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A Few Notes on the Bloch Sphere A Few Notes on the Bloch Sphere David Meyer [email protected],uoregon.edug Last update: August 13, 2021 1 Introduction The fundamental building block of classical computational devices is the two-state system. We typically think of these systems as being built from bits which can take values in f0; 1g. Quantum mechanics, on the other hand, tells us that any such system can exist in a superposition of states. A qubit is a quantum system in which the Boolean states 0 and 1 are represented1 by a prescribed pair of normalized and mutually orthogonal quantum states labeled as fj0i ; j1ig. More specifically, a qubit is defined as a 2-dimensional Hilbert space H2 and we label an orthonormal basis of H2 by fj0i ; j1ig. The state of the qubit is an associated unit length vector in H2. If a state is equal to a basis vector then we say it is a pure state. If a state is any other linear combination of the basis vectors we say it is a mixed state, or that the state is a superposition of j0i and j1i. The state of a qubit bit, j i, is described by j i = α j0i + βj j1i 1 0 where α; β 2 , j0i = , j1i = and jαj2 + jβj2 = 1. C 0 1 Here jxj2 = xx∗, where x∗ is the complex conjugate of x [8]. The complex plane is shown in Figure 1. Note: the Wikipedia uses "bar" (¯x) to denote the complex conjugate as opposed to "star" (x∗). 1I use Dirac or braket notion [6] in this document. 1 Figure 1: The Complex Plane (Image courtesy Wikipedia [9]) More generally, a quantum system is an n-dimensional Hilbert Space Hn. We label an orthonormal basis on Hn by fjx1i ; jx2i ;:::; jxnig such that xi 2 X for some finite X. The associated state of the system is a unit length vector in Hn. If the state S is jφi at some moment in time, then we say that S has state jφi or S is in state jφi. We can also see that a general form for the state of a quantum system is n n X X 2 αi jxii where αi 2 C and jαij = 1 i=1 i=1 Remark: We can already see an important technical problem that must be solved if quan- tum computers are ever to be practical. A superposition of states j0i and j1i is known as a coherent state. A coherent state is extremely "unstable"; it inevitably interacts with its environment and collapses into a pure state (see discussion below). This process is known as decoherence. Algorithms that exploit quantum effects such as superposition are known as quantum algorithms. To apply these quantum algorithms in the real world, the decoherence time must be longer than the time to run the algorithm. As a result methods for increasing the decoherence time are of great interest (see, for example [4]). 2 The Bloch Sphere One question that we might ask, given the complex plane representation of the state of a qubit, is "how does the real, 3-dimensional space in which we live correspond to the 2- 2 dimensional representation of a qubit's state (as represented in the complex plane)"? The Bloch Sphere gives us an elegant way to think about this question. Figure 2: The Bloch Sphere (Image courtesy Wikipedia [7]) The Bloch Sphere, shown in Figure 2, is named for the physicist Felix Bloch [1] and gives us a beautiful way to think about and visualize the state of a qubit in 3-dimensional space. In the Bloch Sphere the pure state of a qubit j i is represented as a point on the surface of the sphere. Mixed states are represented as points in the interior of the sphere. Somewhat surprisingly, specification of a point on the Bloch Sphere requires only two parameters, θ and φ. This itself is remarkable; you might imagine that describing a vector in 3- dimensional space might take three or more parameters. The representation of a classical bits on the Bloch Sphere is given by the poles of the sphere. The representation of the probabilistic classical bit, that is, a bit that is 0 with probability p and 1 with probability 1 − p, is given by the point in z-axis with coordinate 2p − 1. The interior of the Bloch Sphere is used to describe the states of a qubit in the presence of decoherence. The standard way to write the state of a qubit using the Bloch Sphere is θ θ j i = cos j0i + eiφ sin j1i (1) 2 2 where 0 ≤ θ ≤ π and 0 ≤ φ ≤ 2π. See Section 2.3 for a derivation of this equation. 3 2.1 Bloch Sphere Coordinate System Figure 2 shows the coordinate system of the Bloch Sphere. Consider the x axis. Here we can see that π x : θ = ; φ = 0 −! + 2 π π cos j0i + ei·0 sin j1i # ei·0 = e0 = 1 4 4 1 1 = p j0i + (1)p j1i 2 2 1 1 1 1 = p j0i + p j1i # H j0i = p j0i + p j1i 2 2 2 2 ≡ j+i π x : θ = ; φ = π −! − 2 π π cos j0i + eiπ sin j1i # Euler's Identity [11]: eiπ + 1 = 0; eiπ = −1 4 4 1 1 = p j0i + (−1)p j1i 2 2 1 1 1 1 = p j0i − p j1i # H j1i = p j0i − p j1i 2 2 2 2 ≡ |−i On the y axis of the Bloch Sphere we can see that π π y : θ = ; φ = −! + 2 2 π i π π i π π π cos j0i + e 2 sin j1i # Euler's formula [10]: e 2 = cos + i sin = 0 + i · 1 = i 4 4 2 2 1 1 = p j0i + ip j1i 2 2 1 i = p j0i + p j1i 2 2 4 π 3π y : θ = ; φ = −! − 2 2 π i 3π π i 3π 3π 3π cos j0i + e 2 sin j1i # e 2 = cos + i sin = 0 − i · 1 = −i 4 4 2 2 1 1 = p j0i − ip j1i 2 2 1 i = p j0i − p j1i 2 2 Finally, on the z axis of the Bloch Sphere we can see that z+ : θ = 0; φ = 0 −! 0 0 cos j0i + ei0 sin j1i 2 2 = cos 0 j0i + (1) sin 0 j1i = 1 j0i + 0 j1i = j0i and z− : θ = π; φ = 0 −! π π cos j0i + ei0 sin j1i 2 2 π = cos 0 j0i + (1) sin j1i 2 = 0 j0i + 1 j1i = j1i So we can write the state of a qubit in many different ways. Perhaps the most generic way to write j i is j i = α j0i + β j1i The maximally superimposed state is 1 1 j i = p j0i + p j1i (2) 2 2 5 2 Here the probability of measuring j0i is jaj2 = aa∗ = p1 = 1 . Similarly, the probability 2 2 1 of measuring j1i is 2 . Now, note that 1 1 1 eiθ j i = p j0i + p j1i = p j0i + p j1i (3) 2 2 2 2 Why does the second equality in Equation 3 hold? Well, consider jeiθj2. By Euler's formula2 we can see that 2 jeiθj = j(cos θ + i sin θ)j2 = (cos θ + i sin θ)(cos θ − i sin θ) = cos2 θ − cos θ i sin θ + i sin θ cos θ − i2 sin2 θ = cos2 θ − i2 sin2 θ = cos2 θ + sin2 θ = 1 So jeiθj2 = 1. Now, dividing both sides by 2 and taking the square roots gives s 2 r eiθ 1 eiθ 1 = =) p = p (4) 2 2 2 2 iθ So j i = p1 j0i + p1 j1i = p1 j0i + pe j1i. This implies that the statistics of any mea- 2 2 2 2 surements we could perform on the state eiθ j i would be exactly the same as they would be for the state j i. This explains the claim in Section 2.3.1 that global phases have no physical significance. We have seen that measuring j i in the standard basis (fj0i ; j1ig) yields a probability of 1 2 for measuring either j0i or j1i. Is there any measurement that yields information about the phase θ? Consider a measurement in a different basis, the fj+i ; |−i} basis. As we saw above, j+i ≡ p1 (j0i + j1i) and |−i ≡ p1 (j0i − j1i). What does j i look like in this new basis? 2 2 2Euler's formula: eiθ = cos θ + i sin θ 6 The basic approach is to first represent j0i and j1i in the new basis. Here j0i = p1 (j+i+|−i) 2 and j1i = p1 (j+i − |−i). Then 2 1 eiθ j i = p j0i + p j1i 2 2 1 1 eiθ = p p j+i + |−i + p j+i − |−i 2 2 2 1 eiθ = j+i + |−i + j+i − |−i 2 2 1 + eiθ 1 − eiθ = j+i + |−i 2 2 Since eiθ = cos θ + i sin θ we can write 1 + cos θ + i sin θ 1 − (cos θ + i sin θ = j+i + |−i 2 2 What then is the probability of measuring, say, j+i, in the fj+i ; |−i} basis? Well, we know that the probability is the amplitude squared, so, 2 1 + cos θ + i sin θ P (j+i) = 2 1 = (1 + cos θ + i sin θ)(1 + cos θ − i sin θ) 4 1 = 1 + cos θ − i sin θ + cos θ + cos2 θ − cos θi sin θ + i sin θ + cos θi sin θ − i2 sin2 θ 4 1 = 1 + 2 cos θ + cos2 θ + sin2 θ # cos2 θ + sin2 θ = 1 4 1 = 1 + 2 cos θ + 1 4 1 = 2 + 2 cos θ 4 r 1 θ 1 + cos θ θ = 1 + cos θ # cos = ± =) 1 + cos θ = 2 · cos2 2 2 2 2 1 θ = 2 · cos2 2 2 θ θ = cos2 # P (j+i) = cos2 2 2 7 2 θ Similarly, we know that the probability of measuring |−i, P (|−i) = sin 2 .
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