Schrödinger's Cat: Where Does Superposition Stop?

0. Key points

• Schrödinger’s cat is not a piece of publicity for the strangeness of quantum mechanics. It is a counterexample its author built in 1935 in order to sneer: “if you believe this theory at face value, here is the absurdity you get.”
• The quantum notion of superposition is neither a state in which life and death are blended half and half, nor a state that is already one of the two while we merely fail to know which. These three situations can be told apart by experiment. We show exactly how, using polarizers and elementary vector algebra.
• The substance of the cat paradox is the question: if superposition is allowed at the microscopic scale and forbidden at the macroscopic scale, where is the boundary? This is called the measurement problem.
• The Copenhagen interpretation holds that the state collapses to a single alternative at the moment of observation; the many-worlds interpretation holds that no collapse occurs and that the observer branches along with everything else. Both make the same predictions, so no experiment has settled the matter.
• Decoherence explains, by calculation, why the evidence for superposition disappears for macroscopic bodies. It does not explain why we see a single outcome. That gap is still under debate today.

1. Motivation: the cat entered the stage as an objection

The year 1935 was a noisy one for quantum mechanics. In May, Einstein, Podolsky and Rosen published a paper asking whether the quantum-mechanical description of physical reality can be considered complete. Their answer was no. Provoked by that paper, Erwin Schrödinger published a long essay, “The present situation in quantum mechanics”, in the German journal Naturwissenschaften in November of the same year. The cat appears in its fifth section, in barely a dozen lines.

What deserves emphasis is that Schrödinger was no outsider to quantum mechanics. He wrote the wave equation; he is one of the theory’s co-founders. It was this man who confronted the then-standard interpretation of his own theory with: “take it at face value and you get this — is that really acceptable?” The cat, in other words, was born not as a glossy slogan for quantum mechanics but as a complaint.

In the previous article, Laplace’s demon and determinism, we looked at the classical worldview in which knowing all initial conditions fixes the future uniquely (uniqueness of solutions to the initial value problem（Theorem 3.2）［Laplace's Demon and Determinism］). Quantum mechanics inflicts two wounds on that worldview. One is that measurement outcomes can only be predicted probabilistically. The other, the subject of this article, is that before the measurement the state is not one of the alternatives at all. The first wound alone would be tolerable — a die does as much. The troublesome one is the second.

Note

This experiment has never been performed. There are obvious objections on grounds of animal welfare, but even before that, as we shall see, the very act of building the apparatus destroys any means of confirming a “superposed cat”. The cat remains in its box, of unknown vital status, only inside the thought experiment.

2. Preliminaries: superposition is not indecision

Before turning to the cat, let us fix how quantum mechanics writes down a “state” (for the classical way of writing a state, see states and phase space（Definition 2.1）［Laplace's Demon and Determinism］). High-school mathematics suffices.

Definition 2.1（Superposition of states）

Suppose a system has two states $\lvert 0 \rangle$ and $\lvert 1 \rangle$ that are mutually exclusive and that a measurement can always tell apart. Then for any two numbers $\alpha, \beta$ (complex in general, though in this article we use only real ones),

$$\lvert \psi \rangle = \alpha \lvert 0 \rangle + \beta \lvert 1 \rangle, \qquad \alpha^2 + \beta^2 = 1$$

is likewise one of the states the system can occupy. We call it the superposition of $\lvert 0 \rangle$ and $\lvert 1 \rangle$, and the coefficients $\alpha, \beta$ the probability amplitudes. The condition $\alpha^2 + \beta^2 = 1$ is the normalization condition.

The symbol $\lvert \cdot \rangle$ is no more than a marker saying “this is a vector representing a state”. Think of $\lvert 0 \rangle$ and $\lvert 1 \rangle$ as two perpendicular arrows in a plane; then $\lvert \psi \rangle$ is an arrow of length $1$. The normalization condition is just the Pythagorean theorem asserting that the arrow has unit length.

What, then, do the coefficients mean? The Born rule supplies the answer.

Definition 2.2（Born rule）

For a system in the state $\lvert \psi \rangle = \alpha \lvert 0 \rangle + \beta \lvert 1 \rangle$, a measurement of ”$0$ or $1$” returns $0$ with probability $\alpha^2$ and $1$ with probability $\beta^2$. Immediately after the measurement, the state of the system becomes exactly $\lvert 0 \rangle$ or $\lvert 1 \rangle$, whichever corresponds to the outcome.

The normalization condition $\alpha^2 + \beta^2 = 1$ is there to guarantee that the probabilities sum to $1$.

Note carefully: $\alpha^2$ is a probability, but $\alpha$ itself is not. It is the $\alpha$‘s that are added, and the squaring comes afterwards. This ordering is the decisive point separating quantum mechanics from ordinary probability theory. (Pushing the rule “add the amplitudes, then square” all the way to a sum over paths is exactly Feynman’s formulation; see probability amplitudes and the sum over paths（Definition 6.1）［Einstein and Feynman］.) Let us get our hands dirty and see what changes when the order is swapped.

Example 2.3（Three polarizers: sending light through nothing）

Stack two polarizers with their transmission axes crossed ($0^\circ$ and $90^\circ$) and no light gets through at all. Anyone can check this. Now insert a third polarizer at $45^\circ$ between the two. Naively, we have added another obstacle, so the transmitted light should stay at zero or drop further. In fact light comes through.

Let us compute. Unpolarized light of intensity $I_0$ passing the $0^\circ$ plate emerges with half the intensity, $I_0/2$, polarized at $0^\circ$. Write this state as $\lvert 0^\circ \rangle$. To pass it through the $45^\circ$ plate, choose $45^\circ$ and $135^\circ$ as the new reference directions, so that

$$\lvert 0^\circ \rangle = \frac{1}{\sqrt{2}} \lvert 45^\circ \rangle + \frac{1}{\sqrt{2}} \lvert 135^\circ \rangle$$

(this merely resolves the $0^\circ$ arrow along axes tilted by $45^\circ$). By Definition 2.2, the probability of getting through the $45^\circ$ plate is $(1/\sqrt{2})^2 = 1/2$. The intensity becomes $I_0/4$, and the light is remade into $\lvert 45^\circ \rangle$. Repeating the same step,

$$\lvert 45^\circ \rangle = \frac{1}{\sqrt{2}} \lvert 90^\circ \rangle + \frac{1}{\sqrt{2}} \lvert 0^\circ \rangle$$

so the probability of passing the $90^\circ$ plate is again $1/2$. The final intensity is

$$I_0 \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{I_0}{8}.$$

We added a plate and the transmitted light rose from $0$ to $I_0/8$. That adding a filter increases the throughput cannot be explained by the classical picture of a sieve. It happens because the $45^\circ$ plate does not merely attenuate the light: it rebuilds the state.

Remark 2.4

The state $\lvert 0^\circ \rangle$ in Example 2.3 is not “a half-and-half blend of $45^\circ$ light and $135^\circ$ light”. If it were, then after passing the $45^\circ$ plate, sending the light back through a $0^\circ$ plate should let half of it survive as a leftover. What actually happens is only that $\lvert 45^\circ \rangle$ passes a $0^\circ$ plate with probability $1/2$ again; no original $0^\circ$ component “remained”. A superposition is not a mixture. We make this distinction precise in Section 4.

3. The apparatus: a device translating microscopic ambiguity into macroscopic fact

The apparatus in Schrödinger’s original text is as follows.

flowchart TD
A["a single radioactive atom<br/>50% chance of decaying in one hour"] --> B["Geiger counter<br/>detects the decay and passes a current"]
B --> C["a relay drops the hammer"]
C --> D["the flask of prussic acid shatters"]
D --> E["the cat dies"]
A -. "if no decay, nothing happens" .-> F["the cat is alive"]

The chain that translates a microscopic superposition into macroscopic life and death

The ingenuity of this apparatus is that it acts as an amplifier. It translates the ambiguity of a single atom — a microscopic system that quantum mechanics certainly governs — into a difference in a cat, visible to anyone. Moreover the translation is faithful: whether the atom decayed and whether the cat died correspond one to one.

Let us first write down the state on the atom’s side.

Example 3.1（The atom after one hour, and the cat after two）

Consider a single radioactive atom with half-life $T$. The probability that it has not yet decayed at time $t$ is $2^{-t/T}$. By Definition 2.2, a probability is the square of an amplitude, so the undecayed amplitude is $\sqrt{2^{-t/T}} = 2^{-t/(2T)}$, and the state of the atom at time $t$ is

$$\lvert \text{atom}(t) \rangle = 2^{-t/(2T)} \lvert \text{undecayed} \rangle + \sqrt{1 - 2^{-t/T}}\, \lvert \text{decayed} \rangle .$$

Taking $T = 1$ hour and substituting $t = 1$ hour gives $2^{-1/2} = 1/\sqrt{2}$ and $\sqrt{1 - 1/2} = 1/\sqrt{2}$, so

$$\lvert \text{atom}(1\text{h}) \rangle = \frac{1}{\sqrt{2}} \lvert \text{undecayed} \rangle + \frac{1}{\sqrt{2}} \lvert \text{decayed} \rangle .$$

This is the situation Schrödinger described by writing that within one hour perhaps one atom decays, but with equal likelihood it may not.

While we are here, let us also compute $t = 2$ hours. Since $2^{-2/(2\cdot 1)} = 2^{-1} = 1/2$, the undecayed amplitude is $1/2$ and the probability is $1/4$. The probability that the cat is alive has fallen to $25\%$.

Because the apparatus is connected, the state of the atom is transcribed onto the state of the cat. Let us name this “transcribed joint state”.

Definition 3.2（Entanglement）

When the state of a composite system made of two systems A and B cannot be written as a product $\lvert a \rangle \lvert b \rangle$ of a state of A and a state of B, we say that A and B are entangled. For instance,

$$\lvert \Psi \rangle = \frac{1}{\sqrt{2}} \lvert \text{undecayed} \rangle \lvert \text{alive} \rangle + \frac{1}{\sqrt{2}} \lvert \text{decayed} \rangle \lvert \text{dead} \rangle$$

is an entangled state. Here $\lvert \text{undecayed} \rangle \lvert \text{alive} \rangle$ denotes the single state “the atom has not decayed and the cat is alive”.

Evolving the apparatus in time according to the Schrödinger equation, the total state after one hour is precisely this $\lvert \Psi \rangle$. If the atom is in a superposition and the apparatus reads the atom faithfully, the cat is dragged into the superposition too. Nowhere in the equation is there any machinery that would make an exception.

Schrödinger described this as a state in which the living and the dead cat are “smeared out” (verschmiert) in equal measure, and went on to say that such a thing cannot be accepted as a description of reality. For him this was not a consequence of the theory but evidence of a defect in it.

4. How this differs from “we just don’t know”

At this point many readers think: “surely the cat is either alive or dead, and we simply don’t know which until we open the box. Calling that a superposition is just grandiose language.”

The doubt is reasonable. And physics has a definite answer to it. The “we just don’t know” situation has a name of its own, and it can be told apart from a superposition by experiment.

Definition 4.1（Mixed state）

The situation in which the system is either in $\lvert 0 \rangle$ or in $\lvert 1 \rangle$, we do not know which, and each has probability $1/2$, is called an (equally weighted) mixed state. This is probability in the same sense as for a classical coin, and it is a different thing from the superposition of Definition 2.1.

We need one tool for the distinction. Regard $\lvert \text{alive} \rangle$ and $\lvert \text{dead} \rangle$ as two perpendicular arrows in a plane, and consider two new arrows obtained by rotating them by $45^\circ$ in that plane.

Lemma 4.2（Rewriting in a tilted reference frame）

Define

$$\lvert + \rangle = \frac{1}{\sqrt{2}}\bigl(\lvert \text{alive} \rangle + \lvert \text{dead} \rangle\bigr), \qquad \lvert - \rangle = \frac{1}{\sqrt{2}}\bigl(\lvert \text{alive} \rangle - \lvert \text{dead} \rangle\bigr).$$

Then $\lvert + \rangle$ and $\lvert - \rangle$ are again a pair of mutually orthogonal states of length $1$, and conversely

$$\lvert \text{alive} \rangle = \frac{1}{\sqrt{2}}\bigl(\lvert + \rangle + \lvert - \rangle\bigr), \qquad \lvert \text{dead} \rangle = \frac{1}{\sqrt{2}}\bigl(\lvert + \rangle - \lvert - \rangle\bigr)$$

hold.

Proof（Lemma 4.2）

Orthogonality and length are checked by identifying $\lvert \text{alive} \rangle$ and $\lvert \text{dead} \rangle$ with orthogonal unit vectors $\boldsymbol{e}_1, \boldsymbol{e}_2$ and computing inner products. The squared length of $\lvert + \rangle$ is $(1/\sqrt{2})^2 + (1/\sqrt{2})^2 = 1$, likewise $1$ for $\lvert - \rangle$, and their inner product is $(1/\sqrt{2})(1/\sqrt{2}) + (1/\sqrt{2})(-1/\sqrt{2}) = 1/2 - 1/2 = 0$.

The converse formulas amount to solving a linear system. Adding the two defining equations gives $\lvert + \rangle + \lvert - \rangle = (2/\sqrt{2}) \lvert \text{alive} \rangle = \sqrt{2} \lvert \text{alive} \rangle$; subtracting them gives $\lvert + \rangle - \lvert - \rangle = \sqrt{2} \lvert \text{dead} \rangle$. Dividing both sides by $\sqrt{2}$ yields the stated form.

∎

One and the same arrow, read against different coordinate axes. Reading it against the tilted axes is what exposes the difference between superposition and mixture

Read against the alive/dead axes, the arrow $\psi$ in the figure is “a slanted state that is neither”; read against the dashed tilted axes, it is a state lying exactly along the $+$ axis, with no hesitation about it. This difference in appearance is the content of the next proposition.

Proposition 4.3（Superposition and mixture can be distinguished）

Compare the following two.

• State (a): the superposition $\lvert \psi \rangle = \dfrac{1}{\sqrt{2}}\bigl(\lvert \text{alive} \rangle + \lvert \text{dead} \rangle\bigr)$.
• State (b): the mixed state that is $\lvert \text{alive} \rangle$ or $\lvert \text{dead} \rangle$ with probability $1/2$ each (Definition 4.1).

A measurement of “alive or dead” gives alive and dead with probability $1/2$ each in both cases, so it cannot distinguish them. But a measurement of ”$+$ or $-$” yields $+$ with probability $1$ in case (a), whereas in case (b) it yields $+$ with probability $1/2$ and $-$ with probability $1/2$. Hence repeating this measurement many times distinguishes (a) from (b).

Proof（Proposition 4.3）

First, the “alive or dead” measurement. In (a), by Definition 2.2, the probability of alive is the square of the coefficient $1/\sqrt{2}$, namely $1/2$, and likewise $1/2$ for dead. In (b) it is $1/2$ each directly by definition. The two agree.

Next consider the ”$+$ or $-$” measurement.

Case (a). Straight from the defining formulas of Lemma 4.2, $\lvert \psi \rangle = \lvert + \rangle$. Reading this as coefficients with respect to $\lvert + \rangle, \lvert - \rangle$ gives $(\alpha, \beta) = (1, 0)$, so by Definition 2.2 the probability of $+$ is $1^2 = 1$ and that of $-$ is $0^2 = 0$. The outcome is certainly $+$.

Case (b). When the system is in $\lvert \text{alive} \rangle$, the converse formula of Lemma 4.2 gives $\lvert \text{alive} \rangle = \frac{1}{\sqrt{2}}(\lvert + \rangle + \lvert - \rangle)$, so the probability of $+$ is $(1/\sqrt{2})^2 = 1/2$. When the system is in $\lvert \text{dead} \rangle$, likewise $\lvert \text{dead} \rangle = \frac{1}{\sqrt{2}}(\lvert + \rangle - \lvert - \rangle)$ and the probability of $+$ is $(1/\sqrt{2})^2 = 1/2$. Altogether,

$$P_{(b)}(+) = \frac{1}{2}\cdot\frac{1}{2} + \frac{1}{2}\cdot\frac{1}{2} = \frac{1}{2}.$$

Since $P_{(a)}(+) = 1 \ne 1/2 = P_{(b)}(+)$, the gap in probabilities is unambiguous. For instance, preparing the same state $100$ times and measuring each time, (a) gives $+$ on all $100$ runs, while the probability of $100$ consecutive $+$ outcomes in (b) is $2^{-100}$, roughly $10^{-30}$. In practice there is no way to get this wrong.

∎

Example 4.4（Where the coin analogy breaks）

One often meets the explanation that “a superposition is like a coin spinning in the air”. The analogy is half right. A spinning coin resembles a superposition in that heads or tails is not yet settled. But a spinning coin cannot be read against a different axis: a coin has no state “at $45^\circ$ between heads and tails”.

A quantum system does have one. What Proposition 4.3 showed is that two situations indistinguishable as long as we only look at “heads or tails” behave completely differently the moment we look along the tilted direction. Physicists single out superposition precisely because this response to tilted measurements cannot be reproduced by classical probability.

Applying all this to the cat makes the substance of the paradox clear. If the cat in the box really is in the superposition $\lvert \Psi \rangle$ of Definition 3.2, then in principle a measurement distinguishing “alive $+$ dead” from “alive $-$ dead” exists, and its outcomes should differ from those of the “one of the two, we just don’t know” case. Conversely, if such a measurement is absolutely impossible, then the theory must explain why it is impossible. This is the measurement problem.

5. Interpretations: where to break off the story

The computational rules of quantum mechanics themselves have not been broken by a single experiment in nearly a century. The dispute concerns what those rules say about the world. Here are the main positions.

Position	Does collapse occur?	What is the cat doing?	Weakness
Copenhagen interpretation	Yes, at the moment of measurement	Becomes alive or dead the moment the box is opened	“Measurement” is not defined inside the theory
Many-worlds interpretation	No	Branches into a world with a live cat and a world with a dead cat	Hard to make sense of the probability $\alpha^2$
Bohmian mechanics	No	The cat is always one or the other; the wave remains as a pilot	Blatant nonlocality, uneasy fit with relativity
Spontaneous collapse theories (GRW and the like)	Yes, spontaneously and stochastically	Macroscopic objects collapse instantly	Requires new constants in the theory, constrained by experiment

The Copenhagen interpretation, built chiefly by Bohr and Heisenberg, was for a long time the standard textbook account. It has two ingredients. First, while no measurement is made, the state evolves deterministically according to the Schrödinger equation. Second, at the moment of measurement the state collapses to the single state corresponding to the outcome (this is the second half of Definition 2.2). On this view the cat’s fate is settled the moment the box is opened.

Remark 5.1（The measurement problem）

The weakness of the Copenhagen interpretation is that “measurement” is not defined within the theory. Is the Geiger counter a measuring device? Is the cat? The cat’s retina? The experimenter’s brain? The Schrödinger equation treats apparatus, cat and human alike as collections of atoms, so the equation itself offers no ground for drawing a line saying “measurement begins here”. This problem — that the ground for drawing the line is not internal to the theory — is the measurement problem. The cat thought experiment is designed to strike at exactly this point.

The many-worlds interpretation is the position proposed by Hugh Everett III in his 1957 doctoral thesis. Its idea reverses the picture. Discard the extra rule of collapse and keep applying the Schrödinger equation to every system. Then the state $\lvert \Psi \rangle$ of Definition 3.2 does not collapse; it swallows the observer who opens the box, becoming

$$\frac{1}{\sqrt{2}} \lvert \text{decayed} \rangle \lvert \text{dead} \rangle \lvert \text{I saw the dead cat} \rangle + \frac{1}{\sqrt{2}} \lvert \text{undecayed} \rangle \lvert \text{alive} \rangle \lvert \text{I saw the live cat} \rangle .$$

These two terms cease to interfere with each other, so each “I” experiences only the outcome on its own side. Saying that I branch is closer to the original text than saying that worlds multiply. What draws criticism is probability: if both branches are equally real, what exactly is the number $\alpha^2$ the probability of? Several answers to this question are still being proposed today.

The important point is that these two interpretations predict the same experimental results. So “which one is right” cannot, at present, be decided by experiment. To settle it one must build a theory that changes the predictions themselves, as GRW does, and test it. Indeed the parameters of spontaneous collapse theories are being squeezed into ever narrower ranges by experiments at very low temperatures.

Caution

Be wary whenever you meet the phrase “in quantum mechanics the observer’s consciousness creates reality”. This claim follows neither from the Copenhagen interpretation nor from many-worlds. As we shall see in Section 6, the disappearance of interference is explained purely by interactions with air molecules and photons, with no reference to consciousness.

6. Decoherence: why the cat does not interfere

In Section 4 we showed that superposition and mixture can be told apart by a “tilted measurement”. Why, then, has nobody done this with a real cat? Is it merely that our technology falls short?

There is a powerful answer here, developed in the second half of the twentieth century: decoherence. A cat is not isolated in a vacuum. Air molecules strike it, it radiates infrared at body temperature, photons bounce off it. A live cat and a dead cat affect this “environment” in completely different ways. The state including the environment is therefore

$$\lvert \Psi \rangle = \frac{1}{\sqrt{2}} \lvert \text{alive} \rangle \lvert E_1 \rangle + \frac{1}{\sqrt{2}} \lvert \text{dead} \rangle \lvert E_2 \rangle ,$$

where $\lvert E_1 \rangle, \lvert E_2 \rangle$ are the states in which the environment is left in each case. Introduce the number $r = \langle E_1 \vert E_2 \rangle$ measuring how similar the two environment states are (their overlap: $r = 1$ if $\lvert E_1 \rangle = \lvert E_2 \rangle$, and $r = 0$ if they are perfectly distinguishable; we take it real here). Then the result of a “tilted measurement” on the cat is fixed as follows.

Proposition 6.1（Correlation with the environment destroys interference）

For a system in the state $\lvert \Psi \rangle$ above, if we leave the environment untouched and measure only the cat in the basis $\lvert + \rangle, \lvert - \rangle$ of Lemma 4.2, the probability of the outcome $+$ is

$$P(+) = \frac{1 + r}{2}, \qquad r = \langle E_1 \vert E_2 \rangle .$$

In particular $r = 1$ gives $P(+) = 1$ (the same as state (a) of Proposition 4.3), and $r = 0$ gives $P(+) = 1/2$ (the same as state (b)).

Proof（Proposition 6.1）

Substitute the converse formulas $\lvert \text{alive} \rangle = \frac{1}{\sqrt{2}}(\lvert + \rangle + \lvert - \rangle)$ and $\lvert \text{dead} \rangle = \frac{1}{\sqrt{2}}(\lvert + \rangle - \lvert - \rangle)$ of Lemma 4.2 into $\lvert \Psi \rangle$:

$$\begin{aligned} \lvert \Psi \rangle &= \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}}\bigl(\lvert + \rangle + \lvert - \rangle\bigr)\lvert E_1 \rangle + \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}}\bigl(\lvert + \rangle - \lvert - \rangle\bigr)\lvert E_2 \rangle \\ &= \lvert + \rangle \cdot \frac{\lvert E_1 \rangle + \lvert E_2 \rangle}{2} + \lvert - \rangle \cdot \frac{\lvert E_1 \rangle - \lvert E_2 \rangle}{2}. \end{aligned}$$

The probability of obtaining $+$ for the cat is the squared length of the vector accompanying $\lvert + \rangle$ (this is Definition 2.2 reread for the case where the coefficient is itself a vector). Hence

$$\begin{aligned} P(+) &= \left\lVert \frac{\lvert E_1 \rangle + \lvert E_2 \rangle}{2} \right\rVert^2 = \frac{1}{4}\bigl(\langle E_1 \vert E_1 \rangle + \langle E_2 \vert E_2 \rangle + 2\langle E_1 \vert E_2 \rangle\bigr) \\ &= \frac{1}{4}\bigl(1 + 1 + 2r\bigr) = \frac{1 + r}{2} . \end{aligned}$$

Along the way we used that $\lvert E_1 \rangle, \lvert E_2 \rangle$ have length $1$ (the normalization condition). Substituting $r = 1$ gives $P(+) = 1$, and $r = 0$ gives $P(+) = 1/2$.

∎

Corollary 6.2（A cat whose information has leaked into the environment looks classical）

When the environment distinguishes alive from dead perfectly ($r = 0$), every measurement performed on the cat alone returns outcome probabilities identical to those of the mixed state of Definition 4.1. That is, as long as we look only at the cat, no evidence whatsoever of a superposition can be obtained.

Proof（Corollary 6.2）

For the $+$ and $-$ measurement, setting $r = 0$ in Proposition 6.1 gives $P(+) = P(-) = 1/2$, matching state (b) of Proposition 4.3. Carrying out the same computation for an arbitrary tilted reference state $\lvert \theta \rangle = \cos\theta \lvert \text{alive} \rangle + \sin\theta \lvert \text{dead} \rangle$, the cross terms always carry a factor $\langle E_1 \vert E_2 \rangle = 0$ and vanish, so the probability is $\cos^2\theta \cdot \frac{1}{2} + \sin^2\theta \cdot \frac{1}{2} = \frac{1}{2}$, again matching the prediction of the mixed state. Since Proposition 6.1 said that a difference appears only when $r$ is nonzero, the difference disappears at $r = 0$.

∎

This is the decisive point. For an object the size of a cat, interaction with the environment drives $r$ to $0$ extraordinarily fast. A single air molecule bouncing off already scatters differently from a live cat than from a dead one, so the environment “records” the cat’s fate. Molecules collide an outrageous number of times per second, so the interval during which $r$ is appreciably nonzero is shorter than any time in which we could prepare a measurement. The reason we never see a superposed cat is not a matter of technology: it is that the information has already leaked into the air inside the box before the box is opened. (For another reason why quantum effects are invisible at everyday scales, see why quantum effects are invisible in everyday life（Example 5.4）［Laplace's Demon and Determinism］.)

If, then, we can isolate an object from its environment well enough, superposition should become visible even for large objects. And it does.

Example 6.3（How large can an interfering object be?）

In 1999 Arndt and collaborators in Vienna sent C$_{60}$ — the football-shaped molecule of sixty carbon atoms, about $720$ times the mass of a hydrogen atom — through a diffraction grating in high vacuum and observed interference fringes. Such fringes appear only when a single molecule “goes through several slits at once”, so the addition of arrows in the double slit（Example 6.2）［Einstein and Feynman］ applies verbatim to a molecule. These molecules have many internal vibrational modes and radiate infrared in flight, and later experiments confirmed that the stronger the radiation (that is, the more positional information leaks into the environment), the lower the contrast of the fringes. This makes the decline of $r$ in Proposition 6.1 visible in the laboratory.

The record has kept improving since; in 2019 interference was confirmed for giant molecules exceeding twenty thousand atomic mass units. That is still far from a cat (a few kilograms, roughly $10^{27}$ atomic mass units), but it is certain that the naive line “microscopic, so superposition; macroscopic, so no” has been replaced by a mere question of degree of isolation.

Remark 6.4（Has decoherence solved the measurement problem?）

It has not. What Corollary 6.2 says is that the cat becomes indistinguishable from a mixed state, not that the cat has become either alive or dead. The total state $\lvert \Psi \rangle$ is still a superposition. Decoherence gives a perfect answer to “why is no interference visible?” but none to “why do I experience only one outcome?” Filling the remaining gap with the extra rule of collapse is the Copenhagen interpretation; declining to fill it and reading the situation as “each branched I experiences one outcome apiece” is many-worlds. Decoherence did not dissolve the conflict between interpretations; it narrowed the point at issue.

7. Exercises

Exercise 7.1Easy

In the same setting as Example 2.3, suppose the middle polarizer is set at $30^\circ$ instead of $45^\circ$. Express in terms of $I_0$ the intensity of the light that finally passes the $90^\circ$ plate, and state which case is brighter, this one or $45^\circ$. Assume that the probability of passing a polarizer whose axis differs by an angle $\theta$ is $\cos^2\theta$.

Solution

Immediately after the $0^\circ$ plate the intensity is $I_0/2$. The $30^\circ$ plate differs by $30^\circ$, so the transmission probability is $\cos^2 30^\circ = (\sqrt{3}/2)^2 = 3/4$, and the intensity becomes

$$\frac{I_0}{2} \times \frac{3}{4} = \frac{3I_0}{8}.$$

The final $90^\circ$ plate differs from the preceding $30^\circ$ by $60^\circ$, so its transmission probability is $\cos^2 60^\circ = (1/2)^2 = 1/4$. Hence

$$\frac{3I_0}{8} \times \frac{1}{4} = \frac{3I_0}{32}.$$

In the $45^\circ$ case we had $I_0/8 = 4I_0/32$, and $3I_0/32 < 4I_0/32$, so $45^\circ$ is brighter. Indeed the product of transmission probabilities $\cos^2\theta \cos^2(90^\circ - \theta)$ is maximal at $\theta = 45^\circ$.

Exercise 7.2Standard

In the setting of Example 3.1 (half-life $T = 1$ hour), the box is left alone for two hours.

1. Write the state of the atom after two hours as a superposition of $\lvert \text{undecayed} \rangle$ and $\lvert \text{decayed} \rangle$, with explicit numerical coefficients.
2. Find the probability that the cat is alive at that moment.
3. After how many hours does the probability that the cat is alive drop below $10\%$? Use $\log_{10} 2 = 0.301$.

Solution

1. The probability of no decay is $2^{-t/T} = 2^{-2} = 1/4$. The amplitude is its square root, $\sqrt{1/4} = 1/2$. The probability on the decayed side is $1 - 1/4 = 3/4$, so its amplitude is $\sqrt{3}/2$. Hence

$$\lvert \text{atom}(2\text{h}) \rangle = \frac{1}{2} \lvert \text{undecayed} \rangle + \frac{\sqrt{3}}{2} \lvert \text{decayed} \rangle .$$

The normalization condition holds: $(1/2)^2 + (\sqrt{3}/2)^2 = 1/4 + 3/4 = 1$.

2. Life and death of the cat correspond one to one to the state of the atom, so the probability of being alive equals the probability of no decay, $1/4 = 25\%$.

3. Solve $2^{-t} < 0.1$. Taking common logarithms of both sides gives $-t \log_{10} 2 < -1$, that is,

$$t > \frac{1}{\log_{10} 2} = \frac{1}{0.301} = 3.32\ldots$$

so this happens from about 3.33 hours (3 hours 20 minutes) onward.

Exercise 7.3Standard

In the setting of Proposition 6.1, let the overlap of the environment states be $r$ (a real number with $0 \le r \le 1$).

1. Find the probability $P(-)$ of the outcome $-$.
2. Define the visibility of the interference by $V = P(+) - P(-)$. Express $V$ in terms of $r$.
3. For $r = 0.01$, what is the difference between the probabilities of $+$ and $-$, in per cent? Comment on whether $100$ measurements could resolve this difference.

Solution

1. The probabilities sum to $1$, so $P(-) = 1 - P(+) = 1 - \dfrac{1+r}{2} = \dfrac{1-r}{2}$. Computing instead the squared length of the coefficient $(\lvert E_1 \rangle - \lvert E_2 \rangle)/2$ of $\lvert - \rangle$ obtained in the proof of Proposition 6.1 gives $\frac{1}{4}(1 + 1 - 2r) = \frac{1-r}{2}$, in agreement.

2. $V = \dfrac{1+r}{2} - \dfrac{1-r}{2} = r$. In other words, the overlap of the environment states is itself the visibility of the interference.

3. For $r = 0.01$ we get $P(+) = 0.505$ and $P(-) = 0.495$, a difference of $1\%$. In $100$ measurements the mean number of $+$ outcomes is $50.5$, but the statistical fluctuation has standard deviation $\sqrt{100 \times 0.5 \times 0.5} = 5$ counts. A bias of $0.5$ counts is buried in a fluctuation of $5$, so $100$ measurements cannot resolve it. To make the difference exceed the fluctuation one needs roughly $N \times 0.01 > \sqrt{N}/2$, that is about $N > 2500$ measurements. As $r$ shrinks further the required number grows like $r^{-2}$, so for cat-sized values of $r$ it is utterly out of reach.

Exercise 7.4Hard

A friend insists: “the cat is either alive or dead, and humans simply don’t know which. Superposition is just word play.” Explain what is inadequate about this claim, touching on both Proposition 4.3 and Corollary 6.2. Also give the reason why the friend’s claim can be called “effectively correct for a real cat”.

Solution

What is inadequate is that the claim is stated in a form that experiment can refute. The friend’s “one of the two, but we don’t know” is nothing other than the mixed state of Definition 4.1. By Proposition 4.3, a mixed state and a superposition are distinguished by the ”$+$ or $-$” measurement: the former gives $+$ with probability $1/2$, the latter with probability $1$. So the two are in principle different physical states, not word play. Indeed, in systems isolated well enough, as in Example 6.3, the prediction on the superposition side has been confirmed experimentally.

On the other hand, the friend’s claim works almost perfectly for a real cat. By Corollary 6.2, once the environment has recorded life or death ($r = 0$), every measurement made on the cat alone returns the same probabilities as for a mixed state. As estimated in Exercise 7.3, the number of measurements required grows like $r^{-2}$, an astronomical figure at cat scale. The accurate verdict is therefore: correct for practical purposes, incorrect as a matter of principle.

Stopping there, however, falls into the trap of Remark 6.4. What $r = 0$ says is “indistinguishable”, not “settled on one of the two”. To assert that it is settled, one must either add the rule of collapse (Copenhagen) or read the situation as “I find myself in a branch where it looks settled” (many-worlds), and that choice is not currently decidable by experiment.

References

• E. Schrödinger, “Die gegenwärtige Situation in der Quantenmechanik”, Naturwissenschaften 23 (1935), 807–812, 823–828, 844–849 — the original source in which the cat appears. An English translation by J. D. Trimmer is included in the Proceedings of the American Philosophical Society 124 (1980).
• A. Einstein, B. Podolsky, N. Rosen, “Can Quantum-Mechanical Description of Physical Reality Be Considered Complete?”, Physical Review 47 (1935), 777–780 — the paper that directly prompted the cat.
• H. Everett III, “‘Relative State’ Formulation of Quantum Mechanics”, Reviews of Modern Physics 29 (1957), 454–462 — the original paper on the many-worlds interpretation.
• W. H. Zurek, “Decoherence, einselection, and the quantum origins of the classical”, Reviews of Modern Physics 75 (2003), 715–775 (arXiv:quant-ph/0105127) — the standard review of decoherence.
• M. Arndt et al., “Wave–particle duality of C60 molecules”, Nature 401 (1999), 680–682 — the interference experiment with giant molecules.
• R. P. Feynman et al., Feynman Butsurigaku V: Ryoshi Rikigaku (Japanese edition of The Feynman Lectures on Physics, Vol. III), Iwanami Shoten (in Japanese) — Chapter 1, “Quantum behavior”, explains the addition of probability amplitudes carefully, using the double slit.

Appendix: Rewriting the cat as a two-dimensional vector

In components, the discussion of Section 4 becomes plain matrix algebra. Setting

$$\lvert \text{alive} \rangle = \begin{pmatrix} 1 \\ 0 \end{pmatrix}, \qquad \lvert \text{dead} \rangle = \begin{pmatrix} 0 \\ 1 \end{pmatrix},$$

the superposed state becomes $\lvert \psi \rangle = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ 1 \end{pmatrix}$, a column vector with both components positive. The states $\lvert + \rangle, \lvert - \rangle$ of Lemma 4.2 are

$$\lvert + \rangle = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ 1 \end{pmatrix}, \qquad \lvert - \rangle = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ -1 \end{pmatrix},$$

which amounts to rotating the coordinate axes by $45^\circ$. The probability that measuring some state $\lvert \phi \rangle$ yields $\lvert + \rangle$ is the square of the inner product, $\langle + \vert \phi \rangle^2$. For $\lvert \phi \rangle = \lvert \psi \rangle$,

$$\langle + \vert \psi \rangle = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 & 1 \end{pmatrix} \cdot \frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ 1 \end{pmatrix} = \frac{1}{2}(1 + 1) = 1,$$

so the probability is $1$, reproducing case (a) of Proposition 4.3.

A mixed state cannot be written as a single vector. This is the essential difference between mixture and superposition. To write a mixture one must promote the state to a $2 \times 2$ matrix (a density matrix). The matrices corresponding to the superposition $\lvert \psi \rangle$ and to the mixture are, respectively,

$$\rho_{\text{superposition}} = \frac{1}{2}\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}, \qquad \rho_{\text{mixture}} = \frac{1}{2}\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}.$$

The diagonal entries (the probabilities $1/2$ for alive and for dead) agree exactly; only the off-diagonal entries differ. These off-diagonal entries are called the interference terms. What Proposition 6.1 was doing is multiplying the off-diagonal entries by $r$ through the correlation with the environment,

$$\rho(r) = \frac{1}{2}\begin{pmatrix} 1 & r \\ r & 1 \end{pmatrix},$$

so that the interference terms vanish as $r \to 0$. Decoherence is, quite literally, the process of shaving the corners off a matrix. Density matrices are studied properly in a university quantum mechanics course, but remembering the single sentence “the difference between superposition and mixture lies in the off-diagonal entries” is enough to follow the story of the cat to the end.