Gauge transformation in quantum mechanics; Aharonov-Bohm effct Masatsugu Sei Suzuki Department of Physics, SUNY at Binghamton (Date: April 29, 2015) ___________________________________________________________ David Joseph Bohm (20 December 1917 – 27 October 1992) was an American-born British quantum physicist who made contributions in the fields of theoretical physics, philosophy and neuropsychology, and to the Manhattan Project. http://en.wikipedia.org/wiki/David_Bohm _____________________________________________________________ Yakir Aharonov (born 1932 in Haifa, Israel) is an Israeli physicist specializing in Quantum Physics. He is a Professor of Theoretical Physics and the James J. Farley Professor of Natural Philosophy at Chapman University in California. He is also a distinguished professor in Perimeter Institute. He also serves as a professor emeritus at Tel Aviv University in Israel. He is president of the Iyar, The Israeli Institute for Advanced Research. His research interests are nonlocal and topological effects in quantum mechanics, quantum field theories and interpretations of quantum mechanics. In 1959, he and David Bohm proposed the Aharonov-Bohm Effect for which he co- received the 1998 Wolf Prize. http://en.wikipedia.org/wiki/Yakir_Aharonov _____________________________________________________________ 1 Gauge transformation Aharonov-Bohm effect Young's double slits experiment Vector potential _____________________________________________________________ 1. Gauge transformations in electromagnetism We start with the Maxwell's equations, E 4 1 B E c t B 0 4 1 E B j c c t where B: magnetic field E: electric field : charge density J: current density The Lorentz force is defined as 1 F q[E (v B)] . c The Lorentz force is expressed in terms of fields E and B, which is invariant under the gauge transformation (gauge independent). The magnetic field B and electric field E can be expressed by B A , 1 A E , c t where A is a vector potential and is a scalar potential. When E and B are given, and A are not uniquely determined. If we have a set of possible values for the vector potential A and the scalar potential , we obtain other potentials A’ and ’ which describes the same electromagnetic field by the gauge transformation, A', A 2 1 ', c t where is an arbitrary function of r. We note that B and E are gauge-invariant; B' A' (A ) A B 1 A' 1 (A ) 1 1 A E' ' ( ) E c t c t c t c t 2 Canonical momentum and mechanical momentum We now consider the Lagrangian which is defined by 1 1 L mv 2 q( v A) , 2 c where m and q are the mass and charge of the particle. The Canonical momentum is defined as L q p mv A . v c The mechanical momentum is given by q π mv p A. c Then the Hamiltonian is obtained as q 1 1 q H p v L (mv A) v L mv 2 q ( p A)2 q . c 2 2m c The Hamiltonian formalism uses A and , and not E and B, directly. The result is that the description of the particle depends on the gauge chosen. 3. Change of the wave function under a gauge transformation (by Mathematica) The Schödinger equation contains the vector potential A. It may imply that the wave function may change as the vector potential A and scalar potential is changed according to the gauge transformation, 1 A' A , ' c t 3 The Schrödinger equation in the gauge (A, ) takes the form 1 e 2 [ ( p A) e] i . 2m c t where , A and depends on r and t. We note that the charge of electron is denoted as –e. The Schrödinger equation in another gauge ( A' ,' ) takes the form 1 e 2 [ ( p A') e'] ' i ' 2m c t where ' , A' and ' depends on r and t. The wave function changes as ie ' exp( ) , c under the gauge transformation. The difference between ' and is only the phase factor. We now give a proof for this by using the Mathematica. We show that 1 e e 2 e [ ( p A ) e ]exp(i) i exp(i) 2m c c c t t is equivalent to 1 e 2 [ ( p A) e] i . 2m c t First we assume that ' exp(i) where is just a parameter to be determined. We will show that e . c ((Mathematica)) We assume that the electron charge is denoted by –e1 in the program, which means e1>0. (i) We need to calculate directly 1 e e 2 e eq1 [ ( p A ) e ]exp(i) i exp(i) 2m c c c t t 4 in the Cartesian co-ordinates. This equation reduces to 1 e 2 eq2 exp(i){[ ( p A) e] i } 2m c t e with . c The calculation without Mathematica is too complicated for me. ((Program)) Here is a Mathematica program which I made. Clear "Global`" ; 1 x, y, z, t ;ex 1, 0, 0 ; 1 ey 0, 1, 0 ;ez 0, 0, 1 ; 1 x, y, z, t D 1, t ; c — 1 x, y, z, t ;p: Grad , x, y, z &; A1 Ax x, y, z, t , Ay x, y, z, t , Az x, y, z, t e1 Grad 1, x, y, z ; 1x : ex. p A1 &; c e1 e1 1y : ey. p A1 &; 1z : ez. p A1 &; c c H1 : 1 1x 1x 1y 1y 1z 1z e1 1 &; 2 m s1 H1 1 — D 1, t FullSimplify; rule1 0 1, 2,3, 4Exp 1, 2, 3, 4 & ; rule2 0, 0 1, 2, 3, 4 & ; s2 s1 . rule1 FullSimplify; 5 s3 s1 . rule2 FullSimplify; eq1 x,y,z,t s2 s3 FullSimplify; eq2 Solve eq1 0, 1 e1 c — x,y,z,t s2 . eq2 FullSimplify s3 1 s3 1 e12 Ax x, y, z, t 2 0 x, y, z, t 2c2 m e12 Ay x, y, z, t 2 0 x, y, z, t 2 2 2 e1 Az x, y, z, t 0 x, y, z, t 2c e1 m x, y, z, t 0 x, y, z, t 2 c2 m — 0 0,0,0,1 x, y, z, t ce1— 0 x, y, z, t Az 0,0,1,0 x, y, z, t 2 ce1— Az x, y, z, t 0 0,0,1,0 x, y, z, t c2 —2 0 0,0,2,0 x, y, z, t ce1— 0 x, y, z, t Ay 0,1,0,0 x, y, z, t 2 ce1— Ay x, y, z, t 0 0,1,0,0 x, y, z, t c2 —2 0 0,2,0,0 x, y, z, t ce1— 0 x, y, z, t Ax 1,0,0,0 x, y, z, t 2 ce1— Ax x, y, z, t 0 1,0,0,0 x, y, z, t c2 —2 0 2,0,0,0 x, y, z, t 4 Analogy from Classical mechanics The Newton’s second law indicates that the position and the velocity take on, at every point, values independent of the gauge. Consequently, r' r and v' v , or π' π , q Since π mv p A, we have c 6 q q p', A' p A c c or q q p' p (A'A) p . c c In the Hamilton formalism, the value at each instant of the dynamical variables describing a given motion depends on the gauge chosen. 5. Gauge invariance in quantum mechanics In quantum mechanics, we describe the states in the old gauge and the new gauge as and ' . The analogue of the relation in the classical mechanics is thus given by the relations between average values. ' rˆ ' rˆ (gauge independent) ' πˆ ' πˆ (gauge independent) or equivalently q q 'pˆ A' ' pˆ A c c (we will discuss the proof later). We now seek a unitary operator Uˆ which enables one to go from to ' . ' Uˆ From the condition ' ' , we have Uˆ Uˆ UˆUˆ 1ˆ From the condition, ' rˆ ' rˆ Uˆ rˆUˆ rˆ or 7 ˆ ˆ U [rˆ,U ] 0 i pˆ Uˆ is independent of pˆ . We also get q q Uˆ ( pˆ A')Uˆ pˆ A c c or q q q Uˆ ( pˆ A )Uˆ pˆ A c c c or q q q Uˆ pˆUˆ Uˆ AUˆ Uˆ Uˆ pˆ A c c c q q q A pˆ A c c c q pˆ c Note that [rˆ,Uˆ ] 0 , and A is a function of rˆ . ((Note)) Here we show that ' πˆ ' πˆ is equivalent to q q ',pˆ A' ' pˆ A c c q q where πˆ pˆ A and A'. A c c ((Proof)) 8 q q q ' pˆ A' ' pˆ Uˆ A'Uˆ c c c q q pˆ A' c c q q pˆ A' c c q pˆ (A') c q pˆ A c where [Uˆ, A'] 0ˆ , since A' is a function of rˆ . 6. Expression of the unitary operator We assert that Uˆ iq iq Uˆ exp[ (rˆ)] , Uˆ exp[ (rˆ)]. c c Then we get Uˆ pˆUˆ Uˆ [ pˆ,Uˆ ] pˆ ˆ ˆ U i U pˆ rˆ ˆ ˆ iq U Ui pˆ c q pˆ c which coincides with the expression described above. ((Note)) We use the notation such that iq (rˆ) iq Uˆ Uˆ Uˆ () . rˆ c rˆ c _______________________________________________________________________ So we get the gauge transformation for the wave function; iq ' exp[ (rˆ)] . c 9 or iq r ' exp[ (r)] r c The phase factor of the wave function depends on the choice of the form of in the gauge transformation. potential 7. Hamiltonian under the gauge transformation We consider the Schrödinger equation given by ˆ i H .