Quantum Decoherence

This article covers environment-induced decoherence—the process by which a quantum system evolves as it couples to an environment. For example, measuring a previously “isolated” electron involves coupling it to a measurement apparatus, and hence decoherence. What’s interesting about decoherence is how the “weirdness” of quantum mechanics partially disappears in the process. It gives rise to the world of our everyday experience as opposed to the strange superpositions that apparently underlie it.

The Setup

Throughout this section we’ll focus on a simple system—a qubit. The spin of an electron is an example of a qubit in nature. We’ll use orthogonal basis vectors denoted as \(\{ 0, 1\}\). We’ll start with a qubit in the state \(0 + 1 \), completely decoupled from an environment in state \(E_0 \). The total state of the qubit and environment is therefore \( (0 + 1 )\otimes E_0 = 0E_0 + 1E_0 \).

Interaction

Once the qubit interacts with its environment, the state generally cannot be written as \( (0 + 1 )\otimes E\). Allowing the state to evolve according to the Schrodinger equation, it will become \( 0 E^0 (t) +1 E^1 (t) \), allowing \( E^0 (t) \neq E^1 (t) \).

Density Operator/Matrix

The (reduced) density matrix of the qubit shows the effect of the interaction above and the rise of decoherence. We won’t cover density matrices in detail here, but there are many sources covering this topic, including this article on density operators. A short introduction follows, enough to cover the current application.

The density operator associated with a quantum state \( \psi \) will be denoted \( \rho_{\psi} \) and is defined as \( |\psi \rangle \langle \psi | \). It maps a state \( \phi \) to \( \rho_{\psi} \phi = |\psi \rangle \langle \psi | \phi \rangle \). In other words, the density operator for \( \psi \) maps other states to a multiple of \( \psi \) with magnitude being the inner product of the state with \( \psi \). The density matrix is just a matrix representation of this operator, with respect to a given basis.

Linear combinations of density operators are also density operators. Note that the density operator for the state \( \psi_1 + \psi_2 \) is not the same as the sum of the density operators for \( \psi_1 \) and \( \psi_2 \) (e.g. the former maps \( \psi_2 \) to a multiple of \( \psi_1 + \psi_2 \), but the latter maps \( \psi_2 \) to a multiple of \( \psi_2 \)). More to come on this—it’s important for decoherence.

Reduced Density Operator/Matrix

The density matrix for our example state \( 0 E^0 (t) + 1 E^1 (t) \) would be huge due to the complexity of the environment. To focus on the qubit, physicists use the reduced density matrix, which operates on the qubit state alone (ignorant of the environment). The reduced density matrix maps qubit states to other qubit states and provides all the quantum information about the qubit. To learn more about reduced density matrices see x. The reduced density matrix for this qubit follows (using the basis \( \{0, 1 \} \) still).

$$ \rho (t) = \left( \begin{array}{ccc} \langle E^0 (t) | E^0 (t) \rangle & \langle E^1 (t) | E^0 (t) \rangle \\ \langle E^0 (t) | E^1 (t) \rangle & \langle E^1 (t) | E^1 (t) \rangle \end{array} \right) $$

At time 0 (or anytime before the qubit has impacted the environment) the state was \( (0+1)E_0 = 0 E_0 + 1 E_0 \). In other words, \( E^0 (0) = E^1 (0) = E_0 \). Hence, the matrix above has 1 in the upper right and lower left entries (off-diagonal) (\( \langle E_0 | E_0 \rangle = 1 \)). This is the same as the density matrix for the pure qubit state \( 0+1 \) (without tracing out over any environment). This pure state facilitates interference and the weirdness of quantum mechanics.

As time progresses, and the qubit interacts with the environment, \( E^0 (t) \) will differ from \( E^1 (t) \). This means the inner products \( \langle E^0 (t) | E^1 (t) \rangle \) and \( \langle E^1 (t) | E^0 (t) \rangle \) will generally shrink and the reduced density matrix will have smaller values off-diagonal. Shrinking off-diagonal terms are the key to decoherence.

Measurement

Let’s consider the case where the environment consists of a measurement device to tell us humans the state of the qubit. The number of degrees of freedom and complexity of this environment will be huge. Furthermore, the measuring device will magnify the difference between 0 and 1 to make it “seeable” to our perceptions. In other words, measurement will drive major differences between \( E^0 (t) \) and \( E^1 (t) \) and these will be located in a very high-dimensional Hilbert space. Thinking of these two as very different lines in such a space, you may see that \( \langle E^0 (T) | E^1 (T) \rangle \approx 0 \): the lines become orthogonal and our reduced matrix becomes diagonal.

$$ \rho (T) = \left( \begin{array}{ccc} \langle E^0 (T) | E^0 (T) \rangle & 0 \\ 0 & \langle E^1 (T) | E^1 (T) \rangle \end{array} \right) $$

Mixed Vs Pure State

While the matrix at \( t = T \) gives the same probabilities (50% 0 and 50% 1) as the matrix at \( t=0 \), it’s quite different. This diagonal matrix represents what’s called a mixed state of 0 and 1. The matrix at \( t=0 \), with 1s off-diagonal, represents a pure state. We’ll get into this difference more below.

A mixed state is analogous to the result of a coin flip—it’s either 0 or 1, we just don’t know which. Since it’s one or the other, there will be no interference between 0 and 1, as indeed the math will show on this matrix. (Really true in the bigger space?) The pure state, however, was in some sense both 0 and 1 and the math allows for interference effects between those.