My below answer 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 $$ \psi=\frac{\left(Z_{1} \chi_{1}+Z_{0} \chi_{0}\right)}{\sqrt{2}}\, \tag{3}\label{17.25} . $$
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 \eqref{17.25}: $$ \langle\hat{Q}\rangle=\frac{1}{2} \int_{\tau}\left(Z_{1}^{*} \chi_{1}^{*}+Z_{0}^{*} \chi_{0}^{*}\right) \hat{Q}\left(Z_{1} \chi_{1}+Z_{0} \chi_{0}\right) \mathrm{d} \tau, $$ 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 \eqref{17.27} and \eqref{17.28} 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 \eqref{17.29}. 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 \eqref{17.25}. 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 $Q_{10}=0$.