Suppose a projective measurement is performed on a single qubit in the basis $|+\rangle, |−\rangle$, where $|±\rangle \equiv (|0\rangle\pm |1\rangle)/\sqrt{2}$. In the event that we are ignorant of the result of the measurement, the density matrix evolves according to the equation $$ \rho\to\mathcal{E}(\rho) = |+\rangle\langle+|ρ|+\rangle\langle+| + |−\rangle\langle−|ρ|−\rangle\langle−| $$ Illustrate this transformation on the Bloch sphere.

This is given as Exercise 8.15 in Page 378, Chapter 8, Quantum Computation and Quantum Information by Nielsen and Chuang.

My Attempt

Thanks @GaussStrife for pointing out the mistake.

$|+\rangle\langle+|=\frac{1}{2}\begin{bmatrix}1&1\\1&1\end{bmatrix}$ and $|-\rangle\langle-|=\frac{1}{2}\begin{bmatrix}1&-1\\-1&1\end{bmatrix}$

$$ \rho=\frac{1}{2}[I+\vec{r}.\vec{\rho}]=\frac{1}{2}\begin{bmatrix}1+z&x-iy\\x+iy&1-z\end{bmatrix} $$

$$ \mathcal{E}(\rho)=\frac{1}{8}\begin{bmatrix}1&1\\1&1\end{bmatrix}\begin{bmatrix}1+z&x-iy\\x+iy&1-z\end{bmatrix}\begin{bmatrix}1&1\\1&1\end{bmatrix}+\frac{1}{8}\begin{bmatrix}1&-1\\-1&1\end{bmatrix}\begin{bmatrix}1+z&x-iy\\x+iy&1-z\end{bmatrix}\begin{bmatrix}1&-1\\-1&1\end{bmatrix}\\ =\frac{1}{8}\begin{bmatrix}2+2x&2+2x\\2+2x&2+2x\end{bmatrix}+\frac{1}{8}\begin{bmatrix}2-2x&-2+2x\\-2+2x&2-2x\end{bmatrix}=\frac{1}{2}\begin{bmatrix}1&x\\x&1\end{bmatrix} $$ My understanding is that, in the Bloch sphere representation, an arbitrary trace-preserving quantum operation is equivalent to a map of the form, please check, Affine map of single qubit quantum operations $$ \mathcal{E}(\rho)=\vec{r}\xrightarrow{\mathcal{E}}\vec{r}'=M\vec{r}+\vec{c} $$

and we have $I=\begin{bmatrix}1&0\\0&1\end{bmatrix},X=\begin{bmatrix}0&1\\1&0\end{bmatrix},Y=\begin{bmatrix}0&-i\\i&0\end{bmatrix},Z=\begin{bmatrix}1&0\\0&-1\end{bmatrix}$ $$ \frac{1}{2}\begin{bmatrix}1&x\\x&1\end{bmatrix}=\frac{1}{2}\Big[I+xX\Big] $$ Therefore, under the quantum operation, the Bloch vector is transformed as, $\vec{r}=(x,y,z)\xrightarrow{\mathcal{E}}\vec{r}'=(x,0,0)$

If we were to carry out the projective measurement on the basis $|0\rangle,|1\rangle$ then it would be

$$ \mathcal{E}(\rho)=|0\rangle\langle 0|\rho|0\rangle\langle 0| + |1\rangle\langle 1|\rho|1\rangle\langle 1|\\ =\frac{1}{2}\begin{bmatrix}1+z&0\\0&1-z\end{bmatrix}\\ =\frac{1}{2}[I+zZ] \implies \vec{r}'=(0,0,z) $$

  • $\begingroup$ You can take the line defined by the antipodal points $\{ \vert + \rangle, \vert - \rangle\}$ in the Bloch sphere and represent any point inside this line, because it will represent the convex combination between the two possibilities after measurement. $\endgroup$
    – R.W
    Commented Oct 30, 2022 at 18:46
  • 1
    $\begingroup$ In your attempy, you are missing a factor of $\frac{1}{2}$ for the action of your projectors on the current density operator. So the overall scalar should be $\frac{1}{8}$. $\endgroup$ Commented Oct 31, 2022 at 14:16

2 Answers 2


Let $\rho'=\mathcal{E}(\rho)$. We can find out the elements of the Block vector by calculating $$ \vec{n}'=(\text{Tr}(X\rho'),\text{Tr}(Y\rho'),\text{Tr}(Z\rho')). $$ Consider the $Z$ term first: $$ \text{Tr}(Z\rho')=\text{Tr}(Z|+\rangle\langle +|\langle +|\rho|+\rangle+Z|-\rangle\langle -|\langle -|\rho|-\rangle)=\langle +|Z|+\rangle\langle +|\rho|+\rangle+\langle -|Z|-\rangle\langle -|\rho|-\rangle $$ But you can quickly check that $\langle +|Z|+\rangle=0$. The $Y$ term is similar.

Now consider the $X$ term $$ \text{Tr}(X\rho')=\text{Tr}(X|+\rangle\langle +|\langle +|\rho|+\rangle+X|-\rangle\langle -|\langle -|\rho|-\rangle)=\langle +|\rho|+\rangle-\langle -|\rho|-\rangle=\text{Tr}(X\rho). $$ Thus, if we started with a Bloch vector $\vec{n}=(n_X,n_Y,n_Z)$, it transforms into $\vec{n}'=(n_X,0,0)$. This should be quite straightforward to visualise on the Bloch sphere as a projection onto the $x$ axis.


Observe the following:

  1. Given a state $\rho$, performing a projective measurement with elements $\Pi_i$ and not knowing the outcome amounts to performing the operation $\rho\mapsto \sum_i \Pi_i \rho \Pi_i$. In the case of a single qubit, an example would be $\Pi_0\equiv |0\rangle\!\langle 0|$ and $\Pi_1\equiv|1\rangle\!\langle 1|$, or $\Pi_0\equiv |+\rangle\!\langle +|$ and $\Pi_1\equiv|-\rangle\!\langle -|$, which is the one you mentioned.

  2. The Bloch representation of a mixture of states equals the mixture of their individual Bloch representations. More precisely, if $\boldsymbol r(\rho)$ denotes the Bloch vector associated to a state $\rho$, then for any collection of states $\rho_i$ and probabilities $p_i$ we have $\boldsymbol r(\sum_i p_i \rho_i)=\sum_i p_i \boldsymbol r(\rho_i)$. This happens because the mapping between density matrices and expectation values of observables is always linear.

It follows that states obtained after a projective measurement $\{\Pi_i\}$ are in the convex span of the Bloch vectors representing the possible outcomes $\Pi_i$. In the case of a single qubit, where we have the Bloch sphere, for every projective measurement $\Pi_0$ and $\Pi_1$ correspond to two antipodal points on the sphere. It follows that the states obtained must sit on the line connecting said to points. For example, measuring in the computational basis, they must be on the line connecting $|0\rangle$ and $|1\rangle$. Similarly for other bases.


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