Sunday, 15 July 2018

Question about the Dirac notation for partial trace


I saw the following definition for the partial trace operator:



ρA=kek|ρAB|ek, where ek is basis for the state space of system B.



From what I know, in the Dirac notation, the meaning of v|A|u is the inner product of the vectors |v and A|u, so I have two problems with this notation of the partial trace.



First, how can an inner product be an operator? An inner product should be a complex number, so I guess that inner product represents an operator somhow.


Second, what is the meaning of ρAB|ek? ρAB is a mapping on the space AB, but ek is a vector from the space B. So, I don't really know how to interpret the meaning of this notation.



Answer



I believe that an example will help clarify your confusion about notation (as examples usually do). Consider a system of two qubits, A and B, with Hilbert spaces VA and VB spanned by two orthonormal eigenbasis of σz, |0A and |1A; and |0B and |1B. Now suppose that we have a Bell state, |ΨAB=12(|0A|0B+|1A|1B). This state corresponds to a density matrix, ρAB=|ΨABΨ|AB =12(|0A|0B0|A0|B+|0A|0B1|A1|B +|1A|1B0|A0|B+|1A|1B1|A1|B). Now suppose that we wish to get the reduced density matrix for system A. We use your definition for the partial trace over system B with |e1=|0B and |e1=|1B, together with the fact that ϕ|B(|ψA|ϕB)=(ϕ|B|ϕB)|ψA (which is just the inner product of |ϕB and |ϕB, a number, times |ψA) as well as orthonormality, to get, ρA=12(|0A0|A+|1A1|A), a completely mixed state.


Incidentally, this appearance of a completely mixed state is the reason there is no FTL signalling in Bell experiments - a mixed state is complete ignorance about what is going with B if we only study A locally.


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