Tutorials to Quantum Mechanics II, SS Problem Sheet 5 - Solutions

Transcription

1 Lehrstuhl für Kosmologie, Professor Dr. V. Mukhanov Tutorials to Quantum Mechanics II, SS 29 Problem Sheet 5 - Solutions Exercise Calculate ˆq ˆp 2ˆq ˆp ˆq 2 ˆp. 2. Write the Hamiltonian Ĥ ˆq ˆp2 ˆp in canonical form. 1. ˆqˆp 2ˆq ˆpˆq 2 ˆp (i + ˆpˆq)ˆpˆq ˆpˆq(i + ˆpˆq) 2. Canonical Form: ˆp to the left, ˆq to the right. ˆqˆp 2ˆq ˆp 2ˆq 2 ˆp 2ˆq 2 + ˆqˆp 2ˆq ˆp 2ˆq 2 ˆqˆp 2ˆq (ˆp 2ˆq ˆpˆqˆp + ˆpˆqˆp ˆqˆp 2) ˆq (ˆp[ˆp, ˆq] + [ˆp, ˆq]ˆp) ˆq (ˆp[ˆq, ˆp] + [ˆq, ˆp]ˆp) ˆq i2ˆpˆq ˆqˆp 2ˆq ˆp 2ˆq 2 + 2iˆpˆq Exercise 32 Consider the action S (p q H(p, q)) where p and q are to be taken as independent variables. 1. Derive the Hamiltonian equations of motion from the variation of S. 2. What boundary conditions have to be imposed? 3. Show that if H is not explicitly dependent on time, then dh, H H(p, q). 1

2 1. δs δ (qf,t f ) (qf,t f ) (qf,t f ) (qf,t f ) (qf,t f ) (qf,t f ) [p q H(p, q)] (δp dq dq + pδ H (δp dq + pdδq H [δp dq + d(pδq) [δp dq dp [( dq H dp So we have the Hamiltonian equation of motion 2. The boundary condition we imposed is 3. δp H q δq) δp H q δq) H H δq δp q δq] H H δq δp q δq] + pδq (q f,t f ) )δp (dp + H q )δq] dq H dp H q δq qi δq qf dh(p, q) H dp + H dq q H ( H q ) + H q H Exercise Generalise the Hamiltonian equations of motion to systems with many degrees of freedom p i, q i. 2

3 2. Consider a field theory with the action 1 [ S ( t φ) 2 ( φ) 2] d 3 x, 2 where φ φ( x, t) is a scalar function (scalar field). The generalised coordinates shall be q i (t) φ x (t), p i (t) π x (t), where π x are the canonical momenta. Calculate the Hamiltonian H(φ x, π x ). Derive the Lagrangian and Hamiltonian equation of motion. 1. The action of a system with many degrees of freedom is S [ p i q i H(p i, q i )] i So we have δs δ i (qf,t f ) [ i (qf,t f ) p i q i H(p i, q i )] (δp i dq i + p iδ dq i H i δp i H q i δq i ) For each term in the sum, the following derivation is the same to the system with one degree of freedom. The Hamiltonian equation of motion to systems with many degrees of freedom is dq i H i dp i H q i 2. The Lagrangian of the system is L Ld 3 x The canonical momenta is defined as The Hamiltonian is H 1 2 [( tφ) 2 ( φ) 2 ]d 3 x π L φ φ [π φ L]d 3 x [ 1 2 π ( φ)2 ]d 3 x Hd 3 x H H(φ, π, i φ) 3

4 To derive the Lagrangian equation of motion we use the least action principle δs δ L(φ, t φ, x φ)d 3 x δ L(φ, µ φ)d 4 x [ L φ δφ + L ( µ φ) δ( µφ)]d 4 x { L φ δφ + L µ[ ( µ φ) δφ] L µ[ ( µ φ) ]δφ}d4 x The last term can be turned into a surface integral over the boundary of the 4D spacetime region of integration. We restrict that δφ on the boundary, then we get the Lagrange equation of motion or substitute the Lagrange of the system We get L µ [ ( µ L) ] L φ L t [ ( t L) ] + [ L ( L) ] L φ L 1 2 [( tφ) 2 ( φ) 2 ] 2 t φ 2 φ The Hamiltonian equations are obtained in a similar way. δs δ π φ H(φ, π, i φ)d 4 x d 4 x δπ φ + πδ φ H H H δφ δπ φ π i φ δ iφ ( d 4 x δπ φ H ) ( δφ π + H ) π φ H i i φ where we have performed similiar transformation as for the Lagrangian equations. The Hamiltonian equations of motion then are Remark: δ δψ dφ δh δπ π dπ δh δφ is the functional derivative. It is short for δ δψ ψ x i ψ,i 4

5 Exercise 34 Prove that [ˆp, ˆp nˆq m ] i ˆq (ˆpnˆq m ). [ˆp, ˆp nˆq m ] ˆp n (ˆpˆq m ˆq m ˆp) ˆp n [ˆp, ˆq m ] Ex.21 imˆp nˆq m 1 i ˆq (ˆpnˆq m ) Exercise 35 Show that for analytic operators Â(ˆp, ˆq) the following holds: dâ i [Â, Ĥ] Remark: Generally we have dâ  ˆq ˆq +  ˆp ˆp dâ(p(t), q(t)) ψ ψ H ψ(t) t  ψ(t) S 1 ψ(t) Ĥ ψ(t) i i ψ [Â, Ĥ] ψ H ψ(t) ψ(t)  t S + 1 ψ(t) ÂĤ ψ(t) S i + Therefore we have dâ i [Â, Ĥ] We can also prove it directly from the definition of the operator in the Heisenberg picture. Then Exercise 36  (H) (t) Û (t)â(s) Û(t). Â(H) d (t) Û Â (S) Û + t Û (S) Û Â t Û i ĤÛÛ Â (S) Û Û i  (S) ÛÛ ĤÛ i [Â(H) Ĥ], Does the hermiticity of Ĥ follow from the unitarity of the time evolution operator? S 5

6 Û (e iĥt/ ) e iĥ t/! e iĥt/ Ĥ Ĥ Exercise 37 Interaction picture Show that φ(t) satisfies the following equation: i φ(t) I t ˆV (t) φ(t) I Where ˆV (t) Û ˆV (ˆp, ˆq )Û ˆV (ˆp(t), ˆq(t)), and ˆq(t) Û ˆq Û, ˆp(t) Û ˆp Û. The time evolution operator Û is defined according to the free part Ĥ of the Hamiltonian. Find the solution to the differential equation by formal integration and subsequent iteration. In the interaction picture the wave function is defined as φ(t) I e iĥt/ φ(t) S Therefore i t φ(t) I i t (eiĥt/ φ(t) S ) Ĥe iĥt/ φ(t) S + e iĥt/ (Ĥ + ˆV ) φ(t) S e iĥt/ ˆV e iĥt/ φ(t) I ˆV (t) φ(t) I Introduce the time evolution operator and Û(t) e iĥt/ e iĥt/ tû(t) i eiĥt/ (Ĥ Ĥ)e iĥt/ i eiĥt/ V (e iĥt/ e iĥt/ )e iĥt/ i ˆV (t)û(t) Then φ(t) I Û(t) ψ() I 6

7 Integrate the equation for Û we have So Û(t) I i I i t t 1 ˆV (t1 )U(t 1 ) 1 ˆV (t1 ) + ( ) i 2 t t1 1 2 ˆV (t1 ) ˆV (t 2 ) + φ(t) I Û(t) φ() ( ) i n t 2 1 n tn 1 2 n ˆV (t1 ) ˆV (t 2 ) ˆV (t n ) φ() I 7