How to compute the partial trace of the state $|\psi\rangle = \sum_{k}c_k |k\rangle\otimes|k\rangle\otimes |k\rangle$?

Question

Suppose we have a quantum system defined on a Hilbert space of $H=H_A\otimes H_B\otimes H_C$, and there is a state defined in $H$ of the form: \begin{eqnarray} |\psi\rangle = \sum_{k}c_k |k\rangle\otimes|k\rangle\otimes |k\rangle, \tag{1} \end{eqnarray} i.e. the state is a superposition of product states for the three Hilbert spaces $A,B,C$, but where all product states feature the same state $k$ on each subsystem.

Then we want the reduced density operator for the B subspace as: \begin{eqnarray} \rho_B=\text{Tr}_{A,C}|\psi\rangle\langle\psi|=\text{Tr}_{A,C}\left(\sum_{k,j}c_kc_{j^*}|k\rangle\otimes|k\rangle\otimes |k\rangle\langle j|\otimes \langle j|\otimes \langle j| \right).\tag{2} \end{eqnarray} Is it right that, due to the state separation above, the result is: \begin{eqnarray} \rho_B=\sum_{k,j}c_k c_{j}^* |k\rangle\langle j| \hspace{10pt}(?) \tag{3}. \end{eqnarray} This means that no matter which of the two subsystems we trace out, the reduced density operator is always the same. How is this done in case the partial trace is not correctly taken?

Answer 1

TLDR; No, your answer should be $$ \rho_B=\sum_k|c_k|^2|k\rangle\langle k|. $$

You seem to have arrived at equation (3) from equation 2 essentially just by throwing away all the terms on qubits A and C. This is not how the partial trace works. Instead, you have to take an inner product (on both sides), summing over an orthonormal basis of the systems that you are tracing out. In this instance, it's probably easiest to select $|m\rangle\otimes I\otimes |n\rangle$. So, we have $$ \text{Tr}_{AC}|\psi\rangle\langle\psi|=\sum_{m,n}(\langle m|\otimes I\otimes \langle n|)|\psi\rangle\langle\psi|(|m\rangle\otimes I\otimes |n\rangle). $$ The inner products reduce to $\delta_{m,k}\delta_{n,k}\delta_{j,k}$ so you can do all the sums over $m,n,j$, just leaving the sum over $k$.