# Reduced density matrix of 4 qubits

I am trying to find the reduced density matrix of a four qubit quantum system. The Hamiltonian is $$H = -B(\sigma_z \otimes \mathbb{I}_8 + \mathbb{I}_2 \otimes \sigma_z \otimes \mathbb{I}_4 + \mathbb{I}_4 \otimes \sigma_z \otimes \mathbb{I}_2 + \mathbb{I}_8 \otimes \sigma_z)$$

where the $$\sigma_z$$ are the Pauli matrices $$\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$$ acting on each qubit. My problem is to find the reduced density matrix containing the first and second qubit of the subsystem of eigenstates with eigenvalue 0, given that the system is described by a uniform density operator in this subspace.

I believe that the subspace is any state with exactly two |1>s present, for example |1100>. So does this mean the operator is an equally weighted sum of all of these states, say $$\rho = |\psi><\psi|$$ where $$|\psi>$$ is the sum of the 6 states.

I am on the correct line?

I am trying to find the reduced density matrix of a four qubit quantum system. The Hamiltonian is $$H = -B(\sigma_z \otimes \mathbb{I}_8 + \mathbb{I}_2 \otimes \sigma_z \otimes \mathbb{I}_4 + \mathbb{I}_4 \otimes \sigma_z \otimes \mathbb{I}_2 + \mathbb{I}_8 \otimes \sigma_z)$$

...My problem is to find the reduced density matrix containing the first and second qubit of the subsystem of eigenstates with eigenvalue 0, given that the system is described by a uniform density operator in this subspace.

This description of the problem is a little unclear. I'll answer based on what I think this means, but maybe you could make sure the language is exactly as written in your exercise or textbook...

I believe that the subspace is any state with exactly two |1>s present, for example |1100>. So does this mean the operator is an equally weighted sum of all of these states, say $$\rho = |\psi><\psi|$$ where $$|\psi>$$ is the sum of the 6 states.

You are right that there are 6 eigenstate of $$H$$ with eigenvalue zero. And that these states are states with exactly two qubits set to 1. So the states of the entire system with eigenvalue 0 are: $$|3\rangle = |0011\rangle\;, |5\rangle = |0101\rangle\;, |6\rangle = |0110\rangle\;,$$ $$|9\rangle = |1001\rangle\;, |10\rangle = |1010\rangle\;, |12\rangle = |1100\rangle\;,$$

Now, when you write that you want the reduced density matrix, containing the first and second qubits. I think you mean that you are numbering the qubits as 4, 3, 2, 1, in that order. But you might be numbering them like 3, 2, 1, 0, so this is also a little unclear.

I will again make the caveat that I am not entirely sure of the question, but I will proceed anyways. So, I think now you want to start from a uniform mixed state: $$\rho_{total}=\frac{1}{6}\left( |3\rangle\langle 3|+ |5\rangle\langle 5|+ |6\rangle\langle 6|+ |9\rangle\langle 9|+ |10\rangle\langle 10|+ |12\rangle\langle 12| \right)$$ and then trace out the two leftmost qubits... Again, it is not super clear based on the problem description, but I will do it this way and maybe that will help clear things up.

When I trace out over the leftmost qubit, I get an intermediately reduced matrix: $$\tilde\rho = \frac{1}{6} \left( |011\rangle\langle 011|+ |101\rangle\langle 101|+ |110\rangle\langle 110|+ |001\rangle\langle 001|+ |010\rangle\langle 010|+ |100\rangle\langle 100| \right)\;.$$

When I trace out of the now-currently leftmost qubit, I get $$\tilde{\tilde{\rho}} = \frac{1}{6} \left( |11\rangle\langle 11|+ 2|01\rangle\langle 01|+ 2|10\rangle\langle 10|+ |00\rangle\langle 00| \right)$$ $$=\left( \begin{matrix} 1/6 & 0 & 0 & 0\\ 0 & 1/3 & 0 & 0\\ 0 & 0 & 1/3 & 0\\ 0 & 0 & 0 & 1/6\\ \end{matrix} \right)$$