Consider a normalized wavefunction ψ,
(1) ψ = A1ψ1 + A2ψ2 + A3ψ3
Where A1 is the amplitude of ψ1, A2 is the amplitude of ψ2, etc.
Given the rules of QM, the probability of measuring ψ1 is
(2) P1 = A1 A1* = │ A1│2
See equations 23 and 28 in The Essential Quantum Mechanics
The notion is that we started with ψ initially, then after measuring ψ1, the wavefunction underwent a transition to a new state. Equation (2) is the transition amplitude. Now, the transition amplitude in the canonical quantization approach (see The Essential Quantum Field Theory (EQFT) ) is denoted by
(3) for t± → ±∞, < f|U(t,t0)|i > ≡ < f|S|i > ≡ Sfi ( equation 64 in EQFT)
where U(t,t0) = exp(−iH(t−t0)/ℏ)is the time evolution operator (equation 57 in EQFT)
However there is an elapsed time T between the measurements of the initial state ψi and the final state ψf. So we write,
(4) K(ψi,ψf;T) = < ψf|e-iHT/ℏ|ψi >, where K is the propagator.
If we take position as our eigenstates, then equation 4 is,
(5) K(xi,xf;T) = < xf|e-iHT/ℏ|xi > = < xf| ψ >
(6) Where |ψ > = |e-iHT/ℏ|xi >, is the evolved state
(7) K(xi,xf;T) = ∫ δ(x - xf)ψ(x,T)dx = ψ(xf,T)
(8) and │K(xi,xf;T)│2 = ψ*(xf,T)ψ(xf,T), is the probability density at xf.
Writing the Schrödinger Equation (with ℏ restored),
(9) iℏd|ψ>/dt = H|ψ> = E|ψ> ( equ. 19 in The Essential Quantum Mechanics )
A solution is,
(10) |ψ> = C e-i(Et – p.x)/ℏ ,
where C is a constant to be determined later on.
Now to simplify our notation, we will get rid of the Dirac notation, and simply write,
(11) ψ = Ceiϕ
(12) where ϕ = -(Et – p.x)/ℏ, is the phase angle and eiϕ is the phasor.
The wave packet peak travels at the wave group velocity v, which corresponds to the classical particle velocity (fig.1). The time rate of change of the phase angle at the peak is,
(13) dϕ/dt = -(Et – p.v)/ℏ
(14) But E = T + V, where T is the kinetic energy, and V is the potential energy.
(15) Note: T = ½mv2 and p = mv → p.v = 2T
Substituting equations (14) and (15) into (13), we get
(16) dϕ/dt = (T – V)/ ℏ = L/ ℏ, where L is the Lagrangian.
Integrating equation (16),
(17) ϕ = ∫ Ldt/ℏ = S/ℏ where S is the action.
Therefore equation (11) can now be written as,
(18) ψ = C(T)exp(i/ℏ∫Ldt) = C(T) exp(iS/ℏ),
where we take the more generally case that C might be time-dependent.
The central idea of the Path Integral is that a particle/wave traveling between two events could be considered as traveling all possible paths between those two events.
We want to derive equation (18) from the basic idea of the Path Integral. To do that we need to consider finite slices of time. We also discretize space and consider a small number of paths. In our example we will consider three different paths that we label as a, b and c, each of those over two time intervals, t0 → t1 → t2.
For path a:
(19A) La1 = ½m(x122 – x042)/∆t – V(½(x12 + x04)) (19B) La2 = ½m(x252 – x122)/∆t – V(½(x25 + x12))
For path b:
(20A) Lb1 = ½m(x132 – x042)/∆t – V(½(x13 + x04)) (20B) Lb2 = ½m(x252 – x132)/∆t – V(½(x25 + x13))
For path c:
(21A) Lc1 = ½m(x162 – x042)/∆t – V(½(x16 + x04)) (21B) Lc2 = ½m(x252 – x162)/∆t – V(½(x25 + x16))
In terms of the phasors:
For path a:
(22) exp(i/ℏ ∫(La1 + La2)dt)
= exp(i/ℏ ∫ La1dt) exp(i/ℏ ∫ La2dt)
= exp(i/ℏ ∫ ½m(x122 – x042)/∆t – V(½(x12 + x04))dt)
x exp(i/ℏ ∫ ½m(x252 – x122)/∆t – V(½ (x25 + x12))dt)
≈ exp(iS(x04,x12/ℏ) exp(iS(x12,x25)/ℏ)We get similar expression for paths b and c.
The sum of these three paths is:
(23) = ∑ exp(iS(x04,x1j)/ℏ) exp(iS(x1j,x25)/ ℏ), j=1,2,3
Since the transition amplitude is proportional to the above we can multiply by any constant, in this instance, C’ and ∆x1, where xL < x1 < xR.
≈ C’∑ exp(iS(x04,x1j)/ℏ) exp(iS(x1j,x25)/ℏ)∆x1
≈ C’∫xLxR exp(iS(x04,x1)/ℏ) exp(iS(x1,x25)/ℏ)dx1
≈ C’ ∫xLxR exp(i∫x04x25L/ℏ dx1In the general case, we have intervals dx1, dx2, dx3 … dxN, where N → ∞
(25) K(i,f;T=tf–ti)= C ∫∫∫… ∫ ei∫L/ℏ dx1dx2dx3… dxn
= C(T)∫ ei∫L/ℏDx,where the symbol D implies all paths between i and f.
Derivation of the contant
Consider equ.(5), rewritten below
(26) K(xi,xf;T) = < xf|e-iHT/ℏ|xi >
The bra and ket are Dirac delta functions (see equation 27 in The Essential Quantum Mechanics )
(27) K(xi,xf;T) = ∫ δ(x – xf)e-iHT/ℏδ(x – xi)dx
Mathematically the Dirac Delta function is:
(28) δ(x – xi) = (2π)-1∫eik(x – xi)dk, (for the ket)
= (2πℏ)-1 ∫ eip(x – xi)dpThe second line is obtained with p = ℏk
Similarly for the bra,
(29) δ(x – xf) = (2πℏ)-1 ∫ e-ip'(x – xf)dp'
= (2πℏ)-1 ∫ eip'(xf – x)dp'The second line is obtained by taking the minus sign inside the bracket.
When we substitute equ. (28) and (29) into (27), we will get three exponential functions, and a triple integral. To simplify matters, we will examine the argument of each of the exponential function separately, from left to right, and then combine, taking care of not changing the order since we are dealing with operators.
(30) For the bra, E1 = (ip/ℏ)(x – xi)
(31) For the e-iHT/ℏ, E2 = –iHT/ℏ = – iET/ℏ
Where operating on the initial state, H = E, which is the eigenvalue. Note: we have a number(E) instead of an operator(H).
(32) For the ket, E3 = (ip'/ℏ)(xf – x)
Now we can pass E2 as it is a number. So from exp(E1) exp(E2) exp(E3), we can now write it as exp(E2) exp(E1) exp(E3). Let us examine the last two exponents:
(33) exp(E1) exp(E3) = e(ip/ℏ)(x – xi) e(ip'/ℏ)(xf – x)
= ei(p – p')x/ℏ eip'xf/ℏ e-ipxi/ℏThe first exponent in equ.(33), combine with (2πℏ)-1∫dx, gives,
(34) (2πℏ)-1∫ ei(p – p')x/ℏ dx = δ(p – p')
With this result, equation (27) becomes,
= (2πℏ)-1∫∫e–iET/ℏδ(p – p')eip'xf/ℏe-ipxi/ℏdpdp'
= (2πℏ)-1∫e–iET/ℏeip(xf – xi)/ℏdp
= (2πℏ)-1∫e–i(p2/2m)T/ℏeip(xf – xi)/ℏdp(36) Where the energy E is simply the kinetic energy = p2/2m .
Using appendix C,
(37) ∫-∞∞ e–ay2 + by dy= (π/a)½e b2/4a
Making the following correspondence,
y → p, a → iT/2mℏ, b → (i/ℏ)(xf – xi)
(38) K(xi,xf;T) = (m/(i2πℏT))½eim(xf – xi)2/(2Tℏ)
The probability density is then (see equ.(8)),
(39) │K(xi,xf;T)│2 = m/(2πℏT)
We can deduce from equation(39) that:
(i) For very large T, the probability amplitude decreases. That means that the farther away the path is from the classical path (the longer time it will take to go from xi to xf), the less it contributes to the probability amplitude (fig.3).
(ii) As the mass m increases, so is the height, thus the width must decrease (area under envelope is constant - fig.1) That is, the wave packet approaches the classical behavior of a particle.
(iii) If ℏ were to go to zero, the peak would be infinite, giving an exact location, as it should for a classical particle.(fig.1)
Going back to the central idea of the Path Integral - which is that a particle/wave traveling between two events could be considered as traveling all possible paths between those two events - we now see that the greater deviation from the classical path, the smaller contribution we get to the probability amplitude. In addition, we also find that for large mass, or ℏ = 0, we fall into the classical regime.
(A1) C = ∫e–y2dy , integral is from –∞ to +∞.
Squaring both sides,
(A2) C2 = ∫e–y2dy ∫e–x2dx
In the second term, we've replaced y by x, since these are just dummy variable in the integration,
(A3) C2 = ∫∫e–(y2 + x2)dxdy
Switching to polar coordinates,
(A4) Let x = r cosθ
(A5) And y = r sinθ
(A6) Then y2+ x2 = (r cosθ)2 + (r sinθ)2
= r2 cos2θ + r2 sin2θ
= r2 (cos2θ + sin2θ)
= r2(A7) Therefore, C2 = ∫∫e–r2dxdy
Also , the product dydx is just an element of the area of a small square. In polar coordinates, that area is rdrdθ. So,
(A8) C2 = ∫0∞ e–r2rdr ∫02π dθ
Make another change in variable, u = r2, so that du = 2rdr
(A9) C2 = ½ ∫0∞e–udu ∫02π dθ
= ½ (–) (℮–∞ – ℮–0)(2π – 0)
= ½ (–) (0 – 1)(2π)
(A10) Therefore, C = ∫–∞∞ e –y2dy = π½
(B1) C = ∫–∞∞ e–ay2dy
In the case that a constant "a" multiplies y2, we make the following substitution,
(B2) x2 = ay2
(B3)Then x = a½y
(B4) Taking derivatives,
dx = a½dy → dy = a-½dx
Substituting B4 into B1,
(B5) C = ∫–∞∞ e–ay2dy = a-½∫–∞∞ e–x2dx
= a-½ π½ , from A10
= (π/a) ½
(C1) C = ∫–∞∞ e–ay2 + bydy
(C2) The exponent is –ay2 +by = – a[y2 –(b/a)y] .
(C3) We complete the square:
= –a[y2 – (b/a)y + (b/2a)2 – (b/2a)2]
= –a[(y – b/2a)2 – (b2/4a2]
= –a(y – b/2a)2 + b2/4a(C4) Let x2 = (y – b/2a)2
(C5) x = (y – b/2a)
(C6) dx = dy
(C7) C = ∫–∞∞ e–ay2 + bydy , from C1
= ∫–∞∞ e–a(y – b/2a)2 + b2/4ady, from C3
= eb2/4a ∫–∞∞ e–ax2dx, from C4, C6
= (π/a)½eb2/4a , from B5