Homework assignment 3: due Thursday, 10/26/2017

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1 Homework assignment 3: due Thursday, 10/6/017 Physics 6315: Quantum Mechanics 1, Fall 017 Problem 1 (0 points The spin Hilbert space is defined by three non-commuting observables, S x, S y, S z. These operators satisfy the angular momentum commutation relations, so that simultaneous eigenstates of S = Sx + Sy + Sz and S z exist. According to he general theory of angular momentum, these states can be designated by two quantum numbers, s and m s, where s must be either an integer or half integer, and m s {s, s 1,, s}. The theory of spin says that for a given particle, the value of s is fixed. A spin- 1 particle has s = 1/, so that m s { 1/, 1/}. Since s never changes, we can label the two eigenstates of S z as + z and z, where + z = s = 1, m s = 1 and z = s = 1, m s = 1, so that: S z + z = h + z and S z z = h z S + z = 3 h 4 + z S z = 3 h 4 z 1. Find the four matrix elements of S z in the basis of its own eigenstates. Invert the definitions S + = S x + is y and S = S x is y to express S x and S y in terms of S + and S. 3. Use the equation S ± s, m s = h s(s + 1 m s (m s ± 1 s, m s ± 1 to find the matrix elements of S + and S in the basis { + z, z }. 4. From points ( and (3 derive the matrix representations of S x and S y. 1

2 5. Show explicitly that S = h s(s Based on symmetry, write the matrix representations of S x, S y and S z in the basis of eigenstates of S y. Solution. (1 ( ( + Sz + + S S z = z S z + S z = h ( S x = 1 (S + + S S y = 1 i (S + S (3 We get S + + = 0 S + = h + S + = h S = 0 Therefore: S + = h ( ( 0 0 S = h 1 0 (4 From ( and (3 above we get S x = h ( S y = h ( 0 i i 0 (5 S = S x + S y + S z = h 4 = h 4 ( ( ( ( ( h 0 i 0 i + h i 0 i ( ( ( ( h h 1 0 = 3 h ( ( Problem (0 points Determine explicitly the form of the operator S u, corresponding to the spin projection in the direction determined by a unit vector û, defined through the angles θ and ϕ. For a state with a definite value of the spin z-projection, determine the expectation value of S u. Solution The spin operator S is a vector (or, more precisely, pseudo-vector operator. The u-projection operatore S u may be expressed in terms of its components S x, S y, S z in the same manner as for an ordinary, non-operator vector: S u = û S = (sin θ cos ϕs x + sin θ sin ϕs y + cos θs z.

3 where θ, ϕ are the polar and azimuthal angles of the û-direction. Using the explicit form of the spin matrices seen in class, we obtain: S u = h ( cos θ e iϕ sin θ e iϕ. sin θ cos θ The expectation value of S u in the state + is + S u + = h ( cos θ e (1 0 iϕ sin θ e iϕ sin θ cos θ ( 1 0 = h cos θ. ( Analogously, for the state we find: S u = 1 cos θ. Problem 3 (0 points A beam of spin 1/ particles is sent through series of three Stern-Gerlach measuring devices as shown in Fig. 1. The first SGz device transmits particles with S z = h/ and filters out particles with S z = h/. The second device, an SGn device transmits particles with S n = h/ and filters out particles with S n = h/, where the vector ˆn makes an angle θ in the x z plane with respect to the z-axis. The ϕ angle is zero. A last SGz device transmits particles with S z = h/ and filters out particles with S z = + h/. Figure 1: 1. What fraction of the particles transmitted through the first SGz device will survive the third measurement?. How must the angle θ of the SGn device be oriented so as to maximize the number of particles the at are transmitted by the final SGz device? What fraction of the particles survive the third measurement for this value of θ? 3

4 3. What fraction of the particles survive the last measurement if the SGz device is simply removed from the experiment? Solution (b This is maximized by choosing θ = π/ so that f max = 1/4. (c If there is no third device, the fraction surviving is f = +û ˆM(+ˆn + ẑ = +û cos θ + cos θ sin θ + ẑ = cos 3 θ + θ cos sin θ = cos 4 θ (1 + sin θ. Problem 4 (10 points A beam of spin-1/ particles traveling in the y-direction is sent through a Stern- Gerlach apparatus, which is aligned in the z-direction, and which divides the 4

5 incident beam into two beams with S z = ± h/. The S z = h/ beam is allowed to impinge on a second Stern-Gerlach apparatus aligned along the direction given by ê = sin θˆx + cos θŷ 1. Write S e = S ê, the matrix describing the projection of the spin operator in the ê direction, in the basis in which S z is diagonal. Calculate eigenvalues and normalized eigenvectors of S e 3. Calculate the probabilities that the particles will end up in one of the two beams which emerge from the second Stern-Gerlach apparatus. Solution. 5

6 (ccalculate the probabilities that the particles will end up in one of the two beams which emerge from the second Stern-Gerlach apparatus. The probability for the particle to end up in the beam corresponding to positive spin projection is therefore P (+ = sin θ/( cos θ, and for the negative spin projection we get P ( = sin θ/( + cos θ. 6

7 Problem 5 (0 points The ammonia molecule (NH 3 has two configuration states, such that one state is a mirror reflection of the other. If transitions between the two states were impossible, each state energy would be E 0. In fact, the transition amplitude per unit time between the states is c, namely the hamiltonian reads ( E0 c H = c E 0 1. What are the possible energies of the ammonia molecule? ( 1. Assume that, at t = 0, the molecule is found to be in the + = 0 state, in which the Nitrogen atom is above the plane of the three hydrogen atoms. ( What is the probability P (t that the molecule is found in the state 1 = at the time t? 0 Now, assume that each state of the molecule has an electric dipole moment µ. When the molecule is inserted into electric field ɛ, the geometry does not change. However, the Hamiltonian gets a new term: H 1 = µ ɛ and now reads ( E0 + µɛ c H = c E 0 µɛ 1. What are the current possible energies?. What is the expectation value D of the electric dipole moment (defined as H/ ɛ in the ground state? Solution (1 The Hamiltonian is: ( E0 c H = c E 0 from which we can extract the eigenvalues to be E ± = E 0 ±c. These correspond to the kets: ψ + = 1 [ + + ] ψ = 1 [ + ] ( The initial state is + = 1 [ ψ + + ψ ]. The state at the time t is therefore ψ(t = 1 [e ie + t h ψ+ + e ie t h ψ ] 7

8 The probability of finding the particle in the state at time t is ψ(t = sin (ct. (3 The new energies are E ± = E 0 ± (µɛ + c. (4 The dipole moment operator is defined as In the ground state, we get: ˆµ = Ĥ ɛ. D = ψ ˆµ ψ = E ɛ µ ɛ = (µɛ + c. 8