Unitary Time Evolution

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Unitary Time Evolution Unitary time evolution Time evolution of quantum systems is always given by Unitary Transformations. If the state of a quantum system is ψ , then at a later time | i ψ Uˆ ψ . | i→ | i Exactly what this operator Uˆ is will depend on the particular system and the interactions that it undergoes. It does not, however, depend on the state ψ . This | i means that time evolution of quantum systems is linear. Because of this linearity, if a system is in state ψ or φ or any linear combination, the time evolution is| giveni | byi the same operator: (α ψ + β φ ) Uˆ(α ψ + β φ ) = αUˆ ψ + βUˆ φ . | i | i → | i | i | i | i – p. 1/25 The Schrödinger equation As we have seen, these unitary operators arise from the Schrodinger¨ equation d ψ /dt = iHˆ (t) ψ /~, | i − | i where Hˆ (t) = Hˆ †(t) is the Hamiltonian of the system. Because this is a linear equation, the time evolution must be a linear transformation. We can prove that this must be a unitary transformation very simply. – p. 2/25 Suppose ψ(t) = Uˆ(t) ψ(0) for some matrix Uˆ(t) (which we don’t yet assume| i to be| unitary).i Plugging this into the Schrödinger equation gives us: dUˆ(t) dUˆ †(t) = iHˆ (t)Uˆ(t)/~, = iUˆ †(t)Hˆ (t)/~. dt − dt At t = 0, Uˆ(0) = Iˆ, so Uˆ †(0)Uˆ(0) = Iˆ. We see that d 1 Uˆ †(t)Uˆ(t) = Uˆ †(t) iHˆ (t) iHˆ (t) Uˆ(t) = 0. dt ~ − So Uˆ †(t)Uˆ(t) = Iˆ at all times t, and Uˆ(t) must always be unitary. – p. 3/25 For time-independent Hamiltonians we can easily write down the solution to the Schrödinger equation. Using the spectral theorem, we choose a basis k of eigenvectors of {| i} Hˆ with eigenvalues Ek, Hˆ k = Ek k . We then write ψ(t) in terms of this basis: | i | i | i 2 Hˆ = E k k , ψ(t) = α (t) k , α (t) = 1. k| ih | | i k | i | k | Xk Xk Xk Knowing ψ(t) = exp( iHt/ˆ ~) ψ(0) for any t means know- | i − | i ing the amplitudes αk(t). From the Schrödinger equation, dα k = iE α /~ = α (t) = exp( iE t/~)α (0). dt − k k ⇒ k − k k Each energy eigenstates undergoes a steady phase rotation. – p. 4/25 Bloch sphere rotation Any 2 2 Hermitian operator can be written Hˆ = aIˆ+ bXˆ+ × cYˆ + dZˆ with real a,b,c,d. Spin-1/2 unitaries take the form Uˆ(t) = exp (it/~)(aIˆ+ bXˆ + cYˆ + dZˆ) − We now need to use a very useful and important fact. For general operators Aˆ and Bˆ, usually exp(Aˆ)exp(Bˆ) = exp(Aˆ + Bˆ). 6 The one exception to this is when [A,ˆ Bˆ] = 0. In this case only, exp(Aˆ)exp(Bˆ) = exp(Aˆ + Bˆ). – p. 5/25 Since the identity commutes with everything, it exp aIˆ+ bXˆ + cYˆ + d − ~ iat it = exp Iˆ exp bXˆ + cYˆ + dZˆ − ~ − ~ ~ = e−iat/ exp (it/~)(bXˆ + cYˆ + dZˆ) . − Since an overall phase is meaningless, we can always set a = 0. (This is not just true for spin-1/2; one can add or subtract a term aIˆ to any Hamiltonian.) – p. 6/25 The most general spin-1/2 Hamiltonian is therefore Hˆ = bXˆ + cYˆ + dZˆ = E0~n ~σˆ · where 2 2 2 E0 = b + c + d , ~n =(nx,ny,nzp)=(b/E0,c/E0,d/E0), 2 2 2 ˆ ˆ ˆ ˆ with nx + ny + nz = 1 and ~σ =(X, Y, Z). The unitary is exp( iHt/ˆ ~) = cos(E0t/~)Iˆ i sin(E0t/~)~n ~σ.ˆ − − · In the Bloch sphere picture this corresponds to a rotation around the axis ~n at a rate E0/~. This is the most general unitary transformation possible for spin-1/2. – p. 7/25 Controlling unitaries We generalize from steady rotation by assuming we can turn the Hamiltonian on and off. By turning a Hamiltonian on for a particular length of time, we can “rotate” the state by a particular angle. For a spin-1/2, this means we can perform unitary transformations of the form Uˆ(θ) = cos(θ/2)Iˆ i sin(θ/2)~n ~σ.ˆ − · We can do the same with more complicated systems. For a D-dimensional system with a Hamiltonian Hˆ having eigenvalues E and eigenvectors k , we can do the unitary k | i Uˆ(τ) = exp( iE τ/~) k k − k | ih | Xk for any τ. – p. 8/25 Building up unitaries Unfortunately, we cannot always choose the exact values of the eigenvalues Ek or the eigenvectors k . These are generally given to us by nature. But we| sometimesi can increase the range of our options by combining several different unitaries in a row. The important thing to remember is that any product of unitary operators is also unitary: Uˆ †Uˆ = Vˆ †Vˆ = Iˆ = (UˆVˆ )†(UˆVˆ ) = Vˆ †Uˆ †UˆVˆ = I.ˆ ⇒ Suppose there are two different Hamiltonians we can turn on: Hˆ1 and Hˆ2. Then we can perform the unitaries Uˆ1(τ) = exp( iHˆ1τ/~), Uˆ2(τ) = exp( iHˆ2τ/~). − − But we can do much more than these! – p. 9/25 We can also do the unitaries Uˆ2(τ2)Uˆ1(τ1), and Uˆ2(τ3)Uˆ1(τ2)Uˆ2(τ1), and Uˆ2(τ )Uˆ1(τ −1) Uˆ2(τ2)Uˆ1(τ1). n n · · · Let’s see how this works for the spin-1/2. Suppose we can turn on Hamiltonians Hˆ1 = ExX,ˆ Hˆ2 = EyY.ˆ These produce unitaries Uˆ1(θ) and Uˆ2(θ) which correspond, in the Bloch sphere representation, to rotations by θ about the X and Y axes, respectively. – p. 10/25 There is a theorem in geometry that a rotation by any angle θ around any axis ~n can be done by doing three rotations in a row around the X and Y axes: R~n(θ) = RX(φ3)RY (φ2)RX(φ1) for some φ1,φ2,φ3. Since every 2 2 unitary is equivalent to a Bloch sphere rotation about× some axis ~n, any 2 2 unitary equals × UˆX (τ3)UˆY (τ2)UˆX(τ1) for some τ1,τ2,τ3 (up to an overall phase). When we talk about “turning on” and “turning off” Hamiltonians, what do we really mean? – p. 11/25 Here is a picture to give some idea: For many experimental systems, unitaries are effected by turning precisely-tuned lasers on and off for precise lengths of time. – p. 12/25 Tensor products of unitaries We have seen that Hilbert spaces of composite systems are represented by tensor products of the Hilbert spaces of the component systems: 12 = 1 2. H H ⊗ H Therefore, unitaries on the joint system also act on this larger Hilbert space. Suppose that the Hamiltonian of a system is a sum of terms which act only on the individual subsystems: Hˆ = Hˆ1 Iˆ+ Iˆ Hˆ2. ⊗ ⊗ The two terms represent the Hamiltonian of the first and second subsystems, respectively. In this Hamiltonian, the two subsystems are isolated, and do not interact. – p. 13/25 Hˆ1 Iˆ and Iˆ Hˆ2 commute, so ⊗ ⊗ Uˆ(t) = exp( iHt/ˆ ~) − = exp( i(Hˆ1 Iˆ+ Iˆ Hˆ2)t/~) − ⊗ ⊗ = exp( iHˆ1 It/ˆ ~)exp( iIˆ Hˆ2t/~) − ⊗ − ⊗ = exp( iHˆ1t/~) exp( iHˆ2t/~) − ⊗ − Uˆ1(t) Uˆ2(t). ≡ ⊗ It is a tensor product of unitaries. What if the Hamiltonian is not a sum of terms which act on the individual subsystems, but includes terms which act on both subsystems? – p. 14/25 Interactions and entanglement If the Hamiltonian has the form Hˆ = Hˆ1 Iˆ+ Iˆ Hˆ2 + Hˆint, ⊗ ⊗ the unitary Uˆ(t) = exp( iHt/ˆ ~) will not be a tensor product, in general. In− this case, we say that the two subsystems interact. If a tensor-product unitary Uˆ1 Uˆ2 acts on a product state ψ φ , then it will remain⊗ a product state. If a | i ⊗ | i general Uˆ acts on it, generally the state becomes entangled. Since most states are entangled, to produce them from initial product states requires unitaries which are not products. We must have interactions between the subsystems to produce general unitary transformations. – p. 15/25 Let’s take an example for two spin-1/2s: Hˆint = EintZˆ Zˆ. This yields unitary transformations of the form ⊗ Uˆ(θ) = cos(θ/2)Iˆ i sin(θ/2)Zˆ Z.ˆ − ⊗ Suppose we have an initial product state Ψ = ψ φ | i | i ⊗ | i = α1β1 + α1β2 + α2β1 + α2β2 . | ↑↑i | ↑↓i | ↓↑i | ↓↓i When we transform it by Uˆ(θ) it becomes −iθ/2 iθ/2 Uˆ(θ) Ψ = e α1β1 + e α1β2 | i | ↑↑i | ↑↓i iθ/2 −iθ/2 +e α2β1 + e α2β2 , | ↓↑i | ↓↓i which is no longer a product state for θ = mπ/2. The interaction has produced entanglement.6 – p. 16/25 The no-cloning theorem The restriction of time evolution to unitary operators means that certain kinds of evolution are impossible. One impossible task is quantum cloning. Suppose we have a system in an unknown state ψ , | i and we wish to copy it, i.e., to transform a second system starting in some standard state 0 into the same state | i ψ . Is there a unitary Uˆ such that, for any state ψ , | i | i Uˆ ( ψ 0 ) = ψ ψ ? | i ⊗ | i | i ⊗ | i If this is true, then also for φ = ψ | i 6 | i Uˆ ( φ 0 ) = φ φ . | i ⊗ | i | i ⊗ | i – p. 17/25 Consider now a superposition state χ = α ψ + β φ . By linearity, | i | i | i Uˆ( χ 0 ) = Uˆ(α ψ + β φ ) 0 | i ⊗ | i | i | i ⊗ | i = α ψ ψ + β φ φ | i ⊗ | i | i ⊗ | i = (α ψ + β φ ) (α ψ + β φ ), 6 | i | i ⊗ | i | i so Uˆ( χ 0 ) = χ χ , which is a contradiction.
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