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What is the difference between the density matrix for quantum statistical mechanics and density matrix for quantum information theory?

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  • $\begingroup$ Why do you assume there is a difference? $\endgroup$ – By Symmetry Apr 8 '18 at 10:20
  • $\begingroup$ There isn't one - they are the same. $\endgroup$ – Nathaniel Apr 8 '18 at 14:17
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Conceptually they are same. The only difference between them is the number of particles. In quantum information theory, we deal with single particles (Let's call this particle $m$). In this case, the density matrix ($\rho_m$) encodes probabilities, coherence, and decoherence for that single particle. Quantum statistical mechanics deals with many particles (lets say $N$ particles) so the density matrix ($\rho_t$) is taken to be average of the density matrices of each particle, $\rho_t = \frac{1}{N}\sum_{m=1}^N \rho_m$. However, they are same ($\rho_t = \rho_m$) if all the particles are identical, have same probabilities, coherence and we are only looking at the density matrix of the internal states of the particles. Importantly, you will get the same results if you isolate a particle and perform your measurement, and perform your measurement on all particles at once.

Additionally, as quantum mechanics is probabilistic in nature, it becomes very difficult to distinguish the two cases mathematically and it makes sense to keep the same mathematical structure for both cases. Single isolated particle states can also be described by pure wavefunctions but in order to incorporate the randomness of decoherence, we need to use the density matrix which is statistical in its formulation.

EDIT - The particles are assumed to be uncorrelated. The dimensionality of the density matrix for the ensemble of particles and for a single particle is same in my explanation. If you add entanglement then you need a density matrix for the whole system. This density matrix will have a dimension which is much higher than the individual particle density matrix.

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  • $\begingroup$ Sorry but I don’t understand your answer. QI is full of 2-qubit states and density matrices associated with those. The famous Peres-Horodecki criterion (en.m.wikipedia.org/wiki/Peres%E2%80%93Horodecki_criterion) for entanglement is on the partial trace of ρ in - obviously - a 2 qubit system. Moreover there is no reason to suggest that the density matrix for a system is an average of 1-particle density matrices. $\endgroup$ – ZeroTheHero Apr 8 '18 at 13:10
  • $\begingroup$ Wikipedia definition -> Density matrix is a matrix that describes a quantum system in a mixed state, a statistical ensemble of several quantum states. "Statistical ensemble" is a key word here. For a many-particle system, the measurement of an observable will always give the average value of the same observable on a single particle. For example for spin half particles. The average magnetization will be equal to magnetization from single spin 1/2 particle if all of the particles have same density matrix or else it will be an average of magnetizaton from each particle. $\endgroup$ – Rahul Sawant Apr 8 '18 at 13:45
  • $\begingroup$ Additionally, I don't understand why you are mentioning the peres-horedecki criterion. $\endgroup$ – Rahul Sawant Apr 8 '18 at 13:45
  • $\begingroup$ Sorry but you can have density matrices for many-particle systems that are not separable. A system describing two entangled particles does not have a density matrix of the form you suggest. Correlations in simple systems will not produce density matrices of the form you suggest. You can try if you want to write the density matrix of the Werner state as a sum of two density matrices, but this will simply not work. $\endgroup$ – ZeroTheHero Apr 8 '18 at 13:53
  • $\begingroup$ Ok. Now I understand your point. I forgot to add, the particles are uncorrelated. If you add entanglement then you need a density matrix for the whole system. This density matrix will have a dimension which is higher than individual particle density matrix. $\endgroup$ – Rahul Sawant Apr 8 '18 at 14:05

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