Hamiltonian and Green functions.

[tpm]

Let be the 1-D Hamiltonian:

$$\hat H = -\hat D ^{2} + V(\hat x)$$ (1)

and its associated 'Green function' so:

$$-D^{2} G(x,s)+V(x)G(x,s)=\delta (x-s)$$ (2)

then my question is if there is a relationship between:

$$Tr[exp(-a\hat H ) ]$$ and $$Det[1-aG]$$ (3)

where a >0 and 'G' is the Kernel or Green function (due to the relationship between ODE's and Integral equation..

In case (3) is not exact i would like to know if there is any relationship between the trace of a Hamiltonian and the Determinant of its Fredholm Operator (in this case the Green function or propagator)

 

[eljose79]

Thank you for your question. The relationship between the trace of a Hamiltonian and the determinant of its associated Green function is an interesting topic in quantum mechanics.

First, let's clarify some notation. In equation (1), the hat symbol typically denotes a quantum operator, while the lack of hat in equation (2) suggests that we are dealing with a classical problem. Therefore, we will assume that the Hamiltonian in equation (1) represents a quantum mechanical system, while the Green function in equation (2) represents a classical propagator.

Now, let's consider the trace of the operator exponential, Tr[exp(-a\hat H)], where a>0. This can be written as:

Tr[exp(-a\hat H)] = \int \psi^{*}(x)exp(-a\hat H)\psi(x)dx (4)

where \psi(x) is an eigenfunction of the Hamiltonian \hat H. Using the spectral decomposition of the Hamiltonian, we can rewrite this as:

Tr[exp(-a\hat H)] = \sum_{n} \int \psi_{n}^{*}(x)exp(-aE_{n})\psi_{n}(x)dx (5)

where E_{n} is the energy eigenvalue corresponding to the eigenfunction \psi_{n}. Now, using the definition of the Green function in equation (2), we can write:

Tr[exp(-a\hat H)] = \sum_{n} \int \psi_{n}^{*}(x)exp(-aE_{n})\int G(x,s)\psi_{n}(s)dsdx (6)

Next, we can use the completeness relation for the eigenfunctions to write:

Tr[exp(-a\hat H)] = \int \int G(x,s)\sum_{n} \psi_{n}^{*}(x)exp(-aE_{n})\psi_{n}(s)dsdx (7)

But this sum in the integrand is just the propagator for a quantum system, which we can denote as K(x,s). So, we have:

Tr[exp(-a\hat H)] = \int \int G(x,s)K(x,s)dsdx (8)

Finally, we can recognize that this is just the integral of the product of the Green function and the propagator, which is the same

 

[denian]

I would like to clarify that the Hamiltonian and Green functions are fundamental concepts in quantum mechanics and mathematical physics. The Hamiltonian (equation 1) is an operator that represents the total energy of a quantum system, while the Green function (equation 2) is a mathematical tool used to solve differential equations and integral equations.

In response to your question, there is indeed a relationship between the trace of the exponential of a Hamiltonian and the determinant of its Green function. This relationship is known as the Selberg trace formula, which is a mathematical result that relates the trace of the exponential of a self-adjoint operator (such as the Hamiltonian) to the determinant of its resolvent (which is related to the Green function).

Furthermore, in the case where the Hamiltonian is a Fredholm operator (which is a type of integral operator), the trace of its exponential is related to the determinant of its Fredholm operator. This relationship is known as the Duhamel formula.

In summary, there is a deep mathematical connection between the trace of a Hamiltonian and the determinant of its associated Green function or Fredholm operator. This relationship has important implications in quantum mechanics and mathematical physics, and is a topic of ongoing research and study.