Skip to content

Ch. 4 Solutions

← Back to problems　|　Back to chapter

Table of Contents

Basic

Medium

Advanced

Basic

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

→ Back to problem

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. ✓

B-2. Multiplication of Complex Numbers and Phase

→ Back to problem

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. ✓

B-3. Conversion to Polar Form

→ Back to problem

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 \(z = -\sqrt{3} + i\). ✓

B-4. Multiplication in Polar Form

→ Back to problem

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. \(z_1 = 2(\cos 30° + i\sin 30°) = 2(\frac{\sqrt{3}}{2} + \frac{1}{2}i) = \sqrt{3} + i\). \(z_2 = 3(\cos 60° + i\sin 60°) = 3(\frac{1}{2} + \frac{\sqrt{3}}{2}i) = \frac{3}{2} + \frac{3\sqrt{3}}{2}i\).

\[z_1 z_2 = (\sqrt{3} + i)\left(\frac{3}{2} + \frac{3\sqrt{3}}{2}i\right) = \frac{3\sqrt{3}}{2} + \frac{3 \cdot 3}{2}i + \frac{3}{2}i + \frac{3\sqrt{3}}{2}i^2\]
\[= \frac{3\sqrt{3}}{2} + \frac{9}{2}i + \frac{3}{2}i - \frac{3\sqrt{3}}{2} = 0 + 6i = 6i \quad \checkmark\]

B-5. Calculation of Interference Terms

→ Back to problem

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 \(1\).

B-6. Complex Conjugate and Interference Terms

→ Back to problem

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

\[\phi_1 + \phi_2 = \left(\sqrt{2} + \frac{3\sqrt{2}}{2}\right) + \left(\sqrt{2} - \frac{3\sqrt{2}}{2}\right)i = \frac{5\sqrt{2}}{2} - \frac{\sqrt{2}}{2}i\]
\[|\phi_1 + \phi_2|^2 = \left(\frac{5\sqrt{2}}{2}\right)^2 + \left(\frac{\sqrt{2}}{2}\right)^2 = \frac{50}{4} + \frac{2}{4} = \frac{52}{4} = 13 \quad \checkmark\]

B-7. Dirac Notation and the Third Rule

→ Back to problem

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 (\(60°, 180°, 300°\)). ✓

B-8. Adding Amplitudes and Probability

→ Back to problem

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 \(\phi = 0\). ✓

Medium

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

→ Back to problem

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). ✓

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

→ Back to problem

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 \(P = 4A^2 \cdot 1 = 4A^2\). ✓

M-3. Observation Destroys Interference: A Mathematical Explanation

→ Back to problem

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. ✓

M-4. Calculating Amplitudes Through Two Walls

→ Back to problem

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. ✓

M-5. Relationship Between Phase Difference and Path Difference

→ Back to problem

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). ✓

Advanced

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

→ Back to problem

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 \(\varphi\). ✓

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

→ Back to problem

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. ✓