Does Von Neumann entropy fail to account for projective quantum measurements?

Question

I'm trying to understand if this line of reasoning is correct:

For a two-state pure quantum system $\left|\psi \right\rangle=\frac{1}{\sqrt{2}} (\left|0 \right\rangle+\left|1 \right\rangle)$,

1. The Von Neumann entropy is $0$.

2. The possible results for a projective quantum measurements on $\left|\psi \right\rangle$ are to find the system in $\left|0 \right\rangle$ or $\left|1 \right\rangle$ post-measurement. Before the measurement, the probability is $50\%$ for each state. After the measurement either one or the other is known with $100\%$ probability. Thus, knowing the actual measurement value implies knowing one bit of information. Finally, one bit of information is associated with an entropy reduction of $k_B\ln{2}$.

If the Von Neumann entropy before the projective measurement is $0$, then after the measurement, as the production of information increases by one bit, it thus implies that the post-measurement entropy is $-k_B\ln{2}$. Since a total negative entropy is inadmissible in a closed system, the connected environment must increase its entropy to compensate for the decrease. Thus, the projective quantum measurement must increase the entropy of the environment, and cannot happen unless connected to an environment.

Answer 1

The von-Neumann entropy measures the entropy of a state, but for any state we can measure many possible observables. So while it may be that for a measurement of one observable the result on a given state may be very uncertain, for others it may be completely determined.

So for example considering your state $|\psi\rangle$, a measurement of $\sigma^z$, as you propose in the question, is very uncertain but a measurement of $\sigma^x$ will always give the result 1 and has no uncertainty. The von-Neumann entropy is considering all possible measurements, or alternatively considers the best possible measurement to identify the state, and so the entropy of the state is 0.

In general, for any pure state the projection operator onto that state (for example $|\psi\rangle \langle \psi|$) represents an observable for which the state has no uncertainty and so the entropy of all pure states is 0.

Answer 2

Although the current answers and comments here seem very correct and useful, I feel there are also deeper underlying issues behind the OP question which have not been addressed (at least to my satisfaction).

These issue are excellently discussed in this post and I don't feel I can add to this.

My summary of the issue in relation to the OP question:

Consider the process of the quantum state change discussed in the question:

$$ \frac 1 {\sqrt 2} ( \lvert 0 \rangle + \lvert 1 \rangle ) \rightarrow \lvert 0 \rangle $$

There are two distinct processes in which this can occur:

1. A unitary evolution treating the system as a closed system

2. A von Neumannn measurement process.

In both cases we can calculate the before and after von Neumann entropy and get the same answer in each case. All pure states have entropy $0$ so the entropy in both cases changes from $0$ to $0$ as explained in the previous answers and comments.

That being so, we can also say these two situations are fundamentally different processes. The post referenced above discusses ways we can reconcile this.