**a.**

We define the projection operator onto the subspace generated by $\ket{+}$,

(1)
\begin{align} P_+ = \ket{+}\bra{+} = \tfrac{1}{2}(\ket{0}\bra{0} + \ket{0}\bra{1} + \ket{1}\bra{0} + \ket{1}\bra{1}) \end{align}

as in Example 4.3.2 of Rieffel and Polak. We also know that the identity can be written in operator form as:

(2)
\begin{align} I = \ket{0}\bra{0} + \ket{1}\bra{1} \end{align}

Since we want to measure the first qubit in the Hadamard basis, we will apply the identity to the second qubit, and so our two-qubit measurement operator becomes:

(3)
\begin{align} P_+ \otimes I &= \tfrac{1}{2}(\ket{0}\bra{0} + \ket{0}\bra{1} + \ket{1}\bra{0} + \ket{1}\bra{1}) \otimes (\ket{0}\bra{0} + \ket{1}\bra{1}) \\ &= \tfrac{1}{2}(\ket{00}\bra{00} + \ket{00}\bra{10} + \ket{10}\bra{00} + \ket{10}\bra{10} + \ket{01}\bra{01} + \ket{01}\bra{11} + \ket{11}\bra{01} + \ket{11}\bra{11}) \end{align}

Then, defining an arbitrary two-qubit state as

(4)
\begin{align} \ket{\psi} = a_{00} \ket{00} + a_{01} \ket{01} + a_{10} \ket{10} + a_{11} \ket{11} \end{align}

we can directly calculate the probability of obtaining result $\ket{+}$ from the measurement of the first qubit by simply applying our two-qubit measurement operator to $\ket{\psi}$ and taking the norm squared:

(5)
\begin{align} \Big| (P_+ \otimes I) \ket{\psi} \Big|^2 &= \left| \frac{1}{2}\Big[ a_{00} (\ket{00} + \ket{10}) + a_{01} (\ket{01} + \ket{11}) + a_{10} (\ket{00} + \ket{10}) + a_{11} (\ket{01} + \ket{11}) \Big] \right|^2 \\ &= \frac{1}{4} \Big| (a_{00} + a_{10}) \ket{00} + (a_{01} + a_{11}) \ket{01} + (a_{00} + a_{10}) \ket{10} + (a_{01} + a_{11}) \ket{11} \Big|^2 \\ &= \frac{1}{2} \big( |a_{00} + a_{10}|^2 + |a_{01} + a_{11}|^2 \big) \end{align}

Doing the same for the projection operator $P_-$:

(6)
\begin{align} P_- = \ket{-}\bra{-} = \frac{1}{2}(\ket{0}\bra{0} - \ket{0}\bra{1} - \ket{1}\bra{0} + \ket{1}\bra{1}) \end{align}

(7)
\begin{align} P_- \otimes I = \frac{1}{2}(\ket{00}\bra{00} - \ket{00}\bra{10} - \ket{10}\bra{00} + \ket{10}\bra{10} + \ket{01}\bra{01} - \ket{01}\bra{11} - \ket{11}\bra{01} + \ket{11}\bra{11}) \end{align}

(8)
\begin{align} \Big| (P_- \otimes I) \ket{\psi} \Big|^2 &= \left| \frac{1}{2}\Big[ a_{00} (\ket{00} - \ket{10}) + a_{01} (\ket{01} - \ket{11}) + a_{10} (-\ket{00} + \ket{10}) + a_{11} (-\ket{01} + \ket{11}) \Big] \right|^2 \\ &= \frac{1}{4} \Big| (a_{00} - a_{10}) \ket{00} + (a_{01} - a_{11}) \ket{01} + (-a_{00} + a_{10}) \ket{10} + (-a_{01} + a_{11}) \ket{11} \Big|^2 \\ &= \frac{1}{2} \big( |a_{00} - a_{10}|^2 + |a_{01} - a_{11}|^2 \big) \end{align}

Thus, we have the probabilities of each possible outcome:

(9)
\begin{align} P(\textrm{measuring } \ket{+}_1) &= \Big| (P_+ \otimes I) \ket{\psi} \Big|^2 = \frac{1}{2} \big( |a_{00} + a_{10}|^2 + |a_{01} + a_{11}|^2 \big) \\ P(\textrm{measuring } \ket{-}_1) &= \Big| (P_- \otimes I) \ket{\psi} \Big|^2 = \frac{1}{2} \big( |a_{00} - a_{10}|^2 + |a_{01} - a_{11}|^2 \big) \end{align}

**b.**

For the state $\ket{\Phi^+}$, we have $a_{00} = a_{11} = \frac{1}{\sqrt2}$ and $a_{01} = a_{10} = 0$. Then we directly use the result from part (a):

(10)
\begin{align} P(\textrm{measuring } \ket{+}_1) &= \frac{1}{2} \big( |a_{00} + a_{10}|^2 + |a_{01} + a_{11}|^2 \big) = \frac{1}{2} \\ P(\textrm{measuring } \ket{-}_1) &= \frac{1}{2} \big( |a_{00} - a_{10}|^2 + |a_{01} - a_{11}|^2 \big) = \frac{1}{2} \end{align}

**c.**

After applying our measurement operator $P_+ \otimes I$ to the state $\ket{\psi}$, we apply the measurement postulate to obtain our state after measurement:

(11)
\begin{align} \frac{(P_+ \otimes I) \ket{\psi}}{\left|(P_+ \otimes I) \ket{\psi}\right|} &= \frac{a_{00} (\ket{00} + \ket{10}) + a_{01} (\ket{01} + \ket{11}) + a_{10} (\ket{00} + \ket{10}) + a_{11} (\ket{01} + \ket{11})}{|\textrm{numerator}|} \\ &= \frac{(a_{00} + a_{10}) \ket{00} + (a_{01} + a_{11}) \ket{01} + (a_{00} + a_{10}) \ket{10} + (a_{01} + a_{11}) \ket{11}}{|\textrm{numerator}|} \end{align}

And now specializing to the state $\ket{\Phi^+}$ where $a_{00} = a_{11} = \frac{1}{\sqrt2}$ and $a_{01} = a_{10} = 0$, we see that our state after measuring the first qubit to be $\ket{+}_1$ is

(12)
\begin{align} \frac{(P_+ \otimes I) \ket{\Phi^+}}{\left|(P_+ \otimes I) \ket{\Phi^+}\right|} &= \frac{\frac{1}{\sqrt2}(\ket{00}+\ket{01}+\ket{10}+\ket{11})}{|\textrm{numerator}|} \\ &= \frac{1}{2}(\ket{00}+\ket{01}+\ket{10}+\ket{11}) \\ &= \frac{1}{\sqrt2}(\ket{0}+\ket{1}) \otimes \frac{1}{\sqrt2}(\ket{0}+\ket{1}) \\ &= \ket{+} \otimes \ket{+} \end{align}

Since the second qubit is in the state $\ket{+}$ after measurement of the first qubit, if we measure it in the standard basis, we will get either $\ket{0}$ or $\ket{1}$ with equal probability, and after measurement the two-qubit state collapses to either $\ket{+} \otimes \ket{0}$ or $\ket{+} \otimes \ket{1}$, respectively. (The qubits are no longer entangled, so measuring the second qubit has no effect on the first qubit.)

Similarly, if we measure the first qubit to be $\ket{-}_1$, the state becomes:

(13)
\begin{align} \frac{(P_- \otimes I) \ket{\Phi^+}}{\left|(P_- \otimes I) \ket{\Phi^+}\right|} &= \frac{\frac{1}{\sqrt2}(\ket{00}-\ket{01}-\ket{10}+\ket{11})}{|\textrm{numerator}|} \\ &= \frac{1}{2}(\ket{00}-\ket{01}-\ket{10}+\ket{11}) \\ &= \frac{1}{\sqrt2}(\ket{0}-\ket{1}) \otimes \frac{1}{\sqrt2}(\ket{0}-\ket{1}) \\ &= \ket{-} \otimes \ket{-} \end{align}

And since the second qubit is now in the state $\ket{-}$, when measuring the second qubit in the standard basis we will again get either $\ket{0}$ or $\ket{1}$ with equal probability, and the two-qubit state after measurement collapses to either $\ket{-} \otimes \ket{0}$ or $\ket{-} \otimes \ket{1}$, respectively.

**d.**

From part (c), we know that if we start with state $\ket{\Phi^+}$ and measure the first qubit to be $\ket{+}$, the two-qubit state collapses to $\ket{+} \otimes \ket{+}$. Now, measuring the second qubit in the Hadamard basis will return $\ket{+}$ with 100% probability and will leave the state unchanged.

Likewise, if we start with state $\ket{\Phi^+}$ and measure the first qubit to be $\ket{-}$, the two-qubit state collapses to $\ket{-} \otimes \ket{-}$. Now, measuring the second qubit in the Hadamard basis will return $\ket{-}$ with 100% probability and will leave the state unchanged.

It's worth looking back and gaining some intuition here. We know that the Bell state $\ket{\Phi^+} = \frac{1}{\sqrt2}(\ket{00}+\ket{11})$ is perfectly correlated, meaning that if the two qubits are measured in the same basis, they will always produce the same result. What we have shown here is that if we measure only one of the qubits, both qubits immediately collapse to the same state — that is, whatever state corresponds to the obtained measurement result — and therefore the two qubits become unentangled because their state can be expressed as a simple tensor product.