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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 wave function, which describes the state of a physical system, is reduced or collapsed through measuring and recording the measurement's result. This reduction is not caused by human perception or the classical behaviour of macroscopic objects. 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.

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.

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 $Q_{10}$ is equal to zero, it is impossible for the particle to transition between the states. 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 is a property that is typically attributed to a measuring apparatus: it records the result until it is reset by an external action.

The wave function, which describes the state of a physical system, is reduced or collapsed by the process of measuring and recording the result of the measurement. This reduction is not caused by human perception or the classical behaviour of macroscopic objects. The total system may be described either by a pure form (3) or a mixed form consisting of reduced wave functions $\psi_1$ and $\psi_0$ each, which has probability $1/2$ in this case. However, it is more useful to use reduced wave functions because they allow us to calculate the results of future particle measurements without the need to consider the details of the first measuring apparatus. Reduction allows us to describe an isolated physical system without considering the other systems that have interacted with it irreversibly, although it is not necessary to do so since using pure states yields the same results.

This formalism provides a satisfactory explanation regarding the statistics of repeated experiments. However, it has an unsatisfactory aspect in that if a person does the first measurement, then they, plus the particle, will be in the state given by (3). Suppose you object to a conscious entity being in such a state. In that case, you can get around this by postulating that consciousness somehow causes $\psi$ to collapse to state $\psi_1$ or $\psi_0$ randomly with probability $1/2$ for each case in this particular example.

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 wave function, which describes the state of a physical system, is reduced or collapsed through measuring and recording the measurement's result. This reduction is not caused by human perception or the classical behaviour of macroscopic objects. 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.

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

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

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.

So we see that the reduction of the wave function is induced by the measuring process itself, namely, by the recording of the result; it is not induced by human perception or the classical behaviour of macroscopical objects. Furthermore, we see that reduction is not absolutely necessary; after an indelible recording has been made, the total system may be described either by a pure form (3) or a mixed form consisting of reduced wave functions $\psi_1$ and $\psi_0$ each which has probability $1/2$ in this case. Nevertheless, it is more useful to choose reduced wave functions, since the results of further measurements of the particle can be calculated without knowledge of the complicated details of the first apparatus. Reduction enables us to describe an isolated physical system without considering the other systems which have interacted irreversibly with it; we could do so, but this is unnecessary, since we know that a description with pure states yields the same results.

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 somehow causes $\psi$ to actually collapse to state $\psi_1$ or $\psi_0$ randomly with probability $1/2$ for each case in this particular example.

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

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 $Q_{10}$ is equal to zero, it is impossible for the particle to transition between the states. 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 is a property that is typically attributed to a measuring apparatus: it records the result until it is reset by an external action.

The wave function, which describes the state of a physical system, is reduced or collapsed by the process of measuring and recording the result of the measurement. This reduction is not caused by human perception or the classical behaviour of macroscopic objects. The total system may be described either by a pure form (3) or a mixed form consisting of reduced wave functions $\psi_1$ and $\psi_0$ each, which has probability $1/2$ in this case. However, it is more useful to use reduced wave functions because they allow us to calculate the results of future particle measurements without the need to consider the details of the first measuring apparatus. Reduction allows us to describe an isolated physical system without considering the other systems that have interacted with it irreversibly, although it is not necessary to do so since using pure states yields the same results.

This formalism provides a satisfactory explanation regarding the statistics of repeated experiments. However, it has an unsatisfactory aspect in that if a person does the first measurement, then they, plus the particle, will be in the state given by (3). Suppose you object to a conscious entity being in such a state. In that case, you can get around this by postulating that consciousness somehow causes $\psi$ to collapse to state $\psi_1$ or $\psi_0$ randomly with probability $1/2$ for each case in this particular example.