## 13.3 Decoherence and interference

Consider the following interaction between a qubit and its environment: \begin{aligned} |0\rangle|e\rangle &\longmapsto |0\rangle|e_{00}\rangle \\|1\rangle|e\rangle &\longmapsto |1\rangle|e_{11}\rangle \end{aligned} where |e\rangle, |e_{00}\rangle, and |e_{11}\rangle are the states of the environment, which not need to be orthogonal.259 Now assume that the qubit is initially in some general state |\psi\rangle = \alpha|0\rangle + \beta|1\rangle. The resulting qubit-environment interaction is essentially the environment trying to measure the qubit and, as the result, entangling the two together: \Big( \alpha|0\rangle + \beta|1\rangle \Big) |e\rangle \longmapsto \alpha |0\rangle|e_{00}\rangle + \beta |1\rangle |e_{11}\rangle.

Now we can also write this evolution as \begin{aligned} \Big( \alpha|0\rangle + \beta|1\rangle \Big) |e\rangle \longmapsto & \Big( \alpha|0\rangle + \beta|1\rangle \Big) \frac{|e_{00}\rangle+|e_{11}\rangle}{2} \\+& \Big( \alpha|0\rangle - \beta|1\rangle \Big) \frac{|e_{00}\rangle-|e_{11}\rangle}{2}. \\=& \mathbf{1}|\psi\rangle|e_{\mathbf{1}}\rangle + Z|\psi\rangle|e_Z\rangle, \end{aligned} where |e_{\mathbf{1}}\rangle = \frac{1}{2}(|e_{00}\rangle + |e_{11}\rangle) and |e_Z\rangle = \frac{1}{2}(|e_{00}\rangle - |e_{11}\rangle). We can roughly interpret this expression as saying that two things can happen to the qubit: nothing \mathbf{1} (first term), or phase-flip Z (second term).

This, however, should not be taken literally unless the states of the environment, |e_{\mathbf{1}}\rangle and |e_Z\rangle, are orthogonal.260

This process is what we refer to as decoherence.

Let’s look at something a bit more involved now. Suppose a qubit undergoes the usual single-qubit interference evolution, but, between the phase gate and the second Hadamard gate, it is affected by decoherence: \begin{aligned} \times\colon |0\rangle|e\rangle &\longmapsto |0\rangle|e_{00}\rangle \\|1\rangle|e\rangle &\longmapsto |1\rangle|e_{11}\rangle \end{aligned} as described by Figure 13.1.

Let us step through the circuit in Figure 13.1, keeping track of the state of the environment: \begin{aligned} |0\rangle|e\rangle & \overset{H}{\longmapsto} \Big( |0\rangle + |1\rangle \Big) |e\rangle \\& \overset{\phi}{\longmapsto} \Big( |0\rangle + e^{i\phi}|1\rangle \Big) |e\rangle \\& \overset{\times}{\longmapsto} |0\rangle|e_{00}\rangle + e^{i\phi}|1\rangle|e_{11}\rangle \\& \overset{H}{\longmapsto} |0\rangle\Big( |e_{00}\rangle + e^{i\phi}|e_{11}\rangle \Big) + |1\rangle\Big( |e_{00}\rangle - e^{i\phi}|e_{11}\rangle \Big). \end{aligned} Writing \langle e_{00}|e_{11}\rangle=ve^{i\alpha} for v\in\mathbb{R}_{\geqslant 0} and \alpha\in[0,2\pi), the final probability P_k of measuring the output to be in state |k\rangle oscillates with \phi as \begin{aligned} P_{0}(\phi) &= \frac{1}{2}\big(1 + v\cos(\phi + \alpha)\big), \\P_{1}(\phi) &= \frac{1}{2}\big(1 - v\cos(\phi + \alpha)\big). \end{aligned}

As we can see in Figure 13.2, the interference pattern is suppressed by a factor v, which we call the visibility. As v=|\langle e_{00}|e_{11}\rangle| gets closer and closer to 0, we lose all the advantages of quantum interference. For example, in Deutsch’s algorithm (Section 10.4) we would obtain the correct answer with probability at most \frac{1}{2}(1+v). For \langle e_{00}|e_{11}\rangle = 0, the case of perfect decoherence, the network outputs 0 or 1 with equal probabilities, i.e. it is useless as a computing device.

We want to avoid decoherence, or at least diminish its impact on our computing device.

If we wish to study the evolution of the qubit alone, then we can do so in terms of density operators: it evolves from the pure state |\psi\rangle\langle\psi| to a mixed state, which can be obtained by tracing over the environment. We know that the state vector |\psi\rangle=\alpha|0\rangle+\beta|1\rangle evolves as \left( \alpha|0\rangle +\beta |1\rangle\right)|e\rangle \longmapsto \alpha |0\rangle|e_{00}\rangle +\beta |1\rangle |e_{11}\rangle and we can write this as the evolution of the projector |\psi\rangle\langle\psi|, and then trace over the environment to obtain \begin{aligned} |\psi\rangle\langle\psi| \longmapsto & |\alpha|^2|0\rangle\langle 0| \langle e_{00}|e_{00}\rangle+ \alpha\beta^\star |0\rangle\langle 1|\langle e_{11}|e_{00}\rangle \\+ &\alpha^\star\beta |1\rangle\langle 0|\langle e_{00}|e_{11}\rangle + |\beta|^2|1\rangle\langle 1|\langle e_{11}|e_{11}\rangle. \end{aligned} Written in matrix form, this is \begin{bmatrix} |\alpha|^2 & \alpha\beta^\ast \\\alpha^\ast\beta & |\beta|^2 \end{bmatrix} \longmapsto \begin{bmatrix} |\alpha|^2 & \alpha\beta^\ast \langle e_{11}|e_{00}\rangle \\\alpha^\ast\beta \langle e_{00}|e_{11}\rangle & |\beta|^2 \end{bmatrix}. The off-diagonal elements (originally called coherences) vanish as \langle e_{00}|e_{11}\rangle approaches zero. This is why this particular interaction is called decoherence.

Notice that |\psi\rangle|e\rangle \longmapsto \mathbf{1}|\psi\rangle|e_{\mathbf{1}}\rangle+Z|\psi\rangle|e_Z\rangle, implies |\psi\rangle\langle\psi|\longmapsto \mathbf{1}|\psi\rangle\langle\psi| \mathbf{1}\langle e_{\mathbf{1}}|e_{\mathbf{1}}\rangle +Z|\psi\rangle\langle\psi| Z\langle e_Z|e_Z\rangle, only if \langle e_{\mathbf{1}}|e_Z\rangle=0, since otherwise we would have additional cross terms \mathbf{1}|\psi\rangle\langle\psi|Z and Z|\psi\rangle\langle\psi|\mathbf{1}. In this case (i.e. when \langle e_{\mathbf{1}}|e_Z\rangle=0) we can indeed say that, with probability \langle e_{\mathbf{1}}|e_{\mathbf{1}}\rangle, nothing happens, and, with probability \langle e_Z|e_Z\rangle, the qubit undergoes the phase-flip Z. We can also represent this with the Kraus operators \begin{aligned} E_0 &= \sqrt{|e_\mathbf{1}\rangle\langle e_\mathbf{1}|} \mathbf{1} \\E_1 &= \sqrt{|e_Z\rangle\langle e_Z|} Z \end{aligned} which can be shown to satisfy E_0^\dagger E_0+E_1^\dagger E_1=\mathbf{1}.

1. The reason we use two indices in |e_{00}\rangle and |e_{11}\rangle will become clear in a moment, when we consider more general interaction with the environment.↩︎

2. Exercise. Why not?↩︎