University:
Harvard University
Course:
PHYSICS 19 | Introduction to Theoretical Physics
Academic year:

2019
Views:

346

Pages:

18
Author:

basteishx23y

36 2 Quantum Dynamics 2.1 Consider the spin-procession problem discussed in section 2.1 in Jackson. It can also be solved in the Heisenberg picture. Using the Hamiltonian H = - ((eB)/(mc)) * S_{z} = omega*S_{z} write the Heisenberg equations of motion for the time-dependent operators S_{x}(t) S_{5}(t) and S_{z}(t) Solve them to obtain S x, y, z as func-tions of time Let us first prove the following [A S ,B S ]=C S Rightarrow [A_{H}, B_{H}] = C_{H} . (2.1) Indeed we have [A_{H}, B_{H}] =[ mathcal U ^ dagger A s mathcal U , mathcal U ^ dagger B S mathcal U ]= mathcal U ^ dagger A s B s mathcal U - mathcal U ^ dagger B 8 A s mathcal U = mathcal U ^ dagger [A_{S}, B_{S}] * \mathcal{U} = mathcal U ^ dagger C S mathcal U =C it (2.2) The Heisenberg equation of motion gives dS x dt = 1 i hbar [S x ,H]= 1 i hbar [S_{x}, omega*S_{z}] ^ (S - 1, 4 * 20) omega i hbar (-i hbar S y )=- omega S g , (2.3) (d*S_{y})/(dt) = 1 i hbar [S y ,H]= 1 i hbar [S y , omega S z ]^ (S- 1,4,20) omega i hbar (i hbar S x )= omega S x , (2.4) (d*S_{z})/(dt) = 1 i hbar [S z ,H]= 1 i hbar [S z , omega S z ]^ \ S- 1,4,206 = ^ \ S- 1,4,206 0 Rightarrow S z =, constant. (2.5) Differentiating once more eqs. (2.3) and (2.4) we get (d ^ 2 * S_{x})/(d * t ^ 2) = - omega * (d*S_{y})/(dt) = (2.4) = - omega^ 2 S x Rightarrow S x (t)=A cos omega t+B sin omega t Rightarrow S x (0)=A (d ^ 2 * S_{y})/(d * t ^ 2) = omega * (d*S_{x})/(dt) =- omega^ 2 S y Rightarrow S y (t)=C cos omega t+D sin omega t Rightarrow S y (0)=C. But on the other hand (d*S_{x})/(dt) =- omega S y Rightarrow - A * omega * sin omega*t + B*omega * cos omega*t =-C omega cos omega t-D omega sin omega t Rightarrow A = D C = - B (2.6) 2. QUANTUM DYNAMICS 37 So, finally S_{x}(t) = S_{x}(0) * cos omega*t - S_{y}(0) * sin omega*t (2.7) S_{y}(t) = S_{y}(0) * cos omega*t + S_{z}(0) * sin omega*t S_{z}(t) = S_{z}(0) (2.8) (2.9) 2.2 Let (1) be the coordinate operator for a free particle in one dimension in the Heisenberg picture. Evaluate [x(t), x(0)] The Hamiltonian for a free particle in one dimension is given by H = (p ^ 2)/(2m) (2.10) This means that the Heisenberg equations of motion for the operators z and p will be partial p(t) partial t = 1 i hbar [p(t),H(t)]= 1 i hbar [p(t), (p ^ 2 * (t))/(2m)] =0 Rightarrow p(l) = p(0) partial x(t) partial t = 1 i hbar [x,H]= 1 i hbar [x(t), (p ^ 2 * (t))/(2m)] = 1 2m hbar 2p(t)i hbar= p(t) m (2,11) m p(0) m Rightarrow x(t) = t/m * p(0) + x(0) Thus finally [x(t), x(0)] = [t/m * p(0) + x(0), x(0)] = t/m * [p(0), x(0)] =- i hbar t m . (2.13) (2.11) (2.12) 2.3 Consider a particle in three dimensions whose Hamiltonian is given by H= vec p ^ 2 2m +V( vec x ). 2mw (a²+aaaaa) sin² wt + 2mw ta² cos 2wt + 2mw 2m 2mw -(aa+ala)+ 2m. Yat-a) sin 2ct. in + -a) sin 2wt a cos 2wt which means that = = 1 2mw 2 cos 2ot + at cos Zut+i(at-a) sin 2w x [aa + ata + +ata + a² cos 201 4 [(0)aa+10)+(1/aa/1) + (1aa1)] = 1+2+1]= Ame (2.49) So (()) (244) 2mw 2πω 2mw sinwt. (2.51) 2.7 A coherent state of a one-dimensional simple harmonic oscil-lator is defined to be an eigenstate of the (non-Hermitian) annihi-lation operator : αλ) = λιλ), where A is, in general, a complex number. (a) Prove that (A) = -1/10) is a normalized coherent state. (b) Prove the minimum uncertainty relation for such a state. (2.50