Let $M$ be the $2 \times 2$ matrix corresponding to the observable to measure spin along some arbitrary axis $\vec{v}$. This matrix is given by following formula:

\begin{equation} M = v_x X + v_y Y + v_z Z \end{equation}

where $X, Y, Z$ are $2 \times 2$ Pauli matrices and $\vec{v} = (v_x, v_y, v_z)$. E.g., see this

Now if I have a 2 qubit system, then what is the $4 \times 4$ matrix to measure the spin of the first qubit along $\vec{v}$?

  • 1
    $\begingroup$ It should be $M\otimes I$. $\endgroup$
    – narip
    Jun 17, 2021 at 1:59

1 Answer 1


As @narip pointed out in a comment, it should be $M \otimes I$. Moreover, if you want to measure the second qubit along some axis $\vec{n}$ with corresponding matrix $N$, you can do $M \otimes N$.

Expanding this, we get

$$ \begin{align} M \otimes N &= (v_x X_1 + v_y Y_1 + v_z Z_1) \otimes (n_x X_2 + n_y Y_2 + n_z Z_2) \\ &= v_xn_x X_1 X_2 + v_xn_yX_1Y_2 + v_xn_zX_1Z_2 \\ & \hspace{0.2in} + v_yn_xY_1X_2 + v_yn_yY_1Y_2 + v_yn_zY_1Z_2 \\ & \hspace{0.2in} + v_zn_xZ_1X_2 + v_zn_yZ_1Y_2 + v_zn_zZ_1Z_2 \end{align} $$

where $X_1X_2 = X\otimes X$. (Wrote it like this to avoid writing a lot of $\otimes$s.) As @DaftWullie pointed out in his comment, this way you will get an observable with two eigenvalues and therefore your measurement will have two possible values.


Following with the discussion on the comments, if you want are expecting to get four possible outcomes out of the measurement, you can do something like $M_1 + 2N_2$ given that $M$ and $N$ have eigenvalues $\pm 1$. Since both of these are of the form

$$ \begin{align} \sigma_i &= \begin{pmatrix} \cos\theta & \sin\theta\cos\phi - i\sin\theta\sin\phi \\ \sin\theta\cos\phi + i\sin\theta\sin\phi & -\cos\theta \end{pmatrix} \\ &= \begin{pmatrix} \cos\theta & \sin\theta(\cos\phi - i\sin\phi) \\ \sin\theta(\cos\phi + i\sin\phi) & -\cos\theta \end{pmatrix} \\ &= \begin{pmatrix} \cos\theta & e^{-i\phi}\sin\theta \\ e^{i\phi}\sin\theta & -\cos\theta \end{pmatrix}, \end{align} $$

and therefore their characteristic equation is

$$ \begin{align} \text{det}(\sigma_i - \lambda I) &= -(\cos\theta - \lambda)(\cos\theta + \lambda) - \sin^2\theta \\ &= -\cos^2\theta + \lambda^2 - \sin^2\theta \\ &= \lambda^2 - 1 = 0, \end{align} $$

which gives us $\lambda = \pm 1$, we are safe using $M_1 + 2N_2$.

  • $\begingroup$ $X\otimes X$ as an observable has two eigenvalues and hence the measurement has two outcomes. If you want to measure both qubits, you're probably expecting 4 outcomes (or, at least, you need a more precise statement of what you're expecting). In which case, you might be better with something like $X_1+2X_2$. $\endgroup$
    – DaftWullie
    Jun 17, 2021 at 8:05
  • $\begingroup$ @DaftWullie you're right, thanks for your observation! Just some clarifications. Why is the coefficient of $2$ introduced? Would a observable like $M_1 + 2N_2$ with $M,N$ as defined above also work? $\endgroup$
    – epelaez
    Jun 17, 2021 at 10:50
  • $\begingroup$ That depends on the eigenvalues of your operators. If $M$ and $N$ both have eignevalues $\pm 1$, it'll work fine. The trick is just to make sure that all the eigenvalues are distinct, i.e. $\pm1\pm 2$ doesn't have any repetitions, whereas if I'd just done $X_1+X_2$, the eigenvalues would have been $\pm1\pm 1$, and the 0 eigenvalue would have been repeated. $\endgroup$
    – DaftWullie
    Jun 17, 2021 at 12:00
  • $\begingroup$ @DaftWullie I see, thank you! I didn’t know about this, could you point me to some theorem/concept/whatever where this is introduced? $\endgroup$
    – epelaez
    Jun 17, 2021 at 13:35
  • $\begingroup$ I don't know if there is one. This is just the understanding I've come to over time. $\endgroup$
    – DaftWullie
    Jun 17, 2021 at 14:25

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