- #1
cattlecattle
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Hi, It's great to find this forum.
I'm teaching myself QM using Shankar, it's a great book, I've covered 8 chapters so far.
My question is about the notion of using Path Integral method to calculate the propagator. The recipe given by Shankar says the propagator is
<br /> U(x,t;x')=A\int \exp{\frac{iS[x(t)]}{\hbar}}\mathcal{D}[x(t)]<br />
where the integral sums over all possible paths connecting (x',0) and (x, t)
He went on to prove that if the potential is in the form of
<br /> V=a+bx+cx^2+d\dot{x}+ex\dot{x}<br />
then the propagator has a simpler form:
<br /> U(x,t;x')=e^{iS_{cl}/\hbar}\cdot A(t)<br />
In his proof for this, he simply assumed that given x, t and x', the there exists one and only one classical path - at least the action S is well defined. But I found this part confusing, two counter examples:
1. In a harmonic oscillator of angular frequency ω, let x=x'=0, and t=2π/ω, apparently there are infinite classical paths connecting (x', 0) to (x, t) - no matter what your initial velocity, you always come back to the same spot after one full period.
2. Similarly, if x=x'=a > 0, and t=π/ω, there will be no classical paths (there's no way to come back to the same non trivial spot after half period!
In case 1, luckily all such classical paths have the same action S=0, so we can still find the propagator following U(x,t;x')=e^{iS_{cl}/\hbar}\cdot A(t)
But in case 2, there is no classical path at all, what does this imply for the propagator, does the propagator not exist? In QM, the propagator connects one state to the other, so it should always be well-defined.
I'm sure this is treated more rigorously in non-elementary Path Integral texts, but before diving into those, maybe someone here can explain in simple words what this means.
Thanks
I'm teaching myself QM using Shankar, it's a great book, I've covered 8 chapters so far.
My question is about the notion of using Path Integral method to calculate the propagator. The recipe given by Shankar says the propagator is
<br /> U(x,t;x')=A\int \exp{\frac{iS[x(t)]}{\hbar}}\mathcal{D}[x(t)]<br />
where the integral sums over all possible paths connecting (x',0) and (x, t)
He went on to prove that if the potential is in the form of
<br /> V=a+bx+cx^2+d\dot{x}+ex\dot{x}<br />
then the propagator has a simpler form:
<br /> U(x,t;x')=e^{iS_{cl}/\hbar}\cdot A(t)<br />
In his proof for this, he simply assumed that given x, t and x', the there exists one and only one classical path - at least the action S is well defined. But I found this part confusing, two counter examples:
1. In a harmonic oscillator of angular frequency ω, let x=x'=0, and t=2π/ω, apparently there are infinite classical paths connecting (x', 0) to (x, t) - no matter what your initial velocity, you always come back to the same spot after one full period.
2. Similarly, if x=x'=a > 0, and t=π/ω, there will be no classical paths (there's no way to come back to the same non trivial spot after half period!
In case 1, luckily all such classical paths have the same action S=0, so we can still find the propagator following U(x,t;x')=e^{iS_{cl}/\hbar}\cdot A(t)
But in case 2, there is no classical path at all, what does this imply for the propagator, does the propagator not exist? In QM, the propagator connects one state to the other, so it should always be well-defined.
I'm sure this is treated more rigorously in non-elementary Path Integral texts, but before diving into those, maybe someone here can explain in simple words what this means.
Thanks