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Pre-measurement

Let $\left| \psi \right>$ be a pure state:

$$ \left| \psi \right> = a \left| 0 \right> + b \left| 1 \right> $$

The density matrix of $\left| \psi \right>$ is:

$$ \hat{\rho}=\pmatrix{ aa^* & ab^* \\ ba^* & bb^* } $$

The entropy $S=\text{Tr} \hat{\rho} \ln \hat{\rho}$ is:

$$ S=\text{Tr} \pmatrix{ aa^* & ab^* \\ ba^* & bb^* } \ln \pmatrix{ aa^* & ab^* \\ ba^* & bb^* } $$

It is solved as follows:

$$ \begin{align} S&=\text{Tr} U^\dagger \pmatrix{ aa^* + bb^* & 0 \\ 0 & 0 } U U^\dagger \ln \pmatrix{ aa^* + bb^* & 0 \\ 0 & 0 } U\\ &=\text{Tr} \pmatrix{ aa^* + bb^* & 0 \\ 0 & 0 } \ln \pmatrix{ aa^* + bb^* & 0 \\ 0 & 0 } \\ &=(aa^* + bb^*)\ln(aa^* + bb^*) \end{align} $$

Then since $aa^* + bb^*=1$, $S=0$. (We have taken the convention that $0\ln0=0$.)


Post-measurement

Consider that after a measurement, the density matrix expresses a mixture of states (lets call it $\hat{m}$):

$$ \hat{m}=\pmatrix{ aa^* & 0 \\ 0 & bb^* } $$

The entropy of $\hat{m}$ is:

$$ \begin{align} S&=\text{Tr} \pmatrix{ aa^* & 0 \\ 0 & bb^* } \ln \pmatrix{ aa^* & 0 \\ 0 & bb^* }\\ &=aa^* \ln aa^*+ bb^*\ln bb^* \end{align} $$

which is higher than $0$ (as expected from a measurement)

This entropy is the same as the Shannon entropy: it quantifies the information gained by selecting of one element $\left| 0 \right>$ or $\left| 1 \right>$ from the set $\{ \left| 0 \right>,\left| 1 \right> \}$ according to the probability $\rho(\left| 0 \right>)=aa^*$ and $\rho(\left| 1 \right>)=bb^*$:

$$ \begin{align} H&=-\sum_{i=0}^n \left< i \middle| i \right> \ln \left< i \middle| i \right>\\ &=aa^* \ln aa^*+ bb^*\ln bb^* \end{align} $$

So far so good


Problem

Now, here is the problem I am having.

After applying a unitary transformation $U$ to $\left| \psi \right>$, the Von Neumann entropy of both the state ($S(\hat{\rho})=S(U^\dagger \hat{\rho} U)$) and the mixture ($S(\hat{m})=S(U^\dagger \hat{m} U)$) remains the same, but the Shannon entropy of the measurement changes.

Applying $U$ to $\left| \psi \right>$, we get:

$$ \begin{align} \left| \psi' \right> &= \underbrace{\pmatrix{a & b \\ -b^* & a^*}}_{U} \left| \psi \right>\\ &= \pmatrix{a & b \\ -b^* & a^*} \pmatrix{a \\ b}\\ &= aa+bb \left| 0 \right> + -ab^* + ba^* \left| 1 \right> \end{align} $$

The Shannon entropy of this measurement is :

$$ H = (aa+bb)\ln(aa+bb) + (-ab^* + ba^*)\ln (-ab^* + ba^*) $$

Why the departure between the two definitions of entropy? Which is lying?

To me, it looks like it is the Von Neumann entropy that is lying: implicit in the trace operator, the entropy is defined in reference to the eigenbasis. But, this will only be true if the future upcoming measurement is performed against the eigenbasis. If the future measurement is performed against another basis, the entropy will not be the Von Neumann entropy. The Shannon entropy, however, is always correct.

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  • $\begingroup$ Is the logarithm even well-defined? I believe your matrix is singular. $\endgroup$ – AccidentalFourierTransform Sep 4 at 1:47
  • $\begingroup$ @AccidentalFourierTransform After diagonalization, the $\ln$ refers to an element-by-element $\ln$, and not a $\operatorname{matrix-ln}$. $\endgroup$ – Alexandre H. Tremblay Sep 4 at 2:09
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There is no contradiction, just a convoluted comparison. Think carefully about how many measurements are being done in each case:

  • When you started with a post-measurement state $\hat m$, then applied a unitary transformation $U$, and then calculated the Shannon entropy as you defined it, you implicitly applied another measurement by using only the diagonal terms in $U^\dagger\hat m U$ to calculate the Shannon entropy.

  • When you calculated the von Neumann entropy of $U^\dagger\hat m U$, you didn't discard the off-diagonal terms: you didn't apply the second measurement.

The two entropies gave different answers because you asked them different questions. You calculated the von Neumann entropy after only one measurement, and you calculated the Shannon entropy after two measurements.

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Actually, neither entropy formula is incorrect. What you're misunderstanding is what they're telling you. The Von-Neumann entropy formula is basically telling you that the entropy of any pure state is zero. A pure state is any state that can be written as $\hat{\rho} = |\phi\rangle\langle\phi|$ for some state vector $|\phi\rangle$ or, equivalently, it satisfies $\hat{\rho}^2 = \hat{\rho}$.

Once you've taken a measurement, you're in a mixed state, and how mixed the state is depends on which basis the measurement was in and how it relates to $|\psi\rangle$. For example, if $|\psi\rangle$ is an eigenvector for one of the measurement outcomes, then the state after the measurement will be $|\psi\rangle$ with $100\%$ probability. Thus, no entropy is added by the measurement process and the entropy after the measurement will still be zero.

The more possible outcomes that $|\psi\rangle$ overlaps with, the more entropy the measurement process will add.

This is not a mystery - the measurement process is known to be non-linear on the state vector. Equivalently, after a measurement, $\hat\rho$ is now in a mixed state, and just how mixed depends on which basis the measurement was done in.

Where you're messing up is you're calculating $\hat{m}$ incorrectly after applying the unitary transformation (probably because you're reusing the same symbols in $U$ that you used in $|\psi\rangle$). If you apply $U$ and then measure in the basis, your $\hat{m}$ should be $$\hat{m}\rightarrow \left[\begin{array}{cc} (a^*a^* + b^*b^*)(aa+bb) & 0 \\ 0 & (-a^*b + b^*a)(-ab^* + ba^*) \end{array}\right].$$

I would suggest using different symbols for your $U$, like $\alpha$ and $\beta$. That might clear things up for you.

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