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CS191 – Fall 2014 Lecture 14: Open Quantum Systems: Quantum Process Formulation

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CS191 – Fall 2014 Lecture 14: Open Quantum Systems: Quantum Process Formulation CS191 { Fall 2014 Lecture 14: Open quantum systems: quantum process formulation Mohan Sarovar (Dated: October 27, 2014) I. TWO MISSING PIECES Before we start on the bulk of this lecture, we need to cover two concepts that haven't been covered so far. The first is the concept of the entropy of a quantum state, which is defined as: S(ρ) = tr (ρ log ρ) (1) − This expression is most easily evaluated by diagonalizing ρ, in which case S(ρ) = i λi log λi, where λi are the eigenvalues of ρ. 1 The entropy is a measure of how mixed a state is, or how much classical− uncertainty it contains. Another measure of how mixed a state is, which is actually an approximation of theP inverse of the entropy is the purity, which is defined as: P (ρ) = tr (ρ2) (2) Exercise: What are the entropy and purity of a pure state, S( ) and P ( )? What are the j ihId j Idj ih j entropy and purity of the completely mixed state in d dimensions, S( d ) and P ( d )? Note that as the mixedness or uncertainty in the state increases, the entropy increases and the purity decreases. II. OPEN QUANTUM SYSTEMS AND NOISE So far in this course you've mainly seen what ideal quantum operations looks like. For example, a state trans- Pi formation is a unitary map U , and a projective measurement is the following map: j i . But in j i ! j i j i ! ppi reality when we perform a quantum operation (e.g., rotation, measurement, coupling) there will inevitably be errors due to noise in the experimental apparatus. Some examples of errors that are commonly encountered in quantum information processing are: 1. Uncertainty in the implemented rotation angle. 2. Control pulses with electronic noise. 3. Noise from nearby \junk", which are usually uncontrollable degrees of freedom that couple to qubits. In this lecture and the next four we will discuss how to model these nonidealities and how to deal with them and do quantum information tasks reliably in their presence. A. An example Let's start with an example. Suppose we have a qubit that starts in the state = 1 ( 0 + 1 ), and we want to j i p2 j i j i perform a σz rotation on it. To do so we turn on a Hamiltonian in the form ! H = σ : 2 z Then ideally, iHt 1 i ! t i ! t (t) = e− (0) = (e− 2 0 + e 2 1 ): j i j i p2 j i j i 1 We define 0 log 0 = 0. 2 π This affects a rotation around the Z axis, and if we apply it for t = ! , we get a Z-gate. That is, π 1 1 ( ) = ( 0 + 1 ) ( 0 1 ); ! p2 −| i j i ≡ p2 j i − j i E where the last equivalence is due to the unimportance of the global phase of a quantum state. But now, what if ! is not exactly known? Suppose we have some uncertainty in our calibration and all we can 2 2 says is that ! (!0; σ ). That is, ! is Gaussian distributed around some nominal value with some variance σ . Then after a time∼ Nt all we can say is that we have some distribution over Z rotations. We must average over this uncertainty to get an estimate of the actual state. Note that this is exactly what we do in classical physics also, we average over unknowns to get the state. Let's see what we get when we compute this average. Anytime we average over quantum states, we should use the density matrix formalism. The state at time t, for a given value of ! is 1 i ! t i ! t i ! t i ! t i ! t i ! t i ! t i ! t ρ (t) = (t) (t) = e− 2 0 0 e 2 + e− 2 0 1 e− 2 + e 2 1 0 e 2 + e 2 1 1 e 2 ! j ! ih ! j 2 j ih j j ih j j ih j j ih j 1 1 e i!t h i = 2 2 − 1 ei!t 1 2 2 So then, when we average over the rotation parameters, the average state is: ρ(t) = P (!)ρ!(t)d! ZΩ 1 P (!)d! 1 P (!)e i!td! = 2 Ω 2 Ω − 1 P (!)ei!td! 1 P (!)d! 2 ΩR R2 Ω 1 1 P (!)e i!td! = R 2 2 ΩR − ; 1 P (!)ei!td! 1 2 Ω R 2 where Ω is the support of the probability distributionR for ! (the whole real line for the normal distribution above). We see that the off-diagonal elements are complex conjugates of each other, so let's look at one of them: 2 2 (!−!0) (tσ) i!t 1 1 i!t i!0t P (!)e− d! = e− 2σ2 e− d! = e− 2 − : (3) 2 Ω p2πσ Z Z−∞ Therefore, the average density matrix is: (tσ)2 1 1 2 i!0t 2 2 e− − ρ(t) = (tσ)2 : (4) 1 2 +i!0t 1 2 e− 2 ! Exercise: Perform the generalized Gaussian integral required to get the last equality in Eq. (3). This is one of the most useful integrals in physics and it's usually obtained by completing the square in the exponent. There are several things to note about the state in Eq. (4): 1. tr (ρ2) < 1 for all t > 0. That means the state becomes mixed as a result of the uncertain rotation. 2. The off-diagonal elements oscillate at frequency !0, but also decay exponentially with time and increasing uncertainty (σ2). 3. As t or σ , ! 1 ! 1 1 1 0 ρ ; ! 2 0 1 So if we have too much uncertainty about the rotation angle or apply the Hamiltonian for too long, we just get the completely mixed state. Exercise: Compute the purity and entropy of the state in Eq. (4) as a function of t and σ. 3 Environment System FIG. 1: A system embedded in a larger environment. Most coupling between a quantum system and its environment will result in decoherence of the system. This example shows how uncertainty in the operations one performs can result in an increase in entropy or mixedness of the state. This is also true if there is noise (e.g., if ! is a time-dependent random process). In this example, the uncertainty led to a decay of the off-diagonal element of the density matrix, which contains information about the phase of the state (in the computational basis). Consequently, this process is called dephasing. In general, external influences and uncertainty can lead to more general transformations of the state, and we collectively refer to the wide class of phenomena that cause the state of a quantum system to evolve nondeterministically and/or to increase in entropy as decoherence. B. The operator sum decomposition All decoherence arises from a coupling of the system we are interested in to an environment that is (at least partially) uncontrolled and unobserved. See figure 1. For instance, in the example above, the uncertainty in ! could be due to us not knowing the precise parameters that dictate the physics that determine this rotation frequency. As a result, we refer to the dynamics of a system under such couplings to external environments as open system dynamics. In fact, this perspective gives us a general way to model open system dynamics and decohernce processes mathematically. Suppose we begin with the system and environment in a product state at an initial time: %(0) = ρS φE φE ; (5) 0 ⊗ S E E where ρ0 is the (possibly mixed) initial state of the system and φ φ is the pure initial state of the environment. We can always assume the environment is in a pure state initially since if it is not we can expand our definition of the environment until it is. Then, at some late time t, the joint state of the system and environment is: %(t) = USE%(0)USEy ; (6) where USE is a unitary evolution that captures the evolution of the system, the environment, and any coupling between them. Then the reduced state of the system at t is given by tracing out the environment from the above 4 S state, i.e., ρ (t) = tr E(%(t)). We can formally write down what this resulting state is as so: S S E E ρ (t) = tr U ρ φ φ U y E SE 0 ⊗ SE K h i E S E E E = a U ρ φ φ U y a k SE 0 ⊗ SE k k=1 X K S = Akρ0 Aky ; (7) kX=1 where eE for 1 k K is a complete set of orthonormal states (a basis) in the environment Hilbert space, and k ≤ ≤ A aE U φE , which is an operator acting on the system Hilbert space only. k k SE ≡ Example 1 [From Nielsen & Chuang, p.361] Consider an example where the system and environment are single qubits, and the environment starts off in the state 0E . Then suppose the interaction between them after some time t, is described by the unitary σx σy USE = I + σx: (8) p2 ⊗ p2 ⊗ Then the reduced state of the system at this time is given by Eq. (7), with E E σx σy σx A0 = 0 USE 0 = 0 I 0 + 0 σx 0 = p2 × h j j i p2 × h j j i p2 E E σx σy σy A1 = 1 USE 0 = 1 I 0 + 1 σx 0 = p2 × h j j i p2 × h j j i p2 Explicitly, 1 1 ρS = σ ρSσ + σ ρSσ (9) 2 x 0 x 2 y 0 y Notice that Eq.
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