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 a particle and a computer, is given by
$$
\psi=Z_{1} \chi .
$$
After the computer has measured the particle, 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 a particle and a computer, is given by
$$
\psi=Z_{0} \chi .
$$
After the computer has measured the particle, 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 particle and measuring device specification are contained in the volume element $\mathrm{d} \tau$. Multiplication yields
$$
\begin{aligned}
\langle\hat{Q}\rangle= & \frac{1}{2} \int Z_{1}^{*} \chi_{1}^{*} \hat{Q} Z_{1} \chi_{1} \mathrm{d} \tau+\frac{1}{2} \int Z_{0}^{*} \chi_{0}^{*} \hat{Q} Z_{0} \chi_{0} \mathrm{d} \tau \\
& +\operatorname{Re}\left\{\int Z_{1}^{*} \chi_{1}^* \hat{Q} Z_{0} \chi_{0} \mathrm{d} \tau\right\} .
\end{aligned}
\tag{4}
%\label{17.27}
$$
where 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 $\hat{Q}$. The particle can't transition between the states if $Q_{10}$ is equal to zero. This means that the particle would have irreversibly changed the state of the computer, which can be described as an indelible recording of the event. This property is typically attributed to a measuring apparatus: it records the result until an external action resets it.
The total system can be described either by a pure form (3) or a mixed form consisting of collapsed wave functions $\psi_1$ and $\psi_0$, each with a probability of $1/2$ in this case. However, it is more useful to use collapsed wave functions because they allow us to calculate the results of future particle measurements without considering the details of the first measuring apparatus. Collapse enables us to describe an isolated physical system without considering the other systems interacting with it irreversibly. However, doing so is unnecessary since using pure states yields the same results.
If a conscious observer rather than a computer makes the first measurement, then the conscious observer will be in a superposition state as described by state (3). This feature can be removed by postulating that the collapse of the wave function actually occurs when a conscious observer rather than a computer makes the measurement. However, this does not change the theory's predictions, as a conscious observer will make an indelible recording and therefore have $Q_{10}$ equal to zero. Also, there is no known mechanism that would make the wave function collapse when observed by a conscious entity rather than a computer.
