# The Next Step: Quantum Uncertainty and Bell’s Theorem

In the last post, we took a non-standard look at uncertainty relations. This was the crucial picture:

We’ll call a setup like this a network box. It has exactly the same structure as a certain set of possible experiments on a quantum spin-one system. Given an input (a state, represented here by a hex number) and a choice of triangle (experiment, picked out by the pointer position), exactly one of the lights at the vertices of the triangle comes on when you push the button (one result out of three possible.) We focused on two constraints:

(1) If three vertices form a triangle, their associated probabilities (i.e., the probabilities for the lights to come in a relevant experiment) sum to 1.
(2) If a vertex belongs to more than one triangle, its probability doesn’t depend on which triangle we associate it with.

As we saw, (1) and (2) give us an uncertainty relation:

(3) p(a) + p(m) <= 3/2.

Here we add: (3) tells us that p(a) and p(m) can’t both be 1:

(4) p(a) = 1 if and only if p(m) ≠ 1.

We’ll need (4) later. This post takes a step beyond Pitowsky. The picture we need now is a bit different:

Imagine a source that produces network boxes in pairs. We’ve left out the input screen because the joint state is built in at the source. Once the pairs leave the source, they aren’t connected in any ordinary physical way: no wires, no radio signals… But just as before,  if you pick a triangle and push the button, one of the triangle’s vertices light up.

(1) and (2) still hold and so do (3) and (4) too. But this time, the probabilities for the individual network boxes are simple: each outcome has a probability of 1/3. For example: if you set the dial to 7 and push the button, each of the vertices j, k, and l has a 1/3 probability of lighting up.

Suppose Alice has the left-side box and Bob the right-side. They pick settings that overlap. For example: Alice picks 5 and Bob picks 7. The triangles overlap on vertex: k. The new and important bit is this: quantum theory says that either k will light up for both Alice and Bob or for neither. This extends to the case where they pick the same setting. For example: if Alice and Bob both pick 1, the results will either be (a,a) or (b,b) or (c,c).

There’s a special state of two quantum spin-one systems that predicts correlations exactly like this. We do need to add a qualification: the correlations break if Alice or Bob chooses a new setting that doesn’t include the old vertex and runs the experiment again on the same system. We’ll make a stronger assumption to keep the presentation simple: each network box only works once. To keep experimenting, Alice and Bob need to go back to the source and get a new pair of network boxes. We’ll assume there’s no shortage, and in fact a realistic quantum experiment would almost certainly work this way.

Suppose you want to explain how the results for pairs manage to line up as we’ve described. One hypothesis is that the boxes are connected by a causal link of some sort, but we’re assuming that no obvious such link can be found. Taking that seriously, EPR-style reasoning has a lot of appeal. If Alice picks 1 and gets a, she can predict apparently without influencing Bob’s box that he’ll also get a if he picks 1 or if he picks 4. EPR would say that there must be some “element of reality” that’s already built into Bob’s box and determines his result. The simple explanation seems to be: every pair of boxes comes programmed so that experimental results are already determined, with the same wiring for each box in a pair. For example: a pair might be wired so that the light in the red rectangle comes on if you pick a triangle that contains it:

The appearance of probabilities strictly between 0 and 1 (1/3 in this case) at the surface level is just a matter of sampling: box pairs are always wired alike, but there are many ways to wire a pair that would produce perfect correlations of the sort we’ve described.

This is an appealing story.  If it succeeds, there won’t be any mystery about the correlations. They’ll just be a result of the way the boxes were wired at the source. The quantum analogue would be that the correlated quantum experimental results aren’t a matter of spooky action at a distance; they reflect what EPR would call “elements of reality” that the quantum systems already had when they left the source. But given what we’ve already said, it leads to a prediction.

Let’s use bold-face p for the “hidden” 0-1 probabilities that we hope will explain the correlations. If the diagrams represent the fundamental structure of the “wiring,” then (4) still holds for these probabilities. We can never have p(a) = 1 and p(m) = 1. If p(a) = 1, then p(m) has to be 0. But suppose Alice picks setting 4 and Bob picks setting 3. If our story is correct, there could never be a case where a comes on for Bob and m comes on for Alice. (4) rules that out. But if our networks are good models for quantum mechanics then this will happen sometimes. In fact, depending on the details the probability that it happens could be as high as 1/9. Our story can’t explain the correlations; not for our imaginary boxes and not for real-world quantum systems.

This is a different way of getting at what Bell’s theorem gets at. It’s also testable in principle, thought the experiment would be difficult in practice. The quantum states that produce behavior like this are among the ones we call “entangled,” and we have a simple demonstration that there’s no straightforward, “local” explanation for entanglement.

All this might suggest that the real-life quantum experiment would be a demonstration that what Alice does to her system somehow influences Bob’s system and vice-versa. In fact, I think it would be a mistake to jump to that conclusion, though it’s a mistake that many people (including many physicists) make. But that’s a whole ‘nother subject. For now, we can see this case as another confirmation of the power of looking at quantum probabilities combinatorially.

# Quantum uncertainty: another perspective

In a previous post, I talked about the uncertainty relations. The theme there was the way that geometry undergirds quantum probability. This post and the one that follows introduce a related perspective. Even though I don’t share his Bayesian outlook, the discussion owes much to a paper by the late and much-missed Itamar Pitowsky.*

Imagine a set-up like the one in the picture:

The green circles at the vertices of the triangles are lights. The one labeled “l” is on the others are off. Below the seven linked triangles you see a screen (on the left) a dial with seven positions, and a button below the dial. Here’s the idea.

You type a string of digits and letters; they appear on the screen. Then you choose one of the 7 triangles and set the dial so that it points to the corresponding number. When you push the button, exactly of the lights at the vertices of the triangle lights up. In the picture, the triangle was number 7 and light l came on. For a given input string and triangle, it’s generally not possible to predict which light will light up, but given the input and the choice of triangle, there’s a definite probability for each of the relevant lights to light up. (We can ignore the rule that fixes the probabilities.) For example, it might be that for input B35D68, the probabilities for lights j, k and l are p(j) = .2, p(k) = .35, p(l) = .45.

We need one more stipulation: given a choice of input, the probability for a light to come on doesn’t depend on which triangle pick. For example: we supposed that for input B35D68 and dial setting 7, p(k) = .35. The stipulation says that p(k) would also have been .35 if the dial had been set to 5.

What does this have to do with quantum mechanics?

For quantum spin-one systems (don’t worry about exactly what those are), experiments have three possible outcomes. To connect this post with the previous one, the possible outcomes of any experiment could be represented by one of three mutually perpendicular lines. But in a three-dimensional space, a single line belongs to many different sets of three mutually perpendicular lines. The physical correlate: there are distinct experiments on spin-one systems that “overlap”: there’s an outcome of one experiment represented by the same line as one of the outcomes of the other experiment. This means that no matter what state the system starts in, these two outcomes will have the same probability.

In our set-up, each triangle each can be paired with a three-outcome quantum experiment. Vertices of a triangle correspond to triples of perpendicular lines (“eigenvectors,” for the cogniscenti.) Each code number picks out a quantum state, and the cosine-squared rule gives the probabilities. But we can leave those details in the background. What matters are the relations of overlap among the triangles.

Our set-up could be powered by a quantum system. Each run of an “experiment” could mirror what happens when we pick one of 7 appropriately-chosen possible experiments, on a spin-one quantum system. And what follows rests on these two rules:

(1) If three vertices form a triangle, their associated probabilities sum to 1
(2) If a vertex belongs to more than one triangle, its probability is same no matter which triangle we associate it with.

Applying (1) and (2) to our diagram, we get this conclusion:

For any input,

3) p(a) + p(m) ≤ 3/2.

Here’s why. Think about the triangles (a,f,g) and (i,h,m). (1) tells us that A) p(a) + p(f) + p(g) + p(i) + p(h) + p(m) = 2. Now try to suppose p(a) + p(m) > 3/2. We’ll see that this assumption is impossible. Notice first that if p(a) + p(m) > 3/2, then p(f) + p(g) + p(i) + p(h) < 1/2. Otherwise, the total in A) will be bigger than 2. But if

p(f) + p(g) + p(i) + p(h) ≤ 1/2

then we automatically have

p(g) + p(h) ≤ 1/2.

An exactly similar argument shows that

p(c) + p(d) ≤ 1/2.

Now consider triangles (g,h,k) and (c,d,j). Since p(g) and p(h) don’t depend on the triangle we associate with them, and since they add to less than 1/2, we must have p(k) > 1/2. By an exactly similar argument, p(j) > 1/2. But now the triangle (j,k,l) violates (1):the probabilities already add to more than 1. So our assumption that p(a) + p(m) > 3/2 has to be wrong. We have

3) p(a) + p(m) ≤ 3/2.

As Pitowsky points out, this is a kind of uncertainty relation. Being certain of a constraints the probability of m, and vice-versa. We can’t be certain of a and m at the same time. Notice, however: being certain that a would come on if we picked triangle 1 or 4 doesn’t rule out the possibility that light m would come on if we picked triangle 3 or 6 instead — at least, not as far as 3) itself is concerned.

What’s fascinating, as Pitowsky stresses, is that the probability constraints emerge from the combinatorics. It’s a matter of how outcome sets for different possible experiments fit together. This fits with the theme of the previous post: quantum uncertainty isn’t about measurement disturbance. It’s a classically-unexpected dependence relation among possible experiments. In the next post we’ll see just how surprising this is from a classical point of view.

*”Betting on the Outcomes of Measurements: a Bayesian Theory of Quantum Probability,” Studies in the History and Philosophy of Modern Physics, 34 (2003) 395 ff.

# The quantum uncertainty relations – a simple(r) illustration

Kevin Drum notes that Craig Callender is annoyed with the way the quantum uncertainty relations are usually described. You can read Callender’s post here and KD’s post here.   Callender points out that the uncertainty relations aren’t about the measurement of one quantity disturbing the values of another, nor even about measurement as such. They’re about a relationship between the probabilities of different things happening. He worries that the usual way of presenting the “uncertainty principle” leads to nonsense like the very silly “What the #\$*! Do We Know!?” Callender also thinks (I agree) that we’d be better off dropping the phrase uncertainty principle. But if we can’t use measurement disturbance to understand the uncertainty relations, KD would like some better, less misleading analogy. What I’ll offer isn’t an analogy; it’s a quantum case, looked at in the quantum mechanical way. What I hope is that it’s simple enough to give a nudge toward a better way of thinking.

Callender wants to downplay measurement, but experiments give us especially clear cases of applying uncertainty relations to “something happening.” I think that advantage is worth the risks, and so the uncertainty in what follows will be a relationship between the probabilities for things happening in two possible experiments. We’ll pick two particularly simple ones.*

For the first, suppose a physicist produces a photon — a “light particle” and sends it into a special kind of prism (a Woolaston prism), orientated along the lab’s vertical axis. The experiment has two possible results — two things that can happen: either the photon exits the prism in a direction we’ll call “+” (upward, if you like) or in a direction we’ll call “-” (downward, if you like.) The first result means that the photon is vertically polarized; the second that it isn’t (and so is horizontally polarized.) In general, we can’t predict which will happen, but quantum mechanics gives us probabilities. Here’s how they work. (I’ve stripped the math down; what’s missing won’t matter for our purposes.)

Pick two lines at right angles to one another in an abstract mathematical plane.

(Why “abstract?” Because if we generalized the example, we’d need a lot more than three dimensions for even a moderately complex experiment.) The red line V+ represents the vertically polarized outcome, V- is for the “not vertically polarized” or horizontally polarized outcome.

The black line S represents the state. Its job is to answer probability questions. How to pick S depends on such things as how the photon was produced, but in typical cases there’s a clear way to do it. The probability of the “+” result depends on the angle between V+ and S: the smaller the angle, the higher the probability. In a bit more detail: we take the cosine of the angle θ between S and V+, and we square it. The result is the probability of “+” outcome. If we want the probability for the “-” result, we square the cosine of ϕ, the angle between S and V-. Since θ  and ϕ add to 90 degrees, these two numbers add to one, as good probabilities should. Though the usual presentation is a bit different (vectors instead of lines), this geometrical technique really is what quantum mechanics uses.

For the second experiment, rotate the prism 45 degrees toward the horizontal. Once again, we typically can’t predict what the photon will do, but quantum mechanics gives us a new pair of lines to use in calculating probabilities. This time we take the angle between S and D+ or S and D-, depending on which probability we want.  We square the cosine and the result is the probability. The diagram shows how things look for this new experiment. Keeping S the same, we see that the “+” result is more probable this time than in the previous experiment.

Our two experiments are incompatible in this sense: we can’t perform both at once; we can’t align the prism in two different directions at once. However, we can perform whichever experiment we choose, and given the state, we can calculate the probabilities for the outcomes. We’re also in a position to see an uncertainty relation.  Think about this picture:

First: the more certain we are of the outcome of one experiment, the less certain we are about the outcome of the other. Second, in the particular case we’ve described, the blue axes are at 45 degrees to the red axes. This tells us that if we were certain of the outcome of one experiment we would maximally uncertain of the outcome of the other. For example, suppose S coincided exactly with V+. Then the “+” result for the vertical experiment has probability one (the cosine of 0 degrees is 1). But S is at 45 degrees to each of D+ and D- and so in each case the squared cosine is 1/2 That means both outcomes of the diagonal experiment are equally likely.

Notice we haven’t said anything about one experiment “disturbing” the system. We also haven’t said whether the experiment uncovers a pre-existing fact or brings something into being. And we certainly haven’t given any special role to humans or their minds. We’ve described how quantum mechanics goes about its business in a particularly simple case. KD suggests that’s what’s needed is an example where we have something like “independence,” but that’s just what we don’t want. Quantum uncertainty isn’t about independence;  it’s a novel, non-classical kind of dependence. The relationship between the pairs {V+, V-} and {D+, D-} is where the uncertainty relation lies. That relationship is typical of how quantum mechanics represents things: alternative possibilities of a given sort (outcomes of a polarization experiment in this case) are associated with a set of mutually perpendicular lines (more accurately, vectors.) The geometrical relationship between the lines representing the outcomes of one experiment and the lines representing the outcomes of a different experiment are fixed by quantum theory. Given that geometrical relationship, the cosine rule is constrained; the relationships among the probabilities are fixed.

This geometrical way of dealing with probability is a deep part of quantum theory and it has no counterpart in classical physics. In particular, the fact that different sort of situations call for non-overlapping sets such as {V+,V-} and {D+,D-} is arguably the characteristic peculiarity of quantum theory. It is also the source and soul of the uncertainty relations.

Why does quantum theory do things this way? The short and completely unsatisfying answer is that it works. A better answer would talk about how quantum theory evolved and what the rules are for building quantum mechanical models. If the question is what this strangeness means, however, you have wandered into hotly disputed territory. Einstein thought uncertainty meant that quantum theory is incomplete. Some interpretations agree (Callender mentions an approach called Bohmian mechanics.) Others don’t. The range of options is shockingly broad. But whatever we say about the uncertainty relations, the story will be bound up with what we have to say about the way quantum theory weaves geometry and probability together.

* Our example differs from the classic position/momentum uncertainty in one particularly important way: position and momentum are continuous quantities; they call for a different mathematical treatment than polarization. But the examples here point to what’s really important: the geometrical underpinnings of quantum probability.

# Time travel and suspiciously super computers

This is a post about time travel and paradox, but  before we get to that, we need to recall what a function is. Here’s a simple example: multiply by 1/2 and then add 1. This is a function because for any input there’s exactly one output. If we apply the function to 5, we get 3 1/2 as output. If we apply if to 10, we get 6. And so on.

Many (not all) functions have fixed points. If p is a fixed point of a function f, then applying f to p gives us p itself. Our sample function illustrates. If we multiply 2 by 1/2 and then add 1, we get 2 itself; 2 is a fixed point of our function.

Typically, finding a fixed point is a lot harder than just computing the output for some input. Take a more restricted case. Suppose  f is a function that takes strings of 5 0s and 1s as input and gives strings of five  0s and 1s as output. For example: f(00110) might be 10110. The function f might have a fixed point, and there is at least one general procedure for finding it. In the worst case, compute the values of all 32 possible inputs and see if there are any cases where input equals output. Suppose there’s only one fixed point. You might get lucky and find it immediately. Or there might be some relatively simple way to produce it. But in general, finding the fixed point of a function will be a lot more work than finding the output for a particular input.

On to time travel. In a now-classic article, David Deutsch argues that “it is a fundamental principle of the philosophy of science that solutions of problems do not spring fully-formed into the universe.” Deutsch has much more faith than I do in the power of “philosophy of science,” but let that pass. Deutsch would insist that in particular, there shouldn’t be a machine that can find a fixed point for f in one step. But if time travel as we usually think of it is possible, that’s exactly what would happen. To see why, consider this diagram, from Deutsch’s paper.

The box labeled Gf is a calculating device. It takes two string as input. The string that goes with the x on the left consists (in our example) of five bits — “0”s and “1”s — an input for our function f. The string that goes with the 0 on the right is simply a string of five “0”s. The box leaves the first five bits alone and replaces the zeros that make up the second five bits by the value of the function. Ignoring the arrow that loops around though the -1, things might work this way: if the the two strings are 00101 (on the left) and 00000 (on the right), and if f(00101) = 11011, then the input

00101, 00000

would be converted into

00101, 11011.

But now it’s time to look at the full diagram. Imagine that the string of  “0”s passes through the machine, transformed by its operations, and then travels backward in time; that’s what the path through the “-1” represents. Once the information returns to the past, it enter the machine on the left, where x is, along with the original string of “0”s on the right. But we are assuming that other than traveling through time, the string that enters at point x is the same string of “0”s and “1”s as the string that exited and looped back in time. The time-loop has simply moved the information around through time, back before the machine created it, to serve as input to the function f. So the output of the “0” arrow equals the input of the “x” arrow. In other words, the output of the “0” arrow is also the value that f assigns to the input. The only way this setup can work consistently is if it computes in one “simple” step a fixed point of the function f.

There’s no problem supposing that there’s a box like Gf. It’s essentially just a calculator. And if we assume that we can send information around a loop in time, then the box must cough up a fixed point of f. But as Deutsch sees it, this arrangement is actually worse than stories in which a time traveller kills Grandad before Grandad meets Grandma. Stories like that aren’t genuine paradoxes. They’re merely inconsistent. In the case we’ve just described, we have a simple computational gadget that can produce answers to very hard computational problems. (Imagine that n is much bigger than 5; things get very hard very quickly.)  In Deutsch’s view, this is uncomfortably close to the “unproved theorem paradox”: a mathematician publishes the proof of a major theorem. Years later, someone comes across the proof in a journal, gets in a time machine, takes the proof back to the past and presents it to the mathematician himself, who “plagiarizes” it for his publication. Though not logically impossible, this story seems offensively absurd.  In Deutsch’s view, the story we’ve studied here is too close to this absurdity for comfort. Deutsch concludes that classical time travel is therefore physically impossible.

Why “classical” time travel? Because Deutsch thinks that if we bring quantum theory into the story, we can avoid the absurdities in a theoretically appealing way. I’m not convinced he’s right, but that’s another story. My question is whether the Fixed Point Finder above is really as shocking as it seems. In the case of the unproved theorem paradox, complex knowledge appears out of nothing, utterly unexplained. In the case of the Fixed Point Finder, things seem different. We have an explanation. We combine a humble but perfectly intelligible computing device with a remarkable but consistent physical assumption: a closed path in time. When we put the two together, we see that the output of the device must be a fixed point; we can prove it. One person’s absurdity is another’s “What did you expect?” And indeed, Deutsch himself points out that even on his preferred approach, time-loops make for very super computers, though somehow not as super as the one we’ve just described.

And so: if there are closed time-loops, they come with formidable computing power. (There’s a whole intellectual cottage industry built around this fact.) What’s less clear is where the line between peculiar and absurd lies, and whether Deutsch can stay on the right side of it by his own lights. But that’s a much bigger story.

# Three Recent Twitterstorms: some contrarian thoughts

In the last few days there have been three twitterstorms where I find myself at odds with a lot of folks I otherwise tend to agree with. Maybe I’m missing something.

First, the Trayvon Martin verdict. If George Zimmerman had stayed in his car when the dispatcher told him to, Trayvon Martin would still be alive. If he couldn’t bring himself to do that but had left his gun behind, Trayvon Martin would still be alive. (Yes; he was carrying it legally. No; that doesn’t make it okay. America’s gun laws are worse than an embarrassment.) I’m angry as hell at George Zimmerman, and his bone-headed comment that he doesn’t wish he’d done anything differently that night makes me even angrier. But all this is a different matter from thinking he was guilty of second-degree murder or manslaughter under Florida law. I don’t have much reason to believe that the jury got the law wrong, and I haven’t heard anything to suggest that anyone else does either.

There are plenty of smart people on whom this is not lost; Ta-Nehisi Coates, for example. But others seem to think that because something awful happened, there was something this jury could properly have done about it. I’m not convinced.

Item two: juror B37. I’m pretty sure the world doesn’t need any book she might have written. What I’ve been able to glean don’t suggest that this is someone who thinks clearly, writes well or has much insight. But one objection I kept coming across was that it would be wrong from the start for her to write the book, whether or not she could have done it well. The thought seemed to be that that no one should profit from public service nor from a tragedy. Let’s look at both.

The bit about not profiting from public service doesn’t wash. Do we really want to say that ex-Presidents shouldn’t publish their memoirs? And if we don’t want to say that, what makes jurors different?

What about the idea that it would be wrong for the juror to profit from tragedy? The thought has a superficial appeal, but suppose a smart journalist interviewed the jurors and published a book about their deliberations. Would that be wrong? I’d bet no one would object if it happened.

It’s easy to imagine that a well-written book by a thoughtful juror could provide valuable insight. If such a juror wrote such a book, do we really want to say that she shouldn’t publish it? Or that if she did, she should give all the earnings away so as not to profit? You might; I”m not convinced.

The third item is Rolling Stone’s cover photo of Dzhokhar Tsarnaev. I admit: when I look at the photo, I have a very uncomfortable reaction. But I think this is for an instructive reason.  As many people have pointed out, Tsarnaev looks like a young rock star — the kind that might feed a teenage girl’s dreams. He doesn’t fit the stereotype that comes to mind unbidden when I think of an Islamic terrorist. Something went terribly wrong in this young man’s life — something hard to understand. The bold type words on the cover make clear that this is what the story is about. If you can get past your first reaction, you just might find the cognitive dissonance worth sitting with. This New Yorker piece seems to me to get it pretty much right.

So there we are. Maybe I’m missing something in one or another of these cases. If so, I hope you’ll help me see what that might be.

# The Anti-Abortion Movement: a Case of Legislative Arrogance?

Here are some things a sane, thoughtful person could think:

• A fertilized ovum is the first step on the road to full-blown humanity, but it’s not a full-fledged human being.
• If a woman miscarries at two months, this may be sad but it’s not a tragedy in the way it would be if she lost a two-month-old infant.
• Suppose a woman in her 6th month of pregnancy will die if she delivers the baby. The woman is entitled to choose her own life. And if she’s incapacitated, anyone acting on her behalf should save her rather than the fetus unless she has explicitly said otherwise.
• If fetuses can feel pain at 20 weeks, this isn’t morally trivial, but it’s not enough to settle whether a woman at this stage of pregnancy should be able to have an abortion.

It’s not that what’s on the list is beyond reasonable disagreement. It’s that these are all things someone could reasonably believe and that many people do believe. The assumptions behind a good deal of anti-abortion agitation are not obviously true. That means that many  legislators face a question of political morality: given that their views are so controversial, what makes it right to impose them on others?

The first item on the list is worth a few more words, since some of the legislative efforts are on behalf of “personhood” amendments. A fertilized ovum is alive and biologically human, but the questions about abortion aren’t biological questions. Some people think the burgeoning clump of cells is the moral equivalent of a conscious human being. Other people (I’m one) find this all but incomprehensible. People who disagree don’t have anything like a decisive argument to the contrary. The usual arguments either trade on confusions about the difference between moral and biological questions or else appeal to not always well-disguised religious premises.

Does this mean that people like me think the conceptus is no more important than a cluster of skin cells? It certainly needn’t. But agreeing that the conceptus isn’t just a clump of cells doesn’t amount to thinking it has the same standing as the mother.

Maybe you see things differently. That’s fine. But recent anti-abortion legislation rides roughshod over reasonable disagreement. Conservative legislators are so sure they’re right that they’re willing to impose real burdens on people with views at least as defensible as their own.

Ronald Reagan once defended his anti-abortion position by pointing out that if we found someone in a ditch and weren’t sure whether she was alive, we’d err on the side of caution. The thought is that if we aren’t sure of the status of the fetus, we should err on the side of treating it as a full-fledged person.

The analogy isn’t as good as it may seem. The apparently unconscious woman in the ditch is without doubt a full-fledged person if she’s alive; that’s why moral caution is called for. The question about the fetus isn’t whether it’s alive or dead. It’s what to make of its position on a complicated continuum.

But set messy metaphysics aside. Here’s another analogy. Many people think it’s wrong to raise animals to kill and eat them. I’m not a vegetarian, but I’m a carnivore with an uneasy conscience; I don’t think I have good replies to the arguments of my vegetarian friends. Raising animals to consume them is morally questionable and may be very wrong. And yet you’re unlikely find to backers of anti-abortion bills arguing for laws to shut down the slaughterhouse. On the contrary: my bet is that they’d see it as an outrageous infringement on their freedom.

The issue isn’t hypocrisy. It’s what should be a matter of law and what should be left to conscience. There’s a serious case for vegetarianism; the arguments are at least as good as most of the anti-abortion arguments you’ll to hear. Nonetheless, things are murky enough that most of us — conservatives not least — think it would be wrong to outlaw eating meat. Why believe that harsh anti-abortion laws are more obviously justified?

The question of when the law can legitimately impose itself on people who disagree isn’t an easy one. What’s disturbing about conservative legislative efforts on abortion is their complete indifference to the issue — all the more striking since conservatives are otherwise so inclined to argue that government should mostly leave us alone. Some ways of framing the liberal view on abortion make their own mistakes: seeing women’s right to control their bodies as the only relevant issue is just as dogmatic as the kind of conservative view we’ve been discussing. But at the moment, the legislative winds are at the back of the anti-choice movement. It would be easier to disagree respectfully if conservatives showed a bit of recognition that neither their view nor their legislative efforts are obviously right.