Classical limit of the propagator

In summary, the classical correlator vanishes when the Hamiltonian is such that the particle can't be at a certain position starting from another.
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crises
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TL;DR Summary
When taking $$ \hbar \rightarrow 0 $$ shouldnt the propagator result in a delta distribution?
I am currently starting with my first qft lectures and i am trying to see for the free particle that the propagator $$ <x_i | e^{-i\frac{p}{2m} T|x_f}>$$ will equal to one if x_f = 1, x_i=0 m=1 u=1 p=1, T=1 and $$\hbar \rightarrow 0$$ or 0 otherwise. I understand that this limit will result in an exponential term, but it is not one.

Any help, explanation why
 
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  • #2
crises said:
Summary: When taking $$ \hbar \rightarrow 0 $$ shouldn't the propagator result in a delta distribution?
Why do you think so? You can think of propagator as a correlation of the positions of a single particle at two different times. In classical physics the position at one time depends on the position of an earlier time, so they are correlated. So the classical correlation at different times shound not vanish.
 
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What if the Hamiltonian is such that for a given time difference the particle can not be at position $x_2$ starting from position $x_1$ ?

For example, a particle with $p=1$ and $m=1$ has a Hamiltonian $H=\frac{1}{2}$. Suppose 1-D. Then if $x_1=0$ and $x_2=1$ $ \Delta T=1$. So the integral is a constant up to normalizaton factor. However, if T is not as before then in the classical limit there is no contribution, because there is no path that can correspond to those B.C
 
  • #4
crises said:
What if the Hamiltonian is such that for a given time difference the particle can not be at position $x_2$ starting from position $x_1$ ?
Then for those particular values the classical correlator vanishes. In fact, the classical correlator vanishes at all points except those at the classical trajectory, so it is a kind of a delta function. And I guess you want to understand why. Here is a hint.

What is ##f(\varphi)=\lim_{\hbar\rightarrow 0}e^{i\varphi/\hbar}##?
Can you draw the real part of ##f(\varphi)##?
From this drawing can you conclude what is ##\int_{\varphi_1}^{\varphi_2}d\varphi\, f(\varphi)##?
 
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FAQ: Classical limit of the propagator

What is the classical limit of the propagator?

The classical limit of the propagator refers to the behavior of the propagator in the limit of large quantum numbers or in the limit of small Planck's constant. In this limit, the quantum behavior of the system is suppressed and the propagator approaches the classical solution.

How is the classical limit of the propagator calculated?

The classical limit of the propagator is typically calculated using the semiclassical approximation, which involves expanding the propagator in powers of Planck's constant and neglecting higher order terms. This approximation is valid in the limit of small Planck's constant.

What is the importance of the classical limit of the propagator?

The classical limit of the propagator is important because it allows us to connect the quantum behavior of a system to its classical behavior. It also provides a useful tool for solving problems in quantum mechanics, as the classical solutions are often easier to calculate and interpret.

Can the classical limit of the propagator be applied to all systems?

No, the classical limit of the propagator is only applicable to systems that have a well-defined classical limit. This means that the system must have a large number of quantum states or a small value of Planck's constant in order for the classical limit to be valid.

How does the classical limit of the propagator relate to the uncertainty principle?

The classical limit of the propagator is related to the uncertainty principle in that it represents the limit where the uncertainty in the position and momentum of a particle becomes negligible. In this limit, the particle behaves classically and its position and momentum can be known with high precision.

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