Physics 550. Problem Set 6: Kinematics and Dynamics

Transcription

1 Physics 550 Problem Set 6: Kinematics and Dynamics Name: Instructions / Notes / Suggestions: Each problem is worth five points. In order to receive credit, you must show your work. Circle your final answer. Unless otherwise specified, answers should be given in terms of the variables in the problem and/or physical constants and/or cartesian unit vectors (ê x, ê x, ê z ) or radial unit vector ˆr. Staple your work.

2 Problem #1: The Kronecker product of two matrices: C = A B is defined in terms of their matrix elements: C (ik)(jl) = A ij B kl Show that: Tr (C) = Tr (A) Tr (B) The trace of an n n square matrix is defined to be the sum of the elements on its main diagonal. For C: Tr (C) = ik = ik C (ik)(jl) δ (ik)(jl) jl C (ik)(ik) = ik = i A ii B kk A ii k B kk = Tr (A) Tr (B)

3 Problem #2: The Hamiltonian H of a two-component, non-interacting system may be written: H = H 1 I + I H 2 where H 1 and H 2 are the Hamiltonians of the two components, and I is the identity operator. Show that the corresponding time evolution operator U(t) has the form: U(t) = U 1 (t) U 2 (t) where: U 1 (t) = exp( ith 1 / ) U 2 (t) = exp( ith 2 / ) The time-evolution operator for the two-component system is: U(t) = exp( ith/ ) = exp( it(h 1 I + I H 2 )/ ) = exp( it(h 1 H 2 )/ ) = exp( ith 1 / ) exp( ith 2 / ) = U 1 (t) U 2 (t) where is the Kronecker sum: A B = A I m + I n B (where A is n n and B is m m), and the following formula for the matrix exponential: exp(a B) = exp(a) exp(b) has been used.

4 Problem #3: Show that a pure state cannot evolve into a non-pure state, and vice versa. Let ρ be the state operator. The time-evolution of ρ (e.g., in the Schrödinger picture) can be written: ρ(t) = Uρ(t 0 )U where U := U(t, t 0 ) = exp ( i(t t 0 )H/ ). The time-evolution of ρ 2 is therefore: ρ 2 (t) = ( Uρ(t 0 )U ) ( Uρ(t 0 )U ) = Uρ(t 0 ) ( U U ) ρ(t 0 )U = Uρ(t 0 )ρ(t 0 )U = Uρ 2 (t 0 )U The trace of ρ 2 (t) is: Tr ( ρ 2 (t) ) = Tr ( Uρ 2 (t 0 )U ) = Tr ( ρ 2 (t 0 )U U ) = Tr ( ρ 2 (t 0 ) ) Therefore, a pure state will remain a pure state (and similar for a non-pure state).

5 Problem #4: Consider an arbitrary physical system. Denote the Hamiltonian of this system by H 0 (t), and the corresponding evolution operator by U 0 (t, t 0 ): i t U(t, t 0) = H 0 (t)u 0 (t, t 0 ) U 0 (t 0, t 0 ) = I Now suppose that the system is perturbed in such a way that its Hamiltonian becomes: H(t) = H 0 (t) + H 1 (t) The so-called interaction picture (I) may be obtained by the following transformation of states and dynamical variables of the Schrödinger picture (S): ψ I (t) = U 0(t, t 0 ) ψ S (t) ρ I (t) = U 0(t, t 0 )ρ S (t)u 0 (t, t 0 ) R I (t) = U 0(t, t 0 )R S U 0 (t, t 0 ) (a) Find the equations of motion for the state vector ψ I (t) and state operator ρ I (t). (b) Explain why (qualitatively), when H 1 (t) is much smaller than H 0 (t), the motion of ψ I (t) is much slower than that of ψ S (t). (c) Determine the time dependence of the average of the observable represented by the operator R. Compare this to that in the Schrödinger and Heisenberg pictures.

6 Problem #5: The unitary operator: U(v) = exp(iv G) describes the instantaneous (t = 0) effect of a transformation to a frame of reference moving at the velocity v with respect to the original reference frame. The effect of U(v) on the position q and momentum p operators are: UqU 1 = q UpU 1 = m (v vi) where m is the mass (of the quantum state). Find an operator G(t) such that the unitary operator: U(v, t) = exp(iv G(t)) will yield the full Galilei transformation: UqU 1 = q tvi UpU 1 = p mvi Consider an important lemma to the Baker Campbell Hausdorff formula: e A Be A = B + [A, B] +... Applying this to the left-hand side of the above expressions gives: UqU 1 = q + [iv G(t), q] +... UpU 1 = p + [iv G(t), p] +...

7 Noting that G(t) is the generator of an infinitesimal transformation v, all high-order commutator brackets above may be neglected. Therefore, we seek a Hermitian operator G(t) such that: [iv G(t), q] = tv I [iv G(t), p] = mv I Considering the canonical commutation relations, one can see that: G(t) = mq tp (where = 1 has been taken for simplicity) would satisfy the above relations.