenstates focused around $E=\hbar\omega$, and that when it collides with the atoms collapse occurs and only one of the energy eigenstates is selected.  In either case, the energy was already there, and came from the photon gun, which was plugged into the wall.  But to think about this as each photon truly being in the $E=\hbar\omega$ eigenstate, but then spontaneously gaining energy $\Delta E$ from nowhere is not warranted, and even contradicts what has been said, about the source not being a pure planewave.

Professor Shankar then goes on to describe nuclear processes.  Shankar is a professor of nuclear physics at Yale, and I'm a guy with an internet connection, so I'm not going to presume to correct his understanding of nuclear forces.  I'm sure that there is in fact a characteristic range of the nuclear force, and that it can be estimated in that manner.  What I am going to say, however, is that his estimation does not actually follow from the uncertainty principle.  He lists the uncertainty principle as
$$\Delta E\Delta t \geq \hbar.$$
Therefore,
$$\Delta t \geq \hbar/\mu c^2$$
and $$r \geq \hbar/\mu c,$$
according to the uncertainty principle.  This does not put an upper bound on the range of the nuclear force, but actually a lower bound.  If the nuclear force in fact has an upper bound (which I'm sure that it does), then it cannot be derived by merely appealing to the time-energy uncertainty relation.  At the least, it would be necessary to establish that pions, for instance, always move in minimum uncertainty wave-packets.  And maybe they do; Shankar may have left that out because it was too advanced, even though it is true.  If pions are not constrained to have strictly minimum-uncertainty wave functions, then the above argument does not provide an upper bound on range, but a lower bound on range.

Leaving the range issue aside, the real question is if conservation of energy is obeyed.  Again, if $\Delta E$ is to be the amount of energy we can borrow from the universe, then $\Delta E$ can very well be infinite and satisfy the uncertainty principle, so already this cannot be correct.  Rather, $\Delta E$ represents how imprecisely we know the energy of a particular nucleon, and more or less corresponds to the standard deviation of energies.  If $\Delta E=\mu c^2$, this means that not all of the nucleons have exactly the average energy; indeed, some might have more, and some might have as much as $\mu c^2$ more energy.  Those that do can turn that energy in to a pion and exchange it with another nucleon.

To go back to the kids iceskating, suppose I want to know whether they do or do not have snowballs.  And suppose my method of finding this out is as follows: I throw a snowball at them; the more snowballs they have, the more likely my pegging them with a snowball is to make them drop one; the more they drop, the more snowballs it is likely they still have.  So if $A$ is carrying 30 snowballs with him, and I hit him, then he might drop five or so.  Whereas if $B$ only drops one snowball, or zero, then he most likely isn't carrying very many.  This is a terrible way to measure this, by the way; it has a high uncertainty.  Further, if $B$ only has 1 snowball, then he might actually catch the one I threw at him, and I would observe zero.  Therefore my uncertainty is at least 1 snowball, and probably much, much worse.  Say then that I peg both $A$ and $B$ and ascertain by my terrible measurement process that they both have zero snowballs, with uncertainty $\Delta S \geq 1$.  Just after this, I observe the interaction described by Shankar.  Should I conclude that $A$ has miraculously pulled a snowball from thin air, or that he already had one and I didn't know because of imprecision in my measurement?  (This is just an analogy, but the typical method of measurement on the quantum scale is by scattering processes, so it is not too far off.)

I guess my overall point is, the time-energy uncertainty relation cannot be interpreted to render the uncertainty in energy $\Delta E$ as being the amount by which a quantum system can violate the conservation of energy, if for no other reason than the uncertainty principle places no upper bound on this quantity.  While this is a popular interpretation, and probably a useful one in nuclear and atomic physics, it is not a necessary one, or really a logical one.

When a quantum process crosses an energy barrier, $\Delta E$, rather than interpret this as a violation of conservation of energy, we should probably rather interpret this as revelation of our lack of information (uncertainty) in the energy of the system.

If you think I've made some error in my reasoning, then please, point it out.  I love a good debate on quantum mechanics, and I'd hate to be saying anything factually wrong.