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Hi guys, I have a few questions regarding Feynman's formulation of quantum mechanics.
Given the propagator K(x',t';x,t), when is this propagator equal to:
$$K(x',t';x,t)=Ae^{\frac{i}{\hbar}S_{cl}(x',t';x,t)}$$
Where S_cl is the classical action evaluated along the classical path of motion (i.e. where δS=0), and A is some normalization factor (which we we will get to in a moment)? I have seen arguments from method of stationary phase of the path integral:
$$K(x',t';x,t)=\mathcal{N}\int_x^{x'} e^{\int_t^{t'}\mathcal{L}dt}\mathcal{D}x$$
Where ##\mathcal{N}## is another normalization constant (let's not worry too much about convergence of this integral at this point, since I have only basic questions). To show that the most contributions will come about for paths along stationary action if the action is large compared to ##\hbar##. But this is an approximation.
We know; however, from direct integration of the Schrodinger equation that the first equation is identically true (not just as an approximation) for e.g. the free particle propagator. So, this prompts me to ask, when is the first equation true in general and not just as an approximation?
I have seen an argument that the first equation will be identically true, and not just approximately true, given that the potential V(x) is at most quadratic in x (so it would work for e.g. the Harmonic oscillator as well). Is this so?
My next question is in regards to the normalization constant A. I have seen it given, without proof, as:
$$A=\sqrt{\frac{i}{2\pi\hbar}\frac{\partial^2 S_{cl}(x',t';x,t)}{\partial x'\partial x}}$$
How general is this formula? Is this also for cases where the potential is at most quadratic in x?
Given the propagator K(x',t';x,t), when is this propagator equal to:
$$K(x',t';x,t)=Ae^{\frac{i}{\hbar}S_{cl}(x',t';x,t)}$$
Where S_cl is the classical action evaluated along the classical path of motion (i.e. where δS=0), and A is some normalization factor (which we we will get to in a moment)? I have seen arguments from method of stationary phase of the path integral:
$$K(x',t';x,t)=\mathcal{N}\int_x^{x'} e^{\int_t^{t'}\mathcal{L}dt}\mathcal{D}x$$
Where ##\mathcal{N}## is another normalization constant (let's not worry too much about convergence of this integral at this point, since I have only basic questions). To show that the most contributions will come about for paths along stationary action if the action is large compared to ##\hbar##. But this is an approximation.
We know; however, from direct integration of the Schrodinger equation that the first equation is identically true (not just as an approximation) for e.g. the free particle propagator. So, this prompts me to ask, when is the first equation true in general and not just as an approximation?
I have seen an argument that the first equation will be identically true, and not just approximately true, given that the potential V(x) is at most quadratic in x (so it would work for e.g. the Harmonic oscillator as well). Is this so?
My next question is in regards to the normalization constant A. I have seen it given, without proof, as:
$$A=\sqrt{\frac{i}{2\pi\hbar}\frac{\partial^2 S_{cl}(x',t';x,t)}{\partial x'\partial x}}$$
How general is this formula? Is this also for cases where the potential is at most quadratic in x?