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[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.