My below answer, except for the last paragraph, is based on Sec. 17.10 of Greiner's "Quantum Mechanics" textbook. 

The three possible states of the computer are $\chi$ before registering the particle, $\chi_{1}$ if the particle is in state $Z_{1}=\left|1\right>$ and $\chi_{0}$ if the particle is in state $Z_0=\left|0\right>$.

First we consider the case of the particle being in state $Z_1$. The wave function of the total system, consisting of particle and computer, is given by
$$
\psi=Z_{1} \chi .
$$
After the particle has been measured by the computer, the total wave function is
$$
\psi_{1}=Z_{1} \chi_{1} \,. \tag{1}
$$

Similarly if the particle is in state $Z_0$. The wave function of the total system, consisting of particle and computer, is given by
$$
\psi=Z_{0} \chi . 
$$
After the particle has been measured by the computer, the total wave function is
$$
\psi_{0}=Z_{0} \chi_{0} \,.\tag{2}
$$

Now we look at the case of the particle being in the state
$$
\frac{\left(Z_{1}+Z_{0}\right) }{\sqrt{2}}.
$$
The initial state of the system is then
$$
\psi=\frac{\left(Z_{1}+Z_{0}\right) \chi}{\sqrt{2}}\, .
$$
After the measurement, the total wave function is
$$
\psi=\frac{\left(Z_{1} \chi_{1}+Z_{0} \chi_{0}\right)}{\sqrt{2}}\, 
%\label{17.25}
 \tag{3} 
.
$$

Let us consider a second measurement by a person or computer. If the system is in a pure state, the expectation value of the measurement described by the operator $\hat{Q}$ follows from (3):
$$
\langle\hat{Q}\rangle=\frac{1}{2} \int_{\tau}\left(Z_{1}^{*} \chi_{1}^{*}+Z_{0}^{*} \chi_{0}^{*}\right) \hat{Q}\left(Z_{1} \chi_{1}+Z_{0} \chi_{0}\right) \mathrm{d} \tau,
$$
where all variables necessary for the specification of particle and measuring device are contained in the volume element $\mathrm{d} \tau$. Multiplication yields
$$
\begin{aligned}
\langle\hat{Q}\rangle= & \frac{1}{2} \int Z_{1}^{*} \chi_{1}^{*} \hat{Q} Z_{1} \chi_{1} \mathrm{d} \tau+\frac{1}{2} \int Z_{0}^{*} \chi_{0}^{*} \hat{Q} Z_{0} \chi_{0} \mathrm{d} \tau \\
& +\operatorname{Re}\left\{\int Z_{1}^{*} \chi_{1}^* \hat{Q} Z_{0} \chi_{0} \mathrm{d} \tau\right\} . 
\end{aligned}
\tag{4}
%\label{17.27}
$$

Here we have taken into account the Hermiticity of $\hat{Q}$.

To calculate the properties of a mixed state, we have to consider that the expectation value of $\hat{Q}$ in a mixed state is equal to the average of the expectation values, which are calculated by separate measurements with the wave functions $Z_{1} \chi_{1}$ and $Z_{0} \chi_{0}$. Since the number of particles is the same in both states, it holds that

$$
\langle\hat{Q}\rangle^{\prime}=\frac{1}{2} \int Z_{+}^{*} \chi_{+}^{*} \hat{Q} Z_{+} \chi_{+} \mathrm{d} \tau+\frac{1}{2} \int Z_{-}^{*} \chi_{-}^{*} \hat{Q} Z_{-} \chi_{-} \mathrm{d} \tau .
\tag{5}
%\label{17.28}
$$

A comparison of (4) and (5) shows both expectation values to be identical if $$
Q_{10}=\int Z_{1}^{*} \chi_{1}^{*} \hat{Q} Z_{0} \chi_{0} \mathrm{d} \tau=0 . \tag{6} 
%\label{17.29}
$$

Now we want to consider the conditions under which the two states $\psi_{1}$ and $\psi_{0}$ yield a vanishing integral (6). The quantity $\left|Q_{10}\right|^{2}$ can be interpreted as being proportional to the probability of a transition between the states $Z_{1} \chi_{1}$and $Z_{0} \chi_{0}$, caused by the action of the measurement operator $\hat{Q}$. If $Q_{10}$vanishes, a transition between the states is impossible, meaning that the particle would have changed the state of the computer irreversibly; we could describe it as an indelible recording of the event. This is just the property we usually ascribe to a measuring apparatus: it registers the result until it is returned to its initial state by an external action.

This formalism provides a satisfactory explanation in terms of the statistics of repeated experiments. However, it has an unsatisfactory aspect of it in that if the first measurement is done by a person, then they plus the particle will be in the state given by (3). If you object to a conscious entity being in such a state, you can get around this by postulating that consciousness picks out one state randomly according to the Born rule so that the system will be either in state $\psi_1$ or $\psi_0$ with probability 1/2 for each case. This will give identical results to assuming that $Q_{10}=0$.