Partial Trace of Werner State

Question

I am trying to trace out the second qubit of the Werner State: \begin{align} W &=\frac{1-s}{4}I_{4}+\frac{s}{2}(|00\rangle\langle{00}|+|11\rangle\langle11|+|11\rangle \langle00|+|00\rangle \langle 11|)\\[0.5em]&= \left( \begin{array}{cccc} (1+s)/4 & 0 & 0 & s/2 \\ 0 & (1-s)/4 & 0 & 0 \\ 0 & 0 & (1-s)/4 & 0 \\ s/2 & 0 & 0 & (1+s)/4 \\ \end{array} \right) \end{align}

I write down $I_{4}=|11\rangle \langle11|+|00\rangle \langle00|+|10\rangle \langle10|+|01\rangle \langle01|$ and the result is: \begin{align} W&=\frac{1+s}{4}|0\rangle \langle0|\otimes|0\rangle \langle0|+\frac{1+s}{4}|1\rangle \langle1|\otimes|1\rangle \langle1|\\ &+\frac{1-s}{4}|0\rangle \langle0|\otimes|1\rangle \langle1|+\frac{1-s}{4}|1\rangle \langle 1|\otimes|0\rangle \langle0| \\ &+\frac{s}{2}|0\rangle \langle1|\otimes|0\rangle \langle1|+\frac{s}{2}|1\rangle \langle0|\otimes|1\rangle \langle0| \end{align}

Thus tracing out the second qubit results to: \begin{align} W_{A}&=\frac{1+s}{4}|0\rangle \langle0|+\frac{1+s}{4}|1\rangle \langle1|+\frac{1-s}{4}|0\rangle \langle0|+\frac{1-s}{4}|1\rangle \langle1|+\frac{s}{2}|0\rangle \langle1|+\frac{s}{2}|1\rangle \langle0|\\ &=\frac{1}{2}|0\rangle \langle0|+\frac{1}{2}|1\rangle \langle1|+\frac{s}{2}|0\rangle \langle1|+\frac{s}{2}|1\rangle \langle 0|\\ &=\frac{1}{2}\left( \begin{array}{cc} 1 & s \\ s & 1 \\ \end{array} \right) \end{align}.

However the result in the notes seems to be:

\begin{equation} \left( \begin{array}{cc} 1/2 & 0 \\ 0 & 1/2 \\ \end{array} \right) \end{equation}.

I am stuck and I try to find my mistake but i can't. Any help?

Answer 1

$\frac{s}{2}|0\rangle\langle1|\otimes|0\rangle\langle1|+\frac{s}{2}|1\rangle\langle0|\otimes|1\rangle\langle0|$ should disappear when you take the trace over them, as $\langle0|1\rangle$ and $\langle1|0\rangle = 0$

Edit: To give a simpler example, if your traced out the second qubit of $$\frac{1}{\sqrt{2}}|00\rangle+|11\rangle$$, then $\frac{1}{2}|00\rangle\langle11|$ and $\frac{1}{2}|11\rangle\langle00|$ would also disappear.

Answer 2

Note that

$$ \mathrm{tr}_B\left(|0\rangle \langle1|\otimes|0\rangle \langle1|\right) =|0\rangle \langle1| \mathrm{tr} (|0\rangle \langle1|) = |0\rangle \langle1| \, \langle1|0\rangle = |0\rangle \langle1| \cdot 0 = 0 $$

so the two terms marked red below disappear when you take the partial trace

$$ W_{A}=\frac{1+s}{4}|0\rangle \langle0|+\frac{1+s}{4}|1\rangle \langle1|+\frac{1-s}{4}|0\rangle \langle0|+\frac{1-s}{4}|1\rangle \langle1|+\color{red}{\frac{s}{2}|0\rangle \langle1|+\frac{s}{2}|1\rangle \langle0|}. $$

Once these terms are gone, the rest of the calculations yields

$$ \begin{align} W_{A}&=\frac{1+s}{4}|0\rangle \langle0|+\frac{1+s}{4}|1\rangle \langle1|+\frac{1-s}{4}|0\rangle \langle0|+\frac{1-s}{4}|1\rangle \langle1|\\ &=\frac{1}{2}|0\rangle \langle0|+\frac{1}{2}|1\rangle \langle1|\\ &=\frac{1}{2}\left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \\ \end{array} \right) \end{align} $$

as expected.

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There is another way of computing the partial trace from the full density matrix that works particularly well in this case

$$ W_A = \mathrm{tr}_B\begin{pmatrix} (1+s)/4 & 0 & 0 & s/2 \\ 0 & (1-s)/4 & 0 & 0 \\ 0 & 0 & (1-s)/4 & 0 \\ s/2 & 0 & 0 & (1+s)/4 \\ \end{pmatrix} = \begin{pmatrix} \mathrm{tr}\begin{pmatrix} (1+s)/4 & 0 \\ 0 & (1-s)/4 \end{pmatrix} & \mathrm{tr}\begin{pmatrix} 0 & s/2 \\ 0 & 0 \end{pmatrix} \\ \mathrm{tr}\begin{pmatrix} 0 & 0 \\ s/2 & 0 \end{pmatrix} & \mathrm{tr}\begin{pmatrix} (1-s)/4 & 0 \\ 0 & (1+s)/4 \end{pmatrix} \end{pmatrix} = \begin{pmatrix} \frac{1}{2} & 0 \\ 0 & \frac{1}{2} \end{pmatrix}. $$