Lecture Notes: Qubit Representations and Rotations

Home , Bloch sphere, Pauli matrices, Rotation matrix, Unitary matrix

Phys 711 Topics in Particles & Fields — Spring 2013 — Lecture 1 — v0.3 Lecture notes: Qubit representations and rotations

Jeﬀrey Yepez Department of Physics and Astronomy University of Hawai‘i at Manoa Watanabe Hall, 2505 Correa Road Honolulu, Hawai‘i 96822 E-mail: [email protected] www.phys.hawaii.edu/∼yepez

(Dated: January 9, 2013)

Contents mathematical object (an abstraction of a two-state quantum object) with a “one” state and a “zero” state: I. What is a qubit? 1 1 0 II. Time-dependent qubits states 2 |qi = α|0i + β|1i = α + β , (1) 0 1 III. Qubit representations 2 A. Hilbert space representation 2 where α and β are complex numbers. These complex B. SU(2) and O(3) representations 2 numbers are called amplitudes. The basis states are or- IV. Rotation by similarity transformation 3 thonormal

V. Rotation transformation in exponential form 5 h0|0i = h1|1i = 1 (2a) VI. Composition of qubit rotations 7 h0|1i = h1|0i = 0. (2b) A. Special case of equal angles 7 In general, the qubit |qi in (1) is said to be in a superpo- VII. Example composite rotation 7 sition state of the two logical basis states |0i and |1i. If References 9 α and β are complex, it would seem that a qubit should have four free real-valued parameters (two magnitudes and two phases): I. WHAT IS A QUBIT? iθ0 α φ0 e |qi = = iθ1 . (3) Let us begin by introducing some notation: β φ1 e

1 state (called “minus” on the Bloch sphere) Yet, for a qubit to contain only one classical bit of infor- 0 mation, the qubit need only be unimodular (normalized |1i = the alternate symbol is |−i 1 to unity) α∗α + β∗β = 1. (4) 0 state (called “plus” on the Bloch sphere)

1 Hence it lives on the complex unit circle, depicted on the |0i = the alternate symbol is |+i. 0 top of Figure 1. Normalization (4) constrains the value of the magnitudes, so we can write a qubit as When you learn about the Bloch sphere (discussed below) √ 1 − f you will see why the alternate symbols |+i and |−i are |qi = √ , (5) f eiϕ used to denote logical states.1 So what is a qubit? A qubit is the fundamental quan- where 0 ≤ f ≤ 1 and where an irrelevant overall phase is tum state representing the smallest unit of quantum in- factored out. The length (or norm) of the qubit is thus formation containing one bit of classical information ac- an invariant quantity cessible by measurement. We simply take a qubit to be a hq|qi = |α|2 + |β|2 = |p1 − f|2 + |pf|2 = 1. (6)

You should understand why an overall phase is irrelevant 1 The names “up” and “down,” and the respective symbols |↑i 1 to the length of the qubit. The quantum property of and |↓i, are reserved for spin- 2 particles.We will see in another lecture how a 2-qubit encoding conforms with the Pauli exclusion measurement follows from identifying the moduli squared principle for particles with half-integer spin. of the amplitude as an occupation probability f and 1 − 2 f for the qubit to occupy its logical states |1i and |0i, Writing our 2-spinor basis states in terms of qubit states, respectively, as follows: we have θ 2 i ϕ cos f = |β| (7) ξ(↑) ≡ e 2 |+i = 2 eiϕ sin θ 1 − f = |α|2. (8) 2 θ θ = cos |0i + sin eiϕ|1i, (14a) There are only two relevant free parameters to specify the 2 2 ϕ −iϕ θ state of a qubit, but upon measurement, the qubit orig- −i −e sin 2 ξ(↓) ≡ e 2 |−i = θ inally in the superposition state (5) is found to occupy cos 2 only one of its logical states θ θ = − sin e−iϕ|0i + cos |1i. (14b) ( 2 2 |1i, with probability f, |qi −−−−−→measure (9) |0i, with probability 1 − f. III. QUBIT REPRESENTATIONS Thus, upon a single measurement, |qi is found to be in either the state |0i or |1i, an outcome that is said to + = 0 È1\ ` be speciﬁed by a single classical bit ∈ {0, 1}. Thus in z actual experiments, the occupation probability f equals J Èq\ ãä j sin » \ » \ the frequency of occurrence of the result 1 obtained from 2 J q many repeated measurements. 1 J È0\ » \ J ` cos y 2 j II. TIME-DEPENDENT QUBITS STATES ` x The state |q(t)i of a time-dependent qubit, as a two- ` energy level quantum mechanical entity, is governed by -z the Schroedinger wave equation - = 1 FIG. 1 A qubit in Hilbert space in its SU(2) representation ∂ ~ω » \ » \ i~ |q(t)i = σz|q(t)i. (10) (top), and the same qubit on the Bloch sphere in its O(3) ∂t 2 representation (bottom). SU(2) and O(3) are homomorphic. The energy eigenvalues are ±~ω/2 and energy eigenstates are

1 0 A. Hilbert space representation |0i ≡ |1i ≡ , (11) 0 1 The space of all possible orientations of |qi on the com- where |0i is the ground state and |1i is the excited state plex unit circle is called the Hilbert space. In the logical of the qubit. In terms of the angular frequency ω (e.g. basis, the two degrees of freedom of the qubit is often 2 θ Rabi frequency), the time-dependent qubit state is expressed as two angles θ and ϕ, where f = sin 2 . So without any loss of generality the Hilbert space represen- −i ω t i ω t tation of a qubit (1) can be written as |q(t)i = A0e 2 |0i + A1e 2 |1i, (12) 2 θ θ iϕ where the complex probability amplitudes satisfy |A0| + |qi = cos |0i + sin e |1i. (15) 2 2 2 |A1| = 1 since the qubit resides on the complex circle in Hilbert space (or the Bloch sphere in spin space). These angles have a well known geometrical interpreta- Now, we can explicitly write out qubit basis states of tion as Euler angles. the Bloch sphere withu ˆ = (sin θ cos ϕ, sin θ sin ϕ, cos θ) as B. SU(2) and O(3) representations θ −i ϕ 2 θ ϕ θ ϕ cos 2 e −i i |+iu = ϕ = cos e 2 |0i + sin e 2 |1i, θ i 2 sin 2 e 2 2 To understand the geometrical interpretation of a (13a) qubit, consider a three-dimensional space with “unit vec- θ −i ϕ tors” σx, σy, and σz chosen as an orthonormal basis. In 2 θ ϕ θ ϕ − sin 2 e −i i |−iu = ϕ = − sin e 2 |0i + cos e 2 |1i. quantum information theory, one represents each basis θ i 2 cos 2 e 2 2 element by a 2 × 2 matrix, a traceless hermitian genera- (13b) tors of two-dimensional special unitary group, SU(2). To 3 do so, one defines the symmetric product (dot product) In this representation, the qubit is expressed as a matrix as element of the SU(2) group. In quantum information, 1 usually 2×2 unitary matrices are considered single-qubit σi · σj ≡ σi · σj + σj · σi . (16a) quantum gates, but such matrices can themselves repre- 2 sent qubits too. Table I gives a summary of the three Furthermore, one defines the anti-symmetric product qubit representations (cross product) as Representations Qubit i Hilbert space |qi = cos ` θ ´ |0i + sin ` θ ´ eiϕ|1i σ × σ ≡ − σ · σ − σ · σ . (16b) 2 2 i j 2 i j j i O(3) group ~q = (sin θ cos ϕ, sin θ sin ϕ, cos θ) „ cos θ e−iϕ sin θ« SU(2) group M = Note that the centered dot symbol on the R.H.S. of (16) q eiϕ sin θ − cos θ denotes matrix multiplication. Thus, a basis that is orthonormal satisfies the following conditions TABLE I Qubit representations. ( 1, for i = j (normal), σ · σ = (17a) i j 0, otherwise (orthogonal), IV. ROTATION BY SIMILARITY TRANSFORMATION and ( Now that we see a qubit as simply a unit vector on 0, for i = j, the complex circle (Hilbert space representation) or a σi × σj = (17b) σk, for cyclic indices. unit vector on the Bloch sphere (O(3) representation), we can consider rotations of the qubit’s state that keep A fundamental matrix representation that satisfies (17) its length (or norm) invariant. Remarkably, such a rota- is the well-known Pauli basis tion of a qubit is conveniently accomplished by employ- 0 1 0 −i 1 0 ing its SU(2) representation as a 2 × 2 unitary matrix. σ = , σ = , σ = . 1 1 0 2 i 0 3 0 −1 Then, the qubit rotation is induced by a similarity trans- (18) formation, which is to say a double-sided transformation Exercise: acting from the left and the right side. The unitary matrix (acting from the left) along with its matrix inverse If it is not obvious already, verify that the (acting from the right) that is customarily employed for Pauli matrices (18) satisfy the orthonormal- such rotations, about the ith principle axis say, is ity conditions (17) which is just the structure −i θ σ θ θ equation for the SU(2) group U (θ) ≡ e 2 i = σ cos − iσ sin , (22) i 0 2 i 2 [S ,S ] = i S , i j ijk k 1 0 σi where the identity matrix is σ0 = . Explicitly, the where Si ≡ 2 and the structure constant ijk 0 1 is the anti-symmetric Levi-Civita symbol. unitary matrices for the principles directions are cos θ −i sin θ U (θ) = 2 2 (23a) 1 −i sin θ cos θ Now we can express the qubit (15) in vector form (i.e. 2 2 cos θ − sin θ with three real components) as follows: 2 2 U2(θ) = θ θ (23b) sin 2 cos 2 ~q = (sin θ cos ϕ, sin θ sin ϕ, cos θ). (19) −i θ e 2 0 U3(θ) = i θ . (23c) (19) is a representation of a qubit on the Bloch sphere 0 e 2 where θ is the elevation angle and ϕ is the azimuthal angle. In this representation, depicted on the bottom of A general rotation of a qubit about axisn ˆ = (nx, ny, nz) Fig. 1, the qubit is considered as a vector element of the is built using the following unitary matrix (along with its three-dimensional orthogonal group, O(3). Defining the inverse) Pauli spin vector (which has matrix components) −i θ nˆ·~σ θ θ U (θ) = e 2 = σ cos − i(ˆn · ~σ) sin , (24) nˆ 0 2 2 ~σ ≡ (σ1, σ2, σ3), (20) 2 a qubit can also be expressed in matrix form where the identity on the R.H.S. follows since (ˆn · ~σ) = σ0 = 1. Mq ≡ ~q · ~σ (21a) Now a qubit rotation by angle θ about the arbitrary

= sin θ cos ϕ σ1 + sin θ sin ϕ σ2 + cos θ σ3 (21b) axisn ˆ is expressed as the similarity transformation men- −iϕ tioned above (18) cos θ e sin θ = iϕ . (21c) † e sin θ − cos θ Mq0 = Unˆ (θ) · Mq · Unˆ (θ). (25) 4

Here again the centered dot symbol represents matrix The goal is to workout the R.H.S. so that we can de- multiplication. The † symbol denotes the matrix adjoint, termine q~0 in terms of the original vector ~q, the axis of i.e. complex conjugate of the components of the matrix rotationn ˆ, and the angular rotation amount θ. We will combined with matrix transposition. Since see that (28) reduces to a particularly simple and useful rotation formula. † −1 U (θ) = U (θ) = Unˆ (−θ), (26) nˆ nˆ As a preliminary task leading toward reducing (28), we we simply compute the rotated qubit (25) as follows will need some helpful identities. Consider two 3-vectors ~a and ~b. The ﬁrst identity that we need is Mq0 = Unˆ (θ) · Mq · Unˆ (−θ). (27)

Of course using Mq = ~q · ~σ we can write this similarity ~ ~ ~ transformation directly in terms of the 3-vector ~q and the (~a · ~σ) · (b · ~σ) = ~a · b + i ~a × b · ~σ. (29) resulting 3-vector q~0

0 q~ · ~σ = Unˆ (θ) · (~q · ~σ) · Unˆ (−θ). (28) The proof is simple enough and goes as follows:

~ (~a · ~σ)(b · ~σ) = (a1σ1 + a2σ2 + a3σ3)(b1σ1 + b2σ2 + b3σ3) (30a) ~ = ~a · b + a1b2σ1σ2 + b1a2σ2σ1 + O.T. (30b) ~ = ~a · b + i(a1b2 − b1a2)σ3 + O.T. (30c) = ~a ·~b + i ~a ×~b · ~σ, (30d)

where O.T. stands for the “other terms.” In turn, it Finally, we have the useful identity follows that (~a · ~σ) · (~b · ~σ) = ~a ·~b + i ~a ×~b · ~σ (31a)

(32a) (~b · ~σ) · (~a · ~σ) = ~a ·~b − i ~a ×~b · ~σ. (31b) (â · ~σ) · (~b · ~σ) · (â · ~σ) = −(~b · ~σ) + 2 (â ·~b)(â · ~σ), (33)

Then, taking the sum and diﬀerence yields the following two identities

2 (~a · ~σ) · (~b · ~σ) + (~b · ~σ) · (~a · ~σ) = 2~a ·~b (32a) which follows since (ˆa · ~σ) = 1. (~a · ~σ) · (~b · ~σ) − (~b · ~σ) · (~a · ~σ) = 2i ~a ×~b · ~σ. We are now in a position to accomplish the reduction (32b) of (28) which goes as follows 5

0 q~ · ~σ = Unˆ (θ) · (~q · ~σ) · Unˆ (−θ) (34a) (22) θ θ θ θ = σ cos − iσ sin · (~q · ~σ) · σ cos − iσ sin (34b) 0 2 i 2 0 2 i 2 θ θ θ θ = cos2 (~q · ~σ) + sin2 (ˆn · ~σ) · (~q · ~σ) · (ˆn · ~σ) + i sin cos −(ˆn · ~σ) · (~q · ~σ) + (~q · ~σ) · (ˆn · ~σ) 2 2 2 2 (32b) θ θ = cos2 (~q · ~σ) + sin2 (ˆn · ~σ) · (~q · ~σ) · (ˆn · ~σ) + sin θ (ˆn × ~r) · ~σ (34c) 2 2 (33) θ θ = cos2 (~q · ~σ) + sin2 −(~q · ~σ) + 2 (ˆn · ~q)(ˆn · ~σ) + sin θ (ˆn × ~q) · ~σ (34d) 2 2 θ = cos θ (~q · ~σ) + 2 sin2 (ˆn · ~q)(ˆn · ~σ) + sin θ (ˆn × ~q) · ~σ (34e) 2 = cos θ (~q · ~σ) + (1 − cos θ) (ˆn · ~q)(ˆn · ~σ) + sin θ (ˆn × ~q) · ~σ (34f) = [ cos θ ~q + (1 − cos θ) (ˆn · ~q)n ˆ + sin θ nˆ × ~q ] · ~σ. (34g)

Since we are free to choose our fundamental representation ~σ, this identity then implies the following general rotation formula

q~0 = cos θ ~q + (1 − cos θ)n ˆ(ˆn · ~q) + sin θ nˆ × ~q. (35)

This is known as Rodrigues’ rotation formula.

Exercise:

3 Consider the tetrahedron in R centered on the origin O with vertices P0 = (1, 1, 1), P1 = (1, −1, −1), 3 3 P2 = (−1, 1, −1), and P3 = (−1, −1, 1). Let T from R to R be a rotation about the axis through points O and P2 that transforms P1 into P3. Find the images of the transformation of the four corners of the 2π tetrahedron under this transformation. Hint: Use (40) with θ = 3 where nˆ is directed along P2 − O.

As a check of the rotation formula (35), consider the two cases whenn ˆ ⊥ ~q andn ˆ k ~q :

( cos θ ~q + sin θ nˆ × ~q forn ˆ ⊥ ~q, q~0 = (36) cos θ ~q + (1 − cos θ)n ˆ(ˆn · ~q) = ~q forn ˆ k ~q, which is correct by inspection.

V. ROTATION TRANSFORMATION IN EXPONENTIAL FORM

0 Writing this in component form q i = Rij(θ) qj we have

0 qi = [cos θ δij + (1 − cos θ) ninj + sin θ nkikj] qj, (37) 6 whence the transformation matrix for rotations aboutn ˆ is immediately identiﬁed

 2  cos θ + (1 − cos θ) n1 (1 − cos θ) n1n2 − sin θ n3 (1 − cos θ) n1n3 + sin θ n2 2 Rnˆ (θ) = (1 − cos θ) n1n2 + sin θ n3 cos θ + (1 − cos θ) n2 (1 − cos θ) n2n3 − sin θ n1 (38a) 2 (1 − cos θ) n1n3 − sin θ n2 (1 − cos θ) n2n3 + sin θ n1 cos θ + (1 − cos θ) n3  2    n1 n1n2 n1n3 0 −n3 n2 2 = cos θ 13 + (1 − cos θ) n1n2 n2 n2n3 + sin θ  n3 0 −n1 (38b) 2 n1n3 n2n3 n3 −n2 n1 0  2 2    1 − n2 − n3 n1n2 n1n3 0 −n3 n2 2 2 = cos θ 13 + (1 − cos θ)  n1n2 1 − n1 − n3 n2n3  + sin θ  n3 0 −n1 (38c) 2 2 n1n3 n2n3 1 − n1 − n2 −n2 n1 0  2 2    −n2 − n3 n1n2 n1n3 0 −n3 n2 2 2 = 13 + (1 − cos θ)  n1n2 −n1 − n3 n2n3  + sin θ  n3 0 −n1 (38d) 2 2 n1n3 n2n3 −n1 − n2 −n2 n1 0 2 = 13 + (1 − cos θ) λ + sin θ λ, (38e) where the generator λij ≡ ikjnk of a rotation aboutn ˆ is the anti-symmetric matrix

  0 −n3 n2 λ ≡  n3 0 −n1 , (39) −n2 n1 0 which is skew tri-idempotent (i.e. λ3 = −λ). Thus, we can write the rotation matrix in exponential form

θλ Rnˆ (θ) = e . (40)

The determinant det[Rnˆ (θ)] = 1, so the traceless hermitian matrices spaning (39)

0 −1 0  0 0 1 0 0 0  1 0 0 ,  0 0 0 , 0 0 −1 , 0 0 0 −1 0 0 0 1 0 are generators of the SO(3) group. This leads one to deﬁne three traceless hermitian matrices

0 −i 0  0 0 i 0 0 0  J1 = i 0 0 ,J2 =  0 0 0 ,J3 = 0 0 −i , (41) 0 0 0 −i 0 0 0 i 0

3 which are tri-idempotent (Ji = Ji). Thus, the rotation matrix (40) takes on a manifestly unitary form

−iθ(n1J3+n2J2+n3J1) Rnˆ (θ) = e . (42)

The labeling of the components of the rotation axis is arbitrary. By relabeling to match the generators, n1 → A3, n2 → A2, and n3 → A1, we can write (42) as a rotation about Aˆ

−iθ(Aˆ·J~) RAˆ(θ) = e . (43)

With this particular choice of generators, (41) is a 3 × 3 adjoint representation of SU(2)

[Jl,Jm] = i lmnJn. (44)

(41) is also an irreducible spin-1 representation of SU(2). 7

VI. COMPOSITION OF QUBIT ROTATIONS

A rotation through angle β1 aboutn ˆ1, followed by a rotation through angle β2 aboutn ˆ2 is induced by a similarity transformation using a unitary matrix that is the composition

β2 β1 −i 2 nˆ2·~σ −i 2 nˆ1·~σ Unˆ2 (β2)Unˆ1 (β1) = e e (45a) β β β β = cos 2 − i(ˆn · ~σ) sin 2 cos 1 − i(ˆn · ~σ) sin 1 (45b) 2 2 2 2 1 2 β β β β = cos 1 cos 2 − sin 1 sin 2 (ˆn · ~σ)(ˆn · ~σ) 2 2 2 2 1 2 β β β β − i cos 1 sin 2 (ˆn · ~σ) + sin 1 cos 2 (ˆn · ~σ) . (45c) 2 2 2 2 2 1

We need to expand the double dot product term in (45c). To do this, we use the identity (29) that we already derived above. Thus, β β β β U (β )U (β ) = cos 1 cos 2 − sin 1 sin 2 nˆ · nˆ (46) nˆ2 2 nˆ1 1 2 2 2 2 1 2 β β β β β β − i sin 1 cos 2 nˆ + cos 1 sin 2 nˆ − sin 1 sin 2 nˆ × nˆ · ~σ. 2 2 1 2 2 2 2 2 1 2

The overall unitary matrix used to induce a rotation through β12 aboutn ˆ12 is

−i β12 nˆ ·~σ β12 β12 U (β ) = e 2 12 = cos − i sin nˆ · ~σ. (47) nˆ12 12 2 2 12 Finally, equating our two expressions for the unitary matrices, (46)=(47), which is

Unˆ12 (β12) = Unˆ2 (β2)Unˆ1 (β1), (48) the real and imaginary parts lead to the following two identities β β β β β cos 12 = cos 1 cos 2 − sin 1 sin 2 nˆ · nˆ (49a) 2 2 2 2 2 1 2 β β β β β β β sin 12 nˆ = sin 1 cos 2 nˆ + cos 1 sin 2 nˆ − sin 1 sin 2 nˆ × nˆ . (49b) 2 12 2 2 1 2 2 2 2 2 1 2

A. Special case of equal angles

β2 β1 ∆t −i 2 nˆ 2·σ −i 2 nˆ 1·σ For equal angles β1 = β2 = 2 , and making the fol- Unˆ 2 (β2)Unˆ 1 (β1) = e e (52a) lowing identiﬁcations for the rotation axes β1 β2 β1 β2 = cos cos − sin sin nˆ 1 · nˆ 2 nˆ = ψˆ andn ˆ =z, ˆ (50) 2 2 2 2 1 2 h β β β β − i sin 1 cos 2 nˆ + cos 1 sin 2 nˆ (49) reduces to an identity that we will use to prove the 2 2 1 2 2 2 eﬃciency of Grover’s quantum search algorithm β β i − sin 1 sin 2 nˆ × nˆ · σ, (52b) β ∆t ∆t 2 2 1 2 cos 12 = cos2 − sin2 ψˆ · zˆ (51a) 2 2 2 ˆ β ∆t ∆t ∆t where σ = (σx, σy, σz) is a vector of Pauli matrices, n1 sin 12 nˆ = sin cos ψˆ +z ˆ − sin2 ψˆ × z.ˆ and nˆ are unit vectors specifying the respective principal 2 12 2 2 2 2 axes of rotation, and β1 and β2 are real-valued rotation (51b) β2 z −i 2 nˆ 2·σ angles. Let us take US = e as our stream oper- β1 −i 2 nˆ 1·σ ator and UC = e as our collision operator. Let VII. EXAMPLE COMPOSITE ROTATION us choose a reference frame where the particle motion occurs along the zˆ Consider a local evolution operator as a composition of β2 β1 z −i β2 σ −i 2 nˆ 2·σ −i 2 nˆ 1·σ 2 z “qubit rotations” Unˆ 2 = e and Unˆ 1 = e : US = e . (53a) 8

In this frame a general collision operator is that we established above with the collision operator (53b). For the sake of simplicity, let us start with a spe- β1 −i 2 (ασx+βσy +γσz ) UC = e , (53b) cialized construction whereby nˆ 1 is perpendicular to nˆ 2. The solution of (58) in this special case is where α, β, and γ are real valued components subject 2 2 2 to the constraint α + β + γ = 1. Furthermore, let β2 β2 α = cos β = − sin γ = 0. (59) us suppose that the unitary operators (53) are applied 2 2 locally and homogeneously at all the points in the sys- Inserting (59) into (58a) gives tem. So, here we consider a construction whereby the two principal unit vectors specifying the axes of rotation β m c2τ sin 1 = ◦ , (60) are 2 ~

nˆ 1 = (α, β, γ) nˆ 2 = (0, 0, 1). (54) and in turn (58c) is s With this choice, nˆ × nˆ = (β, −α, 0) and nˆ · nˆ = γ, 2 2 1 2 1 2 m◦c τ β2 c pzτ so (52) is a quite general representation of a quantum 1 − sin = − . (61) 2 lattice gas evolution operator ~ ~ In turn, we have z (54) β1 β2 β1 β2 U UC = cos cos − γ sin sin (55a) S s v 2 2 2 2 2 2 2u c pz τ β1 β2 m◦c τ u β β β β ~ 1 2 1 2 cos cos = 1 − u1 − 2 − i α sin cos − β sin sin σ 2 2 t 2 2 2 2 2 x ~ 1 − m◦c τ ~ β1 β2 β1 β2 − i β sin cos + α sin sin σy (62a) 2 2 2 2 s 2 2 2 β β β β m◦c τ c pzτ − i γ sin 1 cos 2 + cos 1 sin 2 σ = 1 − − (62b) 2 2 2 2 z ~ ~ 2 s 2 ic pzτ im◦c τ Eτ 7→ 1 + σz − σx, (55b) = 1 − , (62c) ~ ~ ~ where the last line is chosen as a construction. The rea- with E2 = (m c2)2 + (c p )2. Therefore, the quantum son for choosing this construction is that the quantum ◦ z 0 z lattice gas evolution operator (55a) is algorithm ψ = US UCψ is s 2 2 2 0 ic pzτ im◦c τ z Eτ ic pzτ im◦c τ ψ (z) = 1 + σz − σx ψ(z), (56) US UC = 1 − + σz − σx ~ ~ ~ ~ ~ which is a time-difference representation of the equation (63a) s of motion of a single free Dirac particle with a 2-spinor 2 2 Eτ iEτ c pz m◦c quantum state ψ(z) = (ψ (z), ψ (z))T defined over the = 1 − + σ − σ . L R E z E x set of points {z} in a 1+1 dimensional spacetime. That is, ~ ~ for small τ and for momentum operator pz = −i~∂z, (56) (63b) represents the Dirac equation for a relativistic quantum This result leads us to define the rotation axis particle of mass m◦ 2 m◦c c pz 2 nˆ ≡ − xˆ + zˆ. (64) i~∂tψ = −c pzσzψ + m◦c σxψ. (57) 12 E E 2 To establish a correspondence between (55a) and Since (nˆ 12 · σ) = 1 (an involution), we can em- (55b), we simply choose the real-valued components of ploy Euler’s√ identity and the trigonometric identity −1 2 nˆ 1 to satisfy the following three conditions: sin(cos 1 − x ) = x, so we are free to write (63b) −i β12 nˆ ·σ 2 in a manifestly unitary form e 2 12 as follows: β1 β2 β1 β2 m◦c τ α sin cos − β sin sin = (58a)  s   2 2 2 2 ~ 2 z −1 Eτ β1 β2 β1 β2 U U = exp i cos 1 − nˆ · σ β sin cos + α sin sin = 0 (58b) S C    12  2 2 2 2 ~ β β β β c p τ γ sin 1 cos 2 + cos 1 sin 2 = − z . (58c) (65a) 2 2 2 2 ~  q 2  cos−1 1 − Eτ Additionally, we should respect the reality condition that (64) ~ 2 = exp i σzc pz − σx m◦c  . nˆ 1 have unit norm E α2 + β2 + γ2 = 1 (58d) (65b) 9

The hermitian generator governing the dynamical behav- where in the last line we made the identification ior of the 2-spinor field ψ is the Dirac Hamiltonian s 2 2 E` Eτ hD = −σzc pz + σx m◦c (65c) cos = 1 − , (66) ~c ~ This is a remarkable finding because nowhere in the derivation of (65b) did we invoke the continuum limit or expressing τ in terms of the grid size ` where τ → 0. That is, τ may be taken to be a small but finite quantity, not necessarily infinitesimal. Thus, E` because of the form of (65c), Lorentz invariance would τ = ~ sin . (67) apply to the quantum dynamics even though the space- E ~c time is discrete, albeit there are unexpected deviations from relativistic quantum mechanics (Yepez, 2010). The rotation angle in (65b) is a real scalar quantity, so we may denote this as ` and write References

z ∼ −i ` hD/(~c) Yepez, J., 2010, arXiv:1106.0739 [gr-qc] . US UC = e , (65d)