DOes the trace determine the Hamiltonian

if we define Z as:

[tex] Z(s)=Tr[exp(-sH)] [/tex]

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 [tex] Tr[exp(-sH)]\ne Tr[exp(-sK) [/tex]

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

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

 

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) ??