# DOes the trace determine the Hamiltonian

tpm
if we define Z as:

$$Z(s)=Tr[exp(-sH)]$$

my 2 questions are..

a) is the trace unique and define the Hamiltonian completely? i mean if

we have 2 Hamiltonians H and K then $$Tr[exp(-sH)]\ne Tr[exp(-sK)$$

and if we use the 'Semiclassical approach' then $$Z(s)=Tr[exp(-sH)]\sim As^{-1/2}\int_{-\infty}^{\infty}dx exp(-sV(x))$$

then given Z(s) we could calculate approximately V(x) by solving an integral equation.

StatMechGuy
1 + 2 + 3 = 6
2 + 2 + 2 = 6

Both of these could be the diagonal elements of a hamiltonian. Clearly the trace alone is not sufficient to categorize the entire hamiltonian. You get no information about the off-diagonal elements, too, which is relevant since you have not necessarily taken the trace in a basis which diagonalizes the hamiltonian.

Staff Emeritus