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A form of entropy for pure states?

Physics Asked by Jlee523 on October 9, 2020

A previous question that I asked made me realize I had some major misconceptions about entropy. If I now understand this correctly, given the density matrix of a system, we can calculate Von Neumann Entropy, Renyi Entropy, or whatever other type of entropy we would like. These measures of entropy are all some sort of measure of how uncertain or disordered a system is. We expect that, for a pure state, the entropy should be 0. (This can be seen mathematically since we often return something along the lines of $1 times log(1) = 0$.

However, this definition of entropy works for ensembles, not for single states.

I’m wondering if there is some notion of entropy (or any sort of disorderedness) wavefunctions. For example, if I have two wavefunctions:

$$ |psi rangle = sum_{n=0}^{N-1} frac{1}{N}|n rangle$$ $$ | phi rangle = |l rangle$$

(where $l$ is some non-negative integer), I might glance at $|psi rangle$ and say that it is "more disordered" than $| phirangle$. However, if we were to calculate Von Neumann entropy for these pure states, we would get:

$$ rho_{|psirangle} = |psi rangle langle psi |$$ $$ rho_{|phirangle} = |phi rangle langle phi | $$
So therefore, for both of these cases:

$$S_{VN} = – Tr(rho space log (rho)) = -Tr( rho space log(1) |psi rangle langle psi|) = -Tr(0) = 0 $$

So Von Neumann entropy (and also Renyi entropy) don’t seem to be a good measure for this.

Is there some sort of correlation function that would tell me how disordered or "spread out" a pure state is?

One Answer

If $|psirangle$ and $|phirangle$ had different entropies, then the entropy would depend on the basis, that is, if I transform $|psirangle$ to a new basis:

$$ |krangle equiv frac 1 A sum_{n=0}^{N-1}frac{2^{nk}} N|nrangle $$

then

$$ |psirangle = |0rangle $$

is a pure state (still), and has entropy 0.

Answered by JEB on October 9, 2020

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