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The Quantum Liouville Equation is [itex]\dot{\rho} = \frac{i}{\hbar}[\rho, H][/itex] where the dot denotes the partial derivative with respect to time [itex]t[/itex]. We take [itex]\hbar = 1[/itex] hereafter for convenience.

[tex]

Tr(\dot{\rho}) = 0

[/tex]

Consider [itex]Tr(\rho^2)[/itex] Differentiating with respect to time,

[tex]

\frac{\partial}{\partial t}Tr(\rho^2) = Tr(2 \dot{\rho}\rho) = Tr(2 i[\rho, H]\rho) = 2i\,Tr((\rho H - H\rho)\rho) = 2i\,Tr(\rho H \rho - H\rho\rho) = 0

[/tex]

where we have used [itex]Tr(A B) = Tr(B A)[/itex] to arrive at the last equality, assuming that we are dealing with finite dimensional operators. Hence [itex]Tr(\rho^2)[/itex] is conserved.} Next, we consider [itex]Tr(\rho^3)[/itex]. Differentiating with time, as above, we get

[tex]

\frac{\partial}{\partial t}Tr(\rho^3) = Tr(3 \rho^2 \dot{\rho})= 3i\,Tr(\rho^2 [\rho, H])= 3i\,Tr(\rho^2 \rho H - \rho^2 H\rho) = 0

[/tex]

More generally,

[tex]

\frac{\partial}{\partial t}Tr(\rho^k) = Tr(k \rho^{k-1} \dot{\rho}} = 0

[/tex]

which holds for arbitrary integer [itex]k \geq 1[/itex]. Hence [itex]Tr(\rho^k)[/itex] is conserved for [itex]k \geq 1[/itex].

My analysis suggests that the trace of [itex]\rho^k[/itex] is invariant under evolution even for k > N, where N is the dimension of [itex]\rho[/itex].

Does this seem correct? I read somewhere that [itex]Tr(\rho^k)[/itex] is invariant for k = 1, 2, ...., N-1, where N is the dimension of [itex]\rho[/itex], and further that if [itex]H[/itex] is time-independent (we didn't use this above) then, [itex]Tr(\rho^k H^l)[/itex] is invariant for [itex]k,l = 0, 1, \ldots N-1[/itex]. How does this arise? I know it has to do with Cayley Hamilton theorem, but I don't understand why there ought to be an upper bound on the powers?

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# Trace of higher powers of Density Matrix

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