Checkpoint: Compute Entropies

This checkpoint pulls the module together. You will take a pure bipartite state, read off its reduced density matrix, and compute its entanglement entropy — the same number whether you call it the von Neumann entropy of the part or the entanglement of the whole.

What you have learned

By now you can:

• compute the Shannon entropy $H(\{p_i\}) = -\sum_i p_i \log_2 p_i$ of a distribution;
• compute the von Neumann entropy $S(\rho) = -\sum_k \lambda_k \log_2 \lambda_k$ from the eigenvalues of a density operator;
• distinguish pure ($S = 0$) from mixed ($S > 0$) states, with the maximally mixed state of $d$ levels reaching $S = \log_2 d$;
• form a reduced state by partial trace and read off the entanglement entropy of a pure bipartite state as $S(\rho_A) = S(\rho_B)$;
• find Schmidt coefficients as the singular values of the coefficient matrix, so that the squared coefficients are exactly the eigenvalues of $\rho_A$.

The checkpoint state

Consider the pure two-qubit state already written in Schmidt form,

Because the $A$-labels $\{0,1\}$ pair one-to-one with the $B$-labels $\{0,1\}$, the Schmidt coefficients are $s_1 = \sqrt{1/3}$ and $s_2 = \sqrt{2/3}$. Tracing out qubit $B$,

The entanglement entropy is the von Neumann entropy of this reduced state:

It lies between $0$ (product state) and $1$ (maximally entangled Bell state), confirming partial entanglement — closer to maximal than the $h(1/4)$ example from the entanglement-entropy lesson, because the Schmidt coefficients here are more nearly equal.

Try it

For $|\psi\rangle = \sqrt{1/3}\,|00\rangle + \sqrt{2/3}\,|11\rangle$, form the reduced state $\rho_A$, and return its von Neumann entropy — the entanglement entropy $E$ in bits.

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