3
$\begingroup$

I'm trying to compute the derivative with respect to an inverse temperature parameters $\beta$ of a density matrix that has the following form:

$$\rho(\beta,\mathbf{A}) = \frac{e^{-\beta \mathbf{A}}}{\mathrm{Tr}[e^{-\beta \mathbf{A}}] }$$

where the Hamiltonian is specified by an Hermitian matrix $\mathbf{A}$. How can I compute the following derivative?

$$ \frac{\partial}{\partial \beta} \left( \frac{e^{-\beta \mathbf{A}}}{\mathrm{Tr}[e^{-\beta \mathbf{A}}] } \right) $$

$\mathrm{Tr}$ is the trace operator, and the exponential is here meant as matrix exponential. I think that the derivative of the numerator is simply:

$$\frac{\partial}{\partial \beta}e^{-\beta \mathbf{A}} = -\mathbf{A}e^{-\beta \mathbf{A}}$$

but, I'm not sure this is correct. Is there available at least some book where the basic rules of calculus of matrix functions are available?

$\endgroup$
5
  • 5
    $\begingroup$ What makes you think that $A$ and $e^A$ don't commute? $\endgroup$
    – Omry
    Mar 6, 2017 at 15:24
  • $\begingroup$ Good observation, in fact they commute. math.stackexchange.com/questions/81386/… $\endgroup$
    – linello
    Mar 6, 2017 at 15:28
  • 1
    $\begingroup$ By functional calculus, $e^{-\beta A}$ is a densely defined operator for any self-adjoint $A$. The derivative can be taken (again on a dense domain), and yields the sought result. If both $e^{-\beta A}$ and $-Ae^{-\beta A}$ are trace class, then you can use the product rule to take the derivative of the product as well. $\endgroup$
    – yuggib
    Mar 6, 2017 at 15:40
  • $\begingroup$ Therefore is my computation correct? I'm also interested in computing the trace (the denominator), is correct to say that $\frac{\partial}{\partial \beta} \mathrm{Tr}[e^{-\beta \mathbf{A}}] = \mathrm{Tr}[\frac{\partial}{\partial \beta} e^{-\beta \mathbf{A}}] = \mathrm{Tr}[-\mathbf{A}e^{-\beta \mathbf{A}}]$? $\endgroup$
    – linello
    Mar 6, 2017 at 16:08
  • 1
    $\begingroup$ Yes, provided that you are careful with domains and everything is trace class. The fact that you can put the derivative inside the trace is essentially a consequence of linearity of the trace, and dominated convergence. $\endgroup$
    – yuggib
    Mar 6, 2017 at 16:18

1 Answer 1

6
$\begingroup$

A Hermitian matrix is normal and hence diagonalizable, so choose a diagonalizing eigenbasis where simultaneously $\mathbf{I} = \sum_{n=0}^{\infty} |a_n\rangle\langle a_n|$ and $\mathbf{A} |a_n\rangle = a_n|a_n\rangle.$ In this eigenbasis, $$e^{-\beta\mathbf{A}} = \sum_{n=0}^{\infty} e^{-\beta a_n} ~|a_n\rangle\langle a_n|,$$and the trace operator works out, by its linearity and cyclicity properties, to be simply$$\operatorname{Tr}\mathbf{M} = \operatorname{Tr}(\mathbf{I}~\mathbf{M}) = \sum_{n=0}^\infty\operatorname{Tr} \big(|a_n\rangle\langle a_n|~\mathbf{M} \big) = \sum_{n=0}^\infty\operatorname{Tr} \big(\langle a_n|~\mathbf{M}|a_n\rangle \big) = \sum_{n=0}^\infty\langle a_n|~\mathbf{M}|a_n\rangle.$$So your density matrix can be rewritten in terms of non-matrix quantities as, $$\rho = \frac{e^{-\beta\mathbf{A}}}{\operatorname{Tr}e^{-\beta\mathbf{A}}} = \left(\sum_{m=0}^\infty e^{-\beta a_m}\right)^{-1}~\sum_{n=0}^\infty {e^{-\beta a_n}}|a_n\rangle\langle a_n|,$$and by the normal product rule you get $$\begin{array}{rl}\frac{\partial\rho}{\partial \beta} &=~ \left(\sum_m e^{-\beta a_m}\right)^{-2}~~\sum_\ell e^{-\beta a_\ell} a_\ell~~\sum_n {e^{-\beta a_n}}|a_n\rangle\langle a_n| \\ &-~~~ \left(\sum_m e^{-\beta a_m}\right)^{-1}~\sum_n a_n e^{-\beta a_n}|a_n\rangle\langle a_n|.\end{array}$$You can then rewrite this in basis-independent notation as $$\frac{\partial\rho}{\partial \beta} = - \frac{\operatorname{Tr}(e^{-\beta\mathbf{A}}) ~\mathbf{A}~e^{-\beta\mathbf{A}} ~-~ \operatorname{Tr}(\mathbf{A} e^{-\beta\mathbf{A}})~e^{-\beta\mathbf{A}}}{\left(\operatorname{Tr} e^{-\beta\mathbf{A}}\right)^2},$$so that it looks like the quotient rule, if you wish.

If you like, this is also why the rules aren't very expressly laid out anywhere. There exists a sort of analogy where matrix products work like products, traces work like sums, and so forth, which can be made explicit by choosing a basis: once you have understood this analogy there is not much more to teach.

$\endgroup$
1
  • $\begingroup$ Cool that's exactly what I've obtained with a more naive derivation. This instead is a more rigorous, thank you! $\endgroup$
    – linello
    Mar 6, 2017 at 16:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.