Angular Momentum - set 1

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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 = 5 Y, + 5 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 ] = iz ˆx, ˆL z ] = x, xp y yp x ] = x, yp x ] = yx, p x ] = iy 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, x]p z = ip 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, x]p y = ip y Problem Jacobi identity Recall the cyclic relationship ˆL x, ˆL y ] = iˆl z ˆL y, ˆL z ] = iˆl x ˆL z, ˆL x ] = iˆl 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 ˆL x, ˆL x ] + ˆL y, ˆL y ] + ˆL z, ˆL z ] = 0 Observe that Jacobi identity is also a cyclic relationship. Problem Angular Momentum in Spherical Coordinates ˆL z, cos φ]fφ = i ] φ, cos φ fφ = i cos φ cos φ fφ φ φ = i cos φfφ] cos φf = i sin φfφ = ˆL z, cos φ] = i sin φ φ φ ˆL z, sinφ]fφ = i ] φ, sinφ fφ = i φ = i = ˆL z, sinφ] = i cosφ = isin φ cos φ fφ sinφ sinφ φ sinφfφ] sinφf φ φ = i 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 mψ = ˆL z ψ = i ψ ψ φ = mφ = iψ ψ = m φ = i ψ = ψ L z φ = Ae imφ where A C. Normalising ψ Lz we have π = ψ Lz ψ Lz = ψl z ψ Lz dφ = A 0 π 0 dφ = π A = A = π

5 taking A R without loss of generality. Thus the normalise wavefunction is ψ Lz φ = eimφ π where m Z. This wave function describes the motion of a particle moving in uniform circular motion in the xy-plane. Observe that since m Z, the angular momentum in the z direction is quantised. c To find the momentum eigenket in position basis, kψ = pψ = ˆP z ψ = i ψ z = ikz = ψ ψ = ik z = ψ ψ = ln ψ = ikz + C = ψ p z = Ae ikz = Ae ipz/ where A, C C are some constants of integration and p = k R are the momentum eigenvalues. Attempting to normalising this wavefunction gives = ψ p ψ p = ψ pψ p dz = A dz Hence, the linear momentum eigenket is not normalisable in position space. Suppose we chooose A = π and consider p p = ψ p ψ p = ψ p ψ pdz = A e i p p z dz = A e ik k z dz = A p p = π A δk k = π A δ = π A δ p p = δ p p = δ p p = In the above calculation, we have use the Dirac-delta function π { e ik k z dz = δk k if k = k = 0 if k k e ik k z dz { if p = p 0 if p p and its property δαx = δx α of the Kronecker-delta function where α is some constant. Dirac-delta function is the infinite version δ ij = { if i = j 0 if i j Thus the wave function ψ p z = eipz/ π describes the motion of a freely moving particle with uniform linear momentum p along z-axis in position basis. Consider a linear momentum eigenket p with eigenvalue p, i.e. ˆp p = p p It s position space representation is given by p = ψ p z = eip z/ π. To obtain its momentum space representation, we perform a fourier transform p = φ p p = π ψ p ze i p z dz = π e i p z e i p z dz = π e i p p = π π δ z dz p p = δp p for all p R. Attempting to normalise the momentum eigenket p in momentum space we have = p p = φ p φ p dp = δp pδp pdp = fpδp pdp = fp = δ0 5

6 where fp = δp p. Hence the linear momentum eigenket is not normalisable in momentum space also and we conclude that The property of a state is independent of its representation. The table below gives a summary of position, momentum operators and its eigenkets representation in position and momentum space. Position, Momentum Operators and its Eigenkets in different representation Representation ˆX X ˆP P Position space x R x Ψ x x = δx x i x Ψ p x = eip x/ π Momentum space p R i p Φ x p = e ipx / π p Φ p p = δp p Observe that i In position space ˆX X = x X = X is an eigenket of operator ˆX with eigenvalue x ˆP P = i x Ψ e p x] ip x/ = i = i x π = P is an eigenket of operator ˆP with eigenvalue p. ii In momentum space ˆP P = p P = P is an eigenket of operator ˆP with eigenvalue p ˆX X = i p Φ x p] = i p e ipx / π = i = X is an eigenket of operator ˆX with eigenvalue x. ip ix e ip x/ π = p Ψ p x = p P e ipx / π = x Φ x p = x X Notice that although the operators and eigenkets representation are different in different spaces, its eigenvalues possible observed values are the same in different spaces. Eigenvalues are independent of the representation. Observe that statement is a special case of statement. In general, eigenvalues being the possible observed values should be independent of the frame of reference which they are measured. 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. 6

7 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 = 5 Y, + 5 Y, 5 5 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 = = 8π 5 π 5 + c L z = = 5 = 6 π 5 5 Since l = for both Y, and Y,, L = + = 6. 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 Therefore L x = 0 = L y ] L + + L ] L + L = 0 = 0 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 + L L + + L = L +L + L L + 7

8 L+ L L y = = ] L i + L + L L L + + L = L +L + L L + So L x = L y = L +L + L L +. Recall that L = L x + L y + L z = L = L x + L y + L z = ll + = L x + m = L x = L y = ll + m ] since L z Y l,m = m Y l,m and L Y l,m = ll + Y l,m b L x = L x L x = ll + m ] ll + m = = L y and ll + m L x L y = ] m 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 Y l,m m L z = m P L z = m b Immediately after the measurement of L z =, the state collapse to ψθ, φ = Y,. Using Problem 6 result, we have L x = 0 = L y and L x = L y = + ] =. Hence L x = L y = = = L x L y = = 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. b ˆL z = ] = 8 Since all the spherical harmonics in ψθ, φ has l =, ˆL =. 8

9 c ˆL z ψ = Φ ˆL z ψ = ] 8 ˆL z Y, + 8 ˆL z Y,0 + ˆLz Y, = 8 Y, ˆLz Y, ] ] 8 5 Y, + 5 Y,0 + 5 Y, 8 Y, ˆLz Y, = = 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 = sin θ Ψ = EΨ = 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 Ψ mr φ = EΨ = Ψ φ + mr E 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 = IE 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 = E n = n mr = n 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. 9