Trace of higher powers of Density Matrix

[maverick280857]

Hi,

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

<br /> Tr(\dot{\rho}) = 0<br />

Consider Tr(\rho^2) Differentiating with respect to time,

<br /> \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<br />

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

<br /> \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<br />

More generally,

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

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

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

Does this seem correct? I read somewhere that Tr(\rho^k) is invariant for k = 1, 2, ..., N-1, where N is the dimension of \rho, and further that if H is time-independent (we didn't use this above) then, Tr(\rho^k H^l) is invariant for k,l = 0, 1, \ldots N-1. 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?

 

[Avodyne]

I believe it's correct. It's definitely correct for a time-independent H. In that case, you can write rho in the energy basis, and it's easy to see that the diagonal elements of rho (and any power of rho) are time independent.

 

[maverick280857]

I believe it's correct. It's definitely correct for a time-independent H. In that case, you can write rho in the energy basis, and it's easy to see that the diagonal elements of rho (and any power of rho) are time independent.

Thanks Avodyne. But why are there only (N-1) conserved quantities, where N is the dimension of the density matrix? My calculation suggests that the trace of all higher powers of the density matrix should be conserved.

Also, I don't get the part about Tr(\rho^k H^m) being conserved for k, m = 0, 1, ..., N-1.

 

[Avodyne]

Thanks Avodyne. But why are there only (N-1) conserved quantities, where N is the dimension of the density matrix? My calculation suggests that the trace of all higher powers of the density matrix should be conserved.

Since we require Tr(\rho)=1, there are only N-1 independent eigenvalues of \rho. These can be taken to be the N-1 independent conserved quantities.

Also, I don't get the part about Tr(\rho^k H^m) being conserved for k, m = 0, 1, ..., N-1.

Again it's true for all k,m, by the same argument: in the energy basis, the diagonal elements of \rho^k are time independent. Also, H^m is diagonal and time independent, and hence the diagonal elements of \rho^k H^m are time independent.