Quantum computing, partial trace

AI Thread Summary
The discussion focuses on calculating the partial trace of a density matrix in quantum computing. The density matrix for the combined system is provided, along with the reduced density matrix for subsystems A and B. The process of taking a partial trace over subsystem C is explained mathematically, emphasizing the use of basis states and tensor products. It is noted that when calculating the partial trace, terms from the subsystem being traced out do not interact with the remaining subsystems, simplifying the computation. The explanation concludes that one can effectively ignore the traced-out subsystem to expedite the calculation.
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Homework Statement
Hello, I have to calculate density matrices ## \rho _{AB}, \rho _{AC}, \rho _{BC}## of this state:
##| \psi \rangle _{ABC} = |100 \rangle + | 001 \rangle##
Relevant Equations
##| \psi \rangle _{ABC} = |100 \rangle⟩ + | 001 \rangle##
\begin{align*}
\rho_{AB} &= \text{Tr}_C \, \rho_{ABC} \\
\rho_{AC} &= \text{Tr}_B \, \rho_{ABC} \\
\rho_{BC} &= \text{Tr}_A \, \rho_{ABC}
\end{align*}
I've calculated density matrix
$$ rho_{ABC} = \frac{1}{2} \left( \left| 100 \right\rangle \left\langle 100 \right| + \left| 100 \right\rangle \left\langle 001 \right| + \left| 001 \right\rangle \left\langle 100 \right| + \left| 001 \right\rangle \left\langle 001 \right| \right)$$
and ##
\rho_{AB} = \frac{1}{2} \left( \left| 10 \right\rangle \left\langle 10 \right| + \left| 00 \right\rangle \left\langle 00 \right| \right)

)##. I'm not sure how to calculate other matrices. I don't understand how to calculate partial trace. Could some explain how to do it?
 
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I'm not very well-versed in this subject, but you must already be close? Taking a "partial trace" over one of the subsystems, e.g. C, means to say:$$\mathrm{Tr}_{C}[ \rho] = \sum_c \langle c | \rho | c \rangle $$where in this case the ##|c\rangle## are a basis for C (and similarly for a, b). This notation "##| abc \rangle##" means the tenor product, ##| abc \rangle := |a\rangle \otimes |b \rangle \otimes |c\rangle## (or just ##|a\rangle |b \rangle |c\rangle##) which has its own special properties.

If you look at just this term, for example:$$\sum_c \langle c|100 \rangle \langle 100| c \rangle$$then what you are doing is actually:$$\sum_c \langle c| \bigg{(} |a=1\rangle |b=0\rangle |c=0\rangle \langle a=1 | \langle b=0| \langle c=0| c \rangle \bigg{)}$$The C basis states don't hit the A and B basis states in the tensor product, so you are left with this:$$|a=1\rangle |b=0\rangle \sum_c \langle c|c=0\rangle + \langle a=1 | \langle b=0| \sum_c \langle c=0| c \rangle $$Those sums are both 1 (e.g. ##\sum_c \langle c| c=0 \rangle = \langle 0 | 0 \rangle + \langle 1 | 0 \rangle = 1 + 0 = 1##), leaving you with: $$|a=1\rangle |b=0\rangle + \langle a=1 | \langle b=0|$$

and the same for all the other terms. I.e. in practice I think you would just "ignore" the other two positions/subsystems to take the trace quickly).
 
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