The weirdest thing about QM

There are many spooky things about QM, but I want to share with you what I think is the spookiest. You might want to read the QM for dummies post to get a slightly deeper feel for the math, but this post should stand on its own.

Okay, suppose we have a particle that's in a superposition of spin-up and spin-down (I'm leaving out the normalization term $\frac{\sqrt{2}}{2}$ for simplicity):
$$|\phi\rangle = |p_{up}\rangle + |p_{down}\rangle$$
Don't be scared by the bra-ket notation. $|x\rangle$ is just another way of writing the vector $\vec{x}$.

But of course we never see a particle in a superposition state. Whenever we check, the particle is either spin-up ($|p_{up}\rangle$) or spin-down. The other term disappears:
$$|\phi\rangle = |p_{up}\rangle$$
The question is, when does it disappear? When has it been measured?

Suppose we try to "measure" it by bouncing a second particle (q) off of it. (To measure spin we don't actually "bounce" anything, but the idea is the same.) But instead of that "collapsing" the first particle's state, we discover that the two are now entangled in a joint superposition, written:
$$|\phi\rangle = (|p_{up}\rangle \otimes |q_{up}\rangle) + (|p_{down}\rangle \otimes |q_{down}\rangle)$$
In one of the sum terms, p is spin-up and q "agrees," and in the other term, p is spin-down and q agrees. This doesn't look at all like the earlier "collapsed" state, and indeed we can tell it apart by some clever interference experiment. On the other hand, if we check whether particle p by itself is still exhibiting interference, the answer will be no. So in a sense, p behaving classically, but the p-q system as a whole is "still quantum."

What if we introduce a third particle r? Will it collapse the first two? No, it will just join in on the fun (I'll drop the $\otimes$ symbol for cleanliness):
$$|\phi\rangle = (|p_{up}\rangle |q_{up}\rangle |r_{up}\rangle) + (|p_{down}\rangle |q_{down}\rangle  |r_{down}\rangle)$$
So then what does collapse it? As the system encounters more particles -- a few more in a controlled lab setting, or trillions more as it contacts the messy environment -- they should all just join in. If we lump q and r into the environment (env):
$$|\phi\rangle = (|p_{up}\rangle |env_{up}\rangle) + (|p_{down}\rangle |env_{down}\rangle)$$
In the first term, p is spin-down and the environment reflects this (and vice versa for the other term). So we still haven't answered the question: why do I only see one of those two outcomes?

There are two broadly popular answers to that question:

1) The orthodox ("Copenhagen") interpretation. It says, essentially, that somewhere on the way from one particle to trillions, it suddenly "collapses" into one state or the other. This was by far the favored interpretation in the early days, and roughly 39% of physicists still prefer it even though it's ill-defined[3].

2) The Many Worlds Interpretation (MWI), which says that no collapse ever happens. Instead, the universes keep splitting, ad infinitum. Every time you encounter the result of some indeterminate experiment (which is nonstop, since your environment is filled with tiny particles), your universe splits and you do, too.

The problem with Copenhagen (other than the fact that nobody seems to agree on what it is) is that it doesn't really tell us when or why this special "collapse" (which, by the way, disobeys what is known as unitary evolution which applies in all other cases) happens. We should just quietly assume it does. And for any practical experiment, that's good enough.

But as technology improves and we are able to isolate larger and larger quantum systems, we haven't run into this magic wall where the unitary laws of physics suddenly break down. We keep getting bigger entanglements rather than a sudden collapse. We're starting to question why we needed such an odd assumption to begin with. As a result, more and more physicists are switching to MWI.

But the problem with MWI is in how it answers the fundamental question "why do I only see one outcome?" The answer is that you don't -- you "actually" see all of them. You see, when you encounter the system, you become entangled with it. Each branch of the superposition becomes a parallel universe, and each universe has one copy of you. One copy of you sees "up" and the other copy sees "down." Asking why you you see "up" is nonsense, because in a parallel universe, the "other you" is asking why she sees "down." There isn't one you. In fact, as the universes keep splitting and splitting, there become infinitely many of you.

Yeah.

Now consider what happens after we run 100 iterations of the up/down experiments above. According to Copenhagen, since each time there is a 50%-50% chance of each outcome, you should expect that roughly half of the experiments come out heads. If you get a "lucky streak" of all ups, then as that lucky streak gets longer, you should increasingly suspect that something is wrong with either your setup or QM itself.

What about in MWI?

Well, in MWI there will be $2^{100}$ (~$10^{30}$) universes, each with its own string of outcomes ("UUDDUDD..."). On average, across all of them, there will be as many ups as downs. But since the question of where "you" end up is meaningless, there's no reason "you" shouldn't find yourself in a universe with all up, for example. If there were just one of you, and on each split you followed a path randomly, then we could make a statement like "the expected number of ups is 50, and it's very unlikely that you'll end up with all ups," but that doesn't apply here. You didn't end up with all ups; only "one copy of you" did.

On average, the $2^{100}$ copies of you are fulfilling their quantum (Born rule) obligation to see only half ups, but one of those gals had to be the one who saw all ups... so why shouldn't that be this you? What prevents you you from being the lucky one? Nothing. It doesn't contradict the theory or interpretation in any way. The quantum police won't come knocking just because you keep ending up astronomically lucky.

But the weirdness has just begun.

In MWI, you do indeed model the world as:
$$|\phi\rangle = (|p_{up}\rangle |env_{up}\rangle) + (|p_{down}\rangle |env_{down}\rangle)$$
Right up until the moment you see the result yourself (that is, encounter p or env in any way that would reveal the outcome). But env here includes the bodies and brains of any experimenters that encountered the system before you.

On the other hand, once you encounter the system, you can no longer treat it as a superposition. You cannot do any interference experiment, even in principle, that would suggest that it's still in a superposed state. You are definitively in either one or the other branch. Not only that, but everybody else in that universe will see the same thing you do. If you see 100 ups, they do as well -- even if they saw the outcome long before you.

This should give you a bit of pause. There's nobody else in "your universe" that is capable of collapsing the superposition like that. Each one of them just gets swept up into the entanglement, permitting you (in principle, but certainly not in practice -- probably ever) to do an interference experiment that would demonstrate this. But once you encounter it, all bets are off.

You're the only one who can do this.

Of course, the situation is symmetric for all other conscious observers (assuming there are any). From their perspective, in their universe they are the only one able to do this. And there's no contradiction, because you never inhabit the same universe. All the other people you see in your universe are your copies of them. And from your perspective, they are simply globs of matter obeying mindless physical laws.

Somehow, in your universe (this one!) the buck stops with you. Don't take my word for it. This is well-accepted.

Why isn't this mystical interpretation off-putting to believers of MWI? Well, if you could take the "inside perspective" of any object, you indeed would be the sole observer. But since consciousness is just an illusion, you can't take such a perspective, so there's no problem!

Which is all well and good, but I'm one of those weirdos who does believe he's conscious and inhabiting just one universe. And so maybe I have to suck it up and grudgingly accept my fate as the only conscious observer here. On the other hand, nothing seems to stop me (in principle) from "choosing" which of those universes I want to be "birthed into" -- although nothing explicitly permits me to do this either. In fact, there seems to be nothing in physical reality that can have control over which universe I "end up in," period, given that all of physical reality is (by definition) embedded in the universe. So maybe instead of begrudging acceptance, I should be filled with awe and gratitude.

This is your universe, and your playground. Not "you" as in your body or brain (which, being physical objects, are just part of the environment, just like other experimenters), but that which is looking out your eyes, so to speak.

Is there something looking out your eyes? Something non-physical? Obviously physics says "no." But instead of taking anyone's word for it, I invite you to check for yourself. If the answer is "yes," you have a whole lot of responsibility.

Footnotes

1. You will often hear it said that there's nothing special about conscious observers in QM. For example, famed physicist Sean Carroll:

“Observation” in quantum mechanics is just a suggestive word meaning “interaction between a quantum system and another quantum system, especially in the case where the second system includes many environmental degrees of freedom that are not explicitly accounted for.”

In other words, when q interacted with p, it was an "observation" of sorts. But because it's easy to still see quantum effects with two particles (because we can "explicitly account for" them), we should maybe hold off on calling it an observation until we have so many other particles in on it that we can no longer do so. And since what we call "conscious observers" are themselves made of lots of particles, they qualify as "many environmental degrees of freedom" instead of consciousness being something special.

As we saw, we should indeed treat things this way... with respect to other experimenters. It does not address the question of why I only see one outcome. Indeed, it doesn't even try to address that question, because strictly speaking, it belongs to metaphysics.


2. Many times physicists will explain that the consciousness thing is a red herring. There's nothing interesting to consider about it, and it has nothing to do with QM. If you don't agree, you have good company. Here's the greatest physicist of modern times agreeing with you.


3. Sean Carroll again: "I think Copenhagen is completely ill-defined, and shouldn’t be the favorite anything of any thoughtful person"

Certainty threshold for a convnet in Keras

Suppose you want to fine-tune a convnet such as VGG16 in Keras. You normally replace the final dense layer with your own dense (+softmax) layer that has as many categories as you need, and freeze all the other layers (assuming they are preloaded with useful weights). Then you train the net on your dataset.

 t = Dense(num_categories, activation='softmax', name='predictions')(vgg16.layers[-1].output)  
 classifier = Model(inputs=vgg16.input, outputs=classifier_output)  

Now suppose you want to know not just the relative certainty of each output (which is what a softmax gives you), but something like an "absolute certainty" for each neuron? You should be able to look at the values before the softmax activation. In that case, you can split out the activation layer:

 t = Dense(num_categories, activation=None, name='predictions')(vgg16.layers[-1].output)  
 classifier_output = Activation('softmax')(t)  
 classifier = Model(inputs=vgg16.input, outputs=classifier_output)  

And now suppose we want to train a single binary neuron that will fire only if one of those dense neurons' values exceeds some threshold. You don't know the threshold in advance, but you suspect that for faces belonging to any of the trained categories, its respective neuron should generate a high value, and for other faces none of them will be very high. So you generate training data where faces from the existing categories are true and all other faces are false.

A first guess would be to use a MaxPooling1D layer to collect the maximum of the previous layer, and then feed it to a single dense neuron with sigmoid, so that it can be scaled and produce an output in [0, 1].

Even simpler is to use GlobalMaxPool1D, which instead of using a sliding window just gets the max of all input values. But that layer still requires a 3D input of (batch_size, steps, features). It's not clear to me from the doc what "steps" and "features" mean, but if you look at the impl it's just taking the max along dimension 1 (steps). Presumably features is something like layers.

Our 'predictions' layer is only 2D (batch_size, steps), so we need to make it 3D. First guess:

 global_max = GlobalMaxPool1D()(K.expand_dims(t))  
 yesno_output = Dense(1, activation='sigmoid')(global_max)  
 yesno = Model(inputs=vgg16.input, outputs=yesno_output)  

Alas, that produces a strange error:

AttributeError: 'Tensor' object has no attribute '_keras_history'

It turns out you have to use a Lambda:

 expanded = Lambda(lambda x: K.expand_dims(x))(t)  
 global_max = GlobalMaxPool1D()(expanded)  
 yesno_output = Dense(1, activation='sigmoid')(global_max)  
 yesno = Model(inputs=vgg16.input, outputs=yesno_output)  

And voila, everything works!

QM for dummies

"I will take just this one experiment, which has been designed to contain all of the mystery of quantum mechanics, to put you up against the paradoxes and mysteries and peculiarities of nature one hundred per cent. Any other situation in quantum mechanics, it turns out, can always be explained by saying, 'You remember the case of the experiment with the two holes? It's the same thing.'" -- Richard Feynman

Intro

Quantum Mechanics is hard. To use it expertly, you're going to have to understand math that makes my eyes bleed:

Your only other option seems to be reading popular renditions about half-dead zombie cats which give you no insight at all. I hope to bridge the gap and give you a useful understanding with minimal math.

credit thelifeofpsi.com. their site seems pretty rad.

Quantum states

Suppose you have a particle and some experiment you can run on it that can yield only two possible outcomes. For example, an electron can be spin-up or spin-down. A photon can be vertically polarized or horizontally polarized (it turns out that other forms of polarization, such as circular or elliptical, can be written in terms of those states). We'd like a mathematical model to describe the state of the particle and help us make predictions about it.

It turns out that we can represent a particle's state as a vector. Since there are two possible outcomes, we use a 2-dimensional space. And for reasons we'll see later, we want the states representing those two outcomes to be orthogonal (perpendicular).

Let's stick with the example of spin, which can be measured in any of the spatial directions (x, y, z). If we measure it along the z-axis, the possible outcomes are spin-up ($z_+$) or spin-down ($z_-$). We'll assign those states their own orthogonal vectors (which we write as letters with arrows on top):

$\vec{z_+}$

$\vec{z_-}$

In actuality, QM requires complex numbers instead of real numbers, and for some properties the number of dimensions is infinite. But my hands got tired trying to draw infinitely many complex dimensions, so you'll have to settle for two real ones. Surprisingly, it turns out to be enough to understand most of the important stuff.

What's important to understand is that this is just an abstract, mathematical description of our state. The fact that the states are orthogonal has nothing to do with orthogonality in physical space. For example, spin-up and spin-down would are represented like this even though "up" and "down" aren't perpendicular in physical space. Similarly, you may be used to thinking of the above lines as the x- and y- axes, but resist that temptation: here we'll call them the $z_+$ and $z_-$ axes.

This 2D plane is called the "state space" of the particle, and represents all possible values of the states it can have. Do not think of it as physical space in any way.

Superposition

In classical physics, a particle can only be in one state or the other. In QM, it can be in a superposition. It's a fancy word for any vector that's not directly along one axis:

$\frac{\sqrt{2}}{2}(\vec{z_+} + \vec{z_-})$

The $\frac{\sqrt{2}}{2}$ is there just to make the length equal to 1.

To understand what states "mean," we have to understand what we do with states.

Measurement

If we have a particle in the above state ($\frac{\sqrt{2}}{2}(\vec{z_+} + \vec{z_-})$), we can measure it to see if it's "really" in state $\vec{z_+}$ or state $\vec{z_-}$. The probability of that measurement resulting in state $\vec{z_+}$ is given by projecting the state vector onto the $z_+$-axis and squaring the resulting length. You can think of it as "dropping a perpendicular" (technically, it's taking the inner product and taking the square of the norm):

You can see that the green line touches the $z_+$-axis at $\frac{\sqrt{2}}{2}$, which squared gives $\frac{{1}}{2}$. If you do the same with the other axis, you'll get the same result. This tells us that the particle has a 50-50 chance of being measured as either $z_+$ or $z_-$. Once the state has been measured as (say) $z_+$, the state becomes simply $z_+$ no matter what it was before. This is sometimes called a "collapse."

Another thing to notice is that if the particle state is $\vec{z_-}$, then measuring it as $\vec{z_+}$ (or vice versa) will happen 0% of the time, since projecting onto an orthogonal vector gives you a zero vector. This is why we require the corresponding states to be orthogonal: if the particle is definitely in one state, it is definitely not in the other state, and this is how we make the math work out.

A key point is that we chose to measure it with respect to the $z_+$/$z_-$ axis. But as we shall soon see, that's not our only choice.

Change of basis

In a 2-dimensional plane, there are many choices of axes. For example, we could choose the blue and green lines:

A different axis

What shall we call this axis? Well, it turns out that we got lucky: just like in our previous example, this particular axis has a physical interpretation. The blue line corresponds to spin-up in the (physical) x-direction, and the green line to spin-down.

Why do I say that? The answer, as always, is that it produces a model that agrees with physical experiments. Specifically, if we have an electron that's just been measured to be $x_+$ (i.e., spin-up in the x-direction, corresponding to the blue line), and we measure it in the $z_+$/$z_-$ basis, we actually get each result with 50% probability -- just as the result from the previous section suggests. You can see that the model predicts the same result (50%) for each of the following experiments:

1. Particle starts in $\vec{x_+}$. What are the odds of it being measured as $z_+$?
2. Particle starts in $\vec{z_+}$. What are the odds of it being measured as $x_+$?
3. Particle starts in $\vec{x_-}$. What are the odds of it being measured as $z_+$?
4. Particle starts in $\vec{z_+}$. What are the odds of it being measured as $x_-$?
5. Particle starts in $\vec{x_+}$. What are the odds of it being measured as $z_-$?
6. Particle starts in $\vec{z_-}$. What are the odds of it being measured as $x_+$?
7. Particle starts in $\vec{x_-}$. What are the odds of it being measured as $z_-$?
8. Particle starts in $\vec{z_-}$. What are the odds of it being measured as $x_-$?

And sure enough, these predictions match up with experimental observations. And that is why we call the blue-green axis the $x_+$/$x_-$ axis: because it makes the math match the experiments, and that's what we want out of a model!

As an aside, the above is a rough demonstration of the Heisenberg Uncertainty Principle: knowing the value of a property along one axis makes it uncertain with respect to other axes in the same state space. Momentum and position are similarly just different bases for the same infinite-dimensional space, but that's harder to see visually.

In general, we could pick any axis we want, and it may correspond to some interesting physical property. This just happens to be an easily drawn case with easily interpreted physical significance.
(If you're wondering which axis corresponds to the $y_+$/$y_-$ states, the answer is that we'd have to introduce complex numbers (i.e., numbers with imaginary parts), which can't be drawn on our real 2D plane.)

Interference

The interesting thing about the two-slit experiment is the interference pattern, which we sometimes see and sometimes don't. It turns out that we can model interference with our greatly simplified system, even though it won't look anything like the pretty pattern we see on the screen.

Recall from above that we can write out a state vector in different ways depending on the basis we choose. For example, $\vec{x_+}$ can also be rewritten as $\frac{\sqrt{2}}{2}(\vec{z_+} + \vec{z_-})$. It's the same state, just rewritten.

Now, we can ask about the odds of measuring that state as $x_-$. From the former description ($\vec{x_+}$) it should be immediately obvious that the probability is zero: the state in question is orthogonal to the outcome we want. A particle that's spin-up in the x direction is by definition not spin-down in the same direction.

But if we write it the second way, it may not be so obvious:

What is the probability that a particle in state $\frac{\sqrt{2}}{2}(\vec{z_+} + \vec{z_-})$ is measured as $x_-$?

The answer is the same, because rewriting a vector in a different basis doesn't actually change anything. But if we're not being careful, we might try to reason as follows:

Hmm... well the particle is either in state $\vec{z_+}$ or $\vec{z_-}$. In both cases, the probability of being measured  as $x_-$ is 50%. So on average, we still have a 50% chance!

The above reasoning would be correct in classical mechanics, but it fails in quantum mechanics! Somehow the two "components" ($\vec{z_+}$ and $\vec{z_-}$) have "destructively interfered" to give a result of zero! Similarly, if the question had been to calculate the probability of getting result $x_+$ (for which the result is 100%) we might say that they "constructively interfere."

And that is the essence of the difference between classical and quantum physics: the either/or reasoning fails, because the rules involve these (generally complex-valued) vectors that can represent superpositions whose behavior differs from anything we know in the classical world. Interference is the manifestation of this fact.

Also note that in the first way of writing it ($\vec{x_+}$), it's not a "superposition" of anything. This means that "superposition" is not an intrinsic feature of a system, but of our way of thinking about it. Similarly, the word "interference" implies that the "components" in the superposition are interfering, and is thus just one way of thinking about things.

(For a more mathematical treatment, see details in footnote [1].)

Entanglement

Another interesting property of quantum systems is entanglement. If you want to understand it properly, you have to become familiar with tensor products.

"If you really want to impress your friends and confound your enemies, you can invoke tensor products… People run in terror from the  symbol." -- some Stanford professor, apparently

They're really not that scary, but they're beyond the scope of this guide. So we'll have to make do with another simplification.

Suppose we have two systems: on the left, a particle whose z-spin we care about, and on the right, a detector that's going to measure the z-spin of the particle. Let's call the detector states $\vec{dz_+}$ and $\vec{dz_-}$. To represent the joint state of the two systems, we introduce this operator ($\otimes$) that is vaguely analogous to multiplication:

Joint state = particle state $\otimes$ detector state

So, for example, we might have:
$$(\vec{z_+} + \vec{z_-}) \otimes (\vec{dz_+} + \vec{dz_-})$$

Just like regular multiplication, it distributes over addition, so we can rewrite this:

$$(\vec{z_+} \otimes \vec{dz_+}) + (\vec{z_+} \otimes \vec{dz_-}) + (\vec{z_-} \otimes \vec{dz_+}) + (\vec{z_-} \otimes \vec{dz_-})$$

We can interpret the above in the following way. There are four possible outcomes, corresponding to each term in the sum:

1. $\vec{z_+} \otimes \vec{dz_+}$: particle is spin-up and detector reads spin-up
2. $\vec{z_+} \otimes \vec{dz_-}$: particle is spin-up and detector reads spin-down
3. $\vec{z_-} \otimes \vec{dz_+}$: particle is spin-down and detector reads spin-up
4. $\vec{z_-} \otimes \vec{dz_-}$: particle is spin-down and detector reads spin-down

If the particle has not yet encountered the detector, it's okay that we have some states (2 and 3) where the particle and detector don't agree. But if we have a functioning detector, then after the measurement, we will get only two components in our joint state:

$$(\vec{z_+} \otimes \vec{dz_+}) + (\vec{z_-} \otimes \vec{dz_-})$$

It turns out that the distributive law doesn't help us reduce this state to one of the form:

(some state for the particle) $\otimes$ (some state for the detector)

So the two states are said to be entangled. The physical interpretation is that knowing something about the particle's state gives us information about the detector's state (and vice versa) -- which is just what you'd expect out of a working detector.

Loss of interference

In the previous section, we saw an example where the systems were perfectly unentangled (knowing information about one system told you nothing about the other) and perfectly entangled (knowing the state of one system told you exactly the state of the other). Given different weights for the four terms we looked at, a system can also be partially entangled.

It turns out that the more entangled the system is, the less interference we can exhibit within the individual systems. When it is unentangled, and it is as though the particle has never interacted with the detector, it is free to exhibit interference. When it becomes perfectly entangled, it shows no interference. And with partial entanglement, there will be partial loss of interference.

This is all well understood and easy to show mathematically (though I won't do it here). For "spookiness" we will have to look elsewhere.

Collapse

You'll recall that earlier I said that if a particle is in $\vec{z_+} + \vec{z_-}$ and we measure it, it collapses to one of those two basis states. But this presents two difficulties.

First, this process is not deterministic: there seems to be no way of knowing, a priori, whether the result of the collapse is going to be $\vec{z_+}$ or $\vec{z_-}$.

Second, it is not invertible. Whatever the state was before the collapse, the state is now (say) $\vec{z_+}$. You can't tell where it came from; some information is lost.

These are not just aesthetic concerns. Everything else we know about the dynamical evolution of quantum systems is deterministic and invertible (among other properties of unitary operators), so we're left with the question: what constitutes a collapse? After all, if our theory is complete, it needs to tell us when we should apply the general unitary evolution rules, and when we have to invoke this special one-off rule.

Did the particle collapse when it met the detector, or did it just become entangled (which is a unitary operation)? With small examples (for example, two particles), we can do experiments that show that the system is in a superposition. In fact, the whole field of quantum computing is based on this kind of behavior. And yet in the macroscopic world we never see superpositions -- whatever that would even mean.

So then when does the collapse really happen?

The heart of the matter

Let's back up and figure out why we're even talking of a "collapse."

Recall the two-slit experiment. In one case, the light was spread out more like a wave, and in the other, it behaved more like a particle -- as though the broad wave function "collapsed" into just two possible paths (one for each slit). In that latter case, the interference pattern disappeared.

From physics stackechange

But this fact can be explained through entanglement: the particle becomes entangled with the detector, and so the particle (by itself) will not show interference. As we said, that's easily shown with the math. So is that it? Is the problem solved?

Not quite. The particle still could have taken either of two paths. The same math describes the system as a whole as being in a superposition of states -- one in which the particle went left (and the detector says left) and one in which it went right (and the detector says right):
$$(\vec{particle_{left}} \otimes \vec{detector_{left}}) + (\vec{particle_{right}} \otimes \vec{detector_{right}})$$

So how come we only see one state? This fact can surely still be called a "collapse," and so the question remains: when does it happen?

Founders of QM to the rescue!

According to the Copenhagen Interpretation -- which for most of the 20th century was by far the most accepted, and continues to be the preferred school of thought -- the answer is basically a giant shrug.

The next best thing: a giant slug

Niels Bohr (one of the founders of QM and the Copenhagen school) never mentioned "collapse". Heisenberg (the other main dude who developed the interpretation) basically just said that it's whenever an apparatus measures it.

So much for that excursion.

Decoherence

Unsurprisingly, some in the physics community found this giant question mark a little troubling. One of the key developments in the latter half of the century was decoherence. If there's anything surprising about it, it's that it wasn't developed much sooner.

Basically, in any realistic system, the particles are going to encounter lots of other particles in the surrounding environment. Each interaction only leads to partial entanglement, since any one random particle is unlikely to give you perfect information about the rest of the system. In aggregate, however, trillions of particles will cause enough entanglement that any hope of exhibiting interference will be effectively lost. This process is known as decoherence. Even if it doesn't happen in well-controlled lab settings, your own body and brain are going to cause it.

This explains why the macroscopic world doesn't look spooky. It also has a big advantage over the "collapse" interpretation, which, being binary (collapsed vs not-collapsed), has a hard time accounting for partial loss of interference. But in order to answer the core question of why you see only one outcome, it has to do something a little funny.

You see, dear reader, when you observe the system, you become entangled with it:

$$(\vec{z_+} \otimes \vec{dz_+} \otimes \vec{reader_+}) + (\vec{z_-} \otimes \vec{dz_-} \otimes \vec{reader_-})$$

In other words, no collapse ever "really" takes place. Instead, there are now two copies of you, inhabiting roughly parallel universes. In one, the particle is in $\vec{z_+}$, and in the other, it's in $\vec{z_-}$.

Infinite parallel universes!

The explanation is known as the Many Worlds Interpretation, and it's very much in vogue:

The Many-Worlds Interpretation (MWI) of quantum mechanics holds that there are many worlds which exist in parallel at the same space and time as our own. The existence of the other worlds makes it possible to remove randomness and action at a distance from quantum theory and thus from all physics.

According to this explanation, each time there is a measurement with multiple possible outcomes, the universe splits so that all of them happen. When you encounter the system, you become entangled with it, and thus you also split into two copies, each inhabiting universes with different results, that effectively cannot interfere with each other.

This may leave you feeling a little dissatisfied. If there are two (actually, infinitely many!) copies of you inhabiting parallel universes, which one is "the real you?" Why are you you inhabiting this one and not some other one where things went differently? Have we really solved anything?

Such questions belong to philosophy, and not physics, they say. A common explanation is that there's no reason to privilege "this you" over all the "other you"s: in each parallel universe, there's probably another you is asking why she inhabits that universe. So, you see, there's nothing very strange going on after all! Problem solved!

An experimental test

That's all well and good, but it still feels meaningful to ask why "I" ended up in this universe and not that one, doesn't it? Sure, all those copies of me may technically also be "me," but not in the way that really seems to matter. And so the original itch isn't really scratched.

So maybe you start thinking: do I have any freedom of choice over which path I take? And so I leave you with the following challenge.

Suppose you are sitting in front of a machine that is generating electrons in the $\frac{\sqrt{2}}{2}(\vec{z_+}+ \vec{z_-})$ state once a second. Each time, you measure it in the $z_+ / z_-$ basis, and the universe splits:

The solid red circle represents the "real you" (from your perspective), and the empty ones are all the "other yous." (Ignore the fact that it splits in three in the upper-right. I stole this picture.)

Looking at the diagram from bottom to top, you might ask, at any given branching point: which path will the "real me" take? And such a question might be discarded as meaningless, since as we said, maybe in some funny sense they are all really you.

On the other hand, the "you" that is reading this can meaningfully ask, looking from the top down: what happened along the path that "this me" took? In other words, what does my history look like? How many times did the experiment yield $z_+$ as opposed to $z_-$? That's well-defined physically.

If the above description of QM is true, you should expect that in the long run, the results become arbitrarily close to 50-50. Conversely, if your history looks like it was basically a bunch of random and unsurprising events -- not just for electron experiments, but also macroscopic ones, which after all started out in the quantum domain -- then maybe it's not so interesting to know "why this path?" Because randomness, that's why.

But if not -- if the odds diverge significantly -- then you might start to wonder: why did "this me" end up in a relatively unexpected path? Was it pure chance? Or do I have some control over the path?

Such a question is meaningless within the structure of QM and the Many Worlds Interpretation. The fact that "this" you keeps getting lucky is neither here nor there; they're all you, and on average they didn't beat house odds. If you happen to have some way to choose your path and the quantum police come a-knockin', you can just plead ignorance. Some of "you" had to get lucky, after all.

So, that's it! Get crackin' on your psychic, reality-shifting abilities! If you do manage to somehow choose your path, you've got nobody to answer to!

Footnotes

[1] Recall that to determine the probability of some state $\psi$ being measured as some outcome $\phi$, we take the inner product (and then square its absolute value). With real vectors we call this the dot product, and if you write out the vectors in component form it works like so:
$$(x_1, y_1) \cdot (x_2, y_2) = x_1 * x_2 + y_1 * y_2$$
So to determine the probability of transition from $(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2})$ to $(\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2})$ we get:
$$(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}) \cdot (\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}) = \frac{1}{2} + \frac{-1}{2} = 0$$
This canceling of positive and negative terms can be considered interference. Why? Because we're deciding to think of the components $(a, b)$ as representing two distinct possibilities, one of which produces a positive term ($\frac{1}{2}$) and the other a negative ($\frac{-1}{2}$) which cancel.

Note, however, that if we had written those vectors in the $x_+$/$x_-$ basis, it would have looked like this:
$$(1, 0) \cdot (0, 1) = 0 + 0 = 0$$
Same result, but no reason to call it "interference" here.