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Ch. 5 Solutions

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Table of Contents

Basic

Medium

Advanced

Basic

B-1. Verifying the Normalization Condition

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Solution strategy: Calculate the squared absolute values of \(c_+ = \frac{1+i}{2}\) and \(c_- = \frac{\sqrt{2}}{2}\) respectively, and verify that their sum equals 1.

Detailed calculation:

\[|c_+|^2 = \left|\frac{1+i}{2}\right|^2 = \frac{(1)^2 + (1)^2}{2^2} = \frac{1+1}{4} = \frac{2}{4} = \frac{1}{2}\]
\[|c_-|^2 = \left|\frac{\sqrt{2}}{2}\right|^2 = \frac{(\sqrt{2})^2}{2^2} = \frac{2}{4} = \frac{1}{2}\]

Final answer:

\[|c_+|^2 + |c_-|^2 = \frac{1}{2} + \frac{1}{2} = 1 \quad \checkmark\]

The normalization condition is satisfied.

Verification: As an alternative method, we can confirm that \(|c_+|^2 = c_+^* c_+ = \frac{1-i}{2} \cdot \frac{1+i}{2} = \frac{(1-i)(1+i)}{4} = \frac{1+1}{4} = \frac{1}{2}\).

B-2. Calculating Inner Products

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Solution strategy: Compute inner products using the orthonormality relations \(\langle+|+\rangle = 1\), \(\langle+|-\rangle = 0\), \(\langle-|+\rangle = 0\), \(\langle-|-\rangle = 1\).

Detailed calculation:

\[|\psi\rangle = \frac{1}{\sqrt{3}}|+\rangle + \sqrt{\frac{2}{3}}\,e^{i\pi/4}|-\rangle\]
\[\langle+|\psi\rangle = \frac{1}{\sqrt{3}}\langle+|+\rangle + \sqrt{\frac{2}{3}}\,e^{i\pi/4}\langle+|-\rangle = \frac{1}{\sqrt{3}} \cdot 1 + \sqrt{\frac{2}{3}}\,e^{i\pi/4} \cdot 0 = \frac{1}{\sqrt{3}}\]
\[\langle-|\psi\rangle = \frac{1}{\sqrt{3}}\langle-|+\rangle + \sqrt{\frac{2}{3}}\,e^{i\pi/4}\langle-|-\rangle = \frac{1}{\sqrt{3}} \cdot 0 + \sqrt{\frac{2}{3}}\,e^{i\pi/4} \cdot 1 = \sqrt{\frac{2}{3}}\,e^{i\pi/4}\]

Modulus squared:

\[|\langle+|\psi\rangle|^2 = \left|\frac{1}{\sqrt{3}}\right|^2 = \frac{1}{3}\]
\[|\langle-|\psi\rangle|^2 = \left|\sqrt{\frac{2}{3}}\,e^{i\pi/4}\right|^2 = \frac{2}{3} \cdot |e^{i\pi/4}|^2 = \frac{2}{3} \cdot 1 = \frac{2}{3}\]

Final answer:

\[\langle+|\psi\rangle = \frac{1}{\sqrt{3}}, \quad |\langle+|\psi\rangle|^2 = \frac{1}{3}\]
\[\langle-|\psi\rangle = \sqrt{\frac{2}{3}}\,e^{i\pi/4}, \quad |\langle-|\psi\rangle|^2 = \frac{2}{3}\]

Verification: \(|\langle+|\psi\rangle|^2 + |\langle-|\psi\rangle|^2 = \frac{1}{3} + \frac{2}{3} = 1\) ✓ (normalization condition satisfied)

B-3. Action of Outer Products (Projection Operators)

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Solution strategy: Apply \(\hat{P}_+ = |+\rangle\langle+|\) to \(|\psi\rangle\). The action of an outer product is computed by the procedure "take the inner product, then multiply by the ket."

Detailed calculation:

\[\hat{P}_+|\psi\rangle = |+\rangle\langle+|\psi\rangle = |+\rangle\left(\frac{3}{5}\langle+|+\rangle + \frac{4}{5}\langle+|-\rangle\right)\]
\[= |+\rangle\left(\frac{3}{5} \cdot 1 + \frac{4}{5} \cdot 0\right) = \frac{3}{5}|+\rangle\]

Writing this as a linear combination of \(|+\rangle\) and \(|-\rangle\):

\[\hat{P}_+|\psi\rangle = \frac{3}{5}|+\rangle + 0 \cdot |-\rangle\]

Checking normalization:

\[\left|\frac{3}{5}\right|^2 + |0|^2 = \frac{9}{25} \neq 1\]

Final answer:

\[\hat{P}_+|\psi\rangle = \frac{3}{5}|+\rangle\]

This state is not normalized. The squared norm is \(9/25\), which equals the probability \(|c_+|^2 = 9/25\) of finding the original state \(|\psi\rangle\) in \(|+\rangle\). The normalized post-projection state is \(|+\rangle\) itself.

Verification: We confirm the idempotency of the projection operator \(\hat{P}_+^2 = \hat{P}_+\). \(\hat{P}_+(\hat{P}_+|\psi\rangle) = \hat{P}_+\left(\frac{3}{5}|+\rangle\right) = \frac{3}{5}|+\rangle\langle+|+\rangle = \frac{3}{5}|+\rangle = \hat{P}_+|\psi\rangle\) ✓

B-4. Expansion in the \(x\) Basis

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Solution strategy: Solve equations (5.11) and (5.12) as simultaneous equations for \(|+\rangle\) and \(|-\rangle\).

Detailed calculation:

We restate equations (5.11) and (5.12):

\[|+\rangle_x = \frac{1}{\sqrt{2}}|+\rangle + \frac{1}{\sqrt{2}}|-\rangle\]
\[|-\rangle_x = \frac{1}{\sqrt{2}}|+\rangle - \frac{1}{\sqrt{2}}|-\rangle\]

Adding the two equations:

\[|+\rangle_x + |-\rangle_x = \frac{1}{\sqrt{2}}|+\rangle + \frac{1}{\sqrt{2}}|-\rangle + \frac{1}{\sqrt{2}}|+\rangle - \frac{1}{\sqrt{2}}|-\rangle = \frac{2}{\sqrt{2}}|+\rangle = \sqrt{2}\,|+\rangle\]
\[\therefore \quad |+\rangle = \frac{1}{\sqrt{2}}|+\rangle_x + \frac{1}{\sqrt{2}}|-\rangle_x\]

Final answer:

\[a = \frac{1}{\sqrt{2}}, \quad b = \frac{1}{\sqrt{2}}\]
\[|+\rangle = \frac{1}{\sqrt{2}}|+\rangle_x + \frac{1}{\sqrt{2}}|-\rangle_x\]

Verification: As an alternative method, we directly compute \(a = {}_x\langle+|+\rangle\). Since \({}_x\langle+| = \frac{1}{\sqrt{2}}\langle+| + \frac{1}{\sqrt{2}}\langle-|\), we get \(a = {}_x\langle+|+\rangle = \frac{1}{\sqrt{2}}\langle+|+\rangle + \frac{1}{\sqrt{2}}\langle-|+\rangle = \frac{1}{\sqrt{2}}\) ✓. Similarly, \(b = {}_x\langle-|+\rangle = \frac{1}{\sqrt{2}}\langle+|+\rangle - \frac{1}{\sqrt{2}}\langle-|+\rangle = \frac{1}{\sqrt{2}}\) ✓. The normalization also checks out: \(|a|^2 + |b|^2 = \frac{1}{2} + \frac{1}{2} = 1\) ✓.

B-5. Orthogonality of the \(y\) basis

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Solution strategy: Construct the bra \({}_y\langle+|\) from equations (5.13) and (5.14), and compute the inner product with \(|-\rangle_y\).

Detailed calculation:

From equations (5.13) and (5.14):

\[|+\rangle_y = \frac{1}{\sqrt{2}}|+\rangle + \frac{i}{\sqrt{2}}|-\rangle\]
\[|-\rangle_y = \frac{1}{\sqrt{2}}|+\rangle - \frac{i}{\sqrt{2}}|-\rangle\]

When forming the bra, we take the complex conjugate of the coefficients:

\[{}_y\langle+| = \frac{1}{\sqrt{2}}\langle+| + \left(\frac{i}{\sqrt{2}}\right)^*\langle-| = \frac{1}{\sqrt{2}}\langle+| - \frac{i}{\sqrt{2}}\langle-|\]

Computing the inner product:

\[{}_y\langle+|-\rangle_y = \left(\frac{1}{\sqrt{2}}\langle+| - \frac{i}{\sqrt{2}}\langle-|\right)\left(\frac{1}{\sqrt{2}}|+\rangle - \frac{i}{\sqrt{2}}|-\rangle\right)\]
\[= \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}}\langle+|+\rangle + \frac{1}{\sqrt{2}} \cdot \left(-\frac{i}{\sqrt{2}}\right)\langle+|-\rangle + \left(-\frac{i}{\sqrt{2}}\right) \cdot \frac{1}{\sqrt{2}}\langle-|+\rangle + \left(-\frac{i}{\sqrt{2}}\right)\left(-\frac{i}{\sqrt{2}}\right)\langle-|-\rangle\]
\[= \frac{1}{2} \cdot 1 + \left(-\frac{i}{2}\right) \cdot 0 + \left(-\frac{i}{2}\right) \cdot 0 + \frac{i^2}{2} \cdot 1\]
\[= \frac{1}{2} + \frac{i^2}{2} = \frac{1}{2} + \frac{-1}{2} = 0\]

Final answer:

\[{}_y\langle+|-\rangle_y = 0\]

\(|+\rangle_y\) and \(|-\rangle_y\) are orthogonal. ✓

Verification: \({}_y\langle+|+\rangle_y = \frac{1}{2}\langle+|+\rangle + \frac{1}{2}\left(-i\right)(i)\langle-|-\rangle = \frac{1}{2} + \frac{1}{2} = 1\) ✓ (normalization also confirmed)

B-6. Finding Probabilities from Probability Amplitudes

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Solution Strategy: Read off the expansion coefficients of \(|+\rangle_y\) in the \(z\) basis, and calculate each probability.

Detailed Calculation:

From Eq. (5.13):

\[|+\rangle_y = \frac{1}{\sqrt{2}}|+\rangle + \frac{i}{\sqrt{2}}|-\rangle\]

The probability amplitude for obtaining \(S_z = +\hbar/2\) is \(c_+ = \frac{1}{\sqrt{2}}\), and the probability amplitude for obtaining \(S_z = -\hbar/2\) is \(c_- = \frac{i}{\sqrt{2}}\).

\[P\left(S_z = +\frac{\hbar}{2}\right) = |c_+|^2 = \left|\frac{1}{\sqrt{2}}\right|^2 = \frac{1}{2}\]
\[P\left(S_z = -\frac{\hbar}{2}\right) = |c_-|^2 = \left|\frac{i}{\sqrt{2}}\right|^2 = \frac{|i|^2}{2} = \frac{1}{2}\]

Final Answer:

\[P\left(S_z = +\frac{\hbar}{2}\right) = \frac{1}{2}, \quad P\left(S_z = -\frac{\hbar}{2}\right) = \frac{1}{2}\]

When a particle with spin up in the \(y\) direction is measured along the \(z\) direction, spin up and spin down occur with equal probability (50% each).

Verification: \(P(+) + P(-) = \frac{1}{2} + \frac{1}{2} = 1\) ✓. Furthermore, similar to the \(x\) direction case (Eq. (5.11)), this is consistent with the physical expectation that measurements along orthogonal directions yield equal probabilities.

B-7. Using the Kronecker Delta

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Solution strategy: Evaluate the summand for each case \(j = +\) and \(j = -\), then add them together.

Detailed calculation:

\[\sum_{j \in \{+,-\}} \delta_{+j}\,\delta_{j-} = \delta_{++}\,\delta_{+-} + \delta_{+-}\,\delta_{--}\]

Values of each Kronecker delta: - \(\delta_{++} = 1\), \(\delta_{+-} = 0\), \(\delta_{--} = 1\)

\[= 1 \cdot 0 + 0 \cdot 1 = 0 + 0 = 0\]

Final answer:

\[\sum_{j \in \{+,-\}} \delta_{+j}\,\delta_{j-} = 0\]

Verification: This sum should equal \(\delta_{+-}\) (by the composition rule for Kronecker deltas: \(\sum_j \delta_{ij}\delta_{jk} = \delta_{ik}\)). Since \(\delta_{+-} = 0\), the result is consistent. ✓

B-8. Components of the Basis Transformation Matrix

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Solution strategy: Read off the expansion coefficients of \(|+\rangle_y\), \(|-\rangle_y\) in the \(z\) basis from Eqs. (5.13) and (5.14), and obtain the matrix elements as inner products.

Detailed calculation:

Eq. (5.13): \(|+\rangle_y = \frac{1}{\sqrt{2}}|+\rangle + \frac{i}{\sqrt{2}}|-\rangle\)

Eq. (5.14): \(|-\rangle_y = \frac{1}{\sqrt{2}}|+\rangle - \frac{i}{\sqrt{2}}|-\rangle\)

Reading off each component:

\[\langle+|+\rangle_y = \frac{1}{\sqrt{2}}, \quad \langle-|+\rangle_y = \frac{i}{\sqrt{2}}\]
\[\langle+|-\rangle_y = \frac{1}{\sqrt{2}}, \quad \langle-|-\rangle_y = -\frac{i}{\sqrt{2}}\]

Final answer:

\[U = \begin{pmatrix} \langle+|+\rangle_y & \langle+|-\rangle_y \\ \langle-|+\rangle_y & \langle-|-\rangle_y \end{pmatrix} = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 & 1 \\ i & -i \end{pmatrix}\]

Verification: We confirm unitarity. \(U^\dagger = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 & -i \\ 1 & i \end{pmatrix}\)

\[U^\dagger U = \frac{1}{2}\begin{pmatrix} 1 & -i \\ 1 & i \end{pmatrix}\begin{pmatrix} 1 & 1 \\ i & -i \end{pmatrix} = \frac{1}{2}\begin{pmatrix} 1+1 & 1+1 \cdot(-1) \\ 1+1\cdot(-1) & 1+1 \end{pmatrix}\]

More carefully:

\((U^\dagger U)_{11} = \frac{1}{2}(1 \cdot 1 + (-i)(i)) = \frac{1}{2}(1 + 1) = 1\)

\((U^\dagger U)_{12} = \frac{1}{2}(1 \cdot 1 + (-i)(-i)) = \frac{1}{2}(1 + i^2) = \frac{1}{2}(1 - 1) = 0\)

\((U^\dagger U)_{21} = \frac{1}{2}(1 \cdot 1 + i \cdot i) = \frac{1}{2}(1 + i^2) = \frac{1}{2}(1 - 1) = 0\)

\((U^\dagger U)_{22} = \frac{1}{2}(1 \cdot 1 + i(-i)) = \frac{1}{2}(1 + 1) = 1\)

\[U^\dagger U = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \mathbf{1} \quad \checkmark\]

Medium

M-1. Deriving the Normalization Condition from the Completeness Relation

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Solution strategy: Insert the completeness relation into \(\langle\psi|\psi\rangle = 1\).

Detailed calculation:

Insert the identity operator \(\mathbf{1} = |+\rangle\langle+| + |-\rangle\langle-|\) into the left-hand side of \(\langle\psi|\psi\rangle = 1\):

\[\langle\psi|\psi\rangle = \langle\psi|\mathbf{1}|\psi\rangle = \langle\psi|\left(|+\rangle\langle+| + |-\rangle\langle-|\right)|\psi\rangle\]
\[= \langle\psi|+\rangle\langle+|\psi\rangle + \langle\psi|-\rangle\langle-|\psi\rangle\]

Here, using the property of the inner product \(\langle\psi|j\rangle = \langle j|\psi\rangle^*\):

\[= \langle+|\psi\rangle^*\langle+|\psi\rangle + \langle-|\psi\rangle^*\langle-|\psi\rangle\]
\[= |\langle+|\psi\rangle|^2 + |\langle-|\psi\rangle|^2\]

Since \(\langle\psi|\psi\rangle = 1\):

\[|\langle+|\psi\rangle|^2 + |\langle-|\psi\rangle|^2 = 1 \quad \blacksquare\]

Verification: Confirm with a specific example. For \(|\psi\rangle = \frac{1}{\sqrt{3}}|+\rangle + \sqrt{\frac{2}{3}}|-\rangle\), we have \(|\langle+|\psi\rangle|^2 + |\langle-|\psi\rangle|^2 = \frac{1}{3} + \frac{2}{3} = 1\) ✓

M-2. Completeness Relation for the \(x\) Basis

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Strategy: Express \(|+\rangle_x\), \(|-\rangle_x\) as column vectors in the \(z\) basis, compute the outer products as matrices, and add them together.

Detailed calculation:

Column vector representation in the \(z\) basis:

\[|+\rangle_x \to \frac{1}{\sqrt{2}}\begin{pmatrix}1\\1\end{pmatrix}, \quad |-\rangle_x \to \frac{1}{\sqrt{2}}\begin{pmatrix}1\\-1\end{pmatrix}\]

Matrix representation of the outer product \(|+\rangle_x\,{}_x\langle+|\):

\[|+\rangle_x\,{}_x\langle+| = \frac{1}{\sqrt{2}}\begin{pmatrix}1\\1\end{pmatrix} \cdot \frac{1}{\sqrt{2}}\begin{pmatrix}1 & 1\end{pmatrix} = \frac{1}{2}\begin{pmatrix}1 & 1\\1 & 1\end{pmatrix}\]

Matrix representation of the outer product \(|-\rangle_x\,{}_x\langle-|\):

\[|-\rangle_x\,{}_x\langle-| = \frac{1}{\sqrt{2}}\begin{pmatrix}1\\-1\end{pmatrix} \cdot \frac{1}{\sqrt{2}}\begin{pmatrix}1 & -1\end{pmatrix} = \frac{1}{2}\begin{pmatrix}1 & -1\\-1 & 1\end{pmatrix}\]

Summing:

\[|+\rangle_x\,{}_x\langle+| + |-\rangle_x\,{}_x\langle-| = \frac{1}{2}\begin{pmatrix}1 & 1\\1 & 1\end{pmatrix} + \frac{1}{2}\begin{pmatrix}1 & -1\\-1 & 1\end{pmatrix}\]
\[= \frac{1}{2}\begin{pmatrix}1+1 & 1+(-1)\\1+(-1) & 1+1\end{pmatrix} = \frac{1}{2}\begin{pmatrix}2 & 0\\0 & 2\end{pmatrix} = \begin{pmatrix}1 & 0\\0 & 1\end{pmatrix} = \mathbf{1}\]

Final answer:

\[|+\rangle_x\,{}_x\langle+| + |-\rangle_x\,{}_x\langle-| = \mathbf{1} \quad \blacksquare\]

Verification: It is obvious that acting on an arbitrary state \(|\psi\rangle = \begin{pmatrix}a\\b\end{pmatrix}\) gives \(\mathbf{1}|\psi\rangle = \begin{pmatrix}a\\b\end{pmatrix} = |\psi\rangle\). Furthermore, the same result is obtained as the completeness relation in the \(z\) basis, \(|+\rangle\langle+| + |-\rangle\langle-| = \begin{pmatrix}1&0\\0&0\end{pmatrix} + \begin{pmatrix}0&0\\0&1\end{pmatrix} = \mathbf{1}\), confirming that the completeness relation holds regardless of the choice of basis. ✓

M-3. Unitarity of the Basis Transformation Matrix

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Solution strategy: Since all components of \(U\) are real, \(U^\dagger = U^T\). We compute \(U^T U\) and show it equals the identity matrix.

Detailed calculation:

\[U = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}\]

Since \(U\) is a real matrix:

\[U^\dagger = U^T = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}\]

(This matrix is symmetric, so \(U^T = U\) as well.)

\[U^\dagger U = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix} \cdot \frac{1}{\sqrt{2}}\begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix} = \frac{1}{2}\begin{pmatrix} 1 \cdot 1 + 1 \cdot 1 & 1 \cdot 1 + 1 \cdot (-1) \\ 1 \cdot 1 + (-1) \cdot 1 & 1 \cdot 1 + (-1)(-1) \end{pmatrix}\]
\[= \frac{1}{2}\begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \mathbf{1} \quad \blacksquare\]

Verification: Computing \(UU^\dagger\) also gives the same result (in this case, since \(U = U^T\), we have \(UU^\dagger = U^\dagger U\)). Furthermore, \(\det U = \frac{1}{2}((-1) - 1) = -1\), so \(|\det U| = 1\), satisfying the necessary condition \(|\det U| = 1\) for a unitary matrix. ✓

M-4. Probability of Sequential Measurements

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Solution Strategy: Track the state and probability amplitudes at each stage in sequence, find the overall amplitude, then calculate the probability.

Detailed Calculation:

Step 1: The \(z\)-direction apparatus selects spin-up. The state is \(|+\rangle\).

Step 2: Pass \(|+\rangle\) through the \(x\)-direction apparatus. The probability amplitude for selecting \(S_x = +\hbar/2\) is:

\[{}_x\langle+|+\rangle\]

From the result of D4 (or directly from equation (5.11)):

\[{}_x\langle+|+\rangle = \frac{1}{\sqrt{2}}\]

Step 3: After \(S_x = +\hbar/2\) is selected, the state is \(|+\rangle_x\).

Step 4: Pass \(|+\rangle_x\) through the \(z\)-direction apparatus again. The probability amplitude for finding \(S_z = -\hbar/2\) is:

\[\langle-|+\rangle_x\]

From equation (5.11), \(|+\rangle_x = \frac{1}{\sqrt{2}}|+\rangle + \frac{1}{\sqrt{2}}|-\rangle\), so:

\[\langle-|+\rangle_x = \frac{1}{\sqrt{2}}\]

Overall Amplitude: Multiply the amplitudes at each stage (the "multiply amplitudes" rule from Ch. 4):

\[A = {}_x\langle+|+\rangle \cdot \langle-|+\rangle_x = \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} = \frac{1}{2}\]

Final Answer:

\[P\left(S_z = -\frac{\hbar}{2}\right) = |A|^2 = \left|\frac{1}{2}\right|^2 = \frac{1}{4}\]

Verification and Discussion: If the \(x\)-direction apparatus were not inserted, the probability of obtaining \(S_z = -\hbar/2\) when measuring a particle in state \(|+\rangle\) along the \(z\)-direction would be \(|\langle-|+\rangle|^2 = 0\). By inserting the \(x\)-direction apparatus in between, the \(z\)-direction information is "disturbed," changing what was originally a zero probability to \(1/4\). This demonstrates the non-trivial effect of measurement in quantum mechanics. ✓

Advanced

A-1. Spin Eigenstates in an Arbitrary Direction

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Given eigenstate:

\[|+\rangle_{\hat{n}} = \cos\frac{\theta}{2}\,|+\rangle + e^{i\phi}\sin\frac{\theta}{2}\,|-\rangle\]

(a) Cases \(\theta = 0\) and \(\theta = \pi\)

When \(\theta = 0\):

\[|+\rangle_{\hat{n}} = \cos 0 \cdot |+\rangle + e^{i\phi}\sin 0 \cdot |-\rangle = 1 \cdot |+\rangle + 0 \cdot |-\rangle = |+\rangle \quad \checkmark\]

When \(\hat{\mathbf{n}}\) points along the \(z\)-axis (north pole), the spin-up eigenstate \(|+\rangle\) of \(S_{\hat{n}} = S_z\) is reproduced.

When \(\theta = \pi\):

\[|+\rangle_{\hat{n}} = \cos\frac{\pi}{2}\,|+\rangle + e^{i\phi}\sin\frac{\pi}{2}\,|-\rangle = 0 \cdot |+\rangle + e^{i\phi} \cdot 1 \cdot |-\rangle = e^{i\phi}|-\rangle\]

Up to an overall phase \(e^{i\phi}\), this equals \(|-\rangle\). When \(\hat{\mathbf{n}}\) points in the \(-z\) direction (south pole), the spin-up eigenstate of \(S_{\hat{n}} = -S_z\) corresponds to the spin-down eigenstate \(|-\rangle\) of \(S_z\). ✓

(b) When \(\theta = \pi/2\), \(\phi = 0\)

\[|+\rangle_{\hat{n}} = \cos\frac{\pi}{4}\,|+\rangle + e^{i \cdot 0}\sin\frac{\pi}{4}\,|-\rangle = \frac{1}{\sqrt{2}}|+\rangle + 1 \cdot \frac{1}{\sqrt{2}}|-\rangle = \frac{1}{\sqrt{2}}|+\rangle + \frac{1}{\sqrt{2}}|-\rangle\]

This matches \(|+\rangle_x\) from Eq. (5.11). Since \(\hat{\mathbf{n}} = (\sin(\pi/2)\cos 0, \sin(\pi/2)\sin 0, \cos(\pi/2)) = (1, 0, 0) = \hat{\mathbf{x}}\), this is consistent. ✓

(c) When \(\theta = \pi/2\), \(\phi = \pi/2\)

\[|+\rangle_{\hat{n}} = \cos\frac{\pi}{4}\,|+\rangle + e^{i\pi/2}\sin\frac{\pi}{4}\,|-\rangle = \frac{1}{\sqrt{2}}|+\rangle + i \cdot \frac{1}{\sqrt{2}}|-\rangle = \frac{1}{\sqrt{2}}|+\rangle + \frac{i}{\sqrt{2}}|-\rangle\]

This matches \(|+\rangle_y\) from Eq. (5.13). Since \(\hat{\mathbf{n}} = (\sin(\pi/2)\cos(\pi/2), \sin(\pi/2)\sin(\pi/2), \cos(\pi/2)) = (0, 1, 0) = \hat{\mathbf{y}}\), this is consistent. ✓

(d) Proof of normalization

\[{}_{\hat{n}}\langle+|+\rangle_{\hat{n}} = \left(\cos\frac{\theta}{2}\,\langle+| + e^{-i\phi}\sin\frac{\theta}{2}\,\langle-|\right)\left(\cos\frac{\theta}{2}\,|+\rangle + e^{i\phi}\sin\frac{\theta}{2}\,|-\rangle\right)\]

Using orthonormality \(\langle+|+\rangle = \langle-|-\rangle = 1\), \(\langle+|-\rangle = \langle-|+\rangle = 0\):

\[= \cos^2\frac{\theta}{2}\,\langle+|+\rangle + \cos\frac{\theta}{2}\,e^{i\phi}\sin\frac{\theta}{2}\,\langle+|-\rangle + e^{-i\phi}\sin\frac{\theta}{2}\cos\frac{\theta}{2}\,\langle-|+\rangle + e^{-i\phi}e^{i\phi}\sin^2\frac{\theta}{2}\,\langle-|-\rangle\]
\[= \cos^2\frac{\theta}{2} \cdot 1 + 0 + 0 + |e^{i\phi}|^2\sin^2\frac{\theta}{2} \cdot 1\]
\[= \cos^2\frac{\theta}{2} + \sin^2\frac{\theta}{2} = 1 \quad \blacksquare\]

(e) Probability of obtaining \(S_z = +\hbar/2\) and geometric meaning of \(\theta\)

The probability of obtaining \(S_z = +\hbar/2\) is:

\[P\left(S_z = +\frac{\hbar}{2}\right) = |\langle+|+\rangle_{\hat{n}}|^2 = \left|\cos\frac{\theta}{2}\right|^2 = \cos^2\frac{\theta}{2} \quad \blacksquare\]

Discussion of the geometric meaning of \(\theta\):

The angle \(\theta\) is the polar angle between the \(\hat{\mathbf{n}}\) direction and the \(z\)-axis.

  • When \(\theta = 0\) (\(\hat{\mathbf{n}}\) points in the same direction as the \(z\)-axis): \(P = \cos^2 0 = 1\). One obtains \(S_z = +\hbar/2\) with certainty.
  • When \(\theta = \pi/2\) (\(\hat{\mathbf{n}}\) is perpendicular to the \(z\)-axis): \(P = \cos^2(\pi/4) = 1/2\). Equal probability.
  • When \(\theta = \pi\) (\(\hat{\mathbf{n}}\) points in the \(-z\) direction): \(P = \cos^2(\pi/2) = 0\). \(S_z = +\hbar/2\) is never obtained.

This result is directly connected to the Bloch sphere picture. On the Bloch sphere, \(|+\rangle\) corresponds to the north pole (\(\theta = 0\)) and \(|-\rangle\) corresponds to the south pole (\(\theta = \pi\)). The state \(|+\rangle_{\hat{n}}\) corresponds to the point at polar angle \(\theta\) and azimuthal angle \(\phi\). The "distance" between two states is characterized by the angle \(\theta\) on the Bloch sphere, and the probability is given by the "half-angle formula" \(\cos^2(\theta/2)\).

It is noteworthy that states on diametrically opposite sides (angle \(\theta = \pi\)) of the Bloch sphere correspond to orthogonal states, while the same point (angle \(\theta = 0\)) corresponds to the identical state. The angle \(\theta\) in physical three-dimensional space appears as \(\theta/2\) in the state space—this is deeply related to the property that spin-1/2 changes sign under a \(2\pi\) rotation and returns to its original state only after a \(4\pi\) rotation.

Verification: The probability of obtaining \(S_z = -\hbar/2\) is \(|\langle-|+\rangle_{\hat{n}}|^2 = |e^{i\phi}\sin(\theta/2)|^2 = \sin^2(\theta/2)\). The total \(\cos^2(\theta/2) + \sin^2(\theta/2) = 1\) ✓

A-2. "Which Path Was Taken" and the Disappearance of Interference

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Setting up the problem: A particle initially in state \(|+\rangle_x\) is separated into \(|+\rangle\) and \(|-\rangle\) by a \(z\)-direction Stern-Gerlach apparatus, then the beams are recombined and measured in the \(x\) direction.

(a) When the paths are indistinguishable

Solution strategy: Since the intermediate states are indistinguishable, following the rules of Ch. 4, we "add the amplitudes first, then take the absolute value squared."

Detailed calculation:

Expanding the initial state \(|+\rangle_x\) in the \(z\) basis (from equation (5.11)):

\[|+\rangle_x = \frac{1}{\sqrt{2}}|+\rangle + \frac{1}{\sqrt{2}}|-\rangle\]

The amplitude for obtaining \(S_x = +\hbar/2\) (state \(|+\rangle_x\)) in the final measurement is the sum of amplitudes passing through intermediate states \(|j\rangle\) (\(j = +, -\)):

\[A = \sum_{j=\pm} {}_x\langle+|j\rangle\langle j|+\rangle_x\]

Using the completeness relation \(\sum_{j=\pm}|j\rangle\langle j| = \mathbf{1}\):

\[A = {}_x\langle+|\mathbf{1}|+\rangle_x = {}_x\langle+|+\rangle_x = 1\]

For verification, computing each term explicitly:

\[{}_x\langle+|+\rangle = \frac{1}{\sqrt{2}}, \quad \langle+|+\rangle_x = \frac{1}{\sqrt{2}}\]
\[{}_x\langle+|-\rangle = \frac{1}{\sqrt{2}}, \quad \langle-|+\rangle_x = \frac{1}{\sqrt{2}}\]
\[A = \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} = \frac{1}{2} + \frac{1}{2} = 1\]

Probability:

\[P_a = |A|^2 = |1|^2 = 1\]

(b) When the paths are distinguishable

Solution strategy: Since the paths are distinguishable, following the rules of Ch. 4, we "add the probabilities."

Detailed calculation:

Computing and summing the probability for each path individually:

\[P_b = \sum_{j=\pm} |{}_x\langle+|j\rangle|^2 \cdot |\langle j|+\rangle_x|^2\]
\[= |{}_x\langle+|+\rangle|^2 \cdot |\langle+|+\rangle_x|^2 + |{}_x\langle+|-\rangle|^2 \cdot |\langle-|+\rangle_x|^2\]
\[= \left|\frac{1}{\sqrt{2}}\right|^2 \cdot \left|\frac{1}{\sqrt{2}}\right|^2 + \left|\frac{1}{\sqrt{2}}\right|^2 \cdot \left|\frac{1}{\sqrt{2}}\right|^2\]
\[= \frac{1}{2} \cdot \frac{1}{2} + \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}\]

Comparison: In (a), \(P_a = 1\); in (b), \(P_b = 1/2\).

(c) Analysis and discussion of the interference term

Identifying the interference term explicitly:

Expanding the probability in (a). Writing the amplitude as \(A = A_+ + A_-\) (where \(A_j = {}_x\langle+|j\rangle\langle j|+\rangle_x\)):

\[P_a = |A_+ + A_-|^2 = |A_+|^2 + |A_-|^2 + A_+^* A_- + A_+ A_-^*\]
\[= |A_+|^2 + |A_-|^2 + 2\,\text{Re}(A_+^* A_-)\]

where:

\[A_+ = {}_x\langle+|+\rangle\langle+|+\rangle_x = \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} = \frac{1}{2}\]
\[A_- = {}_x\langle+|-\rangle\langle-|+\rangle_x = \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} = \frac{1}{2}\]
\[|A_+|^2 + |A_-|^2 = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}\]

Interference term:

\[2\,\text{Re}(A_+^* A_-) = 2\,\text{Re}\left(\frac{1}{2} \cdot \frac{1}{2}\right) = 2 \cdot \frac{1}{4} = \frac{1}{2}\]

Therefore:

\[P_a = \frac{1}{2} + \frac{1}{2} = 1\]

On the other hand, the probability in (b) is:

\[P_b = |A_+|^2 + |A_-|^2 = \frac{1}{2}\]

Difference between the two:

\[P_a - P_b = 2\,\text{Re}(A_+^* A_-) = \frac{1}{2}\]

This \(\frac{1}{2}\) is the interference term (cross term).

Discussion:

This result vividly illustrates the essence of the probability amplitude rules from Ch. 4.

  1. When paths are indistinguishable (a): When it is impossible in principle to know which path (\(|+\rangle\) or \(|-\rangle\)) the particle took, the probability amplitudes for the two paths are coherently superposed. The interference term \(2\,\text{Re}(A_+^* A_-)\) survives, and in this case constructive interference occurs, bringing the probability to 1. This is a direct consequence of the completeness relation \(\sum_j |j\rangle\langle j| = \mathbf{1}\), meaning that if the two beams are perfectly recombined, the original state \(|+\rangle_x\) is restored.

  2. When paths are distinguishable (b): When "which-path" information is obtained by tagging, the two paths become distinguishable mutually exclusive events, and the rule of adding probabilities rather than amplitudes applies. The interference term vanishes, and the probability decreases to \(1/2\).

  3. Physical significance: Acquiring path information destroys interference. This is essentially the same phenomenon as the disappearance of interference fringes when one observes "which slit" the particle passed through in a double-slit experiment. In quantum mechanics, the act of acquiring information itself alters the state of the system. The act of tagging corresponds to entangling the spin degree of freedom with the tagging degree of freedom, and the coherence (phase consistency) is lost when one looks at the spin degree of freedom alone.

Verification: In (a), computing the probability of obtaining \(S_x = -\hbar/2\) similarly gives an amplitude of \({}_x\langle-|\mathbf{1}|+\rangle_x = {}_x\langle-|+\rangle_x = 0\), so the probability is 0. \(P(S_x = +\hbar/2) + P(S_x = -\hbar/2) = 1 + 0 = 1\) ✓. In (b), the probability for \(S_x = -\hbar/2\) is similarly \(1/2\), giving \(1/2 + 1/2 = 1\) ✓.