Ch. 4 Solutions

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Table of Contents

Basic

• B-1. Absolute Value and Complex Conjugate of Complex Numbers
• B-2. Multiplication of Complex Numbers and Phase
• B-3. Conversion to Polar Form
• B-4. Multiplication in Polar Form
• B-5. Calculation of Interference Terms
• B-6. Complex Conjugate and Interference Terms
• B-7. Dirac Notation and the Third Rule
• B-8. Adding Amplitudes and Probability

Medium

• M-1. Derivation of the General Formula for Interference Terms
• M-2. Interference Pattern from \(N\) Equally-Spaced, Equal-Amplitude Slits
• M-3. Observation Destroys Interference: A Mathematical Explanation
• M-4. Calculating Amplitudes Through Two Walls
• M-5. Relationship Between Phase Difference and Path Difference

Advanced

• A-1. Quantum Mechanical Analysis of the Mach–Zehnder Interferometer
• A-2. Interference in a 3-State System and the Principle of the "Quantum Eraser"

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Basic

B-1. Absolute Value and Complex Conjugate of Complex Numbers

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Problem:

For each of the following complex numbers, find (a) the complex conjugate \(z^*\), (b) the absolute value \(|z|\), and (c) \(z \cdot z^*\).

1. \(z = 1 + i\)
2. \(z = 3 - 4i\)
3. \(z = -2i\)
4. \(z = 5\)

Strategy: For \(z = a + bi\), apply \(z^* = a - bi\), \(|z| = \sqrt{a^2 + b^2}\), and \(z \cdot z^* = |z|^2 = a^2 + b^2\).

1. \(z = 1 + i\)

(a) \(z^* = 1 - i\)

(b) \(|z| = \sqrt{1^2 + 1^2} = \sqrt{2}\)

(c) \(z \cdot z^* = 1^2 + 1^2 = 2\)

2. \(z = 3 - 4i\)

(a) \(z^* = 3 + 4i\)

(b) \(|z| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5\)

(c) \(z \cdot z^* = 9 + 16 = 25\)

3. \(z = -2i\) (\(a = 0\), \(b = -2\))

(a) \(z^* = +2i\)

(b) \(|z| = \sqrt{0^2 + (-2)^2} = 2\)

(c) \(z \cdot z^* = (-2i)(2i) = -4i^2 = 4\)

4. \(z = 5\) (\(a = 5\), \(b = 0\))

(a) \(z^* = 5\)

(b) \(|z| = 5\)

(c) \(z \cdot z^* = 25\)

Verification: Confirmed that \(z \cdot z^* = |z|^2\) holds in all cases. ✓

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B-2. Multiplication of Complex Numbers and Phase

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Problem:

Compute the following products in the form \(a + bi\).

1. \((1 + i)(1 - i)\)
2. \((2 + 3i)(1 + 2i)\)
3. \(i \cdot (3 + 4i)\)
4. \((1 + i)^2\)

Strategy: Expand normally and substitute \(i^2 = -1\).

1. \((1 + i)(1 - i)\)

\[= 1 \cdot 1 + 1 \cdot (-i) + i \cdot 1 + i \cdot (-i) = 1 - i + i - i^2 = 1 + 1 = 2\]

2. \((2 + 3i)(1 + 2i)\)

\[= 2 \cdot 1 + 2 \cdot 2i + 3i \cdot 1 + 3i \cdot 2i = 2 + 4i + 3i + 6i^2 = 2 + 7i - 6 = -4 + 7i\]

3. \(i \cdot (3 + 4i)\)

\[= 3i + 4i^2 = 3i - 4 = -4 + 3i\]

4. \((1 + i)^2\)

\[= 1 + 2i + i^2 = 1 + 2i - 1 = 2i\]

Verification: Problem 1 has the form \(z \cdot z^*\), so the result is \(|z|^2 = 2\). ✓ For Problem 4, \(|1+i|^2 = 2\), and \(|(1+i)^2| = |2i| = 2 = (\sqrt{2})^2\), which is consistent. ✓

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B-3. Conversion to Polar Form

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Problem:

Express the following complex numbers in polar form \(z = r(\cos\theta + i\sin\theta)\). Find \(r\) and \(\theta\) (\(0 \leq \theta < 2\pi\)) for each.

1. \(z = 1 + i\)
2. \(z = -\sqrt{3} + i\)
3. \(z = -2\)
4. \(z = 3i\)

Strategy: Find \(r = |z|\) and \(\theta = \arg(z)\). Pay attention to the quadrant to confirm the position on the complex plane.

1. \(z = 1 + i\)

\[r = \sqrt{1^2 + 1^2} = \sqrt{2}\]

In the first quadrant with \(\tan\theta = 1/1 = 1\), so \(\theta = \pi/4\).

\[\boxed{z = \sqrt{2}\left(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}\right)}\]

2. \(z = -\sqrt{3} + i\)

\[r = \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{3 + 1} = 2\]

In the second quadrant (real part negative, imaginary part positive). From \(\tan\alpha = 1/\sqrt{3}\), the reference angle is \(\alpha = \pi/6\). Therefore \(\theta = \pi - \pi/6 = 5\pi/6\).

\[\boxed{z = 2\left(\cos\frac{5\pi}{6} + i\sin\frac{5\pi}{6}\right)}\]

3. \(z = -2\)

\[r = 2, \quad \theta = \pi\]

\[\boxed{z = 2(\cos\pi + i\sin\pi)}\]

4. \(z = 3i\)

\[r = 3, \quad \theta = \pi/2\]

\[\boxed{z = 3\left(\cos\frac{\pi}{2} + i\sin\frac{\pi}{2}\right)}\]

Verification: Convert each result back using \(r\cos\theta + ir\sin\theta\) and confirm it matches the original \(z\). For example, problem 2: \(2\cos(5\pi/6) = 2 \cdot (-\sqrt{3}/2) = -\sqrt{3}\), \(2\sin(5\pi/6) = 2 \cdot (1/2) = 1\). Therefore \(z = -\sqrt{3} + i\). ✓

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B-4. Multiplication in Polar Form

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Problem:

Given \(z_1 = 2(\cos\frac{\pi}{6} + i\sin\frac{\pi}{6})\) and \(z_2 = 3(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3})\),

1. Express the product \(z_1 \cdot z_2\) in polar form.
2. Convert the product \(z_1 \cdot z_2\) into the form \(a + bi\).

Strategy: From equation (4.8), the absolute values multiply and the arguments add.

1. Product in polar form

\[r = r_1 \cdot r_2 = 2 \times 3 = 6\]

\[\theta = \frac{\pi}{6} + \frac{\pi}{3} = \frac{\pi}{6} + \frac{2\pi}{6} = \frac{3\pi}{6} = \frac{\pi}{2}\]

\[\boxed{z_1 \cdot z_2 = 6\left(\cos\frac{\pi}{2} + i\sin\frac{\pi}{2}\right)}\]

2. In the form \(a + bi\)

\[\cos\frac{\pi}{2} = 0, \quad \sin\frac{\pi}{2} = 1\]

\[\boxed{z_1 \cdot z_2 = 6(0 + i \cdot 1) = 6i}\]

Verification: We confirm by direct calculation. \(z_1 = 2(\cos 30° + i\sin 30°) = 2(\frac{\sqrt{3}}{2} + \frac{1}{2}i) = \sqrt{3} + i\). \(z_2 = 3(\cos 60° + i\sin 60°) = 3(\frac{1}{2} + \frac{\sqrt{3}}{2}i) = \frac{3}{2} + \frac{3\sqrt{3}}{2}i\).

\[z_1 z_2 = (\sqrt{3} + i)\left(\frac{3}{2} + \frac{3\sqrt{3}}{2}i\right) = \frac{3\sqrt{3}}{2} + \frac{3 \cdot 3}{2}i + \frac{3}{2}i + \frac{3\sqrt{3}}{2}i^2\]

\[= \frac{3\sqrt{3}}{2} + \frac{9}{2}i + \frac{3}{2}i - \frac{3\sqrt{3}}{2} = 0 + 6i = 6i \quad \checkmark\]

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B-5. Calculation of Interference Terms

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Problem:

Two probability amplitudes \(\phi_1 = \frac{1}{\sqrt{2}}\) and \(\phi_2 = \frac{1}{\sqrt{2}} e^{i\theta}\) are given.

1. Express \(|\phi_1 + \phi_2|^2\) in terms of \(\theta\).
2. Calculate the probability \(P\) for each of \(\theta = 0, \, \pi/2, \, \pi\).
3. What is the result \(P_{\text{cl}} = |\phi_1|^2 + |\phi_2|^2\) when probabilities are added classically?

Strategy: Apply equation (4.14). \(|\phi_1| = |\phi_2| = 1/\sqrt{2}\), phase difference \(\delta = \theta\).

1. Express \(|\phi_1 + \phi_2|^2\) in terms of \(\theta\)

\[P = |\phi_1|^2 + |\phi_2|^2 + 2|\phi_1||\phi_2|\cos\delta\]

\[= \frac{1}{2} + \frac{1}{2} + 2 \cdot \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} \cdot \cos\theta\]

\[\boxed{P = 1 + \cos\theta}\]

2. Probability for each \(\theta\)

\(\theta\)	\(\cos\theta\)	\(P\)
\(0\)	\(1\)	\(\boxed{2}\)
\(\pi/2\)	\(0\)	\(\boxed{1}\)
\(\pi\)	\(-1\)	\(\boxed{0}\)

3. Classical sum of probabilities

\[P_{\text{cl}} = |\phi_1|^2 + |\phi_2|^2 = \frac{1}{2} + \frac{1}{2} = \boxed{1}\]

Verification: When \(\theta = 0\), \(\phi_1 + \phi_2 = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \sqrt{2}\), so \(P = |\sqrt{2}|^2 = 2\). ✓ When \(\theta = \pi\), \(\phi_2 = \frac{1}{\sqrt{2}} e^{i\pi} = -\frac{1}{\sqrt{2}}\), so \(\phi_1 + \phi_2 = 0\), \(P = 0\). ✓ Depending on the value of the interference term \(\cos\theta\), \(P\) varies from \(0\) to \(2\), oscillating around the classical value of \(1\).

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B-6. Complex Conjugate and Interference Terms

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Problem:

Given \(\phi_1 = 2e^{i\pi/4}\) and \(\phi_2 = 3e^{-i\pi/4}\),

1. Calculate \(\phi_1 \phi_2^*\).
2. Calculate the interference term \(\phi_1 \phi_2^* + \phi_1^* \phi_2\) and confirm that it is real.
3. Find the probability \(P = |\phi_1 + \phi_2|^2\).

Strategy: Use properties of polar form. \(\phi_2^* = 3e^{+i\pi/4}\).

1. \(\phi_1 \phi_2^*\)

\[\phi_1 \phi_2^* = 2e^{i\pi/4} \cdot 3e^{i\pi/4} = 6e^{i(\pi/4 + \pi/4)} = 6e^{i\pi/2}\]

Since \(e^{i\pi/2} = i\),

\[\boxed{\phi_1 \phi_2^* = 6i}\]

2. Interference term \(\phi_1 \phi_2^* + \phi_1^* \phi_2\)

Since \(\phi_1^* \phi_2 = (\phi_1 \phi_2^*)^* = (6i)^* = -6i\),

\[\phi_1 \phi_2^* + \phi_1^* \phi_2 = 6i + (-6i) = \boxed{0}\]

This is real (\(0\)), confirming that the interference term is indeed real. ✓

Alternative perspective: From Eq. (4.13), the phase difference between \(\phi_1\) and \(\phi_2\) is \(\delta = \pi/4 - (-\pi/4) = \pi/2\), so

\[2|\phi_1||\phi_2|\cos\delta = 2 \cdot 2 \cdot 3 \cdot \cos(\pi/2) = 12 \cdot 0 = 0 \quad \checkmark\]

3. Probability \(P = |\phi_1 + \phi_2|^2\)

\[P = |\phi_1|^2 + |\phi_2|^2 + (\text{interference term}) = 4 + 9 + 0 = \boxed{13}\]

Verification: Compute directly. \(\phi_1 = 2e^{i\pi/4} = 2(\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}i) = \sqrt{2} + \sqrt{2}\,i\). \(\phi_2 = 3e^{-i\pi/4} = 3(\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}i) = \frac{3\sqrt{2}}{2} - \frac{3\sqrt{2}}{2}i\).

\[\phi_1 + \phi_2 = \left(\sqrt{2} + \frac{3\sqrt{2}}{2}\right) + \left(\sqrt{2} - \frac{3\sqrt{2}}{2}\right)i = \frac{5\sqrt{2}}{2} - \frac{\sqrt{2}}{2}i\]

\[|\phi_1 + \phi_2|^2 = \left(\frac{5\sqrt{2}}{2}\right)^2 + \left(\frac{\sqrt{2}}{2}\right)^2 = \frac{50}{4} + \frac{2}{4} = \frac{52}{4} = 13 \quad \checkmark\]

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B-7. Dirac Notation and the Third Rule

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Problem:

A particle travels from a source \(s\) through slit \(k\) (\(k = 1, 2, 3\)) to reach a detector position \(x\). Given the amplitudes for each stage below, calculate the amplitude for the entire path through slit \(k\): \(\phi_k = \langle x | k \rangle \langle k | s \rangle\) for each slit.

\(k\)	\(\langle k \mid s \rangle\)	\(\langle x \mid k \rangle\)
1	\(\frac{1}{\sqrt{3}}\)	\(e^{i\pi/3}\)
2	\(\frac{1}{\sqrt{3}}\)	\(e^{i\pi}\)
3	\(\frac{1}{\sqrt{3}}\)	\(e^{i5\pi/3}\)

Strategy: From the third rule (multiplication), \(\phi_k = \langle x | k \rangle \cdot \langle k | s \rangle\).

\(k = 1\):

\[\phi_1 = e^{i\pi/3} \cdot \frac{1}{\sqrt{3}} = \boxed{\frac{1}{\sqrt{3}} e^{i\pi/3}}\]

\(k = 2\):

\[\phi_2 = e^{i\pi} \cdot \frac{1}{\sqrt{3}} = \boxed{\frac{1}{\sqrt{3}} e^{i\pi}}\]

\(k = 3\):

\[\phi_3 = e^{i5\pi/3} \cdot \frac{1}{\sqrt{3}} = \boxed{\frac{1}{\sqrt{3}} e^{i5\pi/3}}\]

Verification: The absolute value of each amplitude is \(|\phi_k| = \frac{1}{\sqrt{3}} \cdot 1 = \frac{1}{\sqrt{3}}\). The three phases \(\pi/3, \pi, 5\pi/3\) correspond to angles that divide \(2\pi\) into three equal parts (\(60°, 180°, 300°\)). ✓

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B-8. Adding Amplitudes and Probability

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Problem:

Add the \(\phi_1, \phi_2, \phi_3\) found in D7 using the second rule (addition) to obtain the total amplitude \(\phi = \phi_1 + \phi_2 + \phi_3\). Then apply the first rule to calculate the probability \(P = |\phi|^2\).

Strategy: Convert each \(\phi_k\) to the form \(a + bi\) and sum them.

Values from Euler's formula:

\[e^{i\pi/3} = \frac{1}{2} + \frac{\sqrt{3}}{2}i, \quad e^{i\pi} = -1, \quad e^{i5\pi/3} = \frac{1}{2} - \frac{\sqrt{3}}{2}i\]

Therefore

\[\phi_1 = \frac{1}{\sqrt{3}}\left(\frac{1}{2} + \frac{\sqrt{3}}{2}i\right)\]

\[\phi_2 = \frac{1}{\sqrt{3}}(-1)\]

\[\phi_3 = \frac{1}{\sqrt{3}}\left(\frac{1}{2} - \frac{\sqrt{3}}{2}i\right)\]

Calculating the total amplitude:

\[\phi = \phi_1 + \phi_2 + \phi_3 = \frac{1}{\sqrt{3}}\left[\left(\frac{1}{2} + \frac{\sqrt{3}}{2}i\right) + (-1) + \left(\frac{1}{2} - \frac{\sqrt{3}}{2}i\right)\right]\]

Simplifying the expression inside the brackets:

• Real part: \(\frac{1}{2} + (-1) + \frac{1}{2} = 0\)
• Imaginary part: \(\frac{\sqrt{3}}{2} + 0 + \left(-\frac{\sqrt{3}}{2}\right) = 0\)

\[\boxed{\phi = 0}\]

Probability:

\[\boxed{P = |\phi|^2 = 0}\]

Verification: \(e^{i\pi/3}\), \(e^{i\pi}\), \(e^{i5\pi/3}\) are equal to \(e^{i \cdot 2\pi k/3}\) (\(k = 1, 2, 3\)), and using the primitive cube root of unity \(\omega = e^{i2\pi/3}\), these would be \(\omega, \omega^2, \omega^3 = 1\)... but more precisely, since the phases are \(\pi/3, \pi, 5\pi/3\), the sum can be factored as \(e^{i\pi/3}(1 + e^{i2\pi/3} + e^{i4\pi/3})\). Since \(1 + e^{i2\pi/3} + e^{i4\pi/3} = 0\) (sum of the primitive cube roots of unity), we have \(\phi = 0\). ✓

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Medium

M-1. Derivation of the General Formula for Interference Terms

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Problem:

Suppose there are \(N\) indistinguishable paths, and the amplitude for each path is given by \(\phi_k\) (\(k = 1, 2, \ldots, N\)).

1. Expand the probability \(P = \left|\sum_{k=1}^{N} \phi_k \right|^2\) and show that

\[ P = \sum_{k=1}^{N} |\phi_k|^2 + \sum_{j \neq k} \phi_j \phi_k^*\]

1. Explain why the number of interference (cross) terms is \(N(N-1)\).

2. If all amplitudes have equal magnitude \(|\phi_k| = A\), and the phase differences between \(\phi_k\) and \(\phi_j\) (\(j \neq k\)) are all randomly distributed, explain qualitatively why the contribution of the interference terms averages to zero.

1. Proof of the Expansion:

Strategy: Expand \(P = \left|\sum_k \phi_k\right|^2\) in the form \(z \cdot z^*\).

\[P = \left(\sum_{j=1}^{N} \phi_j\right)\left(\sum_{k=1}^{N} \phi_k\right)^* = \sum_{j=1}^{N}\sum_{k=1}^{N} \phi_j \phi_k^*\]

Separating this double sum into terms with \(j = k\) and terms with \(j \neq k\):

\[P = \sum_{k=1}^{N} \phi_k \phi_k^* + \sum_{\substack{j,k=1 \\ j \neq k}}^{N} \phi_j \phi_k^*\]

Since \(\phi_k \phi_k^* = |\phi_k|^2\),

\[\boxed{P = \sum_{k=1}^{N} |\phi_k|^2 + \sum_{j \neq k} \phi_j \phi_k^*}\]

The first term is the sum of probabilities for each path (the classical contribution), and the second term is the interference term. \(\blacksquare\)

2. Number of Interference Terms:

In the double sum \(\sum_{j \neq k}\), \(j\) and \(k\) each take values from \(1\) to \(N\), with the constraint \(j \neq k\). The number of ways to choose 2 different elements from \(N\) with ordering is

\[N(N-1)\]

terms. (\((j,k) = (1,2)\) and \((j,k) = (2,1)\) are counted as separate terms.)

Remark: Since \(\phi_j \phi_k^*\) and \(\phi_k \phi_j^*\) are complex conjugates of each other, grouping by pairs with \(j < k\) gives \(N(N-1)/2\) pairs, and each pair yields a real interference term \(\phi_j \phi_k^* + \phi_k \phi_j^* = 2|\phi_j||\phi_k|\cos\delta_{jk}\).

3. Why Interference Terms Vanish for Random Phases:

Let all \(|\phi_k| = A\), and suppose the phase differences \(\delta_{jk}\) are randomly distributed. Each interference term is

\[\phi_j \phi_k^* = A^2 e^{i\delta_{jk}}\]

and its real part is \(A^2 \cos\delta_{jk}\). If \(\delta_{jk}\) is uniformly randomly distributed over \([0, 2\pi)\),

\[\langle \cos\delta_{jk} \rangle = 0\]

Therefore, the average contribution of the \(N(N-1)\) interference terms is zero. As a result,

\[P \approx \sum_{k=1}^{N} |\phi_k|^2 = NA^2\]

which reduces to the classical addition of probabilities. This is also the essential reason why, when phase coherence is lost in macroscopic systems, quantum interference disappears and classical behavior is recovered (decoherence).

Verification: For \(N = 2\), the number of interference terms is \(2 \cdot 1 = 2\), giving \(\phi_1\phi_2^* + \phi_2\phi_1^* = 2|\phi_1||\phi_2|\cos\delta\), which is consistent with Eq. (4.14). ✓

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M-2. Interference Pattern from \(N\) Equally-Spaced, Equal-Amplitude Slits

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Problem:

Consider \(N\) slits equally spaced with separation \(d\), and let the particle's momentum be \(p\). Assume that the path difference from each slit to the detector position \(x\) is a constant value \(\Delta r\) between adjacent slits, and define the phase difference as \(\delta = p\Delta r / \hbar\). The absolute value of the amplitude through each slit is equal to \(A\).

1. Explain why the amplitude through the \(k\)-th slit (\(k = 0, 1, \ldots, N-1\)) can be written as \(\phi_k = A e^{ik\delta}\).

2. Using the geometric series formula, write the total amplitude \(\phi = \sum_{k=0}^{N-1} A e^{ik\delta}\) in closed form.

3. Show that the probability \(P = |\phi|^2\) is given by

\[ P = A^2 \frac{\sin^2(N\delta/2)}{\sin^2(\delta/2)}\]

1. Verify that for \(N = 2\), this expression agrees with Eq. (4.14) (in the case \(|\phi_1| = |\phi_2| = A\)).

1. Explanation of the amplitude \(\phi_k = Ae^{ik\delta}\):

The path length from the \(k\)-th slit (\(k = 0, 1, \ldots, N-1\)) to the detector position \(x\) is longer by \(k\Delta r\) compared to the path length from the \(0\)-th slit. From the de Broglie relation, when a particle with momentum \(p\) travels an extra distance \(\Delta r\), its phase increases by \(p\Delta r/\hbar = \delta\). Therefore, the amplitude passing through the \(k\)-th slit is advanced in phase by \(k\delta\) relative to the \(0\)-th slit, and we can write

\[\phi_k = A e^{ik\delta}\]

(The common phase factor is omitted.)

2. Closed form via geometric series:

\[\phi = \sum_{k=0}^{N-1} A e^{ik\delta} = A \sum_{k=0}^{N-1} (e^{i\delta})^k\]

Applying the geometric series formula \(\sum_{k=0}^{N-1} r^k = \frac{1 - r^N}{1 - r}\) (\(r \neq 1\)) with \(r = e^{i\delta}\),

\[\boxed{\phi = A \cdot \frac{1 - e^{iN\delta}}{1 - e^{i\delta}}}\]

3. Derivation of the probability \(P = |\phi|^2\):

First, we show that the following identity holds in general:

\[|1 - e^{i\alpha}|^2 = (1 - e^{i\alpha})(1 - e^{-i\alpha}) = 1 - e^{i\alpha} - e^{-i\alpha} + 1 = 2 - 2\cos\alpha = 4\sin^2\frac{\alpha}{2}\]

Using this,

\[|\phi|^2 = A^2 \cdot \frac{|1 - e^{iN\delta}|^2}{|1 - e^{i\delta}|^2} = A^2 \cdot \frac{4\sin^2(N\delta/2)}{4\sin^2(\delta/2)}\]

\[\boxed{P = A^2 \frac{\sin^2(N\delta/2)}{\sin^2(\delta/2)}}\]

4. Verification for the case \(N = 2\):

\[P = A^2 \frac{\sin^2(\delta)}{\sin^2(\delta/2)}\]

Using the double-angle formula \(\sin\delta = 2\sin(\delta/2)\cos(\delta/2)\),

\[P = A^2 \frac{4\sin^2(\delta/2)\cos^2(\delta/2)}{\sin^2(\delta/2)} = 4A^2\cos^2\frac{\delta}{2}\]

On the other hand, from the half-angle formula \(\cos^2(\delta/2) = \frac{1 + \cos\delta}{2}\),

\[P = 4A^2 \cdot \frac{1 + \cos\delta}{2} = 2A^2(1 + \cos\delta)\]

Setting \(|\phi_1| = |\phi_2| = A\) in Eq. (4.14),

\[P = A^2 + A^2 + 2A^2\cos\delta = 2A^2(1 + \cos\delta)\]

The two expressions agree. ✓

Sanity check: When \(\delta = 0\) (the central bright fringe), \(P = A^2 \cdot N^2\) (by L'Hôpital's rule or direct substitution). For \(N = 2\), \(P = 4A^2\). Setting \(\delta = 0\) in the expression above gives \(P = 4A^2 \cdot 1 = 4A^2\). ✓

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M-3. Observation Destroys Interference: A Mathematical Explanation

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Problem:

In the double-slit experiment, let \(\phi_1\) be the amplitude for passing through slit 1, and \(\phi_2\) be the amplitude for passing through slit 2.

1. Using the second rule, explain why the probability when the path through the slits is not observed is \(P = |\phi_1 + \phi_2|^2\).

2. When the path through the slits is observed, the probability of finding that the particle passed through slit 1 is \(P_1 = |\phi_1|^2\), and the probability of finding it passed through slit 2 is \(P_2 = |\phi_2|^2\). The total probability of arriving at detector \(x\) is

\[ P_{\text{obs}} = P_1 + P_2 = |\phi_1|^2 + |\phi_2|^2\]

Explain why no interference term appears in this expression, using the applicability condition of the second rule from the perspective that "the paths have become distinguishable."

1. For the case \(|\phi_1| = |\phi_2|\), qualitatively explain how the difference between \(P\) and \(P_{\text{obs}}\) (namely the interference term \(2|\phi_1||\phi_2|\cos\delta\)) behaves as a function of detector position \(x\), and state what it means for "the interference pattern to disappear."

1. Without Observation:

When we do not observe which slit the particle passes through, the two paths—via slit 1 and via slit 2—are indistinguishable in principle. The condition for applying the second rule ("add amplitudes for indistinguishable paths") is satisfied, so

\[\phi = \phi_1 + \phi_2\]

From the first rule, the probability is

\[P = |\phi_1 + \phi_2|^2 = |\phi_1|^2 + |\phi_2|^2 + 2|\phi_1||\phi_2|\cos\delta\]

The interference term \(2|\phi_1||\phi_2|\cos\delta\) exists, and since \(\delta\) depends on the detector position \(x\), an interference pattern appears.

2. With Observation:

When we observe which slit the particle passes through, the two paths become distinguishable. The condition for applying the second rule ("indistinguishable") is no longer satisfied, so instead of adding amplitudes, we add probabilities (reverting to the classical addition rule of probabilities).

Specifically, the event "passed through slit 1" and the event "passed through slit 2" are mutually exclusive and distinguishable events, so

\[P_{\text{obs}} = P_1 + P_2 = |\phi_1|^2 + |\phi_2|^2\]

The interference term \(2|\phi_1||\phi_2|\cos\delta\) does not appear. This is because the observation records the "which-path" information in the environment (the measuring apparatus), making the two paths distinguishable in principle.

3. What It Means for the Interference Pattern to Disappear:

For the case \(|\phi_1| = |\phi_2| = A\):

• Without observation: \(P(x) = 2A^2(1 + \cos\delta(x))\). Since \(\delta(x)\) is a function of the detector position \(x\), \(P(x)\) oscillates as a \(\cos\) function with respect to \(x\). This is the interference pattern. At bright fringes (\(\delta = 2n\pi\)), \(P = 4A^2\); at dark fringes (\(\delta = (2n+1)\pi\)), \(P = 0\).

• With observation: \(P_{\text{obs}}(x) = 2A^2\). This is a uniform distribution independent of \(x\).

"The interference pattern disappears" means that the oscillating component \(\cos\delta(x)\) vanishes from the probability distribution as a function of detector position \(x\), and the distribution becomes uniform. Because the interference term \(2A^2\cos\delta(x)\) becomes zero, the contrast between bright and dark fringes is completely lost.

Consistency check: \(P_{\text{obs}} = 2A^2\) coincides with the average of \(P(x)\) over \(x\): \(\langle 2A^2(1 + \cos\delta) \rangle_x = 2A^2\) (since the average of \(\cos\delta\) is 0). Even though the interference pattern disappears due to observation, the overall average arrival probability remains unchanged. ✓

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M-4. Calculating Amplitudes Through Two Walls

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Problem:

There are two walls between a source \(s\) and a detector \(x\). The first wall has slits \(A_1, A_2\), and the second wall has slits \(B_1, B_2, B_3\). Which slit the particle passes through is not observed.

1. List all possible paths for a particle to travel from \(s\) to \(x\) (give the number of paths).

2. Using the third rule, write the amplitude for the path "\(s \to A_j \to B_k \to x\)" in Dirac notation.

3. Using the second rule, write down the total amplitude \(\langle x | s \rangle\).

4. Assuming that the amplitudes at each stage are all real and equal to the same value \(c\) (i.e., \(\langle A_j | s \rangle = \langle B_k | A_j \rangle = \langle x | B_k \rangle = c\)), express the probability \(P = |\langle x | s \rangle|^2\) in terms of \(c\).

1. Enumeration of Paths:

Slits in the first wall: \(A_1, A_2\) (2 options) Slits in the second wall: \(B_1, B_2, B_3\) (3 options)

All possible paths are:

\[s \to A_j \to B_k \to x \quad (j = 1, 2; \; k = 1, 2, 3)\]

\[\boxed{\text{Number of paths} = 2 \times 3 = 6}\]

Specifically: \(A_1B_1\), \(A_1B_2\), \(A_1B_3\), \(A_2B_1\), \(A_2B_2\), \(A_2B_3\).

2. Amplitude for Each Path (Third Rule):

The amplitude for the path "\(s \to A_j \to B_k \to x\)" is the product of the amplitudes for three consecutive stages:

\[\boxed{\phi_{jk} = \langle x | B_k \rangle \langle B_k | A_j \rangle \langle A_j | s \rangle}\]

3. Total Amplitude (Second Rule):

Since we do not observe which slits the particle passes through, all 6 paths are indistinguishable. We add the amplitudes:

\[\boxed{\langle x | s \rangle = \sum_{j=1}^{2}\sum_{k=1}^{3} \langle x | B_k \rangle \langle B_k | A_j \rangle \langle A_j | s \rangle}\]

4. Case Where All Amplitudes Equal \(c\):

The amplitude for each path is \(\phi_{jk} = c \cdot c \cdot c = c^3\).

Since all 6 paths have the same amplitude \(c^3\),

\[\langle x | s \rangle = 6c^3\]

\[\boxed{P = |\langle x | s \rangle|^2 = 36c^6}\]

Verification: Dimensionally, since the amplitude is a product of 3 stages, it is of order \(c^3\); multiplying by the number of paths 6 gives \(6c^3\), and the probability is its square, \(36c^6\). If there were only one wall (with \(n\) slits), we would expect \(P = n^2 c^4\), so for \(n = 6\) with 2 stages, \(36c^4\). Since we have 3 stages here, \(c^6\) is correct. ✓

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M-5. Relationship Between Phase Difference and Path Difference

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Problem:

In a double-slit experiment, let the slit separation be \(d\) and the distance from the slit wall to the detector screen be \(L\) (\(L \gg d\)). For a position \(x\) on the screen (with the origin at the center of the screen),

1. Show geometrically that the difference \(\Delta r = r_1 - r_2\) between the distance \(r_1\) from slit 1 to \(x\) and the distance \(r_2\) from slit 2 to \(x\) is, in the approximation \(L \gg d\),

\[ \Delta r \approx \frac{dx}{L}\]

1. Using Eq. (4.22), express the phase difference \(\delta\) as a function of \(x\).

2. From the constructive interference condition (bright fringes) \(\delta = 2n\pi\) (\(n\) is an integer), find the positions \(x_n\) of the bright fringes.

3. Using the de Broglie relation \(p = h/\lambda\), express the spacing \(\Delta x\) between adjacent bright fringes in terms of the wavelength \(\lambda\).

1. Derivation of Path Difference \(\Delta r \approx dx/L\):

Let the positions of slit 1 and slit 2 be at \(\pm d/2\) relative to the central axis perpendicular to the screen. The distances to the detector position \(x\) are:

\[r_1 = \sqrt{L^2 + (x - d/2)^2}, \quad r_2 = \sqrt{L^2 + (x + d/2)^2}\]

We calculate \(r_1 - r_2\) in the approximation \(L \gg d\) and \(L \gg x\).

\[r_1^2 - r_2^2 = \left[(x - d/2)^2 - (x + d/2)^2\right] = -2xd\]

\[r_1^2 - r_2^2 = (r_1 - r_2)(r_1 + r_2) \approx \Delta r \cdot 2L\]

(where \(r_1 + r_2 \approx 2L\) is the approximation for \(L \gg d, x\).)

Therefore

\[\boxed{\Delta r = r_1 - r_2 \approx \frac{-2xd}{2L} = -\frac{dx}{L}}\]

The sign convention depends on the coordinate definition. As the magnitude of the path difference:

\[|\Delta r| \approx \frac{dx}{L}\]

Alternative derivation (geometric): When \(L \gg d\), the two paths are nearly parallel, making an angle \(\theta \approx x/L\). The path difference is \(\Delta r = d\sin\theta \approx d\theta \approx dx/L\). ✓

2. Phase Difference:

Using the form of equation (4.22) (\(\delta = p \Delta r / \hbar\)),

\[\boxed{\delta = \frac{p}{\hbar} \cdot \frac{dx}{L} = \frac{pdx}{\hbar L}}\]

3. Positions of Bright Fringes:

From the constructive interference condition \(\delta = 2n\pi\) (where \(n\) is an integer),

\[\frac{pdx_n}{\hbar L} = 2n\pi\]

\[\boxed{x_n = \frac{2n\pi \hbar L}{pd}}\]

4. Spacing Between Adjacent Bright Fringes:

\[\Delta x = x_{n+1} - x_n = \frac{2\pi\hbar L}{pd}\]

Substituting the de Broglie relation \(p = h/\lambda = 2\pi\hbar/\lambda\),

\[\Delta x = \frac{2\pi\hbar L}{(2\pi\hbar/\lambda) \cdot d} = \frac{\lambda L}{d}\]

\[\boxed{\Delta x = \frac{\lambda L}{d}}\]

Verification: Dimensional analysis: \([\lambda L / d] = \text{m} \cdot \text{m} / \text{m} = \text{m}\). ✓ Physically, the fringe spacing increases with larger wavelength \(\lambda\) and larger screen distance \(L\), and decreases with larger slit separation \(d\). This is in complete agreement with the results of optical interference experiments (Young's experiment). ✓

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Advanced

A-1. Quantum Mechanical Analysis of the Mach–Zehnder Interferometer

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Problem:

A single photon enters a Mach–Zehnder interferometer. The interferometer consists of the following elements:

• Beam splitter BS1: Splits the incident photon into two paths (path \(A\) and path \(B\)). The reflection amplitude is \(\frac{i}{\sqrt{2}}\), and the transmission amplitude is \(\frac{1}{\sqrt{2}}\).
• Mirrors \(M_A\), \(M_B\): Reflect the photon in each respective path. The reflection amplitude is \(i\) (a phase shift of \(\pi/2\)).
• Beam splitter BS2: Recombines the two paths. It has the same properties as BS1 (reflection \(\frac{i}{\sqrt{2}}\), transmission \(\frac{1}{\sqrt{2}}\)).
• Two output ports: detector \(D_1\) and detector \(D_2\).

Assume there is a phase plate in path \(A\) that can introduce an additional phase shift \(e^{i\varphi}\) (no additional phase in path \(B\)).

1. Using the third rule, calculate the amplitudes for the two paths by which the photon travels from the input port to \(D_1\) (via path \(A\) and via path \(B\)). (List the elements the photon encounters in sequence along each path and multiply the amplitudes together.)

2. Using the second rule, find the total amplitude and calculate the probability \(P_1(\varphi)\) that the photon is detected at \(D_1\).

3. Similarly, calculate the probability \(P_2(\varphi)\) of detection at \(D_2\). Verify that \(P_1 + P_2 = 1\) holds.

4. Find the values of \(P_1\) and \(P_2\) for the cases \(\varphi = 0\) and \(\varphi = \pi\), and discuss their physical meaning.

5. Discuss how the interference changes when a "which-path detector" is inserted in path \(A\), based on the conditions for applying the second rule.

Setting Up the Problem:

Components of the interferometer and amplitude assignments:

• BS1: Reflection → \(\frac{i}{\sqrt{2}}\), Transmission → \(\frac{1}{\sqrt{2}}\)
• Mirror: Reflection → \(i\)
• Phase plate (path \(A\) only): \(e^{i\varphi}\)
• BS2: Reflection → \(\frac{i}{\sqrt{2}}\), Transmission → \(\frac{1}{\sqrt{2}}\)

We adopt the following typical configuration: - Path \(A\): Reflected at BS1 → Phase plate → Reflected at mirror \(M_A\) → Transmitted through BS2 to \(D_1\) (or reflected to \(D_2\)) - Path \(B\): Transmitted through BS1 → Reflected at mirror \(M_B\) → Reflected at BS2 to \(D_1\) (or transmitted to \(D_2\))

1. Amplitudes of the Two Paths Leading to \(D_1\):

Via path \(A\) to \(D_1\): BS1 reflection → Phase plate → \(M_A\) reflection → BS2 transmission

\[\phi_A^{(1)} = \frac{i}{\sqrt{2}} \cdot e^{i\varphi} \cdot i \cdot \frac{1}{\sqrt{2}} = \frac{i \cdot i}{\sqrt{2} \cdot \sqrt{2}} \cdot e^{i\varphi} = \frac{i^2}{2} e^{i\varphi} = \frac{-1}{2} e^{i\varphi}\]

\[\boxed{\phi_A^{(1)} = -\frac{1}{2}e^{i\varphi}}\]

Via path \(B\) to \(D_1\): BS1 transmission → \(M_B\) reflection → BS2 reflection

\[\phi_B^{(1)} = \frac{1}{\sqrt{2}} \cdot i \cdot \frac{i}{\sqrt{2}} = \frac{i^2}{2} = -\frac{1}{2}\]

\[\boxed{\phi_B^{(1)} = -\frac{1}{2}}\]

2. Detection Probability at \(D_1\):

By the second rule, the total amplitude is:

\[\phi^{(1)} = \phi_A^{(1)} + \phi_B^{(1)} = -\frac{1}{2}e^{i\varphi} - \frac{1}{2} = -\frac{1}{2}(e^{i\varphi} + 1)\]

Probability:

\[P_1 = |\phi^{(1)}|^2 = \frac{1}{4}|e^{i\varphi} + 1|^2\]

Computing \(|e^{i\varphi} + 1|^2\):

\[|e^{i\varphi} + 1|^2 = (e^{i\varphi} + 1)(e^{-i\varphi} + 1) = 1 + e^{i\varphi} + e^{-i\varphi} + 1 = 2 + 2\cos\varphi = 4\cos^2\frac{\varphi}{2}\]

\[\boxed{P_1(\varphi) = \frac{1}{4} \cdot 4\cos^2\frac{\varphi}{2} = \cos^2\frac{\varphi}{2}}\]

3. Detection Probability at \(D_2\):

Via path \(A\) to \(D_2\): BS1 reflection → Phase plate → \(M_A\) reflection → BS2 reflection

\[\phi_A^{(2)} = \frac{i}{\sqrt{2}} \cdot e^{i\varphi} \cdot i \cdot \frac{i}{\sqrt{2}} = \frac{i^3}{2} e^{i\varphi} = \frac{-i}{2} e^{i\varphi}\]

Via path \(B\) to \(D_2\): BS1 transmission → \(M_B\) reflection → BS2 transmission

\[\phi_B^{(2)} = \frac{1}{\sqrt{2}} \cdot i \cdot \frac{1}{\sqrt{2}} = \frac{i}{2}\]

Total amplitude:

\[\phi^{(2)} = \phi_A^{(2)} + \phi_B^{(2)} = -\frac{i}{2}e^{i\varphi} + \frac{i}{2} = \frac{i}{2}(1 - e^{i\varphi})\]

Probability:

\[P_2 = \left|\frac{i}{2}\right|^2 |1 - e^{i\varphi}|^2 = \frac{1}{4} \cdot 4\sin^2\frac{\varphi}{2}\]

\[\boxed{P_2(\varphi) = \sin^2\frac{\varphi}{2}}\]

Verification of probability conservation:

\[P_1 + P_2 = \cos^2\frac{\varphi}{2} + \sin^2\frac{\varphi}{2} = 1 \quad \checkmark\]

4. Special Cases:

When \(\varphi = 0\):

\[P_1(0) = \cos^2 0 = 1, \quad P_2(0) = \sin^2 0 = 0\]

The photon is always detected at \(D_1\). The amplitudes from the two paths interfere completely constructively at \(D_1\) and completely destructively at \(D_2\). This corresponds to perfect constructive interference (\(D_1\)) and perfect destructive interference (\(D_2\)).

When \(\varphi = \pi\):

\[P_1(\pi) = \cos^2\frac{\pi}{2} = 0, \quad P_2(\pi) = \sin^2\frac{\pi}{2} = 1\]

The photon is always detected at \(D_2\). The phase plate adds a phase shift of \(\pi\) to path \(A\), reversing the interference conditions: perfect destructive interference occurs at \(D_1\) and perfect constructive interference at \(D_2\).

Physical significance: By continuously varying the phase plate setting \(\varphi\), one can continuously control the photon's output destination between \(D_1\) and \(D_2\). This is interference at the single-photon level and serves as direct evidence that the photon behaves as if it "travels through both paths simultaneously."

5. Inserting a Which-Path Detector:

If a device that detects "which path was taken" is inserted in path \(A\), information about which path the photon traveled becomes available in principle. This makes the two paths distinguishable.

The condition for applying the second rule is that "the paths are indistinguishable in principle." When the paths become distinguishable, instead of adding amplitudes, we add probabilities:

\[P_1^{\text{obs}} = |\phi_A^{(1)}|^2 + |\phi_B^{(1)}|^2 = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}\]

\[P_2^{\text{obs}} = |\phi_A^{(2)}|^2 + |\phi_B^{(2)}|^2 = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}\]

The interference terms vanish, giving \(P_1 = P_2 = 1/2\), and the dependence on the phase \(\varphi\) is completely lost. The photon arrives at \(D_1\) and \(D_2\) with equal probability, and changing the phase plate setting no longer affects the detection probabilities. The function of the interferometer is completely destroyed.

Verification: \(P_1^{\text{obs}} + P_2^{\text{obs}} = 1/2 + 1/2 = 1\). ✓ Also, \(|\phi_A^{(1)}|^2 = |{-\frac{1}{2}e^{i\varphi}}|^2 = 1/4\), \(|\phi_B^{(1)}|^2 = |{-\frac{1}{2}}|^2 = 1/4\), which indeed do not depend on \(\varphi\). ✓

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A-2. Interference in a 3-State System and the Principle of the "Quantum Eraser"

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Problem:

Consider an experiment in which a particle travels from a source \(s\) through 3 slits (\(1, 2, 3\)) to reach a detector at \(x\). The amplitude for passing through each slit is

\[ \phi_k = A e^{i\alpha_k} \quad (k = 1, 2, 3)\]

with \(\alpha_1 = 0\), \(\alpha_2 = 2\pi/3\), \(\alpha_3 = 4\pi/3\) (corresponding to the phase differences at the center of equally-spaced slits).

Part I: Complete Interference

1. Calculate the total amplitude \(\phi = \phi_1 + \phi_2 + \phi_3\) and find \(P = |\phi|^2\).

2. Interpret this result by connecting it to the fact that \(e^{i \cdot 2\pi/3}\) is a "primitive cube root of unity."

Part II: Partial Path Information

A "marker" is attached to slit 1, introducing a device that can determine only whether or not the particle passed through slit 1, but cannot distinguish between slits 2 and 3.

1. Discuss how the probability should be calculated in this case, based on Feynman's three rules. Write down the explicit expression for \(P(x)\).

2. Compare with the result from Part I, and discuss whether the phenomenon of "partial recovery of interference by obtaining partial path information" can occur.

Part III: Quantum Eraser

1. If the marker information from Part II is "erased" (i.e., chosen not to be read out), what happens to the interference pattern? Explain by returning to the conditions for applying the second rule. State that this constitutes the basic principle of the "quantum eraser."

Part I: Complete Interference:

1. Calculating the Total Amplitude and Probability:

\[\phi = \phi_1 + \phi_2 + \phi_3 = A(e^{i \cdot 0} + e^{i \cdot 2\pi/3} + e^{i \cdot 4\pi/3})\]

\[= A(1 + e^{i2\pi/3} + e^{i4\pi/3})\]

Setting \(\omega = e^{i2\pi/3}\), \(\omega\) is a primitive cube root of unity, and

\[1 + \omega + \omega^2 = 0\]

(This corresponds to the sum of roots of the second factor \(z^2 + z + 1 = 0\) from \(z^3 - 1 = (z-1)(z^2 + z + 1) = 0\).)

Therefore

\[\phi = A \cdot 0 = 0\]

\[\boxed{P = |\phi|^2 = 0}\]

2. Relationship to Primitive Cube Roots of Unity:

\(e^{i2\pi/3}\) is the primitive cube root of unity \(\omega\), and \(1, \omega, \omega^2\) are located at the vertices of an equilateral triangle inscribed in the unit circle on the complex plane. When these three vectors are added together, they cancel completely to zero due to symmetry.

Physically, since the amplitudes from the three slits have equal magnitude and phases equally spaced by \(120°\), at this detector position (this particular \(x\)), the three waves undergo completely destructive interference, making the probability of a particle arriving zero. This is also consistent with the result of D8.

Part II: Partial Path Information:

3. Method of Calculating the Probability:

By placing a marker on slit 1, it becomes possible to distinguish whether the particle "passed through slit 1" or "did not pass through slit 1 (passed through slit 2 or 3)."

Calculation method based on Feynman's rules:

• Slit 1 is distinguishable from the other slits, so the amplitude for slit 1 is converted to probability independently.
• Slits 2 and 3 are indistinguishable from each other, so their amplitudes are added (second rule).
• Finally, the probabilities of the two distinguishable groups are added (classical addition rule for probabilities).

\[\boxed{P(x) = |\phi_1|^2 + |\phi_2 + \phi_3|^2}\]

Computing explicitly:

\[|\phi_1|^2 = |A|^2 = A^2\]

\[\phi_2 + \phi_3 = A(e^{i2\pi/3} + e^{i4\pi/3})\]

From \(1 + \omega + \omega^2 = 0\), we have \(\omega + \omega^2 = -1\), so

\[\phi_2 + \phi_3 = A \cdot (-1) = -A\]

\[|\phi_2 + \phi_3|^2 = A^2\]

\[\boxed{P = A^2 + A^2 = 2A^2}\]

4. Comparison with Part I:

Situation	Probability
Part I (no path information, complete interference)	\(P = 0\)
Part II (partial path information)	\(P = 2A^2\)
Complete path information (all distinguishable)	\(P = 3A^2\)

In Part I, \(P = 0\) due to completely destructive interference. When partial path information is obtained through the marker, the complete cancellation of the three amplitudes breaks down, giving \(P = 2A^2 > 0\).

The phrase "partial recovery of interference through partial path information" requires careful attention to context. What is actually happening here is:

• The completely destructive interference of Part I (\(P = 0\)) has been partially destroyed by the introduction of the marker.
• Interference still remains between slits 2 and 3 (\(|\phi_2 + \phi_3|^2 = A^2 \neq |\phi_2|^2 + |\phi_3|^2 = 2A^2\)). Indeed, since \(\phi_2 + \phi_3 = -A\), we have \(|\phi_2 + \phi_3|^2 = A^2\), which is less than \(|\phi_2|^2 + |\phi_3|^2 = 2A^2\). Partial destructive interference remains between slits 2 and 3.
• Overall, the probability increased from complete cancellation (\(P = 0\)) to \(P = 2A^2\). This is a case where the probability increased due to "destruction of interference," which is the opposite phenomenon from "recovery of interference."

Therefore, at this particular detector position, the precise statement is: "obtaining partial path information partially destroys the interference, causing the probability to change from zero to non-zero."

Part III: Quantum Eraser:

5. Erasing the Marker Information:

Consider the case where the marker was installed as in Part II, but its information is not read (erased).

Returning to the conditions for applying the second rule, amplitudes are added when the paths are in principle indistinguishable. Erasing the marker information means returning to a state where information about which slit the particle passed through is in principle unobtainable.

If the marker information is completely erased, the three paths become indistinguishable again, and the second rule is restored:

\[\phi = \phi_1 + \phi_2 + \phi_3 = 0\]

\[P = |\phi|^2 = 0\]

The original complete interference pattern (the result of Part I) is recovered.

This is the fundamental principle of the "quantum eraser." A quantum eraser is an operation that recovers a lost interference pattern by retroactively erasing previously acquired path information (which-path information). The key points are:

1. The presence or absence of path information determines interference: Whether interference occurs depends on whether path information is in principle available.
2. Erasing information is a physical operation: Simply "not looking" is insufficient; one must appropriately manipulate the quantum state of the marker to make path information in principle unrecoverable.
3. Recovery through post-selection: In actual quantum eraser experiments, interference fringes are recovered by post-selecting data based on measurement results of the marker (ancillary system). When all data are summed, no interference fringes are visible, but interference fringes appear in sub-ensembles corresponding to specific measurement outcomes.

Verification: Part I (no information) → \(P = 0\), Part II (partial information) → \(P = 2A^2\), complete information → \(P = 3A^2\). As information increases, interference is destroyed and the probability approaches the classical value \(3A^2\). When information is erased, interference is recovered and \(P\) returns to \(0\). This series of results is physically consistent. ✓

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