# Schrödinger's cat and the difficulty of macroscopic superposition state

The Schrödinger's cat was regarded as peculiar since we seldom encounter a superposition state in macroscopic scale: $$\mid \mathrm{dead \,\,cat} \rangle + \mid \mathrm{alive \,\, cat}\rangle$$

We more often describe an unknown cat as $$\mid \mathrm{dead \,\,cat} \rangle \langle \mathrm{dead \,\,cat} \mid+ \mid \mathrm{alive \,\, cat}\rangle \langle \mathrm{alive \,\, cat}\mid$$

without superposition.

I often heard that it is difficult to prepare and maintain a large-scale superposition state. Similar difficulty also occurs in quantum computing.

My question is, actually what is the reason for the difficulty to prepare and maintain a large-scale superposition state? If it is decoherence, why decoherence happens? Is that because of entropy?

$$\renewcommand{\ket}{\left \lvert #1 \right \rangle}$$ $$\renewcommand{\bra}{\left \langle #1 \right \rvert}$$ We can see how decoherence really works, why it messes up superposition states, and why it's particularly prone to messing up states of large objects all through a very simple example $$^{[a]}$$.

# Single two-level system

Suppose we have a quantum system $$S$$ with two possible states. $$S$$ could be a cat and the states could be $$\left \lvert \text{alive} \right \rangle$$ and $$\left \lvert \text{dead} \right \rangle$$, but for the sake of generality we label the states as

$$\ket{\uparrow} \quad \text{and} \quad \ket{\downarrow} \, .$$

## Coherent case

Now suppose $$S$$ is in state $$\ket{y}$$ defined as

$$\ket{y} \equiv \left( \ket{\uparrow} + i \ket{\downarrow} \right) / \sqrt{2} \, .$$

This is a perfectly happy superposition state. It's density matrix is

$$\rho_S = \ket{y} \bra{y} = \frac{1}{2} \left( \ket{\uparrow} \bra{\uparrow} + \ket{\downarrow} \bra{\downarrow} - i \ket{\uparrow} \bra{\downarrow} + i \ket{\downarrow} \bra{\uparrow} \right) = \frac{1}{2} \left[ \begin{array}{cc} 1 & -i \\ i & 1 \end{array} \right] = \frac{1}{2} \left( \text{Id} + \sigma_y \right) \, ,$$ where in the matrix representation we've ordered the states $$\{ \ket{\uparrow}, \ket{\downarrow} \}$$. We can think of this state as a spin pointed along the $$y$$ axis (hence the symbol $$\ket{y}$$). The first two terms are the classical terms (diagonal in the matrix representation) and the other two are the so-called "coherences" (off-diagonal) which disappear via decoherence processes, as we show below.

If we were to prepare $$S$$ in state $$\ket{y}$$ many times and each time measure it along the $$z$$ axis we would get a random sequence of results where half of them are $$\uparrow$$ and half are $$\downarrow$$. Naively you might think that this means that our preparation procedure is giving us a normal "classical" probability distribution where half of the time we prepared $$\ket{\uparrow}$$ and half of the time we prepared $$\ket{\downarrow}$$. However, we can see that this is not true if we rotate $$S$$ about the $$x$$ axis and then measure it along the $$z$$ axis. The operator for the rotation is $$U = \cos(\theta / 2) \, 1 + i \sin(\theta / 2) \, \sigma_x = \left[ \begin{array}{cc} \cos(\theta/2) & i \sin(\theta/2) \\ i \sin(\theta / 2) & \cos(\theta / 2) \end{array} \right]$$ and the density matrix after the rotation is $$U \rho_S U^\dagger = \frac{1}{2} \left[ \begin{array}{cc} 1 - \sin(\theta) & -i \cos(\theta) \\ i \cos(\theta) & 1 + \sin(\theta) \end{array} \right] = \frac{1}{2} \left( \text{Id} - \sin(\theta) \sigma_z + \cos(\theta) \sigma_y \right) \, .$$

As you can see, for a given angle $$\theta$$, the probability to find the system in $$\ket{\uparrow}$$ is $$(1/2)(1 - \sin(\theta))$$, i.e. it depends on how much we rotated. Another way to say this is that $$\langle \sigma_z \rangle_{U \rho_S U^\dagger} = - \sin(\theta) \, ,$$ i.e. the expectation value of $$\sigma_z$$ oscillates as we rotate the system. This makes perfect sense if you think of the two level system as an arrow oriented in 3D space (e.g. a spin): as we rotate the system about the $$x$$ axis its projection about the $$z$$ axis oscillates. So far, nothing about this example tells us anything about decoherence or why it's hard to make big Schrodinger cat states, so now let's get to that.

## Incoherent case

Suppose $$S$$ interacts with some other two level system $$E$$. The letter $$E$$ stands for "environment" which will make sense later. Suppose the state of the combined $$S + E$$ system is $$^{[b]}$$

$$\left( \ket{\uparrow}\ket{\downarrow} + \ket{\downarrow}\ket{\uparrow} \right) / \sqrt{2}$$

where the first ket labels the state of $$S$$ and the second ket labels the state of $$E$$. Now the critical part: what happens if we now do the rotate-and-measure experiment described above on system $$S$$ without doing anything, including measurement, to $$E$$? Experimentally, when we try this in the lab, we find that there is no oscillation in the probability to find $$S$$ in $$\ket{\uparrow}$$ as a function of $$\theta$$! This is decoherence. To describe this mathematically we look at the density matrix of the $$S+E$$ system. The state of the total system is $$\rho_{S+E} = \left[ \begin{array}{cccc} 0&0&0&0 \\ 0 & 1/2 & 1/2 & 0 \\ 0 & 1/2 & 1/2 & 0 \\ 0&0&0&0 \end{array} \right]$$

where we've ordered the states $$\{ \ket{\uparrow}\ket{\uparrow}, \ket{\uparrow}\ket{\downarrow}, \ket{\downarrow}\ket{\uparrow},\ket{\downarrow}\ket{\downarrow} \}$$. To predict the behaviour of experiments done on $$S$$ alone, we take the trace of $$\rho_{S+E}$$ over the part of the space belonging to $$E$$ $$^{[c]}$$. Doing this gives $$\tilde{\rho}_S \equiv \text{Tr}_E \left( \rho_{S+E}\right) = \frac{1}{2} \left[ \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right] = \frac{1}{2} \left( \ket{\uparrow}\bra{\uparrow} + \ket{\downarrow}\bra{\downarrow} \right) = \frac{1}{2}\text{Id} \, .$$ The off-diagonal terms are gone - we have a purely classical state! If we now rotate $$\tilde{\rho}_S$$ by any rotation operator $$U$$ we find that $$U \tilde{\rho}_S U^\dagger = \frac{1}{2} \left[ \begin{array}{cc} 1&0\\0&1 \end{array} \right] = \tilde{\rho}_S$$ and $$\langle \sigma_z \rangle_{U \tilde{\rho}_S U^\dagger} = \text{Tr}_S (U \tilde{\rho}_S U \sigma_z) = 0 \, .$$ Rotations no longer do anything - there's no oscillation and the expectation value of $$\sigma_z$$ is always zero regardless of rotation angle.

This is really, really interesting. Previously we said that a single isolated two level system can be thought of like a spin particle: it's always pointing in some direction in space, so even if measurements along the $$z$$ axis give half up and half down, if you rotate the spin and measure, you see oscillation. On the other hand, we just showed that if we let the two level system interact with something else ($$E$$), the combined system can be left in a state such that the original two level system ($$S$$) doesn't exhibit that oscillation.

What we have just seen is the essence of quantum decohrence. If a quantum system $$S$$ becomes entangled with its surrounding environment $$E$$, then $$S$$ can lose its quantum nature. Of course, if we don't ignore the environment $$E$$ and instead include it in our measurements, then we would observe the full quantum properties of the combined system. In other words, decoherence is just lack of knowledge of the complete system.

If $$E$$ is really big then keeping track of all its degrees of freedom and measuring them in a controlled way is just impossible. That is the essence of why making big Schrodinger cats is hard; if the system $$S$$ is big it interacts with more environmental degrees of freedom and so observing quantum effects is very difficult.

For something a large as a speck of dust interacting with air molecules, the time it takes for the decoherence to kill any off diagonal elements in the density matrix is incredibly small $$^{[d]}$$. Interestingly though, some fairly large systems can be sufficiently isolated from their environments such that they exhibit quantum properties long enough times to be useful; this is, for example, a large fraction of what goes into building a quantum computer.

# Large system

So far we showed what decoherence is, and in particular how it makes a quantum system appear classical. To recap, decoherence happens when your system $$S$$ interacts with the environment $$E$$; if you don't have access to the environment degrees of freedom, then $$S$$ can lose its quantum interference properties and appear classical. In the example we gave, we saw that a single two level system interacting with another one can appear classical. We will now, via an illustrative extention of the same example, that a large system is more prone to decoherence.

Suppose $$S$$ consists of three two level systems in an initial state $$\ket{\uparrow \uparrow \uparrow}$$ with density matrix $$\rho = \ket{\uparrow \uparrow \uparrow} \bra{\uparrow \uparrow \uparrow} \, .$$ Note that there's a sort of redundancy here: we have three separate spins which can be thought of as collectively representing a single spin up.$$^{[e]}$$ Like in the single particle case, we can measure the projection of the spin along the $$z$$ axis, but in this case we use the three-particle operator $$Z^{(3)} \equiv \left( \sigma_z \otimes \sigma_z \otimes \sigma_z \right) \, .$$

## Coherent case

Like before, if we rotate all three spins and measure the average of $$Z^{(3)}$$ we get a sinusoidal dependence on the rotation angle. In particular, if we rotate each spin by an angle $$\theta$$ about the $$x$$ axis, then we get $$\langle Z^{(3)} \rangle_{U \rho U^\dagger} = \cos(\theta)^3 \, .$$

## Incoherent case

Now consider what happens if just one of our spins interacts with the environment. Suppose the middle spin interacts with the environment such that the initial state $$\ket{\uparrow \uparrow \uparrow} \ket{\downarrow}$$ (here the separate second ket with one arrow represents the environment) becomes $$\left( \ket{\uparrow \uparrow \uparrow}\ket{\downarrow} + \ket{\uparrow \downarrow \uparrow}\ket{\uparrow} \right) / \sqrt{2} \, .$$ Writing out the complete four particle density matrix would be tedious and unenlightening. However, the reduced density matrix of the first three particles is $$\tilde{\rho}_S = \frac{1}{2} \left( \ket{\uparrow \uparrow \uparrow}\bra{\uparrow \uparrow \uparrow} + \ket{\uparrow \downarrow \uparrow}\bra{\uparrow \downarrow \uparrow} \right) \, .$$ Note that we have a diagonal density matrix just like we did in the single particle incoherent case. With this density matrix, the expectation value of $$Z^{(3)}$$ following a rotation of all spins by $$\theta$$ is $$\langle Z^{(3)} \rangle_{U \tilde{\rho}_S U^\dagger} = \frac{1}{2} ( \underbrace{\cos(\theta)^3}_{\text{from } \ket{\uparrow \uparrow \uparrow}} + \underbrace{-\cos(\theta)^3}_{\text{from } \ket{\uparrow \downarrow \uparrow}} ) = 0 \, .$$ Here again we've lost the oscillation, and it only took a single spin interacting with the environment to do it. That is why making large Schrodinger cats is hard.

# Notes

$$[a]$$: This is a simplified version of an example from the introductory chapter of my PhD thesis (pdf).

$$[b]$$: This state can be realized if we start with the system in $$\ket{\uparrow}\ket{\downarrow}$$ and subject the system to the Hamiltonian $$H=\sigma_+ \sigma_- + \sigma_- \sigma_+$$ for the proper amount of time such that the propagator is $$U = \left [ \begin{array}{cccc} 1&0&0&0\\ 0&\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}}&0\\ 0&\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}}&0\\ 0&0&0&1 \end{array} \right ] \, .$$

$$[c]$$: Note that, like any other theoretical description, this procedure is justified because it reproduces the results of experiments.

$$[d]$$: I don't remember the numbers but see Schlosshauer's book for a calculation.

$$[e]$$: This kind of redundancy is critical in classical machines. For example, a memory bit in a classical computer could be represented by current in a large number of conduction channels in a transistor; if any one of those channels were to change state, that's such a tiny fraction of the total current that the logical state of the transistor is preserved. This redundancy gives classical computers their robustness against errors on the microscopic level.

It is because of quantum statistical irreversibility, which is closely related to entropy, as the OP suspected.

Qualitatively it is quite easy to understand this. From the laws of quantum mechanics on the microscopic level emerges a classical behaviour for macroscopic (i.e. many particle objects). Of course this is not sufficient though and does not give a lot of insight, so here is what I know about the topic.

The main insight is that what was mentioned above causes the density matrix to become diagonal, i.e. the coherence to disappear due to many body interactions and resulting dissipation. This has been demonstrated quantitatively for exactly solvable systems.

• Leggett and Caldeira solved a system of coupled simple harmonic oscillators in 1983

We apply the influence-functional method of Feynman and Vernon to the study of Brownian motion at arbitrary temperature. By choosing a specific model for the dissipative interaction of the system of interest with its environment, we are able to evaluate the influence functional in closed form and express it in terms of a few parameters such as the phenomenological viscosity coefficient. We show that in the limit h→0 the results obtained from the influence functional formalism reduce to the classical Fokker-Planck equation. In the case of a simple harmonic oscillator with arbitrarily strong damping and at arbitrary temperature, we obtain an explicit expression for the time evolution of the complete density matrix ϱ(x, x′, t) when the system starts in a particular kind of pure state. We compare our results with those of other approaches to the problem of dissipation in quantum mechanics.

• Followup work in the same spirit is due to Zurek (this is a good resource on a related, but slightly different topic), also illuminating the quantum information side of this problem. In particular Unruh&Zurek showed that decoherence happens when coupling to a quantum field. Other important papers are by Joos&Zeh and others.
• More recently (1990s and 2000s) there has been other exactly solved systems that demonstrate phase transitions of quantum systems that look like what the Copenhagen interpretation calls "collapse", but are actual based on just unitary evolution (so no actual collapse there, don't worry ;) ). This is my summary of this excellent answer by Arnold Neumaier, where further references may be found.

I should add that there are indeed macroscopic objects that display coherence phenomena, superconductors and superfluids are the most prominent example. So the "preparation" of such systems can be achieved, but due to the dissipative nature of most everyday systems their coherence is statistically very unlikely.

• why the downvote? This is a full and historically accurate review with reference of why decoherence happens in macroscopic systems, which answers the question. I would appreciate a comment to understand how I can improve my answer. – Wolpertinger Jul 10 '16 at 11:09
• I didn't downvote, but I'd say that while this answer contains lots of useful reference links and gives the right general idea, it doesn't itself actually answer the question. It's particularly nice that you mention superconductors etc. You could add mention of actual quantum devices too, like trapped ions or superconducting qubits. – DanielSank Jul 11 '16 at 13:21
• @DanielSank that's true, I guess I particularly didn't answer the why the "difficulty to prepare and maintain" arises, but rather why we don't encounter macroscopic superpositions in everyday life. I know less about this branch of the topic, so instead of editing I would like to direct readers to your answer (excellent and my +1 btw) and hope that mine at least answers the "why decoherence happens" question. – Wolpertinger Jul 11 '16 at 14:08
• To be clear, I'm really glad you wrote your answer. The large number of links is quite valuable. – DanielSank Jul 11 '16 at 22:10

Let's consider a qubit that has two "classical states" $\left|0\right>$ and $\left|1\right>$, e.g. a current in a flux qubit that flows in one direction or in the other direction, while superpositions of these states are "non-classical" and will decohere into a mixed state of the classical states. What I'm going to demonstrate now is that a superposition is extremely fragile, the fragility can be exploited to turn the system into an extremely sensitive measuring device. In the following I'll focus on using a single qubit as a dark matter (DM) detector.

Suppose we start out with a qubit initialized in the state $\left|0\right>$ and we apply the Hadamard gate $U$ that acts as follows:

$$\begin{split} U\left|0\right> &= \frac{1}{\sqrt{2}}\left[\left|0\right\rangle + \left|1\right\rangle\right]\\ U\left|1\right> &= \frac{1}{\sqrt{2}}\left[\left|0\right\rangle - \left|1\right\rangle\right] \end{split}$$

Note that $U$ is it's own inverse, so applying $U$ again will bring the qubit back to the state $\left|0\right\rangle$ we started out with. But now consider what happens if during the time the qubit spends it time being a superposition of $\left|0\right\rangle$ and $\left|1\right\rangle$ a DM particle collides with it. Then the qubit will get entangled with the DM particle, qubit-DM particle system will be in a state of the form:

$$\left|\psi\right\rangle = \frac{1}{\sqrt{2}}\left[\left|0\right\rangle \left|D_0\right\rangle + \left|1\right\rangle\left|D_1\right\rangle\right]$$

where the states $\left|D_{i}\right\rangle$ are the DM-particle states after scattering off the qubit in state $\left|i\right\rangle$. Here we assume that the scattering is elastic so that the qubit's state does not change at all. So, you might think that because the qubit was not affected by the interaction at all, we cannot perform a measurement on the qubit's state to find out that it has interacted with a DM particle. But watch what happens if we apply the Hadamard gate again to the qubit:

$$U\left|\psi\right\rangle =\left|0\right\rangle\left|D^{+}\right\rangle+\left|1\right\rangle \left|D^{-}\right\rangle$$

where $D^{\pm} = \frac{1}{2}\left[\left|D_0\right\rangle\pm\left|D_1\right\rangle\right]$

So, had there been no interaction the qubit would have returned to the initial state $\left|0\right\rangle$ but now we end up with an entangled state of the qubit and the DM-particle such that there is now a finite probability to find the qubit in the state $\left|1\right\rangle$, despite the fact that collision with the DM-particle happened in a purely elastic way at low energy such that it did not affect the physical state of the qubit in any way at the time of the collision. The probability to find the qubit in the state $\left|1\right\rangle$ is $\frac{1}{2}\left[1-\operatorname{Re}\left\langle D_0\right|D_1\left.\right\rangle\right]$, so it depends on the overlap between the two dark matter states corresponding to scattering off the qubit in the two states of the superposition.

If the states $\left|D_i\right\rangle$ are orthogonal, then you have 50% probability to find the qubit in the states $\left|0\right\rangle$ and $\left|1\right\rangle$, the density matrix after tracing out the DM-particle state is $\frac{1}{2}\left[\left|0\right\rangle\langle 0| + \left|1\right\rangle\langle 1|\right]$.

The larger we make the qubit, the more likely it will be that during the time it spend being a superposition an interaction will have occurred. As demonstrated above, even a purely elastic scattering of a particle is enough to change the superposition into a mixed state. But a macroscopic object at room temperature will emit a huge number of infrared photon, so a superposition will decay into a mixed state extremely rapidly, as the state of the photons will depend on which part of the superposition they were emitted from and a huge number of these photons are emitted every fraction of a second.

A possible way out of this problem would be to consider superposition for which the two parts differ less and less as we make the qubits larger and larger. The change in the state of the environment due to interactions between the two parts of the superposition will then be smaller for each interaction.

So, in conclusion, this example demonstrates that decoherence of macroscopic superpositions happens due to the extreme fragility of of such superpositions. All that's needed is for information about the state to leak out into the environment. One can relate this to entropy increase, but that in itself doesn't explain why this happens so rapidly. But once information has leaked out, it's going to be irreversible in practice, as the degrees if freedom in the environment that now have "detected" the superposition will themselves leak that information to other degrees of freedom in the environment. You could say that once Schrödinger's cat is out of the bag, it's not going to come back.

• This could be a good answer but it's hard to read. For example, the acronym "DM" is used without being defined. I figured out that it means "dark matter" but you should make that more clear by writing "...dark matter (DM)...". – DanielSank Jul 11 '16 at 13:03
• @DanielSank Yes, I agree, I made some changes to the text. – Count Iblis Jul 11 '16 at 17:17

Decoherence happens because in a macroscopic system you are not able to create a small isolated system. In practice you are deal with statistical mixture and not pure state. There's a good description on Wikipedia.

• I briefly read wikipedia about decoherence. My impression is that is an interpretation of experimental result, but does not answer why decoherence happens... – user26143 Mar 2 '14 at 17:33
• The problem is that in macroscopic system objects are inevitable non-isolated: they are coupled with the external ambient. In particular the system "cat+atom" become immediately entangled with ambient and as a consequence it loses his coherence. – LC7 Mar 2 '14 at 17:59
• Haroche says: the cat is a complex and OPEN (it cannot be isolated) system, it's impossible to describe it with a wave function. – LC7 Mar 2 '14 at 18:04
• I still have a question. If we consider a hydrogen atom in external field, e.g. Stark and Zeeman effects, the hydrogen atom is not isolated. Why we can still write a wavefunction for the hydrogen atom? Is that because the interaction from external field do not depend on the details of the configuration of radiation field? – user26143 Mar 2 '14 at 18:54
• No the fact is that you can consider atom plus fiels as a system. In fact the hamiltonian of the system have both atomic and magnetic terms. Sorry for the bad english i'm using my phone now!! – LC7 Mar 2 '14 at 18:58

In any quantum experiment, as soon as the state of the system can in principal be known (be it because of emitted photons available to the experimenter, an interaction with the environment we may be able to read off the state, or any other means by which the observer can determine the state) it decays and ends up in one of the basis states. That's precisely what the measurement postulate of quantum mechanics predicts.

Since in macroscopic systems the probability for this to happen increases significantly, we're hardly able to maintain coherence.

Interestingly, in certain cases we can delete the information gained from decoherence and obtain again superposition as was found in the quantum eraser.

This description is based on the philosophy of the qm. In de broil interpretation bohmian mechanics is deterministic. But dehorence is normal in systems that can have lower preferred configuration. It is not forbidden so it can be achieved.