# Quantum Mechanics mistake in partial trace

I have a given a density matrix by $\rho:=\frac{1}{2} |\psi_1 \rangle \langle \psi_1|+\frac{1}{8} |\psi_2 \rangle \langle \psi_2|+\frac{3}{8} |\psi_3 \rangle \langle \psi_3|.$

Where $|\psi_1\rangle := \frac{1}{\sqrt{2}}(|0. \rangle -i |1. \rangle )$

$|\psi_2\rangle := \frac{1}{\sqrt{3}}(|0. \rangle - |2. \rangle - |3.\rangle )$

$|\psi_3\rangle := \frac{1}{\sqrt{2}}(i|0. \rangle + |3. \rangle )$

where we have that $|0.\rangle := |0\rangle \otimes |0\rangle$

$|1.\rangle := |0\rangle \otimes |1\rangle$

$|2.\rangle := |1\rangle \otimes |0\rangle$

$|3.\rangle := |1\rangle \otimes |1\rangle$

Now I wanted to write down $\rho_A$, which means that I have to trace out the basis B.

I got: $\rho_A=\frac{1}{2}\left(\frac{1}{2}|0\rangle \langle 0|+ \frac{1}{2}|0\rangle \langle 0|\right)+\frac{1}{8}\left(\frac{1}{3}|0\rangle \langle 0|+ \frac{1}{3}|1\rangle \langle 1|-\frac{1}{3}|0\rangle \langle 1|- \frac{1}{3}|1\rangle +\langle 0|\right)+\frac{3}{8}\left(\frac{1}{2}|0\rangle \langle 0|+ \frac{1}{2}|1\rangle \langle 1|\right)$

The problem is: The trace of $tr(\rho_A)=\frac{1}{2}(\frac{1}{2}+\frac{1}{2})+\frac{1}{8}(\frac{1}{3}+\frac{1}{3})+\frac{3}{8}(\frac{1}{2}+\frac{1}{2}) \neq 1$ and therefore this cannot be a density matrix. What have I done wrong?-I have written this in this extended form so that it is easier for you to see what I have down. Does anybody here know how to do this correctly?

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Of the three terms you have written in the last line, it's the second term that leads to $\mathrm{Tr}\rho \neq 1$. Can you check that term again? –  Abhinav Dec 15 '13 at 13:26
sorry, there was a "+" where no "+" should have been, edited it. the thing is that I might have done something wrong there, although I do not know what. –  Xin Wang Dec 15 '13 at 14:00

Focusing exclusively on this term (leaving aside the $1/8$ factor), we have
$$|\psi_2\rangle \langle \psi_2| = \frac{1}{3}(|00\rangle \langle 00|+ |10\rangle \langle 10|+ |11\rangle \langle 11| - |00\rangle \langle 10|-|10\rangle \langle 00| + \ ...),$$ where I have put the ... to indicate that those terms do not contribute to the reduced density matrix of A. Now taking partial trace, we get the correct answer: $$\rho_{A,2} = \frac{1}{3}(|0\rangle \langle 0|+ 2|1\rangle \langle 1|-|0\rangle \langle 1|-|1\rangle \langle 0|)$$ You may have missed the contribution from $|10\rangle \langle 10|$, I guess. Adding 1/8 of this to the other two (correctly obtained) density matrices will give us a reduced density matrix of trace $1$.
A mistake people generally tend to make (but you have not) is in the last two terms, assuming that they do not contribute. This is erroneous, since even though they are not diagonal matrix elements of the full density matrix $\rho_{\mathsf{AB}}$, they still appear when you take a partial trace since they are outer products of the same 2 elements of B.