Quantum Mechanics II - Solutions to Homework 2 | PHYS 581, Assignments of Quantum Mechanics

Download Quantum Mechanics II - Solutions to Homework 2 | PHYS 581 and more Assignments Quantum Mechanics in PDF only on Docsity! Solutions to Homework 2 Phys581, Spring 2007 1 Charged Particle on a Plane 1) The canonical momentum: !p = ! ṙL = m!̇r + e c !A(!r) 2) H = !̇r · !p" 1 2 m!̇r 2 " e c !̇e · !A(!r) = !̇r · ! !p" e c !A(!r) " " 1 2 m!̇r 2 = 1 2 m!̇r 2 = 1 2m ! !p" e c !A(!e) "2 3) We expand the Hamiltonian: H = 1 2m !p2 + e2 2mc2 !A2(!r)" e mc !p · !A(!r) In the quantum mechanical Hamiltonian the only ambiguity is the term !p · !A(!r) # 12 !̂p · !A(!r) + 1 2 !A(!r) · !̂p. But because ! · !A(!r) = 0 !̂p · !A = !A · !̂p (to understand that better assume that each side is applied in some arbitrary function "(!r)). The quantum Hamiltonian is: H = 1 2m !̂p 2 + e2 2mc2 !A2(!r)" e mc !A(!r) · !̂p = p̂21 + p̂22 2m + e2B2 8mc2 # x̂21 + x̂ 2 2 $ + eB 2mc (x̂2p̂1 " x̂1p̂2) where pi = "i!#i. 4) We are now going to evaluate the path integral by summing over all histories that start and end at the same point. The boundary conditions are thus: !r(ti) = !r(tf ) = !r0 The path integral is: % !rf , tf | !ri, ti & = ' D!pD!reiS/! where S = ' tf ti dt ! !p · !̇r "H(!p,!r) " 1 Solutions to Homework 2 Phys581, Spring 2007 is the classical action for the problem. At first we integrate out the momentum !p which will give us an expression involving !r and !̇r. The relevant integral is: ' ! "! dp1dp2 (2$!)2 e i ! ! “ "p·"̇r" 12m ("p" ec "A) 2 ” = ' ! "! dp1dp2 (2$!)2 e i ! !(("p+ ec "A)·"̇r" 12m "p2) = e i ! ! e c "̇r· "A m! 2$% e i ! m"̇r+ i ! ! e c "̇r· "A 2 where % is the time step. In the second equality we perform a shift !p# !p + ec !A and in the third equality we just use Gauss’ integral. The path integral then becomes: % !rf , tf | !ri, ti & = ' D!re i ! R tf ti dtL("̇r) where L(!̇r) is the classical Lagrangian. 5) In the particular case m = 0 the Lagrangian in the exponent will become just L = ec !A · !̇r and the exponent will become: ' tf ti dtL(!̇r) = ' tf ti dt e c !A · !̇r = e c ( !A · d!r The integral is along a closed path. The path integral will then become: % !rf , tf | !ri, ti & = ' D!re 2!i !q H "A·d"r where the quantum of flux !q = hce has been introduced. This integral in the exponent is gauge invariant because the endpoints of the path are the same. According to Stoke’s theorem, the quantity ) !A · d!r is equal to the magnetic flux ! going through the closed path if this is a positively oriented graph (the positive direction is commensurate to the direction of the magnetic field via the right hand rule). Also it is equal to minus the magnetic flux " # BL2 " ! $ = !"BL2 through the area outside the path, which is now viewed as negatively oriented. This means that the term BL2 should not contribute which is possible if e "2#i BL2!q = 1$ BL2 = n!q That is in order for the ambiguity to be removed the total flux throught the plane has to be a integer multiple of the flux quantum. 2 Path Integral for the three-dimensional harmonic oscillator The 3D harmonic oscillator is nothing but three independent 1D harmonic oscillators. The Hamiltonian can be broken into three parts: H(!p,!r) = H1(p1, x1) + H2(p2, x2) + H3(p3, x3) Hi(pi, ri) = p2i 2m + 1 2 m&2x2i 2