$|\psi\rangle=|\rightarrow\rangle=\frac{1}{\sqrt{2}}(|\uparrow\rangle+|\downarrow\rangle)$, then
$\rho_{x+}=|\psi\rangle\langle\psi|= \frac{1}{2}[(|\uparrow\rangle+|\downarrow\rangle)(\langle\uparrow|-\langle\downarrow|)]= \frac{1}{2} \begin{pmatrix} 1&1\\1&1 \end{pmatrix}$.

But then on the other hand we have

$\rho_{z+}=\begin{pmatrix}1&0\\0&0\end{pmatrix}$ and

projection onto $x+$:

$p_{x+}=\frac{1}{2}\begin{pmatrix}1&1\\1&1\end{pmatrix}$ and

$\rho_{x+}=p_{x+}\rho_{z+}p_{x+}^{\dagger}= \frac{1}{2}\begin{pmatrix}1&1\\1&1\end{pmatrix} \begin{pmatrix}1&0\\0&0\end{pmatrix} \frac{1}{2}\begin{pmatrix}1&1\\1&1\end{pmatrix}= \frac{1}{4}\begin{pmatrix}1&1\\1&1\end{pmatrix}$

which is clearly not equal to the $\rho_{x+}$ we calculated above! Would you please let me know if I am missing a point here?!

  • $\begingroup$ @Dan I don't think so since $p_{x+}=|\rightarrow\rangle\langle\rightarrow|=\frac{1}{\sqrt{2}}\begin{pmatrix}1\\1\end{pmatrix}\frac{1}{\sqrt{2}}\begin{pmatrix}1\\1\end{pmatrix}$. $\endgroup$
    – al pal
    Sep 7, 2021 at 17:17

1 Answer 1


You're simply missing the fact if the system undergoes projective evolution by the projection operator $P$, then the post-measurement density matrix is given by $$\rho' = \frac{P\rho P }{\mathrm{Tr}(P\rho P)}$$


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