C/CS/Phy191 Problem Set 6 Solutions 3/23/05

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1 C/CS/Phy191 Problem Set 6 Solutions 3/3/05 1. Using the standard basis (i.e. 0 and 1, eigenstates of Ŝ z, calculate the eigenvalues and eigenvectors associated with measuring the component of spin along an arbitrary direction ˆn for a spin- 1 system. (S ˆn = Ŝ x n x + Ŝ y n y + Ŝ z n z as in class. Show your work. Answer: We first note that the arbitrary unit vector is expressed in the standard spherical coordinates as n = (n x,n y,n z = (cosφsinθ,sinφsinθ,cosθ. With the spin operators S i just proportional to the Pauli matrices, we see that the operator Ŝ n is expressed in matrix form: Ŝ n = Ŝ x n x +Ŝ y n y +Ŝ z n z h cosφsinθ + h ( 0 i 0 sinφsinθ + h ( 0 1 cosθ (1 This is readily simplified to give: Ŝ n h ( cosθ e φ sinθ e iφ sinθ cosθ ( Our task is now to find the eigenvalues and eigenvectors of this matrix. Nontrivial eigenvalues are the solution to the determinant equation: h cosθ λ e iφ sinθ eφ sinθ cosθ λ = 0 (3 This gives the equation λ = h 4, with solutions λ = ± h. Let s just look at the first eigenvalue λ = h and find its associated eigenvector. ( cosθ λ e φ sinθ e iφ sinθ cosθ λ ( a b ( a = b (4 Just the first row gives the following equation: cosθa+e φ sinθb = a (5 This is readily rearranged to give: ( 1 cosθ b = ae iφ sinθ (6 C/CS/Phys 191, Spring 005 1

2 This equation simplifies further by recalling some trigonometric relations (cosθ = 1 sin θ, sinθ = sin θ cos θ. We then obtain: b = ae iφ sin θ cos θ (7 This doesn t uniquely determine a and b yet, but we also have the constraint a + b = 1. Plugging the above equation into the normalization constraint gives: a + sin θ a = 1 a = cos θ cos θ a = cosθ We are free to pick the phase of a because all that matters physically is the relative phase between a and b. Therefore, we can just choose the phase of a to be zero and we have a = cos θ. b is then given by b = ae iφ sin θ cos θ = e iφ sin θ. Our eigenstate +n with eigenvalue + h/ is thus: (8 ( cos θ +n e iφ sin θ +n = cos θ 0 +eiφ sin θ 1 (9 This state should of course look familiar as the generalized spin vector on the Bloch sphere. We might have guessed! Similar algebra will readily show that the other eigenvector n with eigenvalue h/ is given by: n = sin θ 0 +eφ cos θ 1 (10. Suppose 10,000 electrons are prepared in the ψ = 0 spin state and then shot through a Stern- Gerlach device oriented with North South magnet poles aligned 60 from the +ẑ-direction. About how many electrons go into each of the two resulting beams? Answer: If we want to know the probability for measuring a ket ψ in a state φ, the probability is just given by: P φ = φ ψ (11 In this case, particle is in state ψ = 0 and we are measuring in the basis of θ 60 o. The two eigenstates in this basis are (following problem 1: +n = cos θ 0 +eiφ sin θ 1 = cos(30o 0 +e i(0 sin(30 o 1 = (1 n = (13 These give the probabilities P up = +n 0 = 3 4 and P down = n 0 = 1 4. We should therefore expect 7,500 electrons in the upper beam and,500 in the lower beam. C/CS/Phys 191, Spring 005

3 3. Consider a qubit comprised of a single spin- 1 electron. How would you construct a NOT gate in the laboratory to act on this qubit? If your gate involves building a magnet, then please specify the exact orientation, amplitude, and time duration of any proposed applied magnetic fields. Answer: The NOT gate in matrix mechanics looks like ( development operator that looks like Û(t = e Ŝn h m 0 1 t = as in standard matrix mechanics Ŝ x h.. We want to engineer a time. Ŝ x looks like an attractive option, It is important to note that this time development operator is defined in the standard basis, 0 = +z and 1 = z. These states are not eigenstates of Ŝ x, and the time development operator is really only of use if we can apply it to eigenstates. The eigenstates of Ŝ x are + x = 1 ( and x = 1 ( 0 1. To express our time development we can apply it to basis states: Û(t 0 = e m tŝ x / h 0 = e m tŝ x / h ( +x + x / = 1 ( e m t +i +x +e m t x Expressing this again in the standard basis, (14 Û(t 0 = 1 ( e m t ( e +i m t ( 0 1 e+i m t + e = These coefficients should look familiar. Using Euler s relation: m t e+i m t e 0 m t 0 (15 Û(t 0 = cos eb o m t 0 i sin eb o t 0 (16 m We now want to apply this field for a time such that the coefficient on 0 is zero and the coefficient on 1 has magnitude 1. This first occurs when eb o t m = π, so we should apply the field for a time t = mπ eb o. You may be worried that this has taken our state to i 1 and not 1. This is no problem because one could always apply a field along the z-axis for a certain time (evolving φ and eliminate any relative phase. The important part of the gate is to get the timing/fields right for θ. It should be noted that this solution is hardly unique. An applied field in any direction transverse to ẑ for the same time would accomplish the goal, and resonance techniques would also work just fine. 4. An electron is shot with well-defined momentum at a -slit device. Does it get through to the detector? Consider the details below and explain your answer: In case you are worried about the different spatial trajectories the particles take, let s take a quick look at the full quantum state with the spatial dependence factored in: ψ electron = ψ spatial ψ spin C/CS/Phys 191, Spring 005 3

4 ψ electron (x = x ψ spatial ψ spin = ψ spatial (x ψ spin = e ikx ψ spin We further specify that ψ electron (x = 0 and ψ electron (x = x o = ψ path1 + ψ path, where: ψ path1 = e ikx o 0 ψ path = e ikx o e Ŝz h φ 0 Further, φ = eb o m t and t = x o v = x o hk. Assume that we are clever enough to experimentally engineer m φ = π. Answer: Naively, we might think nothing would happen because normally (i.e. classical vectors in spherical coordinates a π rotation brings you right back to where you started. However, this would be incorrect because we need to note the following: ψ > path = e Ŝz h π + h π 0 >= e h 0 >= eπ 0 > So psi > path = 0 >! Normally we ignore the - sign since it s an overall phase factor, but now we can t since the particle is going to interfere with itself!! So what happens at point B? psi > B = 0 > +( 0 > = 0 How many particles get through? ZERO!!! This is one of the strange features of quantum mechanics. For spin-1/ particles two complete rotations on the Bloch sphere are required to return to the initial state. 5. Consider the total spin of a system having electrons: Ŝ T = Ŝ 1 + Ŝ. Show that the entangled state ψ total = has a net spin of zero. In other words, show that it is an eigenstate of Ŝ T with eigenvalue = 0. Answer: Use Ŝ T = ( Ŝ 1 + Ŝ = Ŝ 1 + Ŝ + Ŝ 1 Ŝ, and recall that Ŝ 1 Ŝ = Ŝ 1x Ŝ x + Ŝ 1y Ŝ y + Ŝ 1z Ŝ z. The raising and lowering operators are also very useful (Ŝ + = Ŝ x + iŝ y, Ŝ = Ŝ x iŝ y. We then have that: ((Ŝ1+ ŜT = Ŝ1 + Ŝ + Ŝ 1 (Ŝ+ + Ŝ (Ŝ1+ Ŝ 1 (Ŝ+ Ŝ + + +Ŝ 1z Ŝ z (17 i i A number of terms cancel out in this expansion, and some algebra gives the following expression: Ŝ T = Ŝ 1 + Ŝ + Ŝ 1z Ŝ z + Ŝ 1 Ŝ + + Ŝ 1+ Ŝ (18 Now let s apply this operator to the state ψ total = : Ŝ T ( = ( Ŝ 1 + Ŝ + Ŝ 1z Ŝ z + Ŝ 1 Ŝ + + Ŝ 1+ Ŝ ( (19 Let s look at the action of each of these terms on the state: Ŝ1 ( = 3 h 4 ( Ŝ ( = 3 h 4 ( Ŝ 1z Ŝ z ( = ( ( h h ( Ŝ 1 Ŝ + ( = h ( Ŝ 1+ Ŝ ( = h ( (0 C/CS/Phys 191, Spring 005 4

5 Putting all of these together we get: ( 3 h ŜT ( = h 4 h h ( = 0( So we see that the state is an eigenstate with an associated eigenvalue of zero. 6. Consider an atomic qubit system made from a single hydrogen atom. The qubit levels are the ground state and first excited state of hydrogen. 0 = ground state, E o 1 = first excited state, E 1 Suppose a laser at the resonant frequency ω = E 1 E o is directed at the atom with an intensity such that h 0 eeẑ 1 = 10 6 ev. How long should the laser be pointed at the atom to act as a Hadamard gate? Answer: Like problem 3, we want to construct a time-development operator that looks like a nice, useful quantum gate. In this case, we want the Hadamard gate: (1 H = 1 ( ( For the case of atomic resonance (see lecture 15, we have that a state evolves as ψ(t = cos ω 1t 0 + e i(ωot+π sin ω 1t 1, with ω o = E 1 E o and ω h 1 = V 1 h. A Hadamard gate will take 0 1 ( 0 + 1, and by the fact that unitary transformations are one to one we can be assured that it will take 1 to 1 ( 0 1 modulo an overall phase factor. Further, we note that the ω o frequency is (a always present and (b too fast for us to control anyway, so the most important thing for this gate is to get equal amplitudes on 0 and 1. This is accomplished with the applied laser field. The time evolved ket ψ(t is half 0, half 1 when cos ω 1τ = sin ω 1τ. This occurs first when ω 1τ = π 4. Therefore τ = π. Plugging in the values, we obtain a time ω o = π h V o C/CS/Phys 191, Spring 005 5