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T2: Quantenmechanik Sheet 3, Course 17019 Professor: H. Ruhl, Exercises: N. Elkina, S. Rykovanov Discussion of solutions: Oct. 25-29, 2010 Problem 1: Projection (Sakurai 1.9) � · n̂; +� such that Construct |S � � � �� � � · n̂��S � · n̂; + = h̄ ��S � · n̂; + S 2 (1) where n̂ is the unit vector with polar angle β and azimuthal angle α with respect to the z axis. Solution 1: Cartesian representation for n̂ is given by (sin β cos α, sin β sin α, cos β) (2) � · n̂ as so we can constract the matrix for the operator S � � � � h̄ h̄ 0 1 0 � S · n̂ = (σx nx + σy ny + σz nz ) = sin β cos α + sin β sin α 1 0 i 2 2 −i 0 � h̄ cos β − λ 2 � + cos β � h̄ cos β = iα e sin β 2 � � � · n̂ − λI = 0, so The eigenvalues of this matrix can be found by solving det S 0= � 1 0 0 −1 �� e−iα sin β − cos β �� � � �2 � �2 h̄ h̄ h̄ h̄ − cos β − λ − sin2 β = − + λ2 → λ = ± 2 2 2 2 = (3) � (4) If we are intersted in the normalized eigenket corresponding to the positive eigenvalue, we can write � � �� �S · n̂; + = A|+� + B|−� Thus we have h̄ 2 � cos β eiα sin β eiα sin β − cos β �� A B � h̄ = 2 � A B � (5) Using first equation in this system, one can get: (cos β − 1)A + (e−iα sin β)B = 0 (6) To fix the actual values of A and B we can use the normalization condition |A|2 + |B|2 = 1 (7) so, |B| = |A| 1 − cos β iα e sin β |A| = cos β 2 (8) sin β = 2 sin β β cos 2 2 β 2 (9) 1 − cos β β A = eiα sin sin β 2 (10) 1 − cos β = 2 sin2 We choose the phase of A to be real and positive, i.e. A = cos β 2 B = eiα The final result is � · n̂; +� = cos |S β β |+� + eiα sin |−� 2 2 (11) Problem 2: Hamiltonian, (Sakurai 1.10) Hamiltonian operator for a two-state system is given by H = a(|1��1| − |2��2| + |1��2| + |2��1|) (12) here a is the number with dimension of energy. Find the energy eigenvalues and the corresponding eigenkets (as linear combination of |1� and |2�) Solution 2: The matrix of H is � � a a H= (13) a −a the sum of the eigenvalues λ1 , λ2 of H equals its trace, and their product equals its determinant, i.e. √ √ λ1 + λ2 = 0, λ1 λ2 = −2a2 → λ1 = a 2 λ2 = −a 2 √ For λ1 = a 2 we can write the eigenket as c1 |1� + c2 |2�, then √ � �� � 1− 2 1√ c1 a =0 c2 1 −1 − 2 (14) (15) so, we have � √ (1 − 2)c1 + c2 = 0 |c1 |2 + |c2 |2 = 1 (16) Choosing c1 to be real and positive, we find for the normalized eigenket � c1 c2 � =� 1 √ 4−2 2 � √ 1 2−1 � (17) √ for λ2 = − 2 we write the eigenket as d1 |1� + d2 |2�, then � �� � √ d1 1+ 2 1 a (18) =0 d2 1 −1 + sqrt2 √ we we have (1 + 2)d1 + d2 = 0, plus normalization condition: |d1 |2 + |d2 |2 = 1. Choosing d1 to be real and positive, we find for the normalized eigenket � � � � 1 1 d1 √ (19) =� √ d2 − 2−1 4+2 2 Problem 3: Dispersion of operator, (Sakurai 1.12) � · n̂ with eigenvalue h̄/2, where n̂ is a unit vector lying A spin 1/2 system is known to be in an eigenstate of S in the xz-plane that makes an angle γ with the positive z-axis 1. Suppose Sx is measured. What is the probability of getting +h̄/2? 2. Evaluate the dispersion in Sx that is, (20) �(Sx − �Sx �)2 � Check special cases γ = 0, π/2, π Solution 3: Since 1 1 |Sx ; +� = √ |+� + √ |−� 2 2 So the probability to get +h̄/2 when Sz is measured is given by � � �� � �2 � � 1 � �2 1 γ γ � � � � · n̂; +�� = � √ �+| + √ �−| cos |+� + sin |−� �� = ��Sx ; +|S � � 2 2 2 2 � �2 � 1 1 γ �� 1 γ 1 γ γ γ 1 1 1 � + √ sin � = cos2 + sin2 + cos sin = + sin γ = (1 + sin γ) �√ γ � 2 cos 2 2� 2 2 2 2 2 2 2 2 2 2 • γ = π/2, |Sx ; +�: (22) 1 2 (23) |�Sx ; +|Sx ; +�|2 = 1 (24) 1 2 (25) |�Sx ; +|Sz ; +�|2 = • γ = π/2, |Sx ; +� (21) • γ = π, |Sz ; −� |�Sx ; +|Sz ; −|2 = (2.) Using �(Sx − �Sx �)2 � = �Sx2 � − (�Sx �)2 using Sx = Sx2 = h̄ (|+��−| + |−��+|) 2 h̄2 (|+��−| + |−��+|)(|+��−| + |−��+|) 4 h̄2 h̄2 Sx2 = (|+��+| + |−��−|) = 4 4 � � h̄ � � γ γ γ γ �Sx � = cos �+| + sin �−| (|+��−| + |−��+|) cos |+� + sin |−� = 2 2 2 2 2 h̄ γ γ h̄ γ γ h̄ cos sin + sin cos = sin γ 2 2 2 2 2 2 2 h̄ 2 (�Sx �) = sin2 γ 4 � � h̄2 � � h̄2 � � h̄2 γ γ γ γ γ γ �Sx2 � = cos �+| + sin �−| cos |+� + sin |−� = cos2 + sin2 = 2 2 4 2 2 4 2 2 4 (26) (27) (28) (29) (30) (31) (32) Substituting in (20) we will have �(Sx − �Sx �)2 � = h̄2 h̄2 (1 − sin2 γ) = cos2 γ 4 4 (33) • h̄2 4 (34) �(∆Sx )2 �γ=π/2;|Sx ;+� = 0 (35) h̄2 4 (36) �(∆Sx )2 �γ=0;|Sz ;+� = • • �(∆Sx )2 �γ=0;|Sz ;−� = Problem 4: Uncertainty principle formomentum and coordinate Using uncertaintly principle for momentum and coordinate, find minimal mean valuer for mean of Hamiltonian of harmonic oscillator p̂2 mω 2 x̂2 + 2m 2 (37) � ˆ 2 � �(∆p) ˆ 2 � ≥ h̄ �(∆x) 2 (38) Ĥ(p, x) = Solution 4: The unceraintly relattions is given by � ˆ = x − �x� and ∆x ˆ = x − �x�. From physical point of view, the mean value for momentum and where ∆x coordinate are equal zero. In this case, one can assume � � ˆ 2 � = �x̂2 � = a �(∆x) (39) � � ˆ 2 � = �p̂2x � = b �(∆p) (40) then, we obtain �Ĥ� = b2 mω 2 2 + a 2m 2 (41) Then, 1 h̄2 mω 2 2 h̄ω �Ĥ� ≥ + a = 2m 4a2 2 4 � h̄ 2mωa2 + 2mωa2 h̄ � (42) this function has extremum (min) at a2 = h̄/2mω. Thus min�Ĥ� = h̄ω 2 (43) Problem 5: Density matrix Let ψ� = � cos θ eiφ sin θ � θ, φ ∈ R (44) Does ρ = |ψ��ψ| define a density matrix? Solution 5: We find the 2 × 2 matrix for ρ ρ= � cos2 θ iφ e cos θ sin θ (45) e−iφ cos θ sin θ sin2 θ � (46) Thus ρ = ρ† T rρ = cos2 θ + sinθ = 1 (47) (48) and �x|ρ|x� = |x1 |2 cos2 θ + x1 x2 e−iφ cos θ sin θ + x1 x2 eiφ cos θ sin θ + |x2 |2 sin2 θ (49) ≥ |x1 | cos θ + 2Re(x1 x2 e ) cos θ sin θ + |x2 | sin θ 2 2 iφ 2 2 ≥ (|x1 | cos θ + |x2 | sin θ)2 ≥ 0 for all x = (x1 , x2 )T ∈ C 2 . Thus ρ defines a density matrix. Problem 6: Mixed states A mixed state is a statistical misture of pure states, i.e., the state is described by pairs of probabilities and pure states. Given a mixture {(p1 , |ψ1 �), ..., (Pn , |ψn �)} we define its density matrix to be the positive hermitian matrix ρ= n � j=1 pj |ψj ��ψj | (50) where the pure states |ψ� are normalized (i.e., �ψj |ψj � = 1), and pj ≥ 0 for j = 1, 2, ..., n with n � pj = 1 (51) j=1 1. Find the probability that measurement in the orthonormal basis {|k1 �, ..., |kn �} will yield |kj � 2. Find the density matrix ρU when the mixture is transformed according to the unitary matrix U Solution 6: (1). From the probability distribution of states in the mixture we have for the probability P (kj ) of measuring the state |kj �: P (kj ) = n � l=1 pl |�kj |ψj �|2 = n � l=1 pl �kj |ψl ��ψl |kj � = �kj |ρ|kj � (52) (2). After applying the transform U to the states in the mixture we have the new mixture {(p1 , U |ψ�), ..., (pn , U |ψn �)} with density matrix ρU = n � j=1 pj U |ψj ��ψj |U ∗ = U n � j=1 pj |ψj ��ψj | U ∗ = U ρU ∗ (53)