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.

 

[vanesch]

As Statmechguy points out, the trace isn't sufficient to know the hamiltonian. However, the function Z(s), being a function of s, will allow you (by derivatives for instance) to find all individual diagonal elements of H.

 

[tpm]

Well..since the trace is just the sum of 'Eigenvalues' of a Hamiltonian, considering every Hamiltonian has an infinite number of 'energies'.. perhaps the trace is unique (in this case since Statsmechguay has only pointed a finite-dimensional case), could it Happen that both H and K had the same energies but different potential V(x) ??