The notation $a_{n*}$ is the vector of all elements in row n.

I would guess there is a shorter proof.

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(1)

\begin{align} (A\ket{x})^\dagger=\left(\begin{pmatrix}a_{11}& \dots & a_{1n} \\ \vdots & & \vdots \\ a_{n1}& \dots & a_{nn}\end{pmatrix}\begin{pmatrix}x_1 \\ \vdots \\ x_n\end{pmatrix}\right)^\dagger=\begin{pmatrix}a_{1*}\cdot x \\ \vdots \\ a_{n*}\cdot x\end{pmatrix}^\dagger=\begin{pmatrix}\braket{a_{1*}}{x} \\ \vdots \\ \braket{a_{n*}}{x}\end{pmatrix}^\dagger=\\ \begin{pmatrix}\braket{\overline{x}}{\overline{a_{1*}}} & \dots & \braket{\overline{x}}{\overline{a_{n*}}}\end{pmatrix} =\begin{pmatrix}\overline{x_1} & \dots & \overline{x_n}\end{pmatrix}\begin{pmatrix}\overline{a_{11}}& \dots & \overline{a_{n1}} \\ \vdots & & \vdots \\ \overline{a_{1n}}& \dots & \overline{a_{nn}}\end{pmatrix}=\bra{x}A^\dagger \end{align}

The notation $a_{n*}$ is the vector of all elements in row n.

I would guess there is a shorter proof.

Andrew Thorp 05 Jun 2019 00:29

in discussion Hidden / Per page discussions » Exercise 3.7 Discussion

in discussion Hidden / Per page discussions » Exercise 3.7 Discussion

How would one come to the solution for c? It doesn't seem obvious to me.

We will use the following representations of the Pauli operators:

(1)\begin{align} \hat{I} = \hat{\sigma}_I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \qquad \hat{X} = \hat{\sigma}_X = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \qquad \hat{Y} = \hat{\sigma}_Y = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix} \qquad \hat{Z} = \hat{\sigma}_Z = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \end{align}

We are asked to show that the Pauli operators form a basis for all linear operators on a two-dimensional space. This means that, given any linear operator $\hat{A}$, we want to show that we can express it as a linear combination of the Pauli operators:

(2)\begin{align} \hat{A} = \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix} = b_{I} \hat{\sigma}_I + b_{X} \hat{\sigma}_X + b_{Y} \hat{\sigma}_Y + b_{Z} \hat{\sigma}_Z = \begin{pmatrix} b_{I} + b_{Z} & b_{X} - i b_{Y} \\ b_{X} + i b_{Y} & b_{I} - b_{Z} \end{pmatrix} \end{align}

for some (complex) coefficients $b_{I}$, $b_{X}$, $b_{Y}$, and $b_{Z}$. Clearly, this relation requires that we satisfy the system of linear equations:

(3)\begin{align} a_{11} = b_I + b_Z \qquad a_{12} = b_X - i b_Y \qquad a_{21} = b_X + i b_Y \qquad a_{22} = b_I - b_Z \end{align}

Solving for $b_{I}$, $b_{X}$, $b_{Y}$, and $b_{Z}$, we find:

(4)\begin{align} b_I = \tfrac{1}{2}(a_{11} + a_{22}) \qquad b_X = \tfrac{1}{2}(a_{12} + a_{21}) \qquad b_Y = \tfrac{i}{2}(a_{12} - a_{21}) \qquad b_Z = \tfrac{1}{2}(a_{11} - a_{22}) \end{align}

Therefore, we can write any arbitrary linear operator $\hat{A}$ as a linear combination of the Pauli operators as follows:

(5)\begin{align} \hat{A} = \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix} = \tfrac{1}{2}(a_{11} + a_{22}) \hat{\sigma}_I + \tfrac{1}{2}(a_{12} + a_{21}) \hat{\sigma}_X + \tfrac{i}{2}(a_{12} - a_{21}) \hat{\sigma}_Y + \tfrac{1}{2}(a_{11} - a_{22}) \hat{\sigma}_Z \end{align}

This confirms that the Pauli operators form a basis for all such linear operators on a two-dimensional space.

Note that from Section 5.4.1, we have the following definitions for $R$ and $T$:

(1)\begin{align} R(\alpha) = \begin{pmatrix} \cos{\alpha} & \sin{\alpha} \\ -\sin{\alpha} & \cos{\alpha} \end{pmatrix} \qquad T(\beta) = \begin{pmatrix} e^{i\beta} & 0 \\ 0 & e^{-i\beta} \end{pmatrix} \end{align}

**a.**

Recall that we define $e^{i(\alpha/2)\hat{\sigma}_y}$ as a rotation of $\alpha$ around the $y$-axis of the Bloch sphere (and likewise, $e^{i\alpha\hat{\sigma}_y}$ as a rotation of $2\alpha$), where $\sigma_y$ is the Pauli matrix:

\begin{align} \sigma_y = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix} \end{align}

Recall the following identity, where $\vec{\sigma} = (\hat{\sigma}_x, \hat{\sigma}_y, \hat{\sigma}_z)$:

(3)\begin{align} e^{i\theta(\hat{n}\cdot\vec{\sigma})} = \hat{I} \cos{\theta} + i(\hat{n}\cdot\vec{\sigma})\sin{\theta} \end{align}

Letting $\hat{n} = \hat{y}$ and $\theta = \alpha$, we get:

(4)\begin{align} e^{i\alpha\hat{\sigma}_y} &= \hat{I} \cos{\alpha} + i \hat{\sigma}_y \sin{\alpha} \\ &= \begin{pmatrix} \cos{\alpha} & 0 \\ 0 & \cos{\alpha} \end{pmatrix} + i \begin{pmatrix} 0 & -i \sin{\alpha} \\ i \sin{\alpha} & 0 \end{pmatrix} \\ &= \begin{pmatrix} \cos{\alpha} & \sin{\alpha} \\ -\sin{\alpha} & \cos{\alpha} \end{pmatrix} \\ &= R(\alpha) \end{align}

So we have shown that $R(\alpha)$ is a rotation of $2\alpha$ about the $y$-axis of the Bloch sphere.

**b.**

Recall that we define $e^{i(\beta/2)\hat{\sigma}_z}$ as a rotation of $\beta$ around the $z$-axis of the Bloch sphere (and likewise, $e^{i\beta\hat{\sigma}_z}$ as a rotation of $2\beta$), where $\sigma_z$ is the Pauli matrix:

\begin{align} \sigma_z = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \end{align}

Using the same identity as in part (a):

(6)\begin{align} e^{i\theta(\hat{n}\cdot\vec{\sigma})} = \hat{I} \cos{\theta} + i(\hat{n}\cdot\vec{\sigma})\sin{\theta} \end{align}

and letting $\hat{n} = \hat{z}$ and $\theta = \beta$, we get:

(7)\begin{align} e^{i\beta\hat{\sigma}_z} &= \hat{I} \cos{\beta} + i \hat{\sigma}_z \sin{\beta} \\ &= \begin{pmatrix} \cos{\beta} & 0 \\ 0 & \cos{\beta} \end{pmatrix} + i \begin{pmatrix} \sin{\beta} & 0 \\ 0 & -\sin{\beta} \end{pmatrix} \\ &= \begin{pmatrix} \cos{\beta} + i \sin{\beta} & 0 \\ 0 & \cos{\beta} - i \sin{\beta} \end{pmatrix} \\ &= \begin{pmatrix} e^{i\beta} & 0 \\ 0 & e^{-i\beta} \end{pmatrix} \\ &= T(\beta) \end{align}

So we have shown that $T(\beta)$ is a rotation of $2\beta$ about the $z$-axis of the Bloch sphere.

Alternatively, we could have derived this by working out the action of $T(\beta)$ on an arbitrary qubit state:

(8)\begin{align} \ket{\psi} = e^{i \gamma}\left( \cos{\frac{\theta}{2}} \ket{0} + e^{i \phi} \sin{\frac{\theta}{2}} \ket{1} \right) = e^{i \gamma} \begin{pmatrix} \cos{\frac{\theta}{2}} \\ e^{i \phi} \sin{\frac{\theta}{2}} \end{pmatrix} \end{align}

Applying $T(\beta)$ to state $\ket{\psi}$ gives:

(9)\begin{align} T(\beta) \ket{\psi} &= \begin{pmatrix} e^{i\beta} & 0 \\ 0 & e^{-i\beta} \end{pmatrix} e^{i \gamma} \begin{pmatrix} \cos{\frac{\theta}{2}} \\ e^{i \phi} \sin{\frac{\theta}{2}} \end{pmatrix} \\ &= e^{i \gamma} \begin{pmatrix} e^{i\beta} \cos{\frac{\theta}{2}} \\ e^{i \phi} e^{-i\beta} \sin{\frac{\theta}{2}} \end{pmatrix} \\ &= e^{i (\gamma+\beta)} \begin{pmatrix} \cos{\frac{\theta}{2}} \\ e^{i (\phi-2\beta)} \sin{\frac{\theta}{2}} \end{pmatrix} \\ &= e^{i (\gamma+\beta)}\left( \cos{\frac{\theta}{2}} \ket{0} + e^{i (\phi-2\beta)} \sin{\frac{\theta}{2}} \ket{1} \right) \end{align}

Since the global phase $e^{i (\gamma+\beta)}$ is ignorable, we can see that the only change from the original state $\ket{\psi}$ is that we have taken $\phi \rightarrow \phi - 2\beta$. Since $\phi$ is the azimuthal angle in the Bloch sphere picture, this corresponds to a rotation of $2\beta$ (actually, $-2\beta$) around the $z$-axis of the Bloch sphere.

**c.**

Recall that we define $e^{i(\gamma/2)\hat{\sigma}_x}$ as a rotation of $\gamma$ around the $x$-axis of the Bloch sphere (and likewise, $e^{i\gamma\hat{\sigma}_x}$ as a rotation of $2\gamma$), where $\sigma_x$ is the Pauli matrix:

\begin{align} \sigma_x = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \end{align}

Using the same identity as in parts (a) and (b):

(11)\begin{align} e^{i\theta(\hat{n}\cdot\vec{\sigma})} = \hat{I} \cos{\theta} + i(\hat{n}\cdot\vec{\sigma})\sin{\theta} \end{align}

and letting $\hat{n} = \hat{x}$ and $\theta = \gamma$, we get:

(12)\begin{align} e^{i\gamma\hat{\sigma}_x} &= \hat{I} \cos{\gamma} + i \hat{\sigma}_x \sin{\gamma} \\ &= \begin{pmatrix} \cos{\gamma} & 0 \\ 0 & \cos{\gamma} \end{pmatrix} + i \begin{pmatrix} 0 & \sin{\gamma} \\ \sin{\gamma} & 0 \end{pmatrix} \\ &= \begin{pmatrix} \cos{\gamma} & i \sin{\gamma} \\ i \sin{\gamma} & \cos{\gamma} \end{pmatrix} \end{align}

This is the matrix representation of an operator that corresponds to a rotation of $2\gamma$ about the $x$-axis of the Bloch sphere.

Let Alice's prepared qubit be in the state $\ket{\psi} = a\ket{0} + b\ket{1}$. Eve's qubit begins in state $\ket{0}$, so the combined two-qubit state is $\ket{\psi} \otimes \ket{0} = a \ket{00} + b \ket{10}$, where the first (left) qubit is from Alice and the second (right) qubit is from Eve.

Now, Eve applies a CNOT with the first qubit as the control qubit and the second qubit as the target qubit. Since CNOT takes $\ket{00} \rightarrow \ket{00}$ and $\ket{10} \rightarrow \ket{11}$, this results in a combined two-qubit state of $a \ket{00} + b \ket{11}$.

Next, the first qubit is sent to Bob, who measures it, and Eve measures the second qubit. (It's not clear from the problem statement whether Bob or Eve actually measures first, but the result works out the same either way.) Recall that in BB84, we know that the measurements will occur in either the standard or Hadamard basis. So Bob and Eve will each (randomly) choose one of these bases to measure in, just as Alice (randomly) chose one to prepare in.

We consider two cases:

**1. Alice prepared in the standard basis.** This means that after Eve's CNOT, the combined two-qubit state is either $\ket{00}$ or $\ket{11}$. If Eve measures in the standard basis, she obtains Alice's original bit with certainty, and if Eve measures in the Hadamard basis, she obtains Alice's original bit only 50% of the time. But because the two qubits are unentangled, Bob's measurement is completely unaffected by Eve's presence. (That is, Eve has not changed the state of the transmitted qubit — it is still in the state $\ket{0}$ or $\ket{1}$ as it was prepared by Alice.)

So in this case, Eve learns 50% of the original bits with certainty, and 0% of Bob's measurements are corrupted.

**2. Alice prepared in the Hadamard basis.** This means that after Eve's CNOT, the combined two-qubit state is either $\ket{\Phi^+}=\tfrac{1}{\sqrt2}(\ket{00}+\ket{11})$ or $\ket{\Phi^-}=\tfrac{1}{\sqrt2}(\ket{00}-\ket{11})$. Now, the two qubits are entangled. Whether Bob measures in the standard basis or the Hadamard basis, he will have equal probability of measuring a 0 or 1, regardless of what Alice's original bit was. For example, let's calculate the probability of Bob's Hadamard-basis measurement in the case that Alice prepared a $\ket{+}$:

\begin{align} P(&\textrm{A sent $\ket{+}$, B measures $\ket{+}$}) \\ &= (P_+ \otimes I) \ket{\Phi^+} \\ &= \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})\tfrac{1}{\sqrt2}(\ket{00}+\ket{11}) \\ &= \tfrac{1}{2} \qquad \textrm{(see solution to Exercise 4.7(a) and 4.7(b) from HW 2)} \end{align}

The same holds for the case where Alice prepares $\ket{-}$. That is, even though Alice and Bob are both using the Hadamard basis, Bob now only gets the same result as Alice half the time, because of the CNOT gate applied by Eve. (Note that exactly the same calculation applies to Eve's measurement as well — she will have a 50% probability of getting the correct result, regardless of which basis she measures in. This follows because both states $\ket{\Phi^+}$ and $\ket{\Phi^-}$ are symmetric in the two qubits.)

So in this case, Eve learns 0% of the original bits with certainty, and 50% of Bob's measurements are corrupted.

**In summary:** Assuming Alice prepares in each basis 50% of the time on average, Eve will learn 25% of the original bits with certainty, and 25% of Bob's measurements will be corrupted (that is, for each bit, there is a 75% chance that Eve goes undetected). So when Alice and Bob compare $s$ bits afterward, the probability that Eve goes undetected is:

\begin{align} P(\textrm{Eve undetected after $s$ bits}) = \left(\tfrac{3}{4}\right)^s \end{align}

This is exactly the same result as the direct measure-and-transmit strategy discussed in section 2.4.

Let's prove the contrapositive: If two qubits are *not* entangled, then there *is* a measurement of one of the qubits that gives a single result with certainty.

If two qubits are not entangled, then we can write their state as a tensor product. In general, we can write an arbitrary unentangled two-qubit state as $\ket{\psi} = (a\ket{0}+b\ket{1}) \otimes (c\ket{0}+d\ket{1})$. Let $\ket{v} = a\ket{0}+b\ket{1}$ be the state of the first qubit. Then, if we measure the first qubit in the orthonormal basis $\{\ket{v}, \ket{v^\perp} \}$, the result will be $\ket{v}$ with certainty. So we have proven the desired result.

Note that Example 4.3.5 in Rieffel and Polak designs a similar measurement for a two-qubit system. Expanding this to three qubits, we can write the Hermitian operator

(1)\begin{align} \hat{O} = 2\Big(\ket{000}\bra{000} + \ket{011}\bra{011} + \ket{101}\bra{101} + \ket{110}\bra{110}\Big) + 3\Big(\ket{001}\bra{001} + \ket{010}\bra{010} + \ket{100}\bra{100} + \ket{111}\bra{111}\Big) \end{align}

which specifies measurement with respect to the decomposition of the vector space into subspaces $S_\textrm{even}$ and $S_\textrm{odd}$, which are generated by $\{ \ket{000},\ket{011},\ket{101},\ket{110} \}$ and $\{ \ket{001},\ket{010},\ket{100},\ket{111} \}$, respectively, where we have (arbitrarily) assigned the eigenvalue 2 to “even” and the eigenvalue 3 to “odd”.

**a.**

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

\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):

\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:

\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.

**a.**

Antipodal points on the surface of the Bloch sphere are those where the signs of the Cartesian coordinates have been flipped. On a unit sphere, we can use $\theta = \arccos{z}$ and $\phi = \arctan{\frac{y}{x}}$ (unless $x=y=0$, in which case we are at a pole, and $\phi$ is undefined and irrelevant). Thus, flipping the signs of all Cartesian coordinates leaves $\phi$ unchanged, and takes $\theta \rightarrow \theta + \pi$, since $\cos \theta = -\cos{(\theta + \pi)}$.

So we want to show that the following two antipodal states are orthogonal:

(1)\begin{align} \ket{\psi_1} &= e^{i \gamma}\left( \cos{\frac{\theta}{2}} \ket{0} + e^{i \phi} \sin{\frac{\theta}{2}} \ket{1} \right) \\ \ket{\psi_2} &= e^{i \gamma}\left( \cos{\frac{\theta + \pi}{2}} \ket{0} + e^{i \phi} \sin{\frac{\theta + \pi}{2}} \ket{1} \right) \end{align}

To do this, we simply need to show that their inner product is zero:

(2)\begin{align} \braket{\psi_1}{\psi_2} &= \left[e^{-i \gamma}\left( \cos{\frac{\theta}{2}} \bra{0} + e^{-i \phi} \sin{\frac{\theta}{2}} \bra{1} \right) \right] \left[e^{i \gamma}\left( \cos{\frac{\theta + \pi}{2}} \ket{0} + e^{i \phi} \sin{\frac{\theta + \pi}{2}} \ket{1} \right)\right] \\ &= \cos{\frac{\theta}{2}} \cos{\frac{\theta + \pi}{2}} + \sin{\frac{\theta}{2}} \sin{\frac{\theta + \pi}{2}} \\ &= \cos{\left(-\frac{\pi}{2}\right)} \\ &= 0 \end{align}

where we have used the identity $\cos{(a-b)} = \cos a \cos b + \sin a \sin b$.

**b.**

If two states $\ket{\psi_1} = e^{i \gamma_1}\left( \cos{\frac{\theta_1}{2}} \ket{0} + e^{i \phi_1} \sin{\frac{\theta_1}{2}} \ket{1} \right)$ and $\ket{\psi_2} = e^{i \gamma_2}\left( \cos{\frac{\theta_2}{2}} \ket{0} + e^{i \phi_2} \sin{\frac{\theta_2}{2}} \ket{1} \right)$ are orthogonal, their inner product $\braket{\psi_1}{\psi_2} = 0$. We use this to find a condition on the angles $\theta_1$, $\theta_2$, $\phi_1$, and $\phi_2$:

\begin{align} 0 &= \braket{\psi_1}{\psi_2} \\ &= \left[e^{-i \gamma_1}\left( \cos{\frac{\theta_1}{2}} \bra{0} + e^{-i \phi_1} \sin{\frac{\theta_1}{2}} \bra{1} \right) \right] \left[e^{i \gamma_2}\left( \cos{\frac{\theta_2}{2}} \ket{0} + e^{i \phi_2} \sin{\frac{\theta_2}{2}} \ket{1} \right)\right] \\ &= e^{i(\gamma_2 - \gamma_1)}\left[ \cos{\frac{\theta_1}{2}} \cos{\frac{\theta_2}{2}} + e^{i(\phi_2-\phi_1)} \sin{\frac{\theta_1}{2}} \sin{\frac{\theta_2}{2}} \right] \end{align}

The term $e^{i(\gamma_2 - \gamma_1)}$ can never be zero, so the expression inside the square brackets must vanish. So we must have:

(4)\begin{align} \cos{\frac{\theta_1}{2}} \cos{\frac{\theta_2}{2}} + e^{i(\phi_2-\phi_1)} \sin{\frac{\theta_1}{2}} \sin{\frac{\theta_2}{2}} = 0 \end{align}

There are three ways to satisfy this equation, and all three require that the points are antipodal:

1. $\cos{\frac{\theta_1}{2}} \cos{\frac{\theta_2}{2}} = \sin{\frac{\theta_1}{2}} \sin{\frac{\theta_2}{2}} = 0$. This would imply that $\theta_1 = 0$ and $\theta_2 = \pi$ (or vice versa), meaning the points are at the poles of the sphere and thus are antipodal, regardless of $\phi$.

2. $e^{i(\phi_2-\phi_1)} = 1$ and $\cos{\frac{\theta_1}{2}} \cos{\frac{\theta_2}{2}} + \sin{\frac{\theta_1}{2}} \sin{\frac{\theta_2}{2}} = 0$. This requires that $\phi_1 = \phi_2$ and $\cos{\left( \frac{\theta_1 - \theta_2}{2} \right)} = 0$, and thus $\theta_1 = \theta_2 + \pi$. This is the same condition for antipodality we derived in part (a) of this problem.

3. $e^{i(\phi_2-\phi_1)} = -1$ and $\cos{\frac{\theta_1}{2}} \cos{\frac{\theta_2}{2}} - \sin{\frac{\theta_1}{2}} \sin{\frac{\theta_2}{2}} = 0$. This requires that $\phi_2 = \phi_1 + \pi$ and $\cos{\left( \frac{\theta_1 + \theta_2}{2} \right)} = 0$, and thus $\theta_2 = \pi - \theta_1$. Letting $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ be the Cartesian coordinates of $\ket{\psi_1}$ and $\ket{\psi_2}$, respectively, we have:

\begin{align} x_1 &= \sin \theta_1 \cos \phi_1 \\ y_1 &= \sin \theta_1 \sin \phi_1 \\ z_1 &= \cos \theta_1 \\ x_2 &= \sin \theta_2 \cos \phi_2 = \sin(\pi - \theta_1) \cos(\phi_1 + \pi) = -\sin \theta_1 \cos \phi_1 = -x_1 \\ y_2 &= \sin \theta_2 \sin \phi_2 = \sin(\pi - \theta_1) \sin(\phi_1 + \pi) = -\sin \theta_1 \sin \phi_1 = -y_1 \\ z_2 &= \cos \theta_2 = \cos(\pi - \theta_1) = -z_1 \end{align}

Since the two points have opposite Cartesian coordinates, they are antipodal.

**a.**

We start from the general form $\ket{\psi} = \alpha \ket{0} + \beta \ket{1}$, where $\alpha, \beta \in \mathbb{C}$ and $|\alpha|^2 + |\beta|^2 = 1$. Since $\alpha$ and $\beta$ are complex numbers, we can separate them into the product of a real part and a phase: $\alpha = r_\alpha e^{i \phi_\alpha}$ and $\beta = r_\beta e^{i \phi_\beta}$. This gives:

\begin{align} \ket{\psi} &= \alpha \ket{0} + \beta \ket{1} \\ &= r_\alpha e^{i \phi_\alpha} \ket{0} + r_\beta e^{i \phi_\beta} \ket{1} \\ &= e^{i \phi_\alpha} \left( r_\alpha \ket{0} + e^{i (\phi_\beta - \phi_\alpha)} r_\beta \ket{1} \right) \end{align}

We now define a global phase $\gamma = \phi_\alpha$ and a relative phase $\phi = \phi_\beta - \phi_\alpha$. By our normalization condition $\braket{\psi}{\psi} = 1$, we require $r_\alpha^2 + r_\beta^2 = 1$, and thus without loss of generality we can set $r_\alpha = \cos{\widetilde{\theta}}$ and $r_\beta = \sin{\widetilde{\theta}}$. So we have:

(2)\begin{align} \ket{\psi} = e^{i \gamma} \left( \cos{\widetilde{\theta}} \ket{0} + e^{i \phi} \sin{\widetilde{\theta}} \ket{1} \right) \end{align}

Now, we want to choose $\widetilde{\theta}$ such that these states map onto the surface of a unit Bloch sphere using standard spherical coordinates (where $\theta$ is the polar angle and $\phi$ is the azimuthal angle), and such that $\ket{0}$ is at the top (north pole, i.e., $\theta = 0$) and $\ket{1}$ is at the bottom (south pole, i.e., $\theta = \pi$).

Using the expression we derived above, $\ket{\phi(\widetilde{\theta} = 0)} = e^{i \gamma} \ket{0}$ is at the north pole, and $\ket{\phi(\widetilde{\theta} = \frac{\pi}{2})} = e^{i (\gamma + \phi)} \ket{1}$ is at the south pole. So to achieve the correct result, we must set $\widetilde{\theta} = \frac{\theta}{2}$. This gives the desired expression:

(3)\begin{align} \ket{\psi} = e^{i \gamma}\left( \cos{\frac{\theta}{2}} \ket{0} + e^{i \phi} \sin{\frac{\theta}{2}} \ket{1} \right) \end{align}

Note that the relative phase $\phi$ is irrelevant at the poles.

**b.**

Recall the definition of the states:

\begin{align} \ket{+} &= \frac{1}{\sqrt2}(\ket{0} + \ket{1}) \qquad \ket{\mathbf{i}} = \frac{1}{\sqrt2}(\ket{0} + i \ket{1}) \\ \ket{-} &= \frac{1}{\sqrt2}(\ket{0} - \ket{1}) \qquad \ket{\mathbf{-i}} = \frac{1}{\sqrt2}(\ket{0} - i \ket{1}) \end{align}

By identifying $\cos{\frac{\theta}{2}}$ as the coefficient of $\ket{0}$ and $e^{i \phi} \sin{\frac{\theta}{2}}$ as the coefficient of $\ket{1}$, we can derive the corresponding values of $\theta$ and $\phi$ for each state:

(5)\begin{align} \ket{+}\,&\Leftrightarrow\, \theta = \frac{\pi}{2},\,\phi = 0 \qquad \ket{\mathbf{i}}\,\Leftrightarrow\, \theta = \frac{\pi}{2},\,\phi = \frac{\pi}{2} \\ \ket{-}\,&\Leftrightarrow\, \theta = \frac{\pi}{2},\,\phi = \pi \qquad \ket{\mathbf{-i}}\,\Leftrightarrow\, \theta = \frac{\pi}{2},\,\phi = \frac{3\pi}{2} \end{align}

Notice that all four of these states lie on the equator of the Bloch sphere and thus have $\theta = \frac{\pi}{2}$.

**a.**

Eve is measuring in a random basis at each step. In order to know a bit value "for sure", i.e., with 100% confidence, she would have to measure in exactly the same basis as the one in which Alice prepared the qubit. But there are infinitely many possible measurement bases, and so the probability of doing so is exactly zero. She will have full confidence about $\boxed{0\%}$ of the bits.

**b.**

We interpret the "final key" here as the classical bit string agreed upon by Alice and Bob after they have thrown away the bits for which they used different bases. Of course, in the BB84 protocol, Alice and Bob would then compare their a portion of their classical bit string and throw it away if they detected Eve's presence, but for the purposes of this problem, we ignore that portion.

Suppose Eve measures in the random basis $\{\ket{v}, \ket{v^\perp}\}$, and Alice has prepared the qubits in the basis $\{\ket{a}, \ket{a^\perp}\}$. When the qubit reaches Eve, it must be in either state $\ket{a}$ or $\ket{a^\perp}$. Without loss of generality, assume that Alice prepared $\ket{a}$, and assume that $\ket{v}$ is the “closer” of Eve's basis vectors to $\ket{a}$, i.e., the one such that the angle between $\ket{a}$ and $\ket{v}$ is $0 \le \theta \le \pi/2$. Then, the probability that Eve's measurement produced the same (classical) bit as Alice's is given by $\left|\braket{v}{a}\right|^2 = \cos^2 \theta$. To find the average probability that Eve obtains the correct bit, we integrate this as a probability density over the range of possible angles, $0 \le \theta \le \pi/2$:

(1)\begin{align} \frac{1}{\pi/2} \int_0^{\pi/2} \cos^2 \theta\,d\theta = \frac{1}{2} \end{align}

So on average, $\boxed{50\%}$ of Eve's bits are correct, as compared to Alice's original classical bit string. (Note that after Alice and Bob throw away the bits for which they used different bases, Eve will still have on average only 50% of the bits correct, since her measurement basis is random with respect to either possible basis that Alice and Bob have used, and the same calculation applies.)

**c.**

Remember that Alice and Bob have agreed on their two choices of bases ahead of time, and they will discard any bits in which they used different bases. So we can assume that, for each bit they choose to keep, Alice and Bob have both measured in the same basis. Call this basis $\{\ket{a}, \ket{a^\perp}\}$. Now, if Eve has made a measurement in her random basis $\{\ket{v}, \ket{v^\perp}\}$, then the qubit Bob receives will be in either state $\ket{v}$ or $\ket{v^\perp}$. Let's assume Alice sent $\ket{a}$, and again let $\theta$ be the angle between $\ket{a}$ and $\ket{v}$. What is the probability that Bob will still measure the correct value? We have two cases to consider:

1. Eve measures $\ket{v}$, and then Bob measures $\ket{a}$. Probability = $\left| \braket{v}{a} \right|^2 \left| \braket{a}{v} \right|^2 = \cos^4 \theta$.

2. Eve measures $\ket{v^\perp}$, and then Bob measures $\ket{a}$. Probability = $\left| \braket{v^\perp}{a} \right|^2 \left| \braket{a}{v^\perp} \right|^2 = \sin^4 \theta$.

So the probability that Bob measures the correct bit is $\cos^4 \theta + \sin^4 \theta$. Averaging this over the range of possible angles:

\begin{align} \frac{1}{\pi/2} \int_0^{\pi/2} (\cos^4 \theta + \sin^4 \theta)\,d\theta = \frac{3}{4} \end{align}

This means that, on average, Eve will go undetected $\frac{3}{4}$ of the time. In order for Alice and Bob to have at least a $90\%$ chance of detecting Eve, we need $(\frac{3}{4})^n \le 0.1$, which means that they need to compare $\boxed{n > 8}$ bits.

**a.**

We measure the state $\ket{\psi} = \frac{\sqrt3}{2} \ket{0} - \frac{1}{2} \ket{1}$ in the standard computational basis $\{\ket{0}, \ket{1}\}$. There are two possible results:

1. Result “$\ket{0}$” is obtained with probability $\left|\braket{\psi}{0}\right|^2 = \frac{3}{4}$. The state after measurement is $\ket{0}$.

2. Result “$\ket{1}$” is obtained with probability $\left|\braket{\psi}{1}\right|^2 = \frac{1}{4}$. The state after measurement is $\ket{1}$.

**b.**

We measure the state $\ket{\psi} = \frac{\sqrt3}{2} \ket{1} - \frac{1}{2} \ket{0}$ in the standard computational basis $\{\ket{0}, \ket{1}\}$. There are two possible results:

1. Result “$\ket{0}$” is obtained with probability $\left|\braket{\psi}{0}\right|^2 = \frac{1}{4}$. The state after measurement is $\ket{0}$.

2. Result “$\ket{1}$” is obtained with probability $\left|\braket{\psi}{1}\right|^2 = \frac{3}{4}$. The state after measurement is $\ket{1}$.

**c.**

We measure the state $\ket{\psi} = \ket{\mathbf{-i}}$ in the standard computational basis $\{\ket{0}, \ket{1}\}$. There are two possible results:

1. Result “$\ket{0}$” is obtained with probability $\left|\braket{\psi}{0}\right|^2 = \frac{1}{2}$. The state after measurement is $\ket{0}$.

2. Result “$\ket{1}$” is obtained with probability $\left|\braket{\psi}{1}\right|^2 = \frac{1}{2}$. The state after measurement is $\ket{1}$.

**d.**

We measure the state $\ket{\psi} = \ket{0}$ in the Hadamard basis $\{\ket{+}, \ket{-}\}$. Note that we can rewrite the state in the desired basis as $\ket{\psi} = \frac{1}{\sqrt2}(\ket{+} + \ket{-})$. There are two possible results:

1. Result “$\ket{+}$” is obtained with probability $\left|\braket{\psi}{+}\right|^2 = \frac{1}{2}$. The state after measurement is $\ket{+}$.

2. Result “$\ket{-}$” is obtained with probability $\left|\braket{\psi}{-}\right|^2 = \frac{1}{2}$. The state after measurement is $\ket{-}$.

**e.**

We measure the state $\ket{\psi} = \frac{1}{\sqrt2}(\ket{0} - \ket{1})$ in the basis $\{\ket{\mathbf{i}}, \ket{\mathbf{-i}}\}$. Note that we can rewrite the state in the desired basis as $\ket{\psi} = \frac{1+i}{2} \ket{\mathbf{i}} + \frac{1-i}{2} \ket{\mathbf{-i}}$. There are two possible results:

1. Result “$\ket{\mathbf{i}}$” is obtained with probability $\left|\braket{\psi}{\mathbf{i}}\right|^2 = \frac{1}{2}$. The state after measurement is $\ket{\mathbf{i}}$.

2. Result “$\ket{\mathbf{-i}}$” is obtained with probability $\left|\braket{\psi}{\mathbf{-i}}\right|^2 = \frac{1}{2}$. The state after measurement is $\ket{\mathbf{-i}}$.

**f.**

We measure the state $\ket{\psi} = \ket{1}$ in the basis $\{\ket{\mathbf{i}}, \ket{\mathbf{-i}}\}$. Note that we can rewrite the state in the desired basis as $\ket{\psi} = \frac{-i}{\sqrt2} \ket{\mathbf{i}} + \frac{i}{\sqrt2} \ket{\mathbf{-i}}$. There are two possible results:

1. Result “$\ket{\mathbf{i}}$'' is obtained with probability $\left|\braket{\psi}{\mathbf{i}}\right|^2 = \frac{1}{2}$. The state after measurement is $\ket{\mathbf{i}}$.

2. Result ``$\ket{\mathbf{-i}}$” is obtained with probability $\left|\braket{\psi}{\mathbf{-i}}\right|^2 = \frac{1}{2}$. The state after measurement is $\ket{\mathbf{-i}}$.

**g.**

We measure the state $\ket{\psi} = \ket{+}$ in the basis $\{\ket{a}, \ket{a^\perp}\}$, where $\ket{a} = \frac{1}{2} \ket{0} + \frac{\sqrt3}{2} \ket{1}$ and $\ket{a^\perp} = \frac{\sqrt3}{2} \ket{0} - \frac{1}{2} \ket{1}$. Note that we can rewrite the state in the desired basis as $\ket{\psi} = \frac{\sqrt3 + 1}{2 \sqrt2} \ket{a} + \frac{\sqrt3 - 1}{2 \sqrt2} \ket{a^\perp}$. There are two possible results:

1. Result “$\ket{a}$” is obtained with probability $\left|\braket{\psi}{a}\right|^2 = \frac{2+\sqrt3}{4}$. The state after measurement is $\ket{a}$.

2. Result “$\ket{a^\perp}$” is obtained with probability $\left|\braket{\psi}{a^\perp}\right|^2 = \frac{2-\sqrt3}{4}$. The state after measurement is $\ket{a^\perp}$.

We use the standard convention:

(1)\begin{align} \ket{+} &= \frac{1}{\sqrt2}(\ket{0} + \ket{1}) \qquad \ket{\mathbf{i}} = \frac{1}{\sqrt2}(\ket{0} + i \ket{1}) \\ \ket{-} &= \frac{1}{\sqrt2}(\ket{0} - \ket{1}) \qquad \ket{\mathbf{-i}} = \frac{1}{\sqrt2}(\ket{0} - i \ket{1}) \end{align}

**a.**

$\ket{+} = \frac{1}{\sqrt2}(\ket{0} + \ket{1})$ is a superposition with respect to the standard basis.

It is not a superposition with respect to the Hadamard $\{\ket{+}, \ket{-}\}$ basis.

**b.**

$\frac{1}{\sqrt2}(\ket{+} + \ket{-}) = \ket{0}$ is not a superposition with respect to the standard basis.

**c.**

$\frac{1}{\sqrt2}(\ket{+} - \ket{-}) = \ket{1}$ is not a superposition with respect to the standard basis.

**d.**

$\frac{\sqrt3}{2} \ket{+} - \frac{1}{2} \ket{-}$ is a superposition with respect to the standard basis.

To find a basis in which this is not a superposition, we need to find a vector which is orthogonal. Let $\ket{v} = \frac{\sqrt3}{2} \ket{+} - \frac{1}{2} \ket{-}$. We are looking for $\ket{v^\perp} = \alpha \ket{+} + \beta \ket{-}$ such that $\braket{v}{v^\perp} = 0$ and $|\alpha|^2 + |\beta|^2 = 1$:

\begin{align} \braket{v}{v^\perp} &= \Bigg[ \frac{\sqrt3}{2} \bra{+} - \frac{1}{2} \bra{-} \Bigg]\Bigg[ \alpha \ket{+} + \beta \ket{-} \Bigg] \\ &= \frac{\sqrt3}{2} \alpha - \frac{1}{2} \beta = 0 \implies \alpha = \frac{1}{2},\ \beta = \frac{\sqrt3}{2} \end{align}

Thus, a basis in which $\ket{v}$ is not a superposition is $\{ \ket{v}, \ket{v^\perp} \}$, where these are defined as above.

**e.**

$\frac{1}{\sqrt2}(\ket{\mathbf{i}} - \ket{\mathbf{-i}}) = i \ket{1}$ is not a superposition with respect to the standard basis.

**f.**

$\frac{1}{\sqrt2}(\ket{0} - \ket{1}) = \ket{-}$ is a superposition with respect to the standard basis.

It is not a superposition with respect to the Hadamard $\{\ket{+}, \ket{-}\}$ basis.

The problem is based on Figure 2.3, in which we have a laser emitting photons with random polarization passing through three polaroids $A$, $B$, and $C$. Polaroid $A$ is horizontally polarized, polaroid $C$ is vertically polarized, and polaroid $B$ has a preferred axis $\ket{v} = \cos{\theta} \ket{\rightarrow} + \sin{\theta} \ket{\uparrow}$. We are looking for the fraction of photons (as a function of $\theta$) that pass through all three polaroids and reach the screen.

We do this by simply calculating the fraction of photons that are transmitted at each stage:

(1)\begin{align} \textrm{through polaroid $A$: } &0.5 &\textrm{random polarization initially} \\ &\textrm{transmitted photons are now in state $\ket{\rightarrow}$} \\ \textrm{through polaroid $B$: } &\left|\braket{\rightarrow}{v}\right|^2 \\ &= \left|\bra{\rightarrow}\big[\cos{\theta} \ket{\rightarrow} + \sin{\theta} \ket{\uparrow}\big]\right|^2 &\textrm{substitute definition of $\ket{v}$} \\ &= \left| \cos{\theta} \braket{\rightarrow}{\rightarrow} + \sin{\theta} \braket{\rightarrow}{\uparrow} \right|^2 &\textrm{linearity of inner product} \\ &= \cos^2{\theta} &\textrm{$\braket{\rightarrow}{\rightarrow} = 1$ and $\braket{\rightarrow}{\uparrow} = 0$} \\ &\textrm{transmitted photons are now in state $\ket{v}$} \\ \textrm{through polaroid $C$: } &\left|\braket{v}{\uparrow}\right|^2 \\ &= \left|\big[\cos{\theta} \bra{\rightarrow} + \sin{\theta} \bra{\uparrow}\big]\ket{\uparrow}\right|^2 &\textrm{substitute definition of $\ket{v}$} \\ &= \left| \cos{\theta} \braket{\rightarrow}{\uparrow} + \sin{\theta} \braket{\uparrow}{\uparrow} \right|^2 &\textrm{linearity of inner product} \\ &= \sin^2{\theta} &\textrm{$\braket{\uparrow}{\uparrow} = 1$ and $\braket{\rightarrow}{\uparrow} = 0$} \\ &\textrm{transmitted photons are now in state $\ket{\uparrow}$} \end{align}

And since the three polaroids are in series, we multiply the probabilities:

(2)\begin{align} \boxed{P = 0.5 \cos^2 \theta \sin^2 \theta} \end{align}

**a**. When measured with respect to the standard basis, the second bit of the state $\ket{00}$ is clearly equal to 0 with probability 1.

With respect to the Hadamard basis, the state $\ket{00}$ is equal to $\frac{1}{2}\left(\ket{+}\ket{+} + \ket{+}\ket{-} +\ket{-}\ket{+} +\ket{-}\ket{-}\right)$. Thus, when measured with respect to the Hadamard basis, the second bit equals $\ket{+}$ with probability $\frac{1}{2}$ and equals $\ket{-}$ with probability $\frac{1}{2}$.

As a shorthand for the vectors of the basis $B$, let us write $\ket{u} = \frac{1}{\sqrt{2}}\left(\ket{0} + \mathbf{i}\ket{1}\right)$ and $\ket{v} = \frac{1}{\sqrt{2}}\left(\ket{0} - \mathbf{i}\ket{1}\right)$. Then $\ket{0} = \frac{1}{\sqrt{2}}\left(\ket{u} + \ket{v}\right)$, and $\ket{00} = \frac{1}{2}\left(\ket{u}\ket{u} + \ket{u}\ket{v} +\ket{v}\ket{u} +\ket{v}\ket{v}\right)$. Much as in the previous answer, therefore, we see that when measured with respect to the basis $B$, the second bit equals $\ket{u}$ with probability $\frac{1}{2}$ and equals $\ket{v}$ with probability $\frac{1}{2}$.

**b**. We noted above that, with respect to the Hadamard basis, the state $\ket{00}$ is equal to $\frac{1}{2}\left(\ket{+}\ket{+} + \ket{+}\ket{-} +\ket{-}\ket{+} +\ket{-}\ket{-}\right)$. Thus, when its second bit is measured with respect to the Hadamard basis, one of two things can happen with equal probability:

- Case 1: the second bit is measured as $\ket{+}$, and the state becomes $\frac{1}{\sqrt{2}}\left(\ket{+}\ket{+} +\ket{-}\ket{+}\right)$ – let’s call this state $X$.
- Case 2: the second bit is measured as $\ket{-}$, and the state becomes $\frac{1}{\sqrt{2}}\left(\ket{+}\ket{-} +\ket{-}\ket{-}\right)$ – let’s call this state $Y$.

We can compute

(1)\begin{align} \ket{+}\ket{+} = \frac{1}{2}\left(-\mathbf{i}\ket{u}\ket{u} + \ket{u}\ket{v} + \ket{v}\ket{u} + \mathbf{i}\ket{v}\ket{v}\right) \end{align}

and

(2)\begin{align} \ket{-}\ket{+} = \frac{1}{2}\left(\ket{u}\ket{u} + \mathbf{i}\ket{u}\ket{v} - \mathbf{i}\ket{v}\ket{u} + \ket{v}\ket{v}\right) \end{align}

and so state $X = \frac{1}{2\sqrt{2}}\left((1-\mathbf{i})\ket{u}\ket{u} + (1+\mathbf{i})\ket{u}\ket{v} + (1-\mathbf{i})\ket{v}\ket{u} + (1+\mathbf{i})\ket{v}\ket{v}\right)$. Thus, if Case 1 occurred on the first measurement, then the second measurement yields $\ket{u}$ with probability $\frac{1}{8}\left({\left|1-\mathbf{i}\right|}^2 + {\left|1-\mathbf{i}\right|}^2\right) = \frac{1}{2}$, and yields $\ket{v}$ with probability $\frac{1}{8}\left({\left|1+\mathbf{i}\right|}^2 + {\left|1+\mathbf{i}\right|}^2\right) = \frac{1}{2}$.

Similarly we can compute

(3)\begin{align} \ket{+}\ket{-} = \frac{1}{2}\left(\ket{u}\ket{u} - \mathbf{i}\ket{u}\ket{v} + \mathbf{i}\ket{v}\ket{u} + \ket{v}\ket{v}\right) \end{align}

and

(4)\begin{align} \ket{-}\ket{-} = \frac{1}{2}\left(\mathbf{i}\ket{u}\ket{u} + \ket{u}\ket{v} + \ket{v}\ket{u} - \mathbf{i}\ket{v}\ket{v}\right) \end{align}

and so state $Y = \frac{1}{2\sqrt{2}}\left((1+\mathbf{i})\ket{u}\ket{u} + (1-\mathbf{i})\ket{u}\ket{v} + (1+\mathbf{i})\ket{v}\ket{u} + (1-\mathbf{i})\ket{v}\ket{v}\right)$. Thus, if Case 2 occurred on the first measurement, then the second measurement yields $\ket{u}$ with probability $\frac{1}{8}\left({\left|1+\mathbf{i}\right|}^2 + {\left|1+\mathbf{i}\right|}^2\right) = \frac{1}{2}$, and yields $\ket{v}$ with probability $\frac{1}{8}\left({\left|1-\mathbf{i}\right|}^2 + {\left|1-\mathbf{i}\right|}^2\right) = \frac{1}{2}$.

**c**. As in question b, when the second bit of $\ket{00}$ is measured with respect to the Hadamard basis, one of two things can happen with equal probability:

- Case 1: the second bit is measured as $\ket{+}$, and the state becomes $\frac{1}{\sqrt{2}}\left(\ket{+}\ket{+} +\ket{-}\ket{+}\right)$ – we called this state $X$.
- Case 2: the second bit is measured as $\ket{-}$, and the state becomes $\frac{1}{\sqrt{2}}\left(\ket{+}\ket{-} +\ket{-}\ket{-}\right)$ – we called this state $Y$.

We can compute

(5)\begin{align} \ket{+}\ket{+} = \frac{1}{2}\left(\ket{0}\ket{0} + \ket{0}\ket{1} + \ket{1}\ket{0} + \ket{1}\ket{1}\right) \end{align}

and

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

and so state $X = \frac{1}{\sqrt{2}}\left(\ket{0}\ket{0} + \ket{1}\ket{1}\right)$. Thus, if Case 1 occurred on the first measurement, then the second measurement yields $\ket{0}$ with probability $\frac{1}{2}$ and yields $\ket{1}$ with probability $\frac{1}{2}$.

Similarly we can compute

(7)\begin{align} \ket{+}\ket{-} = \frac{1}{2}\left(\ket{0}\ket{0} - \ket{0}\ket{1} + \ket{1}\ket{0} - \ket{1}\ket{1}\right) \end{align}

and

(8)\begin{align} \ket{-}\ket{-} = \frac{1}{2}\left(\ket{0}\ket{0} - \ket{0}\ket{1} - \ket{1}\ket{0} + \ket{1}\ket{1}\right) \end{align}

and so state $Y = \frac{1}{\sqrt{2}}\left(\ket{0}\ket{0} - \ket{0}\ket{1}\right)$. Thus, if Case 2 occurred on the first measurement, then again the second measurement yields $\ket{0}$ with probability $\frac{1}{2}$ and yields $\ket{1}$ with probability $\frac{1}{2}$.

For $n=1$, the stated theorem is clearly not true – the distance can be zero. For the rest of this answer we assume $n > 1$.

Let’s label the standard basis vectors $\ket{b_0} \ldots \ket{b_{n-1}}$, and let $\ket{b_n}$ be another name for $\ket{b_0}$ – this will simplify the proof notation.

For any state $\ket{\psi}$, the sum of the distances from $\ket{\psi}$ to the basis vectors is

(1)\begin{align} \displaystyle \sum_{i=0}^{n-1} \left| \ket{\psi} - \ket{b_i} \right| = \frac{1}{2} \displaystyle \sum_{i=0}^{n-1} \left( \left| \ket{\psi} - \ket{b_i} \right| + \left| \ket{\psi} - \ket{b_{i+1}} \right| \right) \end{align}

We can now use the triangle inequality:

(2)\begin{align} \left| \ket{\psi} - \ket{b_i} \right| + \left| \ket{\psi} - \ket{b_{i+1}} \right| \ge \left| \ket{b_i} - \ket{b_{i+1}} \right| \end{align}

It is easy to see that the distance between any two standard basis vectors is $\sqrt{2}$. Hence

(3)\begin{align} \displaystyle \sum_{i=0}^{n-1} \left| \ket{\psi} - \ket{b_i} \right| \ge \frac{1}{2} \times n \times \sqrt{2} = \frac{n}{\sqrt{2}} \end{align}

This is a positive lower bound dependent on $n$. (It is not a *tight* lower bound.)

**a**. Assume $\ket{\psi}$ is not entangled with respect to this decomposition. Then (still using the 0/1/2/3 representation for each two-qubit part), $\ket{\psi}$ can be written as

\begin{align} \ket{\psi} = (a_0\ket{0} + a_1\ket{0} + a_2\ket{2} + a_3\ket{3}) \otimes (b_0\ket{0} + b_1\ket{0} + b_2\ket{2} + b_3\ket{3}) \end{align}

We can see that $a_0 b_0 = \frac{1}{2}$, so $a_0$ is non-zero; and $a_1 b_1 = \frac{1}{2}$, so $b_1$ is non-zero. But we can also see that $a_0 b_1 = 0$. This is impossible. Hence $\ket{\psi}$ must in fact be entangled with respect to this decomposition.

**b**. Let us write $\ket{\psi}$ as $\frac{1}{2}(\ket{0000} +\ket{0101} +\ket{1010} + \ket{1111})$.

First consider the one qubit / three qubit decomposition where the one qubit is the leftmost in this representation. Suppose $\ket{\psi}$ is not entangled with respect to this decomposition. Then $\ket{\psi}$ can be written as

(2)\begin{align} \ket{\psi} = (a_0\ket{0} + a_1\ket{1}) \otimes (b_0\ket{000} + b_1\ket{001} + b_2\ket{010} + b_3\ket{011} + b_4\ket{100} + b_5\ket{101} + b_6\ket{110} + b_7\ket{111}) \end{align}

We can see that $a_0 b_0 = \frac{1}{2}$, so $a_0$ is non-zero; and $a_1 b_2 = \frac{1}{2}$, so $b_2$ is non-zero. But we can also see that $a_0 b_2 = 0$. This is impossible. Hence $\ket{\psi}$ must in fact be entangled with respect to this decomposition.

For each of the other three possible positions for the single qubit, a similar argument can be constructed. Hence $\ket{\psi}$ is entangled with respect to any of these four decompositions.

The talk of “opposite signs” seems to assume that $a_1$, $a_2$, $b_1$ and $b_2$ are real, but in fact they can be complex.

Here is an alternative argument. We see that $a_1 a_2 = a_1 b_2$, and these are non-zero; hence $a_2 = b_2$. Similarly $a_1 = b_1$. But then $b_1 b_2$ must be equal to [[$a_1 a_2]], and we can see that this is not the case. Thus the system has no solutions and the state is indeed entangled.

Assume $\ket{GHZ_n}$ is not entangled and $n > 1$. Then $\ket{GHZ_n}$ can be written as

(1)\begin{align} \ket{GHZ_n} = (a_1\ket{0} + b_1\ket{1}) \otimes (a_2\ket{0} + b_2\ket{1}) \otimes \cdots \otimes (a_n\ket{0} + b_n\ket{1}) \end{align}

From the definition of $\ket{GHZ_n}$, we must therefore have

(2)\begin{align} a_1 \cdots a_n = \frac{1}{\sqrt{2}} \end{align}

so all of the $a_i$ values must be non-zero; and

(3)\begin{align} b_1 \cdots b_n = \frac{1}{\sqrt{2}} \end{align}

so all of the $b_i$ values must be non-zero. But we also have

(4)\begin{align} a_1 (b_2 \cdots b_n) = 0 \end{align}

which is clearly impossible. Thus our assumption was wrong and $\ket{GHZ_n}$ must be entangled.