Feynman Path Integral: Formulation of Quantum Dynamics Masatsugu Sei Suzuki Department of Physics, SUNY at Binghamton (Date: January 31, 2021)
(( Overview )) We consider quantum superpositions of possible alternative classical trajectories, a history being a path in configuration space, taken between fixed points (a and b). The basic idea of the Feynman path integral is a perspective on the fundamental quantum mechanical principle of complex linear superposition of such entire spacetime histories. In the quantum world, instead of there being just one classical ‘reality’, represented by one such trajectory (one history), there is a great complex superposition of all these ‘alternative realities’ (superposed alternative histories). Accordingly, each history is to be assigned a complex weighting factor, which we refer to as an amplitude, if the total is normalized to modulus unity, so the squared modulus of an amplitude gives us a probability. The magic role of the Lagrangian is that it tells us what amplitude is to be assigned to each such history. If we know the Lagrangian L, then we can obtain the action S, for that history (the action being just the integral of L for that classical history along the path). The complex amplitude to be assigned to that particular history is then given by the deceptively simple formula amplitude
i i b exp(S ) exp( Ldt ) . ℏ ℏ a
The total amplitude to get from a to b is the sum of these.
(( Path integral )) What is the path integral in quantum mechanics? The path integral formulation of quantum mechanics is a description of quantum theory which generalizes the action principle of classical mechanics. It replaces the classical notion of a single, unique trajectory for a system with a sum, or functional integral, over an infinity of possible trajectories to compute a quantum amplitude. The basic idea of the path integral formulation can be traced back to Norbert Wiener, who introduced the Wiener integral for solving problems in diffusion and Brownian motion. This idea was extended to the use of the Lagrangian in quantum mechanics by P. A. M. Dirac in his 1933 paper. The complete method was developed in 1948 by Richard Feynman. This formulation has proven crucial to the subsequent development of theoretical physics, because it is manifestly symmetric between time and space. Unlike previous methods, the path-integral allows a physicist to easily change coordinates between very different canonical descriptions of the same quantum system.
((The idea of the path integral by Richard P. Feynman))
1 R.P. Feynman, The development of the space-time view of quantum electrodynamics (Nobel Lecture, December 11, 1965). http://www.nobelprize.org/nobel_prizes/physics/laureates/1965/feynman- lecture.html
Feynman explained how to get the idea of the path integral in his talk of the Nobel Lecture. The detail is as follows. The sentence is a little revised because of typo. ______I went to a beer party in the Nassau Tavern in Princeton. There was a gentleman, newly arrived from Europe (Herbert Jehle) who came and sat next to me. Europeans are much more serious than we are in America because they think that a good place to discuss intellectual matters is a beer party. So, he sat by me and asked, «what are you doing» and so on, and I said, «I’m drinking beer.» Then I realized that he wanted to know what work I was doing and I told him I was struggling with this problem, and I simply turned to him and said, ((listen, do you know any way of doing quantum mechanics, starting with action - where the action integral comes into the quantum mechanics?» «No», he said, «but Dirac has a paper in which the Lagrangian, at least, comes into quantum mechanics. I will show it to you tomorrow» Next day we went to the Princeton Library, they have little rooms on the side to discuss things, and he showed me this paper. What Dirac said was the following: There is in quantum mechanics a very important quantity which carries the wave function from one time to another, besides the differential equation but equivalent to it, a kind of a kernel, which we might call K x x),'( ,which carries the wave function x )( known at time t, to the wave function x )'( at time, t . Dirac points out that this function K was analogous to the quantity in classical mechanics that you would calculate if you took the exponential of i / ℏ , multiplied by the Lagrangian L ɺ xx ),( imagining that these two positions x, x’ corresponded t and t . In other words, K x x ),'( is analogous to x'x exp[i L( , x)], ℏ
x'x K x x),'( exp[i L( , x)]. (1) ℏ
Professor Jehle showed me this, I read it, he explained it to me, and I said, «what does he mean, they are analogous; what does that mean, analogous? What is the use of that?» He said, «you Americans ! You always want to find a use for everything!» I said, that I thought that Dirac must mean that they were equal. «No», he explained, «he doesn’t mean they are equal.» «Well», I said, «let’s see what happens if we make them equal.» So I simply put them equal, taking the simplest example where the Lagrangian is 1 xMɺ 2 V x)( , but soon found I had to put a constant of proportionality A in, suitably 2 adjusted. When I substituted exp( Li ℏ )/ for K to get
x'x ,'( tx ) Aexp[i L( , x)] tx ),( dx , (2) ℏ
2 and just calculated things out by Taylor series expansion, out came the Schrödinger equation. So, I turned to Professor Jehle, not really understanding, and said, «well, you see Professor Dirac meant that they were proportional.» Professor Jehle’s eyes were bugging out-he had taken out a little notebook and was rapidly copying it down from the blackboard, and said, «no, no, this is an important discovery. You Americans are always trying to find out how something can be used. That’s a good way to discover things!» So, I thought I was finding out what Dirac meant, but, as a matter of fact, had made the discovery that what Dirac thought was analogous, was, in fact, equal. I had then, at least, the connection between the Lagrangian and quantum mechanics, but still with wave functions and infinitesimal times. It must have been a day or so later when I was lying in bed thinking about these things, that I imagined what would happen if I wanted to calculate the wave function at a finite interval later. I would put one of these factors exp( Li ℏ )/ in here, and that would give me the wave functions the next moment, t and then I could substitute that back into (2) to get another factor of exp( Li ℏ )/ and give me the wave function the next moment, t + 2 , and so on and so on. In that way I found myself thinking of a large number of integrals, one after the other in sequence. In the integrand was the product of the exponentials, which, of course, was the exponential of the sum of terms like L /ℏ . Now, L is the Lagrangian and is like the time interval dt, so that if you took a sum of such terms, that’s exactly like an integral. That’s like Riemann’s formula for the integral Ldt , you just take the value at each point and add them together. We are to take the limit as 0 , of course. Therefore, the connection between the wave function of one instant and the wave function of another instant a finite time later could be obtained by an infinite number of integrals, (because goes to zero, of course) of exponential (iSℏ )/ where S is the action expression (3),
S L ɺ xx ),( dt . (3)
At last, I had succeeded in representing quantum mechanics directly in terms of the action S. This led later on to the idea of the amplitude for a path; that for each possible way that the particle can go from one point to another in space-time, there’s an amplitude. That amplitude is an exponential of i /ℏ times the action for the path. Amplitudes from various paths superpose by addition. This then is another, a third way, of describing quantum mechanics, which looks quite different than that of Schrödinger or Heisenberg, but which is equivalent to them.
1. Introduction The time evolution of the quantum state in the Schrödinger picture is given by
t)( Uˆ tt )',( t )'( , or
3 x t)( x Uˆ tt )',( t )'( dx' x Uˆ tt x')',( x' t )'( , in the x representation, where K(x, t; x’, t’) is referred to the propagator (kernel) and given by
i K xtx t )',';,( x Uˆ tt x')',( x exp[ Hˆ (t t )]' x' . ℏ
Note that here we assume that the Hamiltonian Hˆ is independent of time t. Then we get the form
x t)( dx' K xtx t )',';,( x' t )'( .
For the free particle, the propagator is described by
m im(x x )' 2 K txtx )',';,( exp[ ]. 2iℏ(t t )' ℏ(2 t t )'
(which will be derived later)
((Note)) Propagator as a transition amplitude
i K txtx )',';,( x exp[ Hˆ (t t )]' x' ℏ i i x exp( ˆtH )exp( ˆtH x')' ℏ ℏ , txtx ','
Here we define
i i ,tx exp( ˆtH ) x , ,tx x exp( ˆtH ) . ℏ ℏ
We note that
i i , atx x exp( ˆtH ) a exp( tE ) ax , ℏ ℏ a where
ˆ H a Ea a .
4
(( Heisenberg picture )) The physical meaning of the ket ,tx : The operator in the Heisenberg's picture is given by
i i xˆ exp( ˆ )xtH ˆ exp( ˆtH ) , H ℏ ℏ
i i i xˆ ,tx exp( ˆ )xtH ˆ exp( ˆtH )exp( ˆtH ) x H ℏ ℏ ℏ i exp( ˆ ) ˆ xxtH ℏ
i exp( ˆ ) xxtH ℏ i x exp( ˆtH ) x ,txx ℏ
This means that ,tx is the eigenket of the Heisenberg operator xˆH with the eigenvalue x. We note that
i t)( exp( ˆtH ) . S ℏ H
Then we get
i x t)( x exp( ˆtH ) ,tx . S ℏ H H
This implies that
,tx ,tx , x x . H S where S means Schrödinger picture and H means Heisenberg picture.
2. Propagator We are now ready to evaluate the transition amplitude for a finite time interval
K txtx )',';,( , txtx ',' i i i x exp[ Hˆt)]exp[ Hˆt)].....exp[ Hˆt)] x' ℏ ℏ ℏ
5 where
t t' t (in the limit of N ) N
t t0 Dt t where t0 = t' in this figure.
We next insert complete sets of position states (closure relation)
K txtx )',';,( , txtx ',' i dx dx .... dx dx x exp( Hˆt) x 1 2 N 2 N 1 ℏ N 1 i i x exp( Hˆt) x ...... x exp( Hˆt) x N 1 ℏ N 2 3 ℏ 2 i i x exp( Hˆt) x x exp( Hˆt x') 2 ℏ 1 1 ℏ
This expression says that the amplitude is the integral of the amplitude of all N-legged paths.
((Note))
x t (),','( x ,t (), x ,t (), x ,t (), x ,t ), 11 2 2 33 4 4 .....(xN4 ,t N4 (), xN 3 ,t N3 (), xN2 ,t N2 (), xN 1,tN 1 tx ),(), with
t' t0 t1 t2 t3 t4 ...... tN 3 tN 2 tN 1 t tN
6
We define
x' x0, t' t0 ,
x xN , t tN .
We need to calculate the propagator for one sub-interval
i x exp[ Hˆt) x , i ℏ i1 where i = 1, 2, …, N, and
pˆ 2 Hˆ V (xˆ) , 2m
Then we have
i i x exp( Hˆt) x dp x p p exp( Ht ˆ ) x i ℏ i 1 ii i i ℏ i1 i dp x p p 1ˆ Ht ˆ x O((t 2 )) ii i i ℏ i 1 i pˆ 2 dp x p p 1ˆ t( V (xˆ)) x O((t 2 )) ii i i ℏ 2m i1 i pˆ 2 i dp x p [ p 1ˆ t( ) x p tV (xˆ) x ] O((t 2 )) ii i i ℏ 2m i 1 i ℏ i1 i p 2 i dp x p [ p 1ˆ t( i ) x p tV (x ) x ] O((t 2 )) ii i i ℏ 2m i 1 i ℏ i1 i1 i p 2 dp x p p 1ˆ t[ i V (x )] x O((t 2 )) ii i i ℏ 2m i1 i 1
7
where pi (i = 1, 2, 3, …, N), or
i i pˆ 2 x exp( Hˆt) x dp x p p x 1[ t( i V (x ))] i ℏ i1 ii i ii 1 ℏ 2m i 1 1 i i dp exp[ p (x x 1)[ tE( p , x ]) 2ℏ i ℏ ii i1 ℏ ii 1 1 i i dp exp[ p (x x )]exp[ tE( p , x ]) 2ℏ i ℏ ii i1 ℏ ii 1 1 i (x x ) dp exp[ {p i i 1 t E( p , x )t}] 2ℏ i ℏ i t ii 1 1 i (x x ) dp exp[ {p i i 1 E( p , x )}t] 2ℏ i ℏ i t ii 1 where
p 2 E( p , x ) i V(x ) . i i1 2m i 1
Then we have
K txtx )',';,( , txtx ','
dp1 dp2 dpN 1 dpN lim dx1 dx2...... dxN 1 ...... N 2ℏ 2ℏ 2ℏ 2ℏ i N (x x ) p 2 exp[ {p i i1 ( i V (x ))}t] ℏ i i1 i1 t 2m
We note that
dp i (x x ) p 2 dp i p 2 i exp[ {p i i1 i }t} i exp[ {p (x x ) i i t} 2ℏ ℏ i t 2m 2ℏ ℏ ii i1 2mℏ
m imt x x exp[ ( i i1 2 ]) 2ℏ ti 2ℏ t
______(( Mathematica ))
8 1 p2 Clear "Global` ∗" ; f1 = Exp p x − ∆t ; @ D 2 π — B — 2 m — F Integrate @f1, 8p, −∞, ∞ 2 π ∆t — m ______Then we have m N 2/ K txtx )',';,( lim dx1 dx2...... dxN 1 ( ) N 2ℏ ti N i m xi xi1 2 exp[ t{ ( ) V (xi1 )}] ℏ i1 2 t Notice that as N and therefore t 0 , the argument of the exponent becomes the standard definition of a Riemann integral N t i m xi xi1 2 i ɺ lim t { ( ) V (xi1 )} dtL xx ),( , N ℏ ℏ t 0 i1 2 t t ' where L is the Lagrangian (which is described by the difference between the kinetic energy and the potential energy) 9 x O t 1 L xx ɺ),( m xɺ)( 2 V x)( . 2 It is convenient to express the remaining infinite number of position integrals using the shorthand notation m N 2/ D tx )]([ lim dx1 dx2 ...... dxN 1 ( ) . N 2ℏit Thus we have i K txtx )',';,( , txtx ',' D tx )]([ exp{ S ([ tx )]}, ℏ where t S ([ tx )] dtL xx ɺ),( . t' The unit of S is [erg sec]. 10 When two points at ( ti, xi) and ( tf, xf) are fixed as shown the figure below, for convenience, we use t f S tx )]([ dtL xx ɺ),( . ti x x2 x1 t t1 t2 This expression is known as Feynman’s path integral (configuration space path integral). S tx )]([ is the value of the action evaluated for a particular path taken by the particle. If one wants to know the quantum mechanical amplitude for a point particle at x’, at time t’ to reach a position x, at time t, one integrates over all possible paths connecting the points with a weight factor given by the classical action for each path. This formulation is completely equivalent to the usual formulation of quantum mechanics. The expression for K xtx t )',';,( , xtx t',' may be written, in some loose sense, as iS(N )0, x ,tx xt ,' tx t' exp[ ] N N 0 0 ℏ all)( path iS iS iS exp( path1 ) exp( path2 ) exp( pathn ) ℏ ℏ ℏ 11 where the sum is to be taken over an innumerably infinite sets of paths. x xcl HtL t (a) Classical case Suppose that ℏ 0 (classical case), the weight factor exp[iS ℏ ]/ oscillates very violently. So there is a tendency for cancellation among various contribution from neighboring paths. The classical path (in the limit of ℏ 0 ) is the path of least action, for which the action is an extremum. The constructive interference occurs in a very narrow strip containing the classical path. This is nothing but the derivation of Euler- Lagrange equation from the classical action. Thus the classical trajectory dominates the path integral in the small ħ limit. In the classical approximation (S ℏ ) iS x ,tx xt ,' tx t' "smooth function" exp( cl ) . (1) N N 0 0 ℏ But at an atomic level, S may be compared with ħ, and then all trajectory must be added in xN ,tx N xt 0 ,' tx 0 t' in detail. No particular trajectory is of overwhelming importance, and of course Eq.(1) is not necessarily a good approximation. 12 (b) Quantum case . What about the case for the finite value of S / ℏ (corresponding to the quantum case)? The phase exp[iS ℏ ]/ does not vary very much as we deviate slightly from the classical path. As a result, as long as we stay near the classical path, constructive interference between neighboring paths is possible. The path integral is an infinite-slit experiment. Because one cannot specify which path the particle choose, even when one knows what the initial and final positions are. The trajectory can deviate from the classical trajectory if the difference in the action is roughly within ħ. ((REFERENCE )) R.P. Feynman and A.R. Hibbs, Quantum Mechanics and Path Integrals (extended edition) Dover 2005. ((Note )) We can use the Baker-Campbell-Haudorff theorem for the derivation of the Feynman path integral. We have a Hamiltonian Hˆ T ˆ V ˆ where Tˆ is the kinetic energy and Vˆ( xˆ ) is the potential energy. We consider i i exp(Htˆ ) exp[ ( TVt ˆ ˆ ) ] ℏ ℏ exp(Pˆ Q ˆ ) where i i Pˆ T ˆ t , Qˆ Vˆ t ℏ ℏ We use the Baker-Campbell-Hausdorff theorem 1 1 1 exp()exp(Pˆ Qˆ ) exp{ PQ ˆˆˆˆ ˆˆ [, PQ ] [,[, PPQ ˆˆˆ ]] [ QPQ ,[, ˆ ]] ...} 2 12 12 with i [,]PQˆˆ ( )(2 tTV )[,] 2 ˆ ˆ ℏ i [,[,PPQˆˆˆ ]] ( )(3 tTTV )[,[, 3 ˆˆˆ ]] ℏ 13 i [QPQˆ ,[,ˆ ˆ ]] ( )(3 tVTV )[,[, 3 ˆˆˆ ]] ℏ In the limit of t 0 , we get i exp(Htˆ ) exp( PQ ˆ ˆ ) ℏ exp(Pˆ )exp( Q ˆ ) i i exp( Ttˆ )exp( Vt ˆ ) ℏ ℏ Using this, we get the matrix element i i i xexp( Htxˆ ) x exp( Tt ˆ )exp( Vtx ˆ ) jℏ j1 j ℏ ℏ j 1 i t i xexp( pxˆ 2 ) exp( Vxt ( ) ) j2mℏ j1 ℏ j 1 or 1/2 (x x ) 2 iˆ m im j1 j xjexp( H t ) x j1 exp{[2 V ()])] xj 1 t ℏ2i ℏ t ℏ 2() t 3. Free particle propagator In this case, there is no potential energy. N m N 2/ i m xi xi1 2 K xtx t )',';,( lim dx1 dx2...... dxN 1( ) exp[ t { ( ) }] , N 2ℏ ti ℏ i1 2 t or K(x,t;x',t') lim dx1 dx2 ...... dxN 1 N m m ( ) N / 2 exp[ {(x x')2 (x x )2 (x x )2 .... (x x )2}] 2ℏit 2ℏit 1 2 1 3 2 N1 We need to calculate the integrals, m m f ( ) 2/2 dx exp[ {(x x )' 2 (x x 2 ]) 1 2ℏ ti 1 2ℏ ti 1 2 1 1 m 2/1 m 2 ( ) exp[ (x2 x ])' 4 ℏ ti 4ℏ ti 14 m m g f ( ) 2/1 exp[ (x x 2 ]) 1 1 2ℏit 2ℏit 3 2 1 m m f g dx ( )1/ 2 exp[ (x x') 2 ] 2 1 2 ℏ ℏ 3 6 it 6 it m m g f ( )1 / 2 exp[ (x x )2 ] 2 2 2ℏit 2ℏit 4 3 1 m m f g dx ( )1/ 2 exp[ (x x')2 ] 3 2 3 ℏ ℏ 4 8 it 8 it ...... m m(x x')2 K(x,t;x',t') lim( )1/ 2 exp[ ], N 2ℏNit 2ℏiNt or m m(x x')2 K(x,t;x',t') ( )1/ 2 exp[ ], 2ℏi(t t') 2ℏi(t t') where we use t t' Nt in the last part. (( Mathematica )) 15 Free particle propagator; e=Dt Clear @"Global` ∗"D; ∗ = . exp_ : exp ê 8Complex @re_ , im_ D Complex @re , −im D<; 2 2 h1 = IHx1 − x'L + Hx2 − x1 L M êê Expand; − 2 2 π — ε 2 −m f0 x1_ := Exp h1 ; @ D m B 2 — ε F f1 = Integrate @f0 @x1 D, 8x1 , −∞, ∞ − 1 2 π — ε 2 −m g1 = f1 Exp x3 − x2 2 Simplify; m B 2 — ε H L F êê ∞ m f2 = g1 x2 Simplify , Im > 0 & ‡−∞ êê B B ε — F F ′ 2 m Hx3 −x L 6 ε — 6 π ε — m − 1 2 π — ε 2 −m g2 = f2 Exp x4 − x3 2 Simplify; m B 2 — ε H L F êê ∞ m f3 = g2 x3 Simplify , Im > 0 & ‡−∞ êê B B ε — F F ′ 2 m Hx4 −x L 8 ε — − m ε — 2 2 π − 1 2 π — ε 2 −m g3 = f3 Exp x5 − x4 2 Simplify; m B 2 — ε H L F êê ∞ m f4 = g3 x4 Simplify , Im > 0 & ‡−∞ êê B B ε — F F ′ 2 m Hx5 −x L 10 ε — 10 π ε — m 16 4. Gaussian path integral The simplest path integral corresponds to the vase where the dynamical variables appear at the most up to quadratic order in the Lagrangian (the free particle, simple harmonics are examples of such systems). Then the probability amplitude associated with the transition from the points (x tii ), to (x ,t ff ) is the sum over all paths with the action as a phase angle, namely, i K(x ,t ; tx ), exp[ S ] (tF t ), , f f ii ℏ cl if where Scl is the classical action associated with each path, t f S dtL(x , xɺ t), , cl cl cl ti with the Lagrangian L ɺ txx ),,( described by the Gaussian form, L ɺ txx ),,( )( xta ɺ 2 )( ɺxxtb )( xtc 2 )( xtd ɺ )( xte tf )( If the Lagrangian has no explicit time dependence, then we get (tF tif ), (tF f ti ) . For simplicity, we use this theorem without proof. i K(x ,t tx ),; exp[ S ] (tF t ) . ff ii ℏ cl f i (( Proof )) R.P. Feynman and A.R. Hibbs, Quantum Mechanics and Path Integrals. Let xcl t )( be the classical path between the specified end points. This is the path which is an extremum for the action S. We can represent tx )( in terms of xcl t )( and a new function ty )( ; tx )( xcl t)( ty )( , where ty i )( (ty f ) 0 . At each t, the variables tx )( and ty )( differ by the constantxcl t )( (Of course, this is a different constant for each value of t). Thus, clearly, dxi dyi , 17 for each specific point ti in the subdivision of time. In general, we may say that Dx t)( Dy t)( . The integral for the action can be written as t f tS t ],[ L ɺ tx ),([ (tx )),t]dt , if ti with L ɺ txx ),,( )( xta ɺ2 )( ɺxxtb )( xtc 2 )( xtd ɺ )( xte tf )( . ɺ ɺ We expand L txx ),,( in a Taylor expansion around xcl ,xcl . This series terminates after the second term because of the Gaussian form of Lagrangian. Then 2 2 2 ɺ ɺ L L ɺ 1 L 2 L ɺ L ɺ 2 L txx ),,( L(x , x t), | y | ɺ y ( y yy y |) ɺ . cl cl x xcl xɺ xcl 2 x2 xxɺ xɺ2 ,xx clcl From here we obtain the action t f L L [tS t ], S [t t ], dt( | y | yɺ) if cl if x xcl xɺ xcl ti t f 2 2 2 1 L 2 L ɺ L 2 dt( y yy y |) ɺ ɺ2 ɺ 2 ,xx clcl 2 t x xx x i t f L L S [t t ], dt( | y | yɺ) cl if x xcl xɺ xcl ti t f dt )([ yta ɺ 2 )( yytb ɺ )( ytc 2 ] ti The integration by parts and use of the Lagrange equation makes the second term on the right-hand side vanish. So we are left with t f [tS t ], S [t t ], dt )([ yta ɺ 2 )( yytb ɺ )( ytc 2 ] . if cl if ti Then we can write 18 t f S tx )]([ S [t t ], )([ yta ɺ 2 )( ɺyytb )( ytc 2 ]dt . cl if ti The integral over paths does not depend on the classical path, so the kernel can be written as t i y0 i f K(x ,t tx ),; exp[ S ] exp{ )([ yta ɺ 2 )( ɺyytb )( ytc 2 ]dt}Dy t)( ff ii ℏ cl ℏ y0 ti i (tF t ), exp[ S ] if ℏ cl where t y 0 i f (tF t ), exp{ )([ yta ɺ 2 )( ɺyytb )( ytc 2 ]dt}Dy t)( . if ℏ y 0 ti It is defined hat (tF tif ), is the integral over all paths from y 0 back to y 0 during the interval (t f ti ) . (( Note )) If the Lagrangian is given by the simple form L ɺ txx ),,( )( xta ɺ 2 )( ɺxxtb )( xtc 2 then (tF tif ), can be expressed by (tF tif ), K(x f ,0 t f ; xi ti ),0 . 4. Evaluation of (tF tif ), for the free particle We now calculate (tF f ,tt i t )' for the free particles, where the Lagrangian is given by the form, 1 L ɺ txx ),,( xmɺ2 . 2 Then we have 19 t y 0 i f m (tF t ), exp{ yɺ 2dt}Dy t)( if ℏ 2 . y 0 ti Replacing the variable y by x, we get t x0 i f m (tF t ), exp{ xɺ 2dt}Dx t)( if ℏ 2 x0 ti In this case, formally (tF tif ), is equal to the propagator K(x f ,0 t f ; xi ti ),0 , (tF tif ), K(x f ,0 t f ; xi ti ),0 lim dx1 dx2...... dxN 1 N m m ( )N 2/ exp[ {x 2 (x x )2 (x x )2 .... (x x )2 x 2}] 2ℏ ti 2ℏ ti 1 2 1 3 2 N 1 N 2 N 1 t t t t' where we put x x' 0 , and x x 0 , and t f i . i f N N We now evaluate the following integral; (tF tif ), lim dx1 dx2...... dxN 1 N m m ( ) N 2/ exp[ {x 2 (x x )2 (x x )2 .... (x x )2 x 2}] 2ℏit 2ℏit 1 2 1 3 2 N1 N 2 N 1 We note that f x 2 (x x )2 (x x )2 .... (x x )2 x 2 1 2 1 3 2 N 1 N 2 N 1 2 2 2 (2 x1 x2 .... xN 1 ) (2 xx 21 xx 32 .... xN 2 xN 1) Using the matrix, f can be rewritten as f X ˆXA (Uˆ) UA ˆ Uˆ ˆUA ˆ T (Uˆ ˆUA ˆ ) , with 20 x 2 1 0 0 0 . 0 1 x2 1 2 1 0 0 . . . 0 1 2 1 0 . . X . , A 0 0 1 2 1 . . . . ...... . . . . . . 2 1 xN 1 . . . . . 1 2 We solve the eigenvalue problems to determine the eigenvalues and the unitary operator, such that 0 0 0 .. 0 1 0 2 0 0 .. 0 0 0 0 .. 0 3 ˆ ˆ ˆ U UA 0 0 0 4 .. 0 , ...... ...... 0 0 0 0 .. n where i is the eigenvalue of A. Then we have 0 0 0 .. 0 1 1 0 2 0 0 .. 0 2 0 0 0 .. 0 . 3 f 1 1 1 1 .. N 1 0 0 0 4 .. 0 . ...... . ...... . 0 0 0 0 .. n N 1 N1 2 ii i1 The Jacobian determinant is obtained as x ,( x ,..., x ) 1 2 N 1 detU 1. (1,2 ,..., N 1) Then we have the integral 21 (tF tif ), ttF )',( m N 2/ m 2 2 2 lim d1 d2...... dN 1( ) exp[ { 11 22 ... N 1N 1 }] N 2ℏ ti 2ℏ ti m 2ℏ ti 2ℏ ti 2ℏ ti ( )N 2/ ... ℏ 2 ti m1 m2 mN 1 m m 1 ( )N 2/ ( )(N 2/)1 ℏ ℏ 2 ti 2 ti 21 ....N 1. m 2ℏ ti (det A) m 2ℏiNt m 2ℏ (ti t )' where det A 321 ...N 1 N (( Mathematica )) Example (N = 6). The matrix A (5 x 5): N 1 5 2 1 0 0 0 1 2 1 0 0 A 0 1 2 1 0 0 0 1 2 1 0 0 0 1 2 The eigenvalues of A: 1 2 3 , 2 3, 3 2 , 4 1, 5 2 3 . The unitary operator Uˆ 22 N1 2 f ii . i1 0 0 0 0 1 0 2 0 0 0 Uˆ ˆUA ˆ 0 0 0 0 . 3 0 0 0 4 0 0 0 0 0 5 det A N 6 . ______ 23 Clear "Global` " ; 2 1 0 0 0 1 2 1 0 0 A1 0 1 2 1 0 ; 0 0 1 2 1 0 0 0 1 2 eq1 Eigensystem A1 2 3 , 3, 2, 1, 2 3 , 1, 3 , 2, 3 , 1 , 1, 1, 0, 1, 1 , 1, 0, 1, 0, 1 , 1, 1, 0, 1, 1 , 1, 3 , 2, 3 , 1 1 Normalize eq1 2, 1 Simplify 1 1 1 1 1 , , , , 2 3 2 3 2 2 3 2 Normalize eq1 2, 2 Simplify 1 1 1 1 , , 0, , 2 2 2 2 24 3 Normalize eq1 2, 3 Simplify 1 1 1 , 0, , 0, 3 3 3 4 Normalize eq1 2, 4 Simplify 1 1 1 1 , , 0, , 2 2 2 2 5 Normalize eq1 2, 5 Simplify 1 1 1 1 1 , , , , 2 3 2 3 2 2 3 UT 1, 2, 3, 4, 5 ;U Transpose UT ; UH UT; U MatrixForm 1 1 1 1 1 2 3 2 3 2 2 3 1 1 0 1 1 2 2 2 2 1 0 1 0 1 3 3 3 1 1 0 1 1 2 2 2 2 1 1 1 1 1 2 3 2 3 2 2 3 25 UH.U Simplify; UH.U MatrixForm 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 1 2 3 ; 4 5 s1 UH .A1 .U Simplify ; s1 MatrixForm 2 3 0 0 0 0 0 3 0 0 0 0 0 2 0 0 0 0 0 1 0 0 0 0 0 2 3 f1 Transpose .s1. FullSimplify 2 3 12 3 22 2 32 42 2 3 52 Det A1 6 K1 eq1 1, 1 eq1 1, 2 eq1 1, 3 eq1 1, 4 eq1 1, 5 Simplify 6 6. Calculation of the average Aˆ In order to understand the above discussion, for the sake of clarity, we discuss the fundamental mathematics in detail. ˆ 6.1 The average A under the original basis { bi } 26 We consider the two bases { bi, a i }, where the new basis { ai } is related to the ˆ original basis {} bi through a unitary operator U , ˆ ˆ ˆ aj U b j , bj U a j , bj a j U , ˆ ˆ ˆ ˆ with U U 1. ai is the eigenket of the Hermitian operator A with the eigenvalue ai . ˆ Aai aa i i . Note that ˆ ˆ baij bUb i j aUa i j , or ˆ ˆˆˆ ˆ aUaiji bUUUb jij bUb . In other word, the matrix element of Uˆ is independent of the kind of basis (this is very important property). We also note that ˆˆ ˆ ˆ aAai j bUAUb i j a iij (diagonal matrix) Here we define the Column matrices for the state of the system, 27 1 1 2 2 . . β α (matrix form) . . . . n n with i b i , i a i . We now consider the average over the state under the original basis { bi }. A A ˆ ˆ bii bAb ji b i, j * ˆ bi bAb i ji b i, j * iA ij i i, j A1112 A. . . A 1n 1 A A. . . A 21 22n 2 * * * ...... 1 2 . . . n ...... . ...... . An1 A n 2 . . . A nn n β Aβ (matrix form), using the closure relation. The relation between β and α is obtained as follow. 28 i a i ai b j b j j ai b j j j ˆ bi U b j j j or α U β (matrix form) ˆ since ai b i U . ˆ 6.2 The average A under the new basis { ai } Next, we now consider the average under the new basis { ai }. A A ˆ ˆ aii aAa j a j i, j * ˆ ai aAa i jj a i, j * ia i ij i i, j a10 ... 0 1 0a ...0 2 2 * * * ...... 1 2 . . . n ...... . ...... . 0 0 ... an n 2 ai i i, j 6.3 The calculation of the average using Mathematica (i) Find eigenvalue and eigenkets of matrix A by using Mathematica 29 Eigensystem[ A] which leads to the eigenvalues ai and eigenkets, ai . The eigenkets ahould be normalized using the program Normalize. When the system is degenerate ( the same eigenvalues but different states), further we need to use the program Orthogonize for all eigenkets obtained by doing the process of Eigensystem[ A] (ii) Determine the unitary matrix U Unitary matrix U is defined as ba1112 ba. . . ba 1 n ba2122 ba. . . ba 2 n ...... U ...... ...... ban1 ba n 2 . . . ba nn u1 u 2 . . . u n where b1 a i b2 a i b3 a i ui (matrix form of eigenkets) . . bn a i Thus, we have α U β , β Uα (matrix form) 30 where U uu1 2 ...... u n and * u1 * u2 . U . . * un Note that β Aβ( α Uˆ AUα )( α Aαɶ ) where a1 0 ... 0 0a ...0 2 ...... Uˆ AU (diagonal matrix) ...... ...... 0 0 ... an 6.4 Example-1 (3x3 matrix) Here we discuss a typical example, A is 3x3 matrix. 31 2 2 2 f 2(1 2 3 12 23 31 ) β Aβ 1 * * * A 12 3 2 3 1 A 123 2 3 Where 1 , 2 , and 3 are real, 1 * * * T β , β β , 2 1 2 3 123 3 2 1 1 A 1 2 1 . 1 1 2 Eigenvalue problem of matrix A (we solve the problem using Mathematica. The system is degenerate ) A1 a 1 1 , A2 a 2 2 , A3 a 3 3 . where the eigenvalues and eigenkets are as follows, 1 1 a 3, 0 , 1 1 2 1 32 1 1 a 3 , 2 , 2 2 6 1 1 1 a 0 , 1 . 3 3 3 1 under the basis {bi } . The unitary matrix can be obtained as 1 U 1 2 3 U 2 1 1 1 3 2 6 3 1 1 , 0 2 1 2 2 0 6 3 1 2 1 1 1 1 6 6 6 2 6 3 1 1 1 3 3 3 3 0 0 U U 1, U AU 0 3 0 0 0 0 111 111 1 2 3 263 263 1 1 21 21 β2 Uα 0 2 2 3 63 63 3 111 3 111 1 2 3 263 263 Thus, we have 33 f β Aβ α( U AU ) α a10 0 1 * * * 0a 0 123 2 2 0 0 a3 3 2 2 2 a11 a 22 a 33 2 2 2 31 3 2 0 3 6.5 Example-2 4x4 matrix We also discuu the second example; A is 4x4 matrix. 2 2 22 f 122 2 3 4 222 12 23 31 β Aβ 1 * * * * A 2 1 2 3 4 3 4 1 A2 1 2 3 4 3 4 Where 1 , 2 , 3 and 4 are real, 1 2 * * * * T β β 1234 1234 β 3 4 34 1 10 0 12 10 A . 0 12 1 0 0 11 Eigenvalue problem of matrix A (using Mathematica ) A1 a 1 1 , A2 a 2 2 , A3 a 3 3 , A4 a 4 4 The eigenvalues and eigenkets are obtained as follows, 1 2 2 2 1 1 1 2 2 a1 2 2 , 1 , 1 1 1 2 2 1 2 2 2 1 2 1 2 a2 2 , 2 , 1 2 1 2 35 1 1 1 2 2 1 1 1 2 2 a3 2 2 , 3 . 1 1 1 2 2 1 1 1 2 2 1 2 1 2 a4 0 , 4 . 1 2 1 2 The unitary matrix: U 1 2 3 4 1 11 11 1 2 2 2 2 22 2 1 1 11 11 1 1 22 22 2 2 , 1 1 11 11 1 1 2 22 2 2 2 1 11 11 1 2 2 2 2 22 2 36 1 U 2 3 4 1 1111 1 1 1 222 22 2 2 222 1 1 1 1 2 2 2 2 1 11 1 1 11 1 1 1 1 1 22 2 2 2 2 2 2 1 1 1 1 2 2 2 2 2 20 0 0 0 2 0 0 U U 1, U AU , 0 02 20 0 0 0 0 1 11 11 1 2 2 2 2 22 2 1 1 1 11 11 1 1 1 22 22 2 2 β2 Uα 2 , 3 1 1 11 11 3 1 1 2 22 2 4 2 2 4 1 11 11 1 2 2 2 2 22 2 or 37 1 11 11 12 1 34 2 2 2 2 22 2 1 1 1 11 1 1 112 1 34 22 22 2 2 2 . 3 1 111 1 1 1 1 212 22 34 2 4 2 2 1 11 11 1 12 34 2 2 2 2 22 2 Thus, we have f β Aβ α( U AU ) α 2 20 0 0 1 0 2 0 0 * * * * 2 1 2 3 4 0 02 20 3 0 0 0 0 4 2 2 2 2 a11 a 22 a 33 a 44 22 22 (2 2)12 2 (2 2) 34 0 7. Equivalence with Schrödinger equation The Schrödinger equation is given by iℏ t)( Hˆ t)( . t For an infinitesimal time interval , we can write i )( )0( Hˆ )0( , ℏ from the definition of the derivative, or 38 i x )( x )0( x Hˆ )0( ℏ i ℏ 2 2 [ V (x)] x )0( ℏ 2m x 2 or i x ),( x )0,( x Hˆ )0( ℏ i ℏ 2 2 [ V (x)] x )0,( ℏ 2m x 2 in the x representation. We now show that the path integral also predicts this behavior for the wave function. To this end, we start with x )( dx' K x x )0,';,( x' )0( , or x ),( dx'K x x )0,';,( x )0,'( , where m i x x' x x' K x x )0,';,( exp[ L( , )] 2iℏ ℏ 2 m i 1 (x x )' 2 x x' exp[ { m V ( )}] 2iℏ ℏ 2 2 2 Then we get m i 1 (x x )' 2 x x' x ),( dx'exp[ { m V ( )}] x )0,'( . ℏ ℏ 2 2i 2 2 We now define x'x . Then we have 39 m im 2 i x ),( d exp[ V (x )] (x )0, . ℏ ℏ ℏ 2i 2 2 The dominant contribution comes from the small limit of . Using the Taylor expansion in the limit of 0 V (x ) V x)( , 2 x )0,( 2 2 x )0,( (x )0, x )0,( , x 2 x 2 we get m im 2 i x )0,( 2 2 x )0,( x ),( d exp( 1)[ V (x)][ x )0,( ] ℏ ℏ ℏ 2 2i 2 x 2 x m im 2 x )0,( 2 2 x )0,( i d exp( )[ x )0,( V (x)] ℏ ℏ 2 ℏ 2i 2 x 2 x i iℏ 2 2 1[ V x)( x )0,(] ℏ 2m x 2 Thus we have i ℏ 2 2 x ),( x )0,( [ V (x)] x )0,( , ℏ 2m x 2 which is the same as that derived from the Schrödinger equation. The path integral formalism leads to the Schrodinger equation for infinitesimal intervals. Since any finite interval can be thought of a series of successive infinitesimal intervals the equivalence would still be true. (( Note )) 2/1 2/1 im 2 2iℏ im 2 iℏ 2iℏ d exp( ) , 2d exp( ) . ℏ ℏ 2 m 2 m m (( Mathematica )) 40 Clear @"Global` "D; = Integral @n_ D : m η2 Integrate ηn Exp , η, −∞, ∞ B B 2 — ε F 8 8. Motion of free particle; Feynman path integral The Lagrangian of the free particle is given by m L xɺ 2 . 2 Lagrange equation for the classical path; d L L ( ) ( ) 0 , dt xɺ x or xɺ a , or x at b . This line passes through ( t', x'), ( t, x); 41 x x' x x' (t t )' , 1 t t' 1 x1 t,x t',x' t1 Then we have x x' x x' (t t )' , 1 t t' 1 and 1 dx1 2 1 x x' 2 tL 1)( m( ) m( ) . 2 dt1 2 t t' which is independent of t1. Consequently, we have t m t x x' m x x' t m (x x )' 2 S tL )( dt ( )2 dt ( )2 dt , cl 1 1 1 1 t ' 2 t ' t t' 2 t t' t' 2 t t' and 42 i m(x x )' 2 K txtx )',',,( Aexp[ S ] Aexp[ ]. ℏ cl 2ℏ (ti t )' ((Approach from the classical limit)) To find A, we use the fact that as t t' 0 , K must tend to (x x )' 1 (x x )' 2 (x x )' lim exp[ ] 0 ( ) 2/12 2 . 1 (x x )' 2 lim exp[ ] 0 2 2 2 where , 2 1 (x x )' 2 f xx ,',( ) exp[ ]. (Gaussian distribution). 2 2 2 So we get 2ℏ (ti t )' , m 1 m A , ( ) 2/12 2ℏ (ti t )' or m m(x x )' 2 K txtx )',',,( exp[ ]. 2ℏ (ti t )' 2ℏ (ti t )' Note that m F (t t )' . freeparticle 2ℏ (ti t )' (( Evaluation of S / ℏ )) From the above discussion, S can be evaluated as 43 m (x)2 m(x) x mv p h S x x x , 2 t 2 t 2 2 2 or S mv 1 p 1 x x kx . ℏ 2ℏ 2 ℏ 2 where p is the momentum, p ℏk . Suppose that m is the mass of electron and the velocity v is equal to c/137. We make a S plot of (radian) as a function of x (cm). ℏ (( Mathematica )) NIST Physics constant : cgs units ∗ Clear @"Global` "D; = → × 10 → −27 rule1 9c 2.99792 10 , — 1.054571628 10 , → −28 me 9.10938215 10 =; me c K1 = . rule1 2 — ê 1.2948 × 10 10 K1 ê 137 7 9.4511 × 10 44 SêÑ Hrad L 10 10 10 8 10 6 10 4 100 1 Dx Hcm L -8 -6 -4 10 10 10 0.01 1 100 S Fig. (radian) as a function of x (cm), where v = c/137. m is the mass of electron. ℏ 9. Evaluation of S for the 1D system (example 8-1, Towmsend, 2 nd edition) We consider the Young’s double slit; mx2 m x 1 1 h ℏ S x px x x . (2 t) 2 t 2 2 The phase difference between two paths is evaluated as S 1 p 1 x kx x , ℏ 2 ℏ 2 45 S If is comparable to , the interference effect can be observed. Such a condition is ℏ satisfied when x . (( Note )) In classical physics, the phase difference is given by 2 x . 10. What is the Lagrangian for photon? (( Landau-Lifshitz )) For a particle, we have the Hamilton equations H H pɺ , v rɺ r p In view of the analogy, we can immediately write the corresponding equation for rays: kɺ , v rɺ r k ɺ In vacuum, ck , so that k 0, v c n ( n is a unit vector along the direction of propagation); in other words, in vacuum the rays are straight lines, along which the light travels with velocity c. Pursuing the analogy, we can establish for geometrical optics a principle analogous to the principle of least action in mechanics. However, it cannot be written in Hamiltonian form as Ldt 0 , since it turns out to be impossible to introduce, for rays, a function analogous to the Lagrangian of a particle. Since the Lagrangian of a particle is related to the Hamiltonian H by the equation H Lp H , p replacing the Hamiltonian H by the angular frequency and the momentum by the wave vector k, we should have to write for the Lagrangian in optics, k . But this k expression is equal to zero, since ck .But this expression is also clear directly from the consideration that the propagation of rays is analogous to the motion of particles with zero mass. 46 As is well, in the case where the energy is constant, the principle of least action for particles can also be written in the form of the so-called principle of Maupertuis: S p d l 0 where the integration extends over the trajectory of the particle between two of its points. In this expression the momentum is assumed to be a function of the energy and the coordinates. The analogous principle for rays is called Fermat’s principle. In this case, we can write by analogy: k d l 0 In vacuum, k n , and we obtain ( dl n dl : c dl 0 which corresponds to rectilinear propagation of the rays. (( REFERENCE )) L.D. Landau and E.M. Lifshitz, The Classical Theory of Fields (Pergamon, 1996). ((Note)) From a view point of the phaser diagram for the interference, photons are one of the best examples. I will use this example for the explanation of photon propagator. It seems to me that theorists hesitate to use the word of photon. Why is that? It is hard to find the i expression of the action S for photon, even if the path integral is expressed by exp(px ) ; ℏ plane-wave form. The Lagragian form of photon is not simple compared to that of particle (such as electron). 11. Single slit experiment In order to understand the path integral method, let us go back to the Young’s double slit experiment. We obtain an interference pattern, independent of whether we use a light source, or particle (electron) source. This can, of course, be explained by saying that there is a probability amplitude associated with each path. Note that a path integral approach offers a road to quantum mechanics for systems that are not readily accessible via Hamiltonian mechanics (Merzbacher, 1998). In his book, Feynman discussed the principle of least action for particles (Feynman, 1964). Is it possible to say intuitively that L T cp 47 where T is the kinetic energy of photon, c is the velocity of light, and p is the momentum.? The momentum p is expressed by 2 ℏ pℏ k , where is the wave length of photon. Then the action S can be evaluated as S Ldt cpdt cp t p x where p is assumed to be constant and x c t . Then we have i i exp(S ) exp( px ) exp( ikx ) ℏ ℏ The phase of change in wave function for photon is k x where k is the wave number. ______Imagine the slit divided into many narrow zones, width y (= = a/N). Treat each as a secondary source of light contributing electric field amplitude E to the field at P. 48 We consider a linear array of N coherent point oscillators, which are each identical, even to their polarization. For the moment, we consider the oscillators to have no intrinsic phase difference. The rays shown are all almost parallel, meeting at some very distant point P. If the spatial extent of the array is comparatively small, the separate wave amplitudes arriving at P will be essentially equal, having traveled nearly equal distances, that is E E r )( E (r ) ... E (r ) E r)( 0 10 20 0 N 0 N The sum of the interfering spherical wavelets yields an electric field at P, given by the real part of E Re[E )( er (kri 1t ) E )( er (kri 2 t) ... E )( er (kri N t) ] 0 0 0 (kri 1t ) ( rrik 12 ) ( rrik 13 ) ( N rrik 1)) Re[E0 )( er 1[ e e ... e ]] (( Note )) When the distances r1 and r2 from sources 1 and 2 to the field point P are large compared with the separation , then these two rays from the sources to the point P are nearly parallel . The path difference r2 – r1 is essentially equal to sin . Here we note that the phase difference between adjacent zone is 49 a (rk r ) k sin k( sin ) 2 1 N (rk 3 r2 ) (rk 4 r3 ) (rk N rN 1) 2 where k is the wavenumber, k . It follows that (rk 2 r1) (rk 3 r1) 2 (rk 4 r1) 3 (rN r1) (N )1 Thus the field at the point P may be written as (kri 1 t ) i i 2 ( Ni )1 E Re[E0 )( er 1[ e e ... e ]] We now calculate the complex number given by Z 1 ei ei2 ... e ( Ni )1 1 eiN 1 ei eiN 2/ (eiN 2/ eiN 2/ ) ei 2/ (ei 2/ ei 2/ ) N sin( ) e (Ni 2/)1 2 sin( ) 2 If we define D to be the distance from the center of the line of oscillators to the point P, that is 1 1 kD (N )1 k sin kr1 (N )1 kr1 2 2 1 k(D r ) (N )1 1 2 50 Then we have the form for E as N sin( ) ~ E Re[E )( er (kDi t) 2 ] Re[ eE ti ] 0 sin( ) 2 The intensity distribution within the diffraction pattern due to N coherent, identical, distant point sources in a linear array is equal to c ~ 2 I S 0 E 2 N sin2 ( ) sin2 ( ) sin 2 ( ) 2 2 2 I I0 I0 Im 2 2 2 sin ( ) sin ( ) 2 2N 2 in the limit of N→∞, where 2 sin 2 ( ) 2N 2N c I 0 [E (r)]2 0 2 0 c c I I N 2 0 [NE (r)]2 0 E 2 m 0 2 0 2 0 N Nk sin kasin k sin where a = N. We make a plot of the relative intensity I/Im as a function of . sin2 ( ) I 2 2 Im 2 Note that 51 sin 2 ( ) 2 2 d 2 2 IêIm 0.06 0.05 0.04 0.03 0.02 0.01 bê2p -4 -2 0 2 4 The numerator undergoes rapid fluctuations, while the denominator varies relatively slowly. The combined expression gives rise to a series of sharp principal peaks separated by small subsidiary maxima. The principal minimum occur in directions in direction m such that ka sin m 2 2 1 asin 2m 2m m m k 2 12. Phasor diagram (i) The system with two paths 52 The phasor diagram can be used for the calculation of the double slilts (Young) interference. We consider the sum of the vectors given by OS and ST . The magnitudes of these vectors is the same. The angle between OS and ST is (the phase difference). Fig. Phasor diagram for the double slit. In this figure, QO QS QT R . SOM STM 2/ . Then we have OT 2OM 2OS cos 2Acos . 2 2 2 The resultant intensity is proportional to OT , 2 I OT 4A2 cos2 2A2 1( cos) . 2 Note that the radius R is related to OS (= A) through a relation A 2Rsin . 2 When A = 1 (in the present case), we have the intensity I as I 4cos2 2 53 The intensity has a maximum ( I = 4) at 2n and a minimum ( I = 0) at 2 (n )2/1 . (ii) The system with 6 paths . Fig. The resultant amplitude of N = 6 equally spaced sources with net successive phase difference φ. = N φ = 6 φ. (iii) The system with 36 paths (comparable to single slit) 54 Fig. The resultant amplitude of N = 36 equally spaced sources with net successive phase difference φ. (iv) Single slit in the limit of N→∞ We now consider the system with a very large N. We may imagine dividing the slit into N narrow strips. In the limit of large N, there is an infinite number of infinitesimally narrow strips. Then the curve trail of phasors become an arc of a circle, with arc length equal to the length E0. The center C of this arc is found by constructing perpendiculars at O and T. The radius of arc is given by 55 E0 R R(N) . in the limit of large N, where R is the side of the isosceles triangular lattice with the vertex angle φ, and is given by N kasin , with the value being kept constant. Then the amplitude Ep of the resultant electric field at P is equal to the chord OT , which is equal to sin E E 2Rsin 2 0 sin E 2 . P 2 2 0 2 Then the intensity I for the single slits with finite width a is given by sin I I ( 2 )2 . m 2 where Im is the intensity in the straight-ahead direction where a The phase difference φ is given by kasin 2 sin 2psin . We make a plot of I/Im as a function of , where p = a/ is changed as a parameter. Fig. The relative intensity in single-slit diffraction for various values of the ratio p = a/. The wider the slit is the narrower is the central diffraction maximum. 13. Gravity: Feynman path integral 56 ((Calculation from the classical limit)) m L xɺ 2 mgx , 2 d L L ( ) ( ) , dt xɺ x xɺɺ g , g x t 2 At B , 1 2 1 1 Initial conditions: x1 = x and t1= t, g x t 2 At B . 2 and x1 x ' and t1 t ' g x' t'2 At'B , 2 A and B are determined from the above two equations. '( tt )[ (tg t )(' t t ) x]2 (2 t t x') x 1 1 1 , 1 (2 t t )' ɺ dx1 (tg t )(' t t 2' t1) (2 x x )' x1 . dt1 (2 t t )' Then we get the expression of the Lagrangian m L(x , ɺ tx ), xɺ 2 mgx . 1 11 2 1 1 The Hamilton’s principle function is 57 t S txtx )',',,( L x ,( ɺ tx ), dt cl 1 11 1 t ' m[g 2 (t t )' 4 (12 x x )' 2 12 (tg t 2 ()' x x )' (24 t t )' i K txtx )',';,( Aexp[ S tx tx )]',':,( . ℏ cl ((Classical limit)) When x x' and t t'T , we have 1 S tx tx )',':,( mg T 32 x' gmT , cl 24 and i ix'mgT K txtx )',';,( Aexp( mg T 32 ) ℏ ℏ 24 ix'mgT A exp( ) 1 ℏ where i A Aexp( mg T 32 ) . 1 24ℏ (( Mathematica )) 58 −1 Clear "Global` ∗" ; eq1 = x g t2 + A t + B; @ D 2 −1 eq2 = x0 g t0 2 + A t0 + B; 2 rule1 = Solve @8eq1, eq2 <, 8A, B −Ht0 − t1 L Hg Ht − t0 L Ht − t1 L + 2 xL + 2 Ht − t1 L x0 2 Ht − t0 L v1 = D@x1 , t1 D êê Simplify g Ht − t0 L Ht + t0 − 2 t1 L + 2 Hx − x0 L 2 Ht − t0 L m L1 = v1 2 − m g x1 FullSimplify 2 êê 1 m g t − t0 t + t0 − 2 t1 + 2 x − x0 2 − 2 IH H L H L H LL 8 Ht − t0 L 4 g Ht − t0 L H−Ht0 − t1 L Hg Ht − t0 L Ht − t1 L + 2 xL + 2 Ht − t1 L x0 LM t K1 = L1 t1 FullSimplify ‡t0 êê m g2 t − t0 4 − 12 x − x0 2 + 12 g t − t0 2 x + x0 − I H L H L H L H LM 24 Ht − t0 L rule1 = 8x → x0 , t → t0 + T<; K2 = K1 ê. rule1 êê Simplify 1 − g m T g T2 + 24 x0 24 I M 14. Simple harmonics: Feynman path integral (( Classical limit )) m 1 L xɺ 2 m 2 x 2 , 2 2 0 d L L ( ) ( ) , dt xɺ x ɺɺ 2 x 0 x , 59 x1 Acos( t10 ) Bsin( t10 ) . Initial conditions: x1 = x and t1= t, x Acos0t Bsin0t , and x1 x' and t1 t ', x' Acos0t'Bsin0t'. A and B are determined from the above two equations. x'sin[0 (t t1)] xsin[0 '( tt 1)] x1 , sin[0 (t t )]' ɺ dx1 x'0 cos[0 (t t1)] x0 cos[0 '( tt 1)] x1 . dt1 sin[0 (t t )]' Then we get the expression of the Lagrangian m 1 L(x ɺ tx ),, xɺ 2 m 2x 2 1 11 2 1 2 0 1 2 m0 2 [2xx'cos[0 (t t 2' t1) 2sin [0 (t t )]' 2 2 x' cos[20 (t t1)] x cos[0 '( tt 1)]) The Hamilton’s principle function is t S txtx )',',,( L(x , ɺ tx ), dt cl 1 11 1 t' m0 2 2 [(x x )' cos(0 (t t ))' 2xx ]' 2sin(0 (t t )' i K txtx )',';,( Aexp[ S tx tx )]',':,( , ℏ cl or 60 im K txtx )',';,( Aexp[ 0 {(x2 x 2 )' cos( (t t ))' 2xx }]' . ℏ 0 2 sin[0 (t t )]' In the limit of t t' 0 , we have im K txtx )',',,( Aexp[ 0 (x x 2 ])' . ℏ 2 sin{0 (t t )}' To find A, we use the fact that as t t' 0 , K must tend to (x x )' , 1 (x x )' 2 (x x )' lim exp[ ] 0 ( ) 2/12 2 . 1 (x x )' 2 lim exp[ ] 0 2 2 2 where , 2 1 (x x )' 2 f xx ,',( ) exp[ ] (Gaussian distribution). 2 2 2 In other words 1 1 m A , 0 . 2 1/ 2 2 ℏ ( ) 2i sin{0 (t t )}' So we get 2iℏsin{ (t t )}' 1 m 0 , A 0 2/12 ℏ m0 ( ) 2 isin{0 (t t )}' or m K txtx )',',,( 0 2ℏisin{ (t t )}' 0 im exp[ 0 [(x2 x 2 )' cos{ (t t )}' 2xx ]]' ℏ 0 2 sin[0 (t t )]' 61 Note that m K(x xt ',,0 t )',0 0 (tF t )' ℏ 2 isin{0 (t t )}' and i K txtx )',',,( (tF t )' exp[ S txtx )]'';,( . ℏ cl (( Mathematica )) Clear @"Global` ∗"D; ∗ expr_ := expr ê. 8Complex @a_ , b_ D Complex @a, −bD< seq1 = x A Cos @ω0 tD + B Sin @ω0 tD; seq2 = x0 == A Cos @ω0 t0 D + B Sin @ω0 t0 D; srule1 = Solve @8seq1 , seq2 <, 8A, B x1 = A Cos @t1 ω0D + B Sin @t1 ω0D ê. srule1 êê Simplify; v1 = D@x1, t1 D êê Simplify; m m L1 = v1 2 − ω02 x1 2 Simplify; 2 2 êê t L1 t1 FullSimplify ‡t0 êê 1 m ω0 −2 x x0 + x2 + x0 2 Cos t − t0 ω0 Csc t − t0 ω0 2 I I M @H L DM @H L D If the initial state of a harmonic oscillator is given by the displaced ground state wave function m (x,0) exp[ 0 (x x ) 2]. 2ℏ 0 When x , with m 0 . ℏ Then we have 62 1 1 )0,( exp[ ( 2 ]) , 4/1 2 0 and t),( K t )0,';,( )0,'( d' 1 i t sin( t) 2 exp( 0 )exp[ 0 e 0ti {( 2 )cot( t) i 2 4/1 2 2 0 0 1 20 }] sin(0t) 1 i t 1 2 exp( 0 )exp[ e 0ti {( 2 )cos( t) i 2 sin( t) 4/1 2 2 0 0 0 20}] 0ti 0ti 1 i t 1 2 e e exp( 0 )exp[ e 0ti {( 2e 0ti ( ) 4/1 2 2 0 2 20}] 1 i t 1 ti 1 2 2 ti exp[ 0 { 2 2 e 0 1( e 0 )}] 4/1 2 2 0 2 0 or 1 i t 1 t),( exp[ 0 { 2 2 (cos t isin t) 4/1 2 2 0 0 0 1 2 1( cos2 t isin2 t)}] 2 0 0 0 1 1 t 1 exp[ ( cos t)2 i( 0 sin t 2 sin2 t)] 4/1 2 0 0 2 0 0 4 0 0 Finally we have 1 | t |),( 2 exp[( cos( t))2 ] 2/1 0 0 (( Mathematica )) 63 Clear @"Global` ∗"D; ∗ exp_ := exp ê. 8Complex @re_ , im_ D Complex @re , −im D<; 1 2 2 KSH @ξ_, t_ , ξ1_ D := Exp B IIξ + ξ1 M Cos @ω0 tD − 2 ξ ξ1MF; 2 π Sin @ω0 tD 2 Sin @ω0 tD 2 −1 4 ξ − ξ0 ϕ0 ξ_ := π ê Exp − H L ; @ D B 2 F ∞ f1 = KSH ξ, t, ξ1 ϕ0 ξ1 ξ1 FullSimplify , Im Cot t ω0 > −1, ω0 t > 0 & ‡−∞ @ D @ D êê @ 8 @ @ DD ξ2+ξ02 Cot t ω0 + ξ ξ+2 ξ0 Csc t ω0 − J N @ D H @ DL 2 +Cot t ω0 H @ DL − Csc @t ω0D 1 4 π ê 1 − Cot @t ω0D ∗ Amp1 = f1 f1 êê FullSimplify 2 −Hξ−ξ0 Cos @t ω0DL π 0 0 . rule1 = 8ω → 1, ξ → 1<; H1 = Amp1 ê rule1; Plot @Evaluate @Table @H1, 8t, 0, 20, 2 0.5 0.4 0.3 0.2 0.1 x -3 -2 -1 1 2 3 15. Gaussian wave packet propagation (quantum mechanics) x t)( x Uˆ tt )',( t )'( dx' x Uˆ tt x')',( x' t )'( , i K txtx )',';,( x Uˆ tt x')',( x exp[ Hˆ (t t ))' x' , ℏ or x t)( dx'K txtx )',';,( x' t )'( . K(x, t; x’, t’) is referred to the propagator (kernel) For the free particle, the propagator is given by 64 m im(x x )' 2 K txtx )',';,( exp[ ]. 2iℏ(t t )' ℏ(2 t t )' Let’s give a proof for this in the momentum space. Hˆ is the Hamiltonian of the free particle. 1 kx eikx , 2 ˆ H k Ek k , with ℏ 2k2 E E , k , k 2m k ℏ i K txtx )',';,( x exp[ Hˆ (t t ))' x' ℏ i dk kx k exp[ Hˆ (t t ))' x' ℏ ℏki 2 dk kx k exp[ (t t ))' x' 2m 1 ℏki 2 dk exp[ik(x x )' (t t )]' 2 2m Note that ℏki 2 iℏ(t t )' m(x x )' m(x x )' 2 exp[ik(x x )' (t t )]' exp[ {k }2 ], 2m 2m ℏ(t t )' 2iℏ(t t )' and dk exp(ik 2 ) , i or m im(x x )' 2 K txtx )',';,( exp[ ]. 2iℏ(t t )' ℏ(2 t t )' 65 ((Quantum mechanical treatment)) Probability amplitude that a particle initially at x’ propagates to x in the interval t-t’. This expression is generalized to that for the three dimensions. m im | r r |' 2 K r t r t )',';,( [ ] 2/3 exp[ ]. 2iℏ(t t )' ℏ(2 t t )' We now consider the wave function: 1 x2 x (t )0 exp(ik x ) . 0 4 2 2 x x where the probability 2 2 1 x P x (t )0 exp( 2 ) , 2 x 2 x has the form of Gaussian distribution with the standard deviation x . The Fourier transform is given by k (t )0 dx xk x (t )0 1 1 x2 dxexp(ikx)exp(ik x ) 2 0 4 2 2 x x 1 1 2 exp[ 2 (k k 2 ]) 2 x x 0 2 x 4/1 2 2 2 x exp[ x (k k0 ]) or 2 2 k (t )0 exp[2 2 (k k 2 ]) x x 0 2 1 1 (k k0 ) exp[ 2 ] . 2 1 1 2 2 x 2 x 2 1 (k k0 ) exp[ 2 ] 2 k 2 k 66 1 which is the Gaussian distribution with the standard deviation k . 2 x Then we get the wave function at time t, x t)( dx' K xtx )0,';,( x' )0( 1 m x'2 im(x x )' 2 dx'exp[ik x' ] 2 ℏti 0 4 2 2ℏt 2 x x 1 m 1 mx(x 4ik 2 ) 2ik 2tℏ 2 2 exp[ 0 x 0 x ] ℏ 2 ℏ 2 2 ti 1 2im 4m x 2it x 2 ℏ x t tℏ (xx 4ik 2 ) 2ik 2 2 1 1 0 x 0 m x exp[ ℏ ] 2 itℏ 2 it x 1 4 x 1( ) 2 2m 2 2m x x 2 2 tℏ 2 itℏ ({ xx 4ik0 x ) 2ik0 x 1}( 2 ) 1 1 m 2m x exp[ ℏ ℏ ] 2 itℏ 2 it it x 1 4 x 1( 1)( ) 2 2m 2 2m 2 2m x x x ℏ 0tk 2 (x ) 4 2 2 4 1 1 m 8( k mx (xt 4k ℏ)) exp[ m i 0 x 0 x ] ℏ 22 ℏ22 itℏ 2 t 2 4 t 2 x 1 2 4 x 1( 2 4 ) 8m x 1( 2 4 ) 2m x 4m x 4m x Since 1 1 itℏ ℏ22 1 t 2 1 2 4 2m x 4m x the probability is obtained as ℏ tk (x 0 )2 2 1 x t)( exp[ m ]. 22 ℏ22 t ℏ 2 t 2 1 2 x 1( ) x 2 4 4m2 4 4m x x 67 which has the form of Gaussian distribution with the standard deviation t ℏ 22 1 . x 2 4 4m x and has a peak at ℏ tk x 0 . m The Fourier transform: 1 k t)( dx xk x t)( dxeikx x t)( 2 tℏ (xx 4ik 2 ) 2ik 2 2 1 1 1 0 x 0 m x dxexp[ikx ℏ ] 2 2 itℏ 2 it x 1 4 x 1( ) 2 2m 2 2m x x 2 2 ℏ 1 (k k0 ) ik t exp[ 2 ] 1 1 2m 2 4 2 x 2 x where tℏ (xx 4ik 2 ) 2ik 2 2 1 1 0 x 0 m x x t)( exp[ ℏ ]. 2 itℏ 2 it x 1 4 x 1( 2 ) 2 2m 2m x x Then the probability is obtained as 2 2 1 (k k0 ) k t)( exp[ 2 ]. 2 k 2 k where 1 k . 2 x Therefore 68 2 2 k t)( k (t )0 . H»yHx,tL»L2 2.0 1.0 0.5 0.2 0.1 x 0 1 2 3 4 5 2 Fig . Plot of x t)( as a function of x where the time t is changed as a parameter. In summary, we have ℏ tk (x 0 )2 2 1 x t)( exp[ m ]. 22 ℏ22 t ℏ 2 t 2 1 2 x 1( ) x 2 4 4m2 4 4m x x and 2 2 1 (k k0 ) k t)( exp[ 2 ]. 2 k 2 k 16. Wave packet for simple harmonics (quamtum mechanics) ((L.I. Schiff p.67-68)) 69 i x t)( x exp( ˆtH ) (t )0 ℏ i x exp( ˆtH x') x' (t )0 dx' ℏ We define the kernel K txx ),',( as i K txx ),',( x exp( ˆtH x') ℏ i nx exp( tE ) xn ' ℏ n n i exp( tE ) x)( * x )'( ℏ n n n n Note that x , with m 0 . ℏ Then we have i tx ),( exp( tE ) dx' * x )'( x )'( x)( . ℏ n n n n We assume that 2/1 1 x)( exp[ 2 (x a 2 ]) , 4/1 2 or 1 1 )( n nx x)( , n n or 1 1 )( exp[ ( 2 ]) , 4/1 2 0 70 with 0 x0 . We need to calculate the integral defined by I dx' * x )'( x )'( n 2/1 1 dx' * x )'( exp[ 2 x'( a 2 ]) n 4/1 2 d 2/1 1 *2/1 )( exp[ ( 2 ]) n 4/1 2 0 1 1 d* )( exp[ ( 2 ]) 4/1 n 2 0 Here 1 2 n 2 2 n () ( 2 n!) e Hn(). Then we get 1 1 2 1 I ( 2n n!) 2 d exp( )H ()exp[ ( )2 ]) . 1 / 4 2 n 2 0 Here we use the generating function: n 2 s exp(2s s ) Hn (). n 0 n! Note that 2 1 sn 2 1 d exp(2s s2 )exp( )exp[ ( 2 ]) d H )( exp( )exp[ ( 2 ]) 0 n 0 2 2 n0 n! 2 2 sn 1 d H )( exp[( 2 2 )] n 0 0 n0 n! 2 The left-hand side is 71 1 2 d exp(2s s2 )exp[( 2 2 )] 2/1 exp(s 0 ) 0 2 0 0 4 2 sn n 2/1 exp( 0 ) 0 4 n0 n! Thus we have 2 sn n s n 1 1/ 2 exp( 0 ) 0 d H ()exp[(2 2 )]. n 0 0 4 n 0 n! n0 n! 2 or 2 1 1/ 2 exp( 0 ) n dH () exp[( 2 2 )]. 0 n 0 0 4 2 Then 1 2 I ( 2n n!) 2 exp( 0 ) n , 4 0 and 1 i 2 (x,t) exp( E t)(2 n n!) 2 exp( 0 ) n (x) . ℏ n 0 n n 4 Since 1 E ℏ (n ) , n 0 2 or i i exp( E t) exp( t in t), ℏ n 2 0 0 and 1 t),( tx ),( , we get 72 1 2 t n 2 0 i 0 n i (,t) (2 n!) exp( )(0e ) exp( 0t)n () , n 0 4 2 or 2 2 1 0t i t),( 2( n n )! 1 exp( 0 )( ei )n exp( )et 2 H )( 4/1 4 0 2 0 n n0 2 2 i0t 1 0 i 1 0e n 4/1 exp( 0t ) ( ) H n )( 4 2 2 n0 n! 2 Using the generating function i 0t 1 0e n 1 2 2i n 0t i 0t ( ) Hn () exp[ 0 e 0e ], n0 n! 2 4 we have the final form 2 2 1 i 1 2 0t 0t t),( exp( 0 t e2i ei ) 4/1 4 2 2 0 4 0 0 1 2 2 i 1 exp[ 0 t 2 (cos2 t isin 2 t) 4/1 4 2 2 0 4 0 0 0 0(cos0t isin0t)] OR 2 2 1 1 (,t) exp[ 0 2 2 cos2 t 2 cos t] 1/ 2 2 2 0 0 0 0 or 2 1 t),( exp[( cos t 2 ]) 2/1 0 0 2 (,t) represents a wave packet that oscillates without change of shape about = 0 with amplitude 0 and angular frequency 0. 17. Mathematica 73 ,t 2 0.6 0.4 0.2 4 2 2 4 2 1 Fig. The time dependence of t),( exp[( cos t 2 ]) , where =1. 2/1 0 0 0 T 2 /0 . The peak shifts from = 0 at t = 0 to = 0 at t = T/4, = 0 at t = T/2, = - 0 at t = 3 T/4, and = 0 at t = T. 74 75 18. Neutron interferometry Suppose that the interferometry initially lies in a horizontal plane so that there are no gravitational effects. We then rotate the plane formed by the two paths by angle about the segment AC. The segment BD is now higher than the segment AC by l2sin . 76 Fig. Dependence of gravity-induced phase on angle of rotation . From R. Colella, A. W. Overhauser, and S. A. Werner, Phys.. Rev. Lett. 34 ( 1975) 1472. ((Note)) mgtt[24 ( ') 12( xx ') 2 12 gtt ( ') 2 ( xx ') S( xtxt , , ', ') cl 24(t t ') l where x is the height Along the path BD, t t' T 1 , and xx' x l sin v 0 1 l Along the path AC, t t' T 1 , and x x' x 0 v 0 2 4 2 m( g T 12 gT 2 x 0 ) Scl ( x0 , T ) 24 T m( g2 T 3 24 Tgx ) 0 24 From the previous discussion we have the propagator for the gravity i K ;,( xtx t ), Aexp( S ) 00 ℏ cl i ix mgT exp( mg T 32 0 ) 24ℏ ℏ ix mgT A exp( 0 ) 1 ℏ For the path ABD and path ACD, we have i i i exp[ S(ABD)] exp[ S(ACD)] exp[ S(ACD)]{1 exp(i)}. ℏ ℏ ℏ The phase difference between the path ABD and the path ACD is given by S()()1 ABD S ACD 1 m2 gl l mgl T sin 1 2 sin , ℏ ℏ2 ℏ p or 1 m2 gl l m 2 gl l 12sin 12 sin sin . ℏ2 ℏ 2 ℏ 2 77 where p is the momentum p mv . T is related to l1 as l ml ml T 1 1 1 . v p 2ℏ The Probability is 2 2 sin 4cos2 ( ) . 2 Note that m2 lgl 21 , 2ℏ2 with mg =1.025 neV/cm. where g = 9.80446 m/s 2 at Binghamton, NY (USA). The energy of neutron is evaluated as 2 ℏ2 ℏ2 2 E k 2 = 40.63 meV 2M n 2M n for =1.419Å. (( Mathematica )) 78 Clear "Global` " ; rule1 1.054571628 10 27 , mn 1.674927211 10 24 , qe 4.8032068 10 10 , meV 1.602176487 10 15 , A0 10 8, neV 1.602176487 10 21 , g 980.446, 1.419 A0 ; mn g . rule1 neV 1.02497 2 2 2 E1 . rule1 2 mn meV 40.6266 19. Quantum-mechanical interference to detect a potential difference A low-intensity beam of charged particles, each with charge q, is split into two parts. each part then enters a very long metallic tube shown above. Suppose that the length of the wave packet for each of the particles is sufficiently smaller than the length of the tube so that for a certain time interval, say from t0 to t, the wave packet for the particle is definitely within the tubes. During this time interval, a constant electric potential V1 is applied to the upper tube and a constant electric potential V2 is applied to the lower tube. The rest of the time there is no voltage applied to the tubes. Here we consider how the interference pattern depends on the voltages V1 and V2. Without the applied potentials, the amplitude to arrive at a particular point on the detecting screen is 79 1 2 . The intnsity is proportional to 2 2 2 * * I0 1 2 1 2 (1 2 21 ) . With the potential, the Lagrangian in the path is modified as L0 L L0 qV . Thus the wave functions are modified as i t i t exp[ (qV )dt] exp[ (qV )dt] 1 ℏ 1 2 ℏ 12 t0 t0 iqV iqV exp[ 1 (t t )] exp[ 2 (t t )] 1 ℏ 0 2 ℏ 0 iqV iqV exp[ 1 t] exp[ 2 t] 1 ℏ 2 ℏ where t t t0 . The intensity of the screen is proportional to I 2 iqV iqV 2 exp[ 1 t] exp[ 2 t] 1 ℏ 2 ℏ iq(V V ) 2 exp[ 1 2 t] 1 2 ℏ We assume that q( V V ) 1 2 t . ℏ Then we have i 2 2 2 *i * i Ie12 1 2( 12 ee 12 ) When 1 2 0 , we have 80 2 2 I 2 (1 cos ) 4 cos 2 0 0 2 The intensity depends on the phase; I becomes maximum when 2n and minimum at 1 (2 n ) . 2 (( Note )) Method with the use of gauge transformation The proof for the expression can also be given using the concept of the Gauge trasnformation. The vector potential A and scalar potential are related to the magnetic field B and electric field E by B A , 1 A E , c t The gauge transformation is defined by A' A , 1 ' , c t where c is an arbitrary function. The new wave function is related to the old wave function through iq '()r exp( ) () r . ℏc Suppose that ' 0 . Then r)(' 0 r )( is the wave function of free particle. Then we get 1 , c dt . c t The wave function r )( is given by 81 iq (r ) exp( ) ( r ) ℏc 0 iq exp( c dt ) (r ) ℏc 0 iq exp( dtV ) (r ) ℏ 0 When V is the electric potential and is independent of time t, we have iq ()r exp[ V ( t t )] () r ℏ 0 0 This expression is exactly the same as that derived from the Feynman path integral. 20. Quantization of magnetic flux and Aharonov-Bohm effect The classical Lagrangian L is defined by 1 q L mv 2 q v A . 2 c in the presence of a magnetic field. In the absence of the scalar potential ( 0 ), we get 1 q e L mv 2 v A L )0( v A , cl 2 c c c where the charge q = -e (e>0), A is the vector potential. The corresponding change in the action of some definite path segment going from (rn1,tn1 ) to (rn1,tn ) is then given by t e n dr S )0( n,( n )1 S )0( ,( nn )1 dt A , c dt tn1 This integral can be written as t r e n dr e n dt A A dr , c dt c t n1 rn1 where dr is the differential line element along the path segment. Now we consider the Aharonov-Bohm (AB) effect. This effect can be usually explained in terms of the gauge transformation . Here instead we discuss the effect using the Feynman’s path integral. In the best known version, electrons are aimed so as to pass through two regions that are free of electromagnetic field, but which are separated from 82 each other by a long cylindrical solenoid (which contains magnetic field line), arriving at a detector screen behind. At no stage do the electrons encounter any non-zero field B. Fig. Schematic diagram of the Aharonov-Bohm experiment. Electron beams are split into two paths that go to either a collection of lines of magnetic flux (achieved by means of a long solenoid). The beams are brought together at a screen, and the resulting quantum interference pattern depends upon the magnetic flux strength- despite the fact that the electrons only encounter a zero magnetic field. Path denoted by red (counterclockwise). Path denoted by blue (clockwise) 83 Reflector C1 Incident Electron beams slit -1 P slit -2 Q Screen B out of page C2 Reflector Fig. Schematic diagram of the Aharonov-Bohm experiment. Incident electron beams go into the two narrow slits (one beam denoted by blue arrow, and the other beam denoted by red arrow). The diffraction pattern is observed on the screen. The reflector plays a role of mirror for the optical experiment. The path1: slit-1 – C1 – S. The path 2: slit-2 – C2 – S. Let 1B be the wave function when only slit 1 is open. ie r)( r)( exp[ dr A(r)] , (1) ,1 B 0,1 ℏc Path1 The line integral runs from the source through slit 1 to r (screen) through C 1. Similarly, for the wave function when only slit 2 is open, we have ie r)( r)( exp[ dr A(r)], (2) ,1 B 0,2 ℏc Path2 The line integral runs from the source through slit 2 to r (screen) through C 2. Superimposing Eqs.(1) and (2), we obtain ie ie r)( r)( exp[ dr A(r)] r)( exp[ dr A(r)] . B 0,1 ℏc Path1 0,2 cℏ Path2 84 The relative phase of the two terms is dr A r)( dr A r)( dr A r)( ( A) da , Path1 Path2 by using the Stokes’ theorem, where the closed path consists of path1 and path2 along the same direction. The relative phase now can be expressed in terms of the flux of the magnetic field through the closed path, e e e e A dr ( A) da B da . cℏ cℏ cℏ cℏ where the magnetic field B is given by B A . The final form is obtained as ie r)( exp[ dr A(r [)] r)( exp(i ) (r)], B ℏc Path2 0,1 0,2 and is the magnetic flux inside the loop. It is required that 2n . Then we get the quantization of the magnetic flux, 2cℏ n , n e where n is a positive integer, n = 0, 1,2,….. Note that 2cℏ =4.1356675 10 -7 Gauss cm 2. e which is equal to 2 0 , where 0 is the magnetic quantum flux, 2cℏ = 2.067833758(46) x 10 -7 Gauss cm 2. (NIST) 0 2e We note that e e 0 ℏ ℏ c0 c 0 85 The intensity is i i II0 (1 e )(1 e ) 2I0 [1 cos( )] 2 4cos 2 2 4cos 0 4 I I0 3 2 1 1 1 2 3 4 5 x 0 Suppose that the area for the region of magnetic field is A 1 mm 2 =10 -2 cm 2. 102B 10 5 B 7 0 2.06783 10 2.06783 4 If B 0.3 G , /0 1.45 10 . If B 1 mG , / 0 48 . If B 0.01 mG , / 0 0.48 . 21. Example-1: Feynman path integral We consider the Gaussian position-space wave packet at t = 0, which is given by 1 x2 x (t )0 exp( ) (Gaussian wave packet at t = 0). 2 2 2 86 The Gaussian position-space wave packet evolves in time as x (t dx K ;,( xtx )0, x )0( 0 0 0 1 m x 2 im(x x )2 dx exp[ 0 0 ] (1) 2 2 ℏti 0 2 2 2ℏt 1 1 x2 exp[ ℏ ] 2 ℏti 2 ti 2 (2 ) m m where m im(x x )2 K ;,( xtx ,t )0 ( ) 2/1 exp[ 0 ] (free propagator) (2) 00 2iℏt 2ℏt (Note) You need to show all the procedures to get the final form of x (t . (a) Prove the expression for x (t given by Eq.(1). 2 (b) Evaluate the probability given by x (t for finding the wave packet at the position x and time t. (a) The Gaussian wave packet: 1 x2 x (t )0 exp( ) . (Gaussian) 2 2 2 The free propagator: m im(x x )2 K ;,( xtx ,t )0 ( ) 2/1 exp[ 0 ]. 00 2iℏt 2ℏt Then we have x (t dx K ;,( xtx )0, x )0( 0 0 0 2/1 2 2 1 m x0 im(x x0 ) dx0 exp[ ] 2 2 ℏti 2 2 2ℏt Here we have 87 x 2 im(x x )2 x 2 im(x2 2 xx x 2 ) 0 0 0 0 0 2 2 2ℏt 2 2 2ℏt 1 im imx imx2 ( )x 2 ( )x 2 2 2ℏt 0 ℏt 0 2ℏt 2 ax0 bx0 c b b2 4ac a(x )2 0 2a 4a where 1 im imx imx2 a , b , c . 2 2 2ℏt ℏt 2ℏt Then we get the integral b b2 dx exp[ax 2 bx c] dx exp[a(x )2 c 0 0 0 0 0 2a 4a b2 b exp(c ) dx exp[a(x )2 0 0 4a 2a b2 exp(c ) a 4a 1 Note that when Re(a) 0 , the above integral can be calculated as 2 2 b 1 dx exp[a(x )2 dy exp( y2 ) , 0 0 2a a a b with the replacement of variable as y a(x ) and dy dx a . Thus we have 0 2a 0 1 m b2 x (t exp(c ) 2 2 ℏti a 4a 2 imx 1 m imx2 ℏt exp[ ] 2 2 ℏti 1 im 2 2ℏt 1 im (4 ) 2 2ℏt 2 2 2ℏt or 88 2 imx 1 m imx2 ℏt x (t exp[ ] 2 1 im 2 2ℏt 1 im 2 ℏti ( ) (4 ) 2 2ℏt 2 2 2ℏt 2 mx 1 m imx2 ℏt exp[ 2 ] 2 im 2 2ℏt ℏt im ℏti 2( ℏti ) (2 ) 2ℏt 2ℏt 2 mx 1 1 imx2 ℏt exp[ ] ℏti ℏ ℏti 2 2 2 t 2 m 2im( m ) 2ℏt 2 mx 2 2 1 1 imx ℏt ℏt ℏ exp[ ℏ ] 2 ti 2 2ℏt 2im 2 ti m m or 1 1 imx2 imx2 2 x (t exp( ) ℏti ℏ ℏ ℏti 2 2 2 t 2 t 2 m m 1 1 imx2 2 exp[ 1( )] ℏti ℏ ℏti 2 2 2 t 2 m m ℏti 1 1 imx2 exp[ ( m )] ℏti ℏ ℏti 2 2 2 t 2 m m 1 1 x2 exp[ ℏ ] 2 ℏti 2 ti 2 (2 ) m m Finally we get 1 1 x2 x (t exp[ ℏ ]. 2 ℏti 2 ti 2 (2 ) m m 89 It is clear that at t = 0, x (t is the original Gaussian wave packet. (b) 1 1 x2 ℏti x (t exp[ ( 2 )] 2 ℏti ℏ t 22 m 2 (2 4 ) m m2 ℏti 2 22 i x 1 1 x m exp[ 22 ]exp[ 22 ] 2 ℏti 4 ℏ t 4 ℏ t 2 (2 ) (2 ) m m2 m2 ℏt 2 22 i x 1 1 x exp[ ]ei exp[ m ] 22 4/1 ℏ t 22 ℏ t 22 2 4 ℏ t 4 4 (2 2 ) (2 2 ) 2 m m m where ℏti ℏ t 22 ℏ t 22 2 4 ei with arctan( ) . m m2 m 22 Then we have 22 2 1 1 x x (t exp[ ] 22 ℏ 22 2 4 ℏ t 4 t ( 2 ) m2 m 2 1 1 The height of x (t is . 2 ℏ t 22 4 m2 1 ℏ t 22 The width is x 4 . m2 22. Example-2: Feynman path integral Suppose that the Gaussian wave packet is given by 1 x2 x (t )0 exp(ik x ) . (Gaussian) 2 0 2 2 90 Here we discuss how such a Gaussian wave packet propagates along the x axis as the time changes. The free propagator: m im(x x )2 K ;,( xtx ,t )0 ( ) 2/1 exp[ 0 ]. 00 2iℏt 2ℏt Then we have x (t dx K ;,( xtx )0, x )0( 0 0 0 2/1 2 2 1 m x0 im(x x0 ) dx0 exp[ ik x00 ] 2 2 ℏti 2 2 2ℏt Here we have x 2 im(x x )2 x 2 im(x2 2 xx x 2 ) 0 0 ik x 0 0 0 ik x 2 2 2ℏt 00 2 2 2ℏt 00 1 im mx imx 2 ( )x 2 (ki )x 2 2 2ℏt 0 0 ℏt 0 2ℏt 2 ax0 bx0 c b b2 4ac a(x )2 0 2a 4a where 1 im mx imx2 a , b (ki ) , c . 2 2 2ℏt 0 ℏt 2ℏt Then we get the integral b b2 dx exp[ax 2 bx c] dx exp[a(x )2 c 0 0 0 0 0 2a 4a b2 b exp(c ) dx exp[a(x )2 0 0 4a 2a b2 exp(c ) a 4a 91 1 Note that when Re(a) 0 , the above integral can be calculated as 2 2 b 1 dx exp[a(x 2 ]) dy exp( y 2 ) , 0 0 2a a a b with the replacement of variable as y a(x ) and dy dx a . Thus we have 0 2a 0 1 m b2 x (t exp(c ) 2 2 ℏti a 4a 2 imx ik 1 m imx2 0 ℏt exp[ ] 2 2 ℏti 1 im 2 2ℏt 1 im (4 ) 2 2ℏt 2 2 2ℏt or 2 imx ik 1 m imx 2 0 ℏt x (t exp[ ] 2 1 im 2 2ℏt 1 im 2 ℏti ( ) (4 ) 2 2ℏt 2 2 2ℏt 2 mx k0 1 m imx2 ℏt exp[ 2 ] 2 im 2 2ℏt ℏt im ℏti 2( ℏti ) (2 ) 2ℏt 2ℏt 2 mx k 1 1 imx 2 ℏt 0 exp[ ] ℏti ℏ ℏti 2 2 2 t 2 m 2im( m ) 2ℏt 2 mx k0 1 1 imx 2 2ℏt ℏt ℏ exp[ ℏ ] ti 2 2ℏt 2im 2 ti 2 m m or 92 ℏ ℏ 0tk 2 2 2 ti 2 im(x ) ( ) 1 1 imx x (t exp( m m ) ℏ ℏ ℏ ℏ 22 2 ti 2 2 t 2 t 4 t m m2 1 1 i 4/1 e 2 ℏ t 22 4 2 m ℏ ℏ 0tk 2 2 0tk 2 2 im(x ) 4 (x ) imx 1 exp( m )exp[ m ] 2ℏt 2ℏt ℏ t 22 2 ℏ t 22 4 4 m2 m 2 where ℏti ℏ t 22 ℏ t 22 2 4 ei with arctan( ) . m m2 m 22 Then we have ℏ tk 2 (x 0 )2 2 1 1 x (t exp[ m ] . 2 ℏ 22 ℏ t 22 4 t 4 ( 2 ) m 2 m This means the center of the Gaussian wave packet moves along the x axis at the constant velocity. 2 1 1 (i) The height of x (t is . 2 ℏ t 22 4 m2 1 ℏ t 22 (ii) The width is x 4 . m2 ℏk (iii) The Group velocity is v 0 . g m 23. Summary: Feynman path integral The probability amplitude associated with the transition from the point tx ii ),( to (x ,t ff ) is the sum over all paths with the action as a phase angle, namely, 93 i Amplitude = exp( S) , ℏ All paths where S is the action associated with each path. So we can write down K(x ,t ff ; tx ii ), x ,t ff ,tx ii t i f exp( dtL) ℏ All t paths i i (tF t )exp( S ) f i ℏ cl where Scl is the classical action associated with each path. If the Lagrangian is given by the simple form L ɺ txx ),,( )( xta ɺ 2 )( ɺxxtb )( xtc 2 , then (tF tif ), can be expressed by (tF tif ), K(x f ,0 t f ; xi ti ),0 . 24. Comment by Roger Penrose on Feynman path integral Roger Penrose: Road to the Reality. A Complete Guide to the Law of the Universe (Jonathan Cape London, 2004). Here is a very interesting comment by Roger Penrose (Nobel laureate, 2020) on the Feynman Path integral. The content is the same, but some sentences are appropriately revised. The Lagrangian is in many respects more appropriate than a Hamiltonian when we are concerned with a relativistic theory. The standard Schrödinger/Hamiltonian quantization procedures lie uncomfortably with the spacetime symmetry of relativity. However, unlike the Hamiltonian, which is associated with a choice of time coordinate, the Lagrangian can be taken to be a completely relativistically invariant entity. The basic idea, like so many of the ideas underlying the formalism of quantum theory, is one that goes back to Dirac, although the person who carried it through as a basis for relativistic quantum theory was Feynman. Accordingly, it is commonly referred to as the formulation in terms of Feynman path integrals or Feynman sum over histories. 94 The basic idea is a different perspective on the fundamental quantum mechanical principle of complex linear superposition. Here, we think of that principle as applied, not just to specific quantum states, but to entire spacetime histories. We tend to think of these histories as ‘possible alternative classical trajectories’ (in configuration space). The idea is that in the quantum world, instead of there being just one classical ‘reality’, represented by one such trajectory (one history). There is a great complex superposition of all these ‘alternative realities’ (superposed alternative histories). Accordingly, each history is to be assigned a complex weighting factor, which we refer to as an amplitude if the total is normalized to modulus unity, so the squared modulus of an amplitude gives us a probability. We are usually interested in amplitudes for getting from a point a to a point b in configuration space. The magic role of the Lagrangian is that it tells us what amplitude is to be assigned to each such history. If we know the Lagrangian L, then we can obtain the action S, for that history (the action being just the integral of L for that classical history). The complex amplitude to be assigned to that particular history is then given by the deceptively simple formula i i Complex amplitude exp(S ) exp[ Ldt ] . ℏ ℏ For each history, there will be some action S, where S is the integral of the Lagrangian along the path. All the histories are supposed to ‘coexist’ in quantum superposition, and each history is assigned a complex amplitude exp(iS /ℏ ) . How are we to make contact with Lagrange’s requirement, perhaps just in some approximate sense, that there should be a particular history singled out for which the action is indeed stationary? The idea is that those histories within our superposition that are far away from a ‘stationary-action’ history will basically have their contributions cancel out with the contributions from neighboring histories. This is because the changes in S that come about when the history is varied will produce phase angles exp(iS /ℏ ) that vary all around the clock, and so will cancel out on the average. Only if the history is very close to one for which the action is large and stationary (so the argument runs), will its contribution begin to be reinforced by those of its neighbors, rather than cancelled by them. because in this case there will be a large bunching of phase angles in the same direction. This is indeed a very beautiful idea. In accordance with the ‘path-integral’ philosophy, not only should we obtain the classical history as the major contributor to the total amplitude—and therefore to the total probability—but also the smaller quantum corrections to this classical behavior, arising from the histories that are not quite classical and give contributions that do not quite cancel out, which may often be experimentally observable. 95 ______REFERENCES L.D. Landau and E.M. Lifshitz, The Classical Theory of Fields (Pergamon, 1996). R.P. Feynman, Nobel Lecture December 11, 1965: The development of the space-time view of quantum electrodynamics http://www.nobelprize.org/nobel_prizes/physics/laureates/1965/feynman- lecture.html R.P. Feynman and A.R. Hibbs, Quantum Mechanics and Path Integrals (extended edition) Dover 2005. J.J. Sakurai and J. Napolitano, Modern Quantum Mechanics , second edition (Addison- Wesley, New York, 2011). A. Das; Lecture Notes on Quantum Mechanics , 2nd edition (Hindustan Book Agency, 2012). John S. Townsend, A Modern Approach to Quantum Mechanics , second edition (University Science Books, 2012). R.P. Feynman, R.B. Leighton, and M. Sands, The Feynman Lectures on Physics vol. II (Basic Book, 2010). J. Mehra, The beat of different drum; The life and science of R. Feynman (Oxford University Press, 1994). H. Rauch and S.A. Werner, Neutron Interferrometry: Lessons in Experimental Quantum Mechanics (Oxford, 2000). K. Gottfried and T.-M. Yan, Quantum mechanics: Fundamentals , 2nd edition (Springer, 2003). M. Reuter, Classical and Quantum Dynamics: From Classical Paths to Path Integral , 3rd edition (Springer, 2001). L.S. Schulman, Techniques and Applications of Path Integration (Wiley-Interscience, 1981). Colella, A. W. Overhauser, and S. A. Werner, Phys.. Rev. Lett. 34 ( 1975) 1472.”Observationof Gravitationally Induced Quantum Interference .” E. Merzbacher, Quantum Mechanics, 3 rd edition (John Wiley & Sons, 1998). Roger Penrose: Road to the Reality. A Complete Guide to the Law of the Universe (Jonathan Cape London, 2004). ______APPENDIX-I Mathematical formula-1 b2 dxexp(ax2 bx c) exp( c) a 4a for Re[a] 0 96 (( Mathematica )) ______APPENDIX-II Action in the classical mechanics We start to discuss the calculus of variations with an action given by the form t f S L ɺ xx ],[ dt , ti dx where xɺ . The problem is to find has a stationary function x x)( so as to minimize dt cl the value of the action S. The minimization process can be accomplished by introducing a parameter . 97 x xf x t xcl t xi ti tf t Fig. tx i )( xi , (tx f ) x f , x t)( xcl t)( t)( , where is a real number and x ticl )( xi , xcl (t f ) x f , ti )( 0 , (t f ) 0 , x x d )( dt , 0 98 t f S[x ] L(x t)( t),( xɺ t)( ɺ t))( dt , cl cl ti has a minimum at = 0. t f S[x ] L[x t)]( dt , 0 cl ti L[x ] L 0 , L | 0 d , 0 t S[x ] f L L { t)( ɺ(t)}dt x xɺ t i t f t f L L t d L t)( dt (t |)} f ( t)() dt x xɺ t i dt xɺ t i t i t f L d L [ ( )] t)( dt x dt xɺ t i (1) The Taylor expansion: t f S[x ] L(x t)( t),( xɺ t)( ɺ t))( dt cl cl ti ((Fundamental lemma)) If t f M t t)()( dt 0 t i for all arbitrary function (t) continuous through the second derivative, then M(t) must identically vanish in the interval ti t t f . ______From this fundamental lemma of variational and Eq.(1), we have Lagrange equation 99 L d L ( ) 0 . (2) x dt xɺ L can have a stationary value only if the Lagrange equation is valid. In summary, x2 S L xx ɺ),( dt , x1 L d L S = 0 ( ) 0 . x dt xɺ 100