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Ch. 4 Problems

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Basic

Medium

Advanced

Basic

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

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. \(z = 5\)
Hint

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

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B-2. Multiplication of Complex Numbers and Phase

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

Expand normally and substitute \(i^2 = -1\). Calculate following the approach of equation (4.2).

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B-3. Conversion to Polar Form

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. \(z = 3i\)
Hint

\(r = |z| = \sqrt{a^2 + b^2}\), \(\theta = \arctan(b/a)\) (but be careful about which quadrant the point is in). It helps to plot the point on the complex plane to verify the angle.

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B-4. Multiplication in Polar Form

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 \(a + bi\).
Hint

Use equation (4.8): the moduli are multiplied, and the arguments are added. Note that \(\frac{\pi}{6} + \frac{\pi}{3} = \frac{\pi}{2}\).

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B-5. Calculation of Interference Terms

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?
Hint

Use Eq. (4.14). We have \(|\phi_1| = |\phi_2| = \frac{1}{\sqrt{2}}\), and the phase difference is \(\delta = \theta\). Substitute into \(P = |\phi_1|^2 + |\phi_2|^2 + 2|\phi_1||\phi_2|\cos\delta\) and compute.

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B-6. Complex Conjugate and Interference Terms

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

Since \(\phi_2^* = 3e^{+i\pi/4}\), we have \(\phi_1 \phi_2^* = 2 \cdot 3 \cdot e^{i\pi/4} \cdot e^{i\pi/4} = 6e^{i\pi/2}\). You may also use Eq. (4.13).

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B-7. Dirac Notation and the Third Rule

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 \(\frac{1}{\sqrt{3}}\) \(e^{i5\pi/3}\)
Hint

Use the third rule (multiplication). \(\phi_k = \langle x | k \rangle \cdot \langle k | s \rangle\). When multiplying a complex number by a real number, the real number changes the magnitude while the phase remains unchanged.

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B-8. Adding Amplitudes and Probability

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 \(P = |\phi|^2\).

Hint

It is helpful to convert each \(\phi_k\) to the form \(a + bi\) before adding. Use \(e^{i\pi/3} = \frac{1}{2} + \frac{\sqrt{3}}{2}i\), \(e^{i\pi} = -1\), and \(e^{i5\pi/3} = \frac{1}{2} - \frac{\sqrt{3}}{2}i\).

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Medium

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

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.

Hint

Separate \(\left|\sum_k \phi_k\right|^2 = \left(\sum_j \phi_j\right)\left(\sum_k \phi_k\right)^* = \sum_j \sum_k \phi_j \phi_k^*\) into the cases \(j = k\) and \(j \neq k\). If the phases are random, the average of \(\cos\delta_{jk}\) is 0.

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

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 \(|\phi_1| = |\phi_2| = A\)).
Hint

Sum of a geometric series: \(\sum_{k=0}^{N-1} r^k = \frac{1 - r^N}{1 - r}\). Substitute \(r = e^{i\delta}\) and use \(|1 - e^{iN\delta}|^2 = 4\sin^2(N\delta/2)\) (it helps to first show that \(|1 - e^{i\alpha}|^2 = 2 - 2\cos\alpha = 4\sin^2(\alpha/2)\)).

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M-3. Observation Destroys Interference: A Mathematical Explanation

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."
Hint

The applicability condition of the second rule is that "the paths are in principle indistinguishable." When observation provides information about which path was taken, the two paths can no longer be called "indistinguishable." In that case, instead of adding amplitudes, one adds probabilities (reverting to the classical addition rule for probabilities).

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M-4. Calculating Amplitudes Through Two Walls

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 \(c\).

Hint

The number of paths is the number of choices at the first wall × the number of choices at the second wall. The amplitude for each path is a product of three stages (third rule). Add the amplitudes for all paths (second rule). If all amplitudes are equal, the sum of amplitudes for \(N\) paths is \(N \times (\text{amplitude for one path})\).

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M-5. Relationship Between Phase Difference and Path Difference

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 \(\lambda\).

Hint

Draw the paths from the two slits to the detector position \(x\), and use the fact that when \(L \gg d\), the two paths are approximately parallel. The path difference becomes \(d\sin\theta \approx d \cdot x/L\). Use \(p/\hbar = 2\pi/\lambda\).

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Advanced

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

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.

Hint

The process of reaching \(D_1\) via path \(A\): reflection at BS1 → phase plate → reflection at mirror \(M_A\) → transmission (or reflection) at BS2. Multiply the amplitudes of each element in sequence. Whether the photon is transmitted or reflected at BS2 to reach \(D_1\) depends on the geometric configuration of the interferometer. In a typical configuration, path \(A\) is transmitted through BS2 to reach \(D_1\), and path \(B\) is reflected at BS2 to reach \(D_1\) (the reverse assignment is also acceptable, but be consistent).

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

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."
Hint

Part I: \(1 + e^{i2\pi/3} + e^{i4\pi/3} = 0\) (the sum of primitive \(n\)-th roots of unity is 0). Part II: Since slit 1 is distinguishable from the others, the amplitude for slit 1 is independently converted to a probability. Since slits 2 and 3 are indistinguishable, their amplitudes are added. \(P = |\phi_1|^2 + |\phi_2 + \phi_3|^2\). Part III: When the marker information is erased, all paths become indistinguishable again, and the rule of adding amplitudes is restored.

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