How do I construct a Density Matrix corresponding to a Hamiltonian?

Question

I have a Hamiltonian and I want to know the corresponding density matrix. The matrix I'm interested in is the one in this question.

Answer 1

There's many different density matrices that can correspond to a given Hamiltonian.

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For the 8x8 matrix in your question, there's 8 different "eigenstate" density matrices that can be obtained, one for each of the 8 eigenvectors. The density matrices are constructed by doing the outer product of the eigenvectors. For the $i^{\rm{th}}$ eigenstate of the Hamiltonian, the density matrix $\rho_i$ is:

$ \rho_i = |\psi_i\rangle_ \langle \psi_i| $.

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A system can also be in a "pure" superposition of eigenstates, for example:

$|\psi \rangle = \frac{1}{\sqrt{2}}|\psi_1\rangle + \frac{1}{\sqrt{2}}|\psi_2\rangle $.

Then the density matrix is once again made by doing the outer product of the pure wave function $|\psi\rangle$ with itself.

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A system can also be in a "mixed" state, which means it's a linear combination of "pure" states.

In this case you would construct the density matrix like this (for example):

$\rho = 0.5 \rho_1 + 0.5\rho_2$,

which descrbes a state which is a 50% mixture of $\rho_1$ and a 50% mixture of $\rho_2$.

Answer 2

Your question remains very unclear as to what it actually is that you want to calculate.

There is no direct correspondence between a system Hamiltonian and the quantum state of the system. No matter what the Hamiltonian, any quantum state is a valid state of the system.

Where a Hamiltonian comes in useful is, if you know the state at some time (say, $t=0$), you can find out what the state is at any later time via the Schroedinger equation $$ i\frac{\partial |\psi\rangle}{\partial t}=H(t)|\psi\rangle. $$ If $H$ does not change in time, you get $$ |\psi(t)\rangle=e^{-iHt}|\psi(0)\rangle $$ or, if your initial state is a mixed state, $$ \rho(t)=e^{-iHt}\rho(0)e^{iHt}. $$

Now, there are two reasonable things that might be relevant in terms of a state derived from a Hamiltonian - the thermal state and the ground state (which is the thermal state at 0 temperature). At temperature $T$, the thermal state is $$ \rho_{\text{thermal}}=\frac{e^{-H/(k_BT)}}{\text{Tr}(e^{-H/(k_BT)})}, $$ while the ground state is simply the eigenstate of $H$ with the smallest energy. You can (crudely) think of the thermal state as the best guess about what the state would be if you cooled it to a temperature $T$.

In one of the comments on another answer, you say

I need it to get the purity of my qubit states and the internal energy of the system vs. the magnetisation factor h

Purity has nothing to do with the Hamiltonian. If you know the density matrix $\rho$ of your system, purity is just $\text{Tr}(\rho^2)$. The Hamiltonian will help you with the expected internal energy: $\text{Tr}(\rho H)$ but, again, the state has to be provided from elsewhere, not from the Hamiltonian.