1
Classical action for the HO
The action is
Z
1
1
dt mẋ2 − mω 2 x2
2
2
.
(1)
δL δL
d2
−
= 2 x + ω 2 x = 0.
δ ẋ
δx
dt
(2)
S=
The EL equation are
∂t
The solution is
x(t) = asin(ωt) + bcos(ωt) , ẋ = aωcos(ωt) − bωsin(ωt).
(3)
Boundary conditions:
x(0) = x0 , x(T ) = xf
(4)
give
b = x0 , a =
xf − x0 cos(ωT )
.
sin(ωT )
(5)
Plug it into the action:
Z
mω 2 2
(a − b2 )(cos2 (ωt) − sin2 (ωt)) − 4absin(ωt)cos(ωt)
S = dt
2
Z
mω 2 2
= dt
(a − b2 )cos(2ωt) − 2absin(2ωt)
2
T
mω 2
=
(a − b2 )sin(2ωt) + 2abcos(2ωt) 0
4
mω 2
(a − b2 )sin(2ωT ) + 2ab (cos(2ωT ) − 1)
=
4
mω 2
=
(a − b2 )sin(ωT )cos(ωT ) − 2absin2 (ωT )
2
(6)
mωcos(ωT ) 2
=
xf + x20 (cos2 (ωT ) − sin2 (ωT )) − 2x0 xf cos(ωT )
2sin(ωT )
− mωsin(ωT )(xf x0 − x20 cos(ωT ))
mωcot(ωT ) 2
mω
1
2 1
3
2
=
xf +
x
cos (ωT ) + sin (ωT )cos(ωT )
2
sin(ωT ) 0 2
2
2
cos (ωT )
− mωx0 xf sin(ωT ) +
sin(ωT )
mωcot(ωT ) 2
mω
=
(xf + x20 ) −
x0 xf .
2
sin(ωT )
The free particle action can be obtained from the ω → 0 limit which is
regular:
m
m
m 2
Sf ree,cl =
(x0 + x2f ) − x0 xf =
(x0 − xf )2 .
(7)
2T
T
2T
Depends only on the difference as should be.
1
2
Double slit
The propagator of the free particle is
i
Kcl (0, 0, T, X) = F e ~ Scl .
(8)
The normalization can be found from demanding
lim K(0, 0; T, X) = δ(X).
(9)
T →0
In the future we will prove this property by connection to the Sch.eq. ThereRfore the factor F can be fixed from integrating over both sides such that
dXK = 1. Since
r
Z
imX 2
2πi~T
(10)
=
dXexp
2T
m
p m
we find that F = 2πi~T
and the propagator is
r
Kcl (0, 0, T, X) =
m
exp
2πi~T
imX 2
2T
.
(11)
For d- dimensional theory, on the same way we find
m d2
~ =
Kcl (0, ~0, T, X)
exp
2πi~T
~2
imX
2T
!
(12)
Consider two slits experiment. The screen is at z = L. The wall is located
at z = D and the two slits on the wall at (x, y, z) = (±a, 0, D). The
particle propagate from (x, y, z, t) = (0, 0, 0, 0) to (±a, 0, D, t0 ) and then to
(xf , yf , L, T ). What will be the amplitude?
Z
i
h
(xf ,yf ,L,T )
(a,y 0 ,D,t0 ) (xf ,yf ,L,T )
(−a,0,D,t0 ) (xf ,yf ,L,T )
= dt0 K(0,0,0,0) K(a,0,D,t
+
K
K
K(0,0,0,0)
0)
(−a,0,D,t0 )
(0,0,0,0)
Z T
m 3
1
=
dt0
×
2πi~ (t0 (T − t0 )) 32
0

 
!
2 + y 2 + (L − D)2
2
2
im
(x
−
a)
f
im a + D
f
exp
+
exp 
0
2t
2(T − t0 )

 !
im (xf + a)2 + yf2 + (L − D)2
im a2 + D2
 .
+exp
exp 
2t0
2(T − t0 )
(13)
2
Use the integral
Z
0
T
 √
√ 2 
√
r √
A
+
B 
B
1
π A+ B
A

√
−
=
exp
−
dt0
exp
−

 .
3
t0
T − t0
T3
T
AB
[t0 (T − t0 )] 2
(14)
In our case
A=−
im(a2
+
2
D2 )
, B± = −
im (xf ∓ a)2 + yf2 + (L − D)2
2
.
(15)
2
In the large L limit, B = − imL
2 , the prefactor is a constant. The exponents
are
√
K∼e
−
A+B+ +2
T
AB+
√
−
+e
A+B− +2
T
AB−
and its absolute value squared is
√
√
A + B− + 2 AB− A + B+ + 2 AB+
+
.
|K|2 ∼ 2 + 2 cos −
iT
iT
(16)
(17)
The integrand is
2maxf
+O
T
1
L
(18)
so we get an interference pattern proportional to
max f
.
cos2
~T
The usual result is
∼ cos
2
πdsin(θ)
λ
(19)
.
(20)
These are the same since d is the distance between slits: 2a = d.
xf
The wavelength is λ = hp = 2π~T
mL and sin(θ) = L at leading order. Therefore
amxf
πdsin(θ)
=
.
λ
~T
3
(21)