Path Intergal Shoichi Midorikawa 1 Introduction The Feynman path integral1 is one of the formalism to solve the Schrödinger equation. However this approach is not peculiar to quantum mechanics, and M. Kac is the one who recognized the applicability to the diffusion equation. Therefore, this formula is nowadays known as Feynman-Kac formula. The purpose of this note is to derive the path integral from the Schrödinger equation in a general way so that it can also be applicable to the diffusion equation. I will also show a somewhat different way to calculate the prefactor of the kernel of the simple harmonic oscillator. Derivation From Schrödinger Equation Consider the one-dimensional Schrödinger equation iħ ( ) t ψ(x, t) = ħ m x + V (x) ψ(x, t). (1) The formal solution is ψ(x, t) = i ( ) ħ ħ m x + V (x) t ψ(x, 0). We can rewrite ψ(x, t) = The kernel is defined by K(x, t; x 0, t 0 ) = i ħ i ħ ( ħ ) m x + V (x) t δ(x x 0 )ψ(x 0, 0) dx 0 ( ħ ) m x + V (x) (t t 0 ) δ(x x 0 ) () 1
Then we have ψ(x, t) = The successive use of eq.(3) leads to ψ(x c, t c ) = = K(x, t; x 0, t 0 )ψ(x 0, t 0 ) dx 0 (3) K(x c, t c ; x b, t b )ψ(x b, t b ) dx b K(x c, t c ; x b, t b ){K(x b, t b ; x a, t a )ρ(x a, t a ) dx a } dx b This leads to the following relationship between the kernels and Let K(x c, t c ; x a, t a ) = Nϵ = t ϵ = t i+1 t i (i = 0, 1,,, t N ) t 0 = 0, x 0 = 0, t N = t x N = x K(i + 1; i) = K(x i+1, t i+1 ; x i, t i ) for abbreviation. Then, K(x, t; 0, 0) = K(x c, t c ; x b, t b )K(x b, t b ; x a, t a ) dx b Furthermore, we have for small ϵ i ) ( ħ ħ m x + V (x) ϵ = iħ V (x)ϵ i ħ K(N; N 1)K(N 1; N ) K(; 1)K(1; 0) dx N 1 dx dx 1 (4) ( ħ m x ) ϵ ϵ ħ ħ m x, V (x) + O(ϵ 3 )
Thus i ħ ( ħ ) m x + V (x) ϵ iħ V (x)ϵ i ħ as ϵ 0. Using the integral representation of the delta function δ(x i x i 1 ) = 1 πħ e ip(x i x i 1 )/ħ dp we can evaluate K(i; i 1) in the limit of ϵ 0: K(i; i 1) = 1 i πħ e = 1 = 1 i πħ e i πħ e ħ V (x i)ϵ ħ V (x i)ϵ ħ V (x i)ϵ The p integral is of the Gaussian form e p σ dp = πσ ) ϵ m x ( ħ dp i ) ( ħ ϵ e ip(x i x i 1 )/ħ ħ m x dp i ( ) p ħ m ϵ p(x i x i 1 ) dp iϵ mħ ( p m(x i x i 1 ) ϵ ) + i m(x i x i 1 ) ϵħ We will apply this formula even in case of imaginary σ. Substitution of σ = mħ iϵ gives { ( ) m K(i; i 1) = πiħϵ i h ϵ m xi x i 1 V (x i )} (5) ϵ By substituiting eq.(5) into eq.(4) and taking the limit of ϵ 0, we obtain ( m ) N/ K(x, t; 0, 0) = lim dx 1 dx N 1 ϵ 0 πiħϵ { N i h ϵ ( ) m xi x i 1 V (x i )} (6) ϵ i=1 3
In a continuum limit N ( ) m xi x i 1 ϵ V (x i ) ϵ i=1 t ( ) 1 dt mẋ V (x) = 0 t 0 dt L(ẋ, x) = S (x(t)) where L(ẋ, x) is a Langrangian and S (x(t)) is the action. Furthermore we define the notation ( m ) N/ Dx(t) = lim dx 1 dx N 1 ϵ 0 πiħϵ Then we have K(x, t; 0, 0) = Dx(t) 3 Harmonic Oscillator i ħ S(x(t)) The Lagrangian of the harmonic oscillator is L = m (ẋ ω x ) We wish to calculate Kb; a = b a Dx(t) i tb dt L (ẋ(t), x(t)) ħ t a the integral over all paths which go from (x a, t a ) to (x b, t b ). We can represent x in terms of classical path x c and quantum fluctuation y around classical path; x(t) = x c (t) + y(t) with y(t b ) = y(t a ) = 0 Then, m (ẋ ω x ) = m = m { (ẋc + ẏ) ω (x c + y) } (ẋ ) c ω x c + m(ẋc ẏ ω x c y) + m (ẏ ω y ) 4
The action is Sb, a = tb t a m (ẋ c ω x c) dt tb t a tb m(ẋ c ẏ ω m x c y) dt+ (ẏ ω y ) dt t a The second term is, using integration by parts, tb tb m(ẋ c ẏ ω x c y) dt = mẋ c y t b ta m(ẍ c + ω x c )y dt t a t a which vanishes because y(t a ) = y(t b ) = 0 and ẍ c + ω x c = 0. Thus Sb, a can be divided into two parts. the classcal and the quantum parts; Sb, a = S cl b, a + S q 0, 0 Notice that S q 0, 0 is a function of the time interval T = t b t a. This means that K(b; a) must be of the form where K(b; a) = F (T )e i ħ S clb,a F (T ) = 0 0 i T ħ 0 m Now we evaluate S cl b, a. We write y(t) = A cos(ωt + φ a ). (ẏ ω y ) dt Dy(t) (7) Then x a = A cos φ a, x b = A cos(ωt + φ a ), where T = t b t a. The classical action represented by x a and x b is S cl b, a = mω ( ) x a x b sin ωt A sin φ a (x b cos ωt x a ) (8) Since x b = A cos(ωt + φ a ) = x a cos ωt A sin φ a sin ωt, we have A sin φ a = x a cos ωt x b sin ωt 5
By substituting this into eq.(8), we have S cl b, a = mω ( x a x b sin ωt (x ) b cos ωt x a ) (x a cos ωt x b ) sin ωt mω = ((x a + x b) ) cos ωt x a x b sin ωt Determination Of The Prefactor We evaluate 0 i T m F (T ) = (ẏ ω y ) dt Dy(t) (9) 0 ħ 0 by discritizing the functional integral Dy(t), namely, ( m ) N/ 0 = lim m N ( ) (yi y i 1 ) ω ϵ yi dy 1 dy dy N 1 ϵ 0 πiħϵ iϵ 0 i=1 Note that y 0 = y N = 0. We ress the sum in the onent in terms of the vector and matrix notation where N ( ) (yi y i 1 ) ω ϵ yi = y T M N 1 y i=1 y T = (y 1, y,, y N, y N 1 ) and M N 1 = ω ϵ 1 0 0 0 0 1 ω ϵ 1 0 0 0 0 1 ω ϵ 0 0 0..... 0 0 0 ω ϵ 1 0 0 0 0 1 ω ϵ 1 0 0 0 0 1 ω ϵ 6
The recursive formula of det M n is det M n = ( ω ϵ ) det M n 1 det M n, and the characteric equation is x ( ω ϵ )x + 1 = 0 The solution of this equation is ) x = (1 ω ϵ ± iωϵ 1 ω ϵ 4 cos θ ± i sin θ Note that, det M 1 = ω ϵ = cos θ = sin θ sin θ det M = ( ω ϵ ) ( ω ϵ ) = det M 3 = cos θ We can show det M N 1 = sin Nθ sin θ sin 3θ sin θ sin θ sin θ sin 3θ sin θ = sin 4θ sin θ by mathematical induction. Hece, m 1 m sin θ F (T ) = lim = lim ϵ 0 πiϵ det M N 1 ϵ 0 πiϵ sin Nθ mω = πi sin T ω where we have used T = Nϵ, lim sin θ = lim θ = ωϵ, and lim sin Nθ = sin Nωϵ = sin T ω. ϵ 0 ϵ 0 ϵ 0 Finally the kernel for the harmonic oscillator is { mω K(b; a) = πi sin T ω i mω ((x a + x b) ) } cos ωt x a x b ħ sin ωt 7
4 The Fokker-Planck Equation 4.1 The solution of the Fokker-Planck equation The Fokker-Planck equation for Orstein-Uhlenbeck process is given by 1 P (v, t) = γ (vp (v, t)) + P (v, t) t v v (10) We can solve it using the integral transform method. Let Then, P (v, t) = 1 π 1 P (v, t) = t π e ikv F (k, t) dk e ikv F (k, t) dk t 1 P (v, t) = k e ikv F (k, t) dk v π The special care must be made in the next calculation. 1 (vp (v, t)) = v π ( ) ve ikv F (k, t) dk v Replacing the v in the parentheses by i gives k 1 ( (vp (v, t)) = i ) v π v k eikv F (k, t) dk = 1 ( i ) π k v eikv F (k, t) dk = 1 ( ( )) ke ikv F (k, t) dk π k The integration by parts finally gives 1 (vp (v, t)) = v π e ikv k F (k, t) dk (11) k 8
Then the Fourier transform of the differential equation (10) is F (k, t) = γ k k F (k, t) + F (k, t) t k Dividing by F (k, t), we obtain ln F (k, t) = γ k k ln F (k, t) + t k (1) If we make the Gaussian anzats ln F (k, t) = ikm(t) k S(t) (13) with the unknown function m(t) and S(t), we have ikṁ(t) k Ṡ(t) = ikγm(t) + k γs(t) k γ Compareing equal power of k, we find ṁ(t) = γm(t) ( Ṡ(t) = γ S(t) 1 ) We choose as the initial condition P (v, 0) = δ(v v 0 ) = 1 π meaning that namely, F (k, 0) = ( ikv 0 ), m(0) = v 0, S(0) = 0. The solution is + e ik(v v 0) dk, m(t) = v 0 e γt, (14) S(t) = 1 ( ) 1 e γt. (15) 9
The substitution of these into equation (1) gives P (v, t) = 1 i (v m(t)) k k π S(t) dk = 1 S(t) ( ) (v m(t)) k i π S(t) After the integration, we obtain 1 P (v, t) = πs(t) (v m(t)) S(t) By substituting equations (14) and (15), we obtain eventually P (v, t) = π (1 e γt ) (v v 0 e γt ) (1 e γt ) (v m(t)) dk S(t) If t = ϵ 1, then 1 e γϵ γϵ and e γϵ 1 γϵ, and P (v, t) becomes P (v, ϵ) 4πγϵ (v v 0(1 γϵ)) (16) 4γϵ 4. The Path Integral Formula The Green s function satisfies t G(v, t; v 0, 0) = γ The formal solution is G(v, ϵ; v 0, 0) = { v v + 1 ϵγ v v + 1 G(v, t; v v 0, 0) } δ(v v v 0 ) The calcuation v vδ(v v 0) must be done as the equation (11) has been 10
derived, namely, v vδ(v v 0) = 1 π = 1 π = 1 π = 1 π = 1 π Therefore we get G(v, ϵ; v 0, 0) = 1 π v veik(v v 0) dk ( ) e ikv 0 v veikv dk ( ) e ikv 0 1 v i k eikv dk ( ) ( ) 1 i k e ikv 0 v eikv dk = 1 π = 1 π = 1 π = 1 π iv 0 ke ik(v v 0) dk { } ϵγ v v + 1 e ik(v v0) dk v { } ϵγ iv 0 k k e ik(v v 0) dk iϵγv 0 k ϵγ k + ik(v v 0 ) dk ϵγ k + ik (v v 0 + ϵγv 0 )) dk ϵγ ( k i (v v 0(1 ϵγ)) ϵγ (v v 0(1 ϵγ)) 4ϵγ We finally obtain after the Gaussian intergral of k G(v, ϵ; v 0, 0) = 4πγϵ (v v 0(1 ϵγ)), 4ϵγ which is the same as equation (16). 11 dk )
In gengeral { } G(v i, t i ; v i 1, t i 1 ) = ϵγ v i + 1 δ(v v i vi i v i 1 ) = 4πγϵ (v i v i 1 (1 ϵγ)) 4ϵγ Thus the path-integral representation of the Fokker-Planck solution is ( ) N/ G(v, t; v 0, 0) = lim ϵ 0 4πγϵ References 4ϵγ N (v i v i 1 (1 ϵγ)) dv 1 dv N 1 i=1 1 R. Feynman, Space-Time Approach to Non-Relativistic Quantum Mechanics (1948) Rev. Modern Physics. 0. R.P. Feynman and A.R. Hibbs, Quantum Mechanics and Path Integrals (McGrawHill, New York, 1965). M.Kac, Transactions of the American Mathematical Society 65(1949)1-13. 1