An alternative - Hamiltonianversion of the Feynman path integral is often useful when one isdealing with non-Cartesian variables or with constrained systemsa. Show thatimx2dpexp2пhiрхSm(6)exp27Tihe2eh2mhb. Using result (a) show that the propagator may be written asdx n-1dpn-1dpod dpidax2Dj (xf,t;Xi, ti)= limn o27Th2тћ2πh(PI( -V(x)E2mexphl=1dtpHa,p)ED(t)Dp(t)] exp(7)h

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Asked Oct 22, 2019
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It's a quantum mechanics problem.

An alternative - Hamiltonian
version of the Feynman path integral is often useful when one is
dealing with non-Cartesian variables or with constrained systems
a. Show that
imx2
dp
exp
2пh
iрх
S
m
(6)
exp
27Tihe
2eh
2mh
b. Using result (a) show that the propagator may be written as
dx n-1dpn-1
dpod dpidax2
Dj (xf,t;Xi, ti)
= lim
n o
27Th
2тћ
2πh
(PI( -
V(x)E
2m
exp
h
l=1
dtpHa,p)
ED(t)Dp(t)] exp
(7)
h
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An alternative - Hamiltonian version of the Feynman path integral is often useful when one is dealing with non-Cartesian variables or with constrained systems a. Show that imx2 dp exp 2пh iрх S m (6) exp 27Tihe 2eh 2mh b. Using result (a) show that the propagator may be written as dx n-1dpn-1 dpod dpidax2 Dj (xf,t;Xi, ti) = lim n o 27Th 2тћ 2πh (PI( - V(x)E 2m exp h l=1 dtpHa,p) ED(t)Dp(t)] exp (7) h

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Expert Answer

Step 1

a)

Using the property of the infinite integral,

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e4a a (a>0) -00

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Step 2

The integral in the given problem can be solved using the above property and taking the limits of integration from negative infinity to positive infinity. The integral can be solved as,

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is pixp 2mTh 1 + S 2mh 2Th 2лй 2лй -0 X т 2is 2Thie im 2ha V2Thie II

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Step 3

b)

The expression for the path integral in...

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Dx (t)Dp(t)e0#0]A

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