Quantum Bayesian Networks

April 17, 2012

Mirror, Mirror, on the Wall, What is the Most Symmetric State of Them All in Quantum SIT?

Filed under: Uncategorized — rrtucci @ 3:50 pm

SIT =Shannon Information Theory

Let H({\underline{a}_1,\underline{a}_2, \ldots, \underline{a}_N}) denote the classical entropy of a probability distribution of the random variables {\underline{a}_1,\underline{a}_2, \ldots, \underline{a}_N}.

Consider a Hilbert space {\cal H}_{\underline{a}_1}\otimes{\cal H}_{\underline{a}_2}\otimes\ldots\otimes{\cal H}_{\underline{a}_N} = {\cal H}_{\underline{a}_1,\underline{a}_2, \ldots, \underline{a}_N}. For any density matrix \rho_{\underline{a}_1,\underline{a}_2, \ldots, \underline{a}_N} that acts on {\cal H}_{\underline{a}_1,\underline{a}_2, \ldots, \underline{a}_N}, let S({\underline{a}_1,\underline{a}_2, \ldots, \underline{a}_N}) be its von Neumann entropy.

Suppose |\psi\rangle_{\underline{a}_1,\underline{a}_2, \ldots, \underline{a}_N}\in {\cal H}_{\underline{a}_1,\underline{a}_2, \ldots, \underline{a}_N} is a ket (pure quantum state) with density matrix given by \rho_{\underline{a}_1,\underline{a}_2, \ldots, \underline{a}_N} = [|\psi \rangle_{\underline{a}_1,\underline{a}_2, \ldots, \underline{a}_N}][h.c.]. Then it is well known (a consequence of the Schmidt decomposition or of the Araki-Lieb Inequality) that for this pure state

S({\underline{a}_1,\underline{a}_2, \ldots, \underline{a}_N})=0

S(\underline{a}_J) = S(\underline{a}_{J^c})

where J and J^c are disjoint sets whose union equals \{1, 2, \ldots, N\}, and where \underline{a}_J = (\underline{a}_j)_{j\in J} and \underline{a}_{J^c} = (\underline{a}_j)_{j\in J^c}. For example, if N=4, then we have

S(\underline{a}_1, \underline{a}_2, \underline{a}_3, \underline{a}_4)=0
S(\underline{a}_1) = S(\underline{a}_2, \underline{a}_3, \underline{a}_4) and permutations
S(\underline{a}_1, \underline{a}_2) = S(\underline{a}_3, \underline{a}_4) and permutations

Pure quantum states thus have a very high degree of symmetry. So much so that we can make a model of their entropy as follows. I’ll call it the RUM (Roots of Unity Model) of pure states.

Define

S(\underline{a}_J) = \left|\sum_{j\in J}  \underline{a}_j \right|

(note that the entropy contributions are summed coherently rather than incoherently. The latter is common for macroscopic classical systems)

where we now interpret {\underline{a}_1,\underline{a}_2, \ldots, \underline{a}_N} as the Nth roots of unity. In other words, we are now setting

\underline{a}_j = {\rm exp}(i\frac{2\pi (j-1)}{N}) for j=1, 2, \ldots, N.

The model works because

\sum_{j=1}^N \underline{a}_j = 0

so

\sum_{j\in J} \underline{a}_j = - \sum_{j\in J^c} \underline{a}_j

S(\underline{a}_J) =|\sum_{j\in J} \underline{a}_j| = |\sum_{j\in J^c} \underline{a}_j| = S(\underline{a}_{J^c}).

Henceforth we will abbreviate \sum_{j\in J}  \underline{a}_j = \sum \underline{a}_J

I bet a lot of people have discovered RUM before, but I myself discovered it on my own yesterday. I find RUM very helpful because (besides being good for numbing pain and inducing forgetfulness🙂 ) it helps me visualize better the entropy of a pure quantum state. For example, it “explains” to me why quantum conditional entropies can be negative. Indeed, suppose J and K are two disjoint subsets of \{1,2,\ldots,N\}. Then

S(\underline{a}_J|\underline{a}_K) = S(\underline{a}_{J\cup K} ) - S(\underline{a}_K) =|\sum \underline{a}_{J \cup K}| - |\sum \underline{a}_K|

Also, the Araki-Lieb and sub additivity inequalities

|S(\underline{a}_J) - S(\underline{a}_K)| \leq S(\underline{a}_J ,\underline{a}_K)\leq S(\underline{a}_J) + S(\underline{a}_K)

become the well known triangle inequalities

\left|\;\;|\sum\underline{a}_J| - |\sum\underline{a}_K|\;\;\right| \leq |\sum\underline{a}_{J \cup K}|\leq |\sum \underline{a}_J| + |\sum \underline{a}_K|

1 Comment »

  1. Someone said to me by email

    I’m not sure what to make of RUM yet. Is it clear that this technique can capture the calculation of an arbitrary von Neumann entropy ? It seems like it is not general enough…

    My reply:

    I totally agree. That’s why I call it a model. I was thinking of something like the Bohr model of the hydrogen atom. A crude picture but it can be used to understand SOME (not all) of the features of the hydrogen spectrum, and it gives some intuition about what is going on.

    Comment by rrtucci — April 25, 2012 @ 4:49 pm


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