Angular Momentum - set 1

Transcription

1 Angular Momentum - set PH0 - QM II August 6, 07 First of all, let us practise evaluating commutators. Consider these as warm up problems. Problem : Show the following commutation relations ˆx, ˆL x = 0, ˆx, ˆL y = i ẑ, ˆx, ˆL z = i ŷ Problem : ˆp x, ˆL x = 0, ˆp x, ˆL y = i ˆp z, ˆp x, ˆL z = i ˆp y Using the commutation relations for the angular momentum operators, prove the Jacobi identity Problem : Prove the following relations ˆL x, ˆL y, ˆL z + ˆL y, ˆL z, ˆL x + ˆL z, ˆL x, ˆL y = 0 ˆL z, cos φ = i sin φ ˆL z, sinφ = i sin φ cos φ where φ is the azimuthal angle. Now, let us see how we can use our commutator calculation skills to some physical problems. Problem : a Prove that one can measure the z-components of the angular momentum and linear momentum simultaneously. b Write down the ˆL z operator in polar co-ordinates, and solve for the wavefuction of a particle moving with constant value of ˆL z = m. Identify the motion of the particle described by ˆL z = m. c Write down the operator ˆP z in differential operator form use cartesian co-ordinates for ˆP z and solve for the wave function of a particle moving with constant momentum k along the z-direction you may take the normalisation constant to be π.

2 d Now write down the wave function of a particle with m and k as the z-components of the angular and linear momenta. e Identify the combined motion of the particle. Problem 5: Consider the angular momentum state described by the wavefunction ψθ, φ = sin θ cos θe iφ cos θe iφ. a Is ψθ, φ an eigenstate of ˆL or ˆL z? b Find the probability of measuring for the z-component of the orbital angular momentum. c Find the expectation values of L and L z in this state. Hint: You may find it convenient to express ψθ, φ in terms of spherical harmonics. Problem 6: a Show that the spherical harmonics satisfy the following expectation values L x = 0 = L y ; L x = L y = ll + m. b Verify the uncertainty relation for the simultaneous measurement of x- and y- components of the orbital angular momentum Problem 7: L x L y m, where L x = L x L x. Consider a particle in the angular momentum state ψθ, φ = Y, + Y, + Y,0 + Y, + Y, a If L z is measured in this state, what values will be obtained and with what probabilities? b If after a measurement of L z, we obtain a result of, calculate the uncertainties L x and L y and their product L x L y and verify that it satisfies the generalised uncertainty relation. Problem 8: Consider a particle in the angular momentum state ψθ, φ = 8 Y, + 8 Y,0 + AY,

3 a Find A such that ψθ, φ is normalised. b Calculate the expectation values of ˆL z and ˆL in this state. c Calculate Φ ˆL z ψ where 8 Φθ, φ = Y, + Y,0 + 5 Y, Problem 9: A particle of mass m is fixed at one end of a rigid rod of negligible mass and length R. The other end of the rod rotates in the xy plane about a bearing located at the origin, whose axis is in the z-direction. a Write the system s total energy in terms of its angular momentum, L. b Write down the time-independent Schrödinger equation of the system. Hint: In polar coordinates, only φ varies. c Solve for the possible energy levels of the system, in terms of m and the moment of inertia I = mr. d Explain why there is no zero-point energy.

4 Problem Set Solution Problem Angular momentum in Cartesian coordinates Commutator of ˆx with the various components of angular momentum ˆL x, ˆL y, ˆL z : ˆx, ˆL x = x, yp z zp y = 0, since x commutes with y, z, p y, p z ˆx, ˆL y = x, zp x xp z = x, zp x = zx, p x = i hz ˆx, ˆL z = x, xp y yp x = x, yp x = yx, p x = i hy Commutator of ˆp x with the various components of angular momentum ˆL x, ˆL y, ˆL z : ˆp x, ˆL x = p x, yp z zp y = 0, since p x commutes with y, z, p y, p z ˆp x, ˆL y = p x, zp x xp z = p x, xp z = p x xp z xp z p x = p x, xp z = i hp z ˆp x, ˆL z = p x, xp y yp x = p x, xp y = p x xp y xp y p x = p x, xp y = i hp y Problem Jacobi identity Recall the cyclic relationship ˆL x, ˆL y = i hˆl z ˆL y, ˆL z = i hˆl x ˆL z, ˆL x = i hˆl y i.e. we cannot measure them simultaneously to arbitrary accuracy. Jacobi identity: ˆL x, ˆL y, ˆL z + ˆL y, ˆL z, ˆL x + ˆL z, ˆL x, ˆL y = i h ˆL x, L ˆ x + ˆL y, ˆL y + ˆL z, ˆL z = 0 Observe that Jacobi identity is also a cyclic relationship. Problem ˆL z, cos φfφ = ˆL z, sinφfφ = Angular Momentum in Spherical Coordinates i h φ, cos φ fφ = i h φ = i h cos φfφ cos φf φ φ i h φ, sinφ fφ = i h φ = i h = ˆL z, sinφ = i h cosφ = i hsin φ cos φ cos φ cos φ fφ φ = i h sin φfφ = ˆL z, cos φ = i h sin φ fφ sinφ sinφ φ sinφfφ sinφf φ φ = i h cosφfφ Problem Linear Momentum vs Angular Momentum a ˆL z, ˆp z = xˆp y yˆp x, ˆp z = 0 as ˆp z commutes with x, y, ˆp y, ˆp z Hence we can measure the z-component of the linear and angular momentum to arbitrary accuracy. b m hψ = ˆL z ψ = i h ψ ψ φ = mφ = iψ ψ = m φ = i ψ = ψ φ = Ae imφ where A C. This wave function describes the motion of a particle of mass m moving in uniform circular motion in the xy-plane.

5 c hkψ = ˆP z ψ = i h ψ z = ikz = ψ ψ = ik z = ψ ψ = ln ψ = ikz = ψ z = Be ikz where B C. This wave function describes the motion of a freely moving particle with uniform momentum along z-axis. Notice that both the wavefunctions in b and c are plane waves which are non-normalizable. However, we can normalised plane waves by working in momentum space via Fourier transform. d ψz, φ = ψ z ψ φ = ABe ikz+mφ = Ce ikz+mφ where C = AB C. e The particle is moving in helical motion - uniform circular motion in the xy plane and constant linear motion along z-axis. Charge particle entering a uniform magnetic field obliquely with constant speed will move in a helical manner. Problem 5 Observe that Angular Momentum on Spherical Harmonics 8π π ψθ, φ = sin θ cos θe iφ cos θe iφ = Y, + Y, where Y, = 8π sin θ cos θeiφ and Y, = π sin θe iφ are the normalised spherical harmonics. a ψθ, φ is an eigenstate for ˆL since l = is the same for both Y, and Y,. ψθ, φ is not an eigenstate for ˆL z since m = for Y, and m = for Y,. b P L z = h = 8π π + c L z = h 6 + h = 5 h = 6 π 5 Since l = for both Y, and Y,, L = + h = 6 h. Problem 6 Uncertainty Relationship in Angular Momentum a First consider L + = Y l,m L + Y l,m = α Y l,m Y l,m+ = 0 for some α R. raising operator L = Y l,m L Y l,m = β Y l,m Y l,m = 0 for some β R. lowering operator as Y i,j Y k,l = δ ik δ jl i.e. Y l,m are normalised orthonormal functions. As a result L+ + L L x = = L+ L L y = = i i L + + L L + L = 0 = 0

6 Therefore L x = 0 = L y Next consider L + = Y l,m L + Y l,m = γ Y l,m Y l,m+ = 0 for some γ R. L = Y l,m L Y l,m = δ Y l,m Y l,m = 0 for some δ R. Now L+ L + L x = = L+ L L y = = i So L x = L y = L +L + L L +. Recall that L + + L + L + L L + + L = L +L + L L + L + L + L L L + + L = L +L + L L + L = L x + L y + L z = L = L x + L y + L z = ll + h = L x + m h = L x = L y = ll + m h since L z Y l,m = m h Y l,m and L Y l,m = ll + h Y l,m b L x = L x L x = ll + m h ll + m = h = L y and ll + m L x L y = h m h as l m = ll + mm + = ll + m m Problem 7 Given Angular Momentum State ψθ, φ = Y, + Y, + Y,0 + Y, + Y, a Since L z Y l,m = m h Y l,m m L z = m h - h - h 0 h h P L z = m h b Immediately after the measurement of L z = h, the state collapse to ψθ, φ = Y,. Using Problem 6 result, we have L x = 0 = L y and L x = L y = + h = h. Hence L x = L y = h = h = L x L y = h h = h m =. The uncertainty relationship is satisfied. Problem 8 Given Angular Momentum State ψθ, φ = 8 Y, + 8 Y,0 + AY, a Normalization: = A = A = Recall that the spherical harmonics Y l,m are orthonormal functions.

7 b ˆL z = h = h 8 Since all the spherical harmonics in ψθ, φ has l =, ˆL = h. c ˆL z ψ = 8 ˆL z Y, + 8 ˆL z Y,0 + ˆLz Y, = 8 Y, ˆLz Y, h Φ ˆL z ψ = 8 Y, + Y,0 + 5 Y, 8 Y, ˆLz Y, h = 8 8 h = h 5 Problem 9 Particle in Circular Motion a Ĥ = ˆL mr = mvr mr = mv Setting the potential energy of the particle to zero in the xy plane b Time independent Schrödinger equation: ĤΨ = EΨ = ˆL where ˆL = h sin θ Ψ = EΨ = h mr mr sin θ + θ θ sin θ φ. sin θ sin θ + θ θ sin θ For circular motion in the xy plane, the polar angle θ = 90 0 is constant. Thus φ Ψ = EΨ h Ψ mr φ = EΨ = Ψ φ + mr E h Ψ = 0 c The above Schrödinger equation is a second order linear homogeneous differential equation. The auxiliary equation is λ + k = 0 where k = mr E h = IE h and I = mr is the moment of inertia of the system. Solving the auxiliary equation we have λ = ±ki which give Ψφ = Ae ikφ + Be ikφ for some A, B C. Now, the particle is in circular motion and thus the wave function must be periodic with period π which imply that Ψφ + π = Ψφ This is the boundary condition for uniform circular motion. As a result e ikφ = e ikφ+π = e ikπ+kφ = k = n Z i.e. k is an integer. Discretization of the energy arises from the boundary conditions. Now n = k = mr E n h = E n = n h mr = n h I where n {0, ±, ±,...} and I = mr d Zero-point energy ZPE or ground state energy is the lowest possible energy that a quantum mechanical system may have. In this question, the ground state energy or ZPE is E 0 = 0. Hence there is no zeropoint energy. This is because the Hamiltonian involve only the azimuthal angle φ. It does not contain any canonical conjugate pair of variables.