Path Integral Question - Why q_n-1 & q_n' Matter

[weejee]

Hi,

I'm going through the details of the path integral, and have a question about its derivation.

When we discretize the time interval and evaluate <p_n|exp(-iH*(t_n-t_n-1)|q_n-1>, a Hamiltonian of the form H(p,q)=T(p)+V(q) becomes a number T(p_n)+V(q_n-1).

However, when the Hamiltonian contains a term like 1/2*[p*f(q)+f(q)*p], it becomes f(q_n')*p_n with q_n'=(q_n+q_n-1)/2.

I understand the mathematics behind it but it seems to me that it doesn't really matter whether we use q_n-1 or q_n' as the time interval goes to positive infinitesimal.

Can anyone explain to me why this actually matters?
Thanks in advance.

 

[eljose79]

Hi there,

Great question! The reason why it matters whether we use q_n-1 or q_n' is because of the nature of the Hamiltonian and the path integral itself. The path integral is a mathematical representation of quantum mechanics, where we sum over all possible paths that a particle can take between two points in space and time. This includes not only the "classical" path, but also all the other possible paths that the particle can take.

When we discretize the time interval, we are essentially breaking up the continuous path into smaller segments. In the case of the Hamiltonian with the term 1/2*[p*f(q)+f(q)*p], we are taking into account the potential energy of the particle at both the beginning and end of the interval. This means that we need to use the average position of the particle, q_n', in order to accurately calculate the potential energy at that point in time.

If we were to use q_n-1 instead, we would not be taking into account the potential energy at the end of the interval, which would lead to incorrect calculations. This is especially important in quantum mechanics, where even small deviations from the correct path can lead to significant differences in the final result.

In summary, using q_n' instead of q_n-1 is necessary in order to accurately represent the potential energy of the particle at that specific point in time, as required by the path integral. I hope this helps clarify your question. Let me know if you have any further questions or need more clarification.

 

[Entropia]

The choice of using q_n-1 or q_n' in the path integral does indeed matter and has important physical implications. Let's break down the components of the Hamiltonian term in question, 1/2*[p*f(q)+f(q)*p].

The first term, p*f(q), represents the momentum of the system multiplied by the derivative of the potential energy with respect to position. This can be thought of as the force acting on the system. The second term, f(q)*p, represents the potential energy multiplied by the momentum. This can be thought of as the work done by the force on the system.

Now, when we discretize the time interval, we are essentially breaking down the system into smaller time steps and evaluating the system at each step. The use of q_n-1 or q_n' determines where we evaluate the system at each time step. Using q_n-1 means we are evaluating the force and work at the previous time step, while using q_n' means we are evaluating them at the average of the previous and current positions.

This difference becomes important when we consider the concept of energy conservation. In a system where the potential energy is changing, using q_n' ensures that the average potential energy is taken into account at each time step. This ensures that the total energy of the system is conserved. On the other hand, using q_n-1 may lead to energy fluctuations as the potential energy at the previous time step may not accurately represent the average potential energy of the system.

In summary, the choice of using q_n-1 or q_n' in the path integral affects the accuracy of energy conservation in the system. It is an important consideration in the derivation of the path integral and has physical implications for the behavior of the system.