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Ch. 3 Problems

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Table of Contents

Basic

Medium

Advanced

Basic

B-1. Calculation of Interference Terms

Two probability amplitudes are given as \(\phi_1 = \frac{1}{\sqrt{2}} e^{i\pi/3}\) and \(\phi_2 = \frac{1}{\sqrt{2}} e^{-i\pi/6}\). Calculate the following.

  1. \(P_1 = |\phi_1|^2\)
  2. \(P_2 = |\phi_2|^2\)
  3. \(P_{12} = |\phi_1 + \phi_2|^2\)
  4. The value of the interference term \(2\mathrm{Re}(\phi_1^* \phi_2)\)
Hint

When calculating \(\phi_1^* \phi_2\), it is convenient to first find the phase difference \(\delta = \theta_1 - \theta_2\) and then use \(2|\phi_1||\phi_2|\cos\delta\). Note that \(\delta = \pi/3 - (-\pi/6) = \pi/2\).

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B-2. Phase Difference and Interference Intensity

In equation (3.2), let \(I_1 = I_2 = I_0\). For each of the following phase differences \(\delta\), find \(I_{12}\) as a multiple of \(I_0\).

  1. \(\delta = 0\)
  2. \(\delta = \pi/2\)
  3. \(\delta = \pi\)
  4. \(\delta = 2\pi/3\)
Hint

When \(I_1 = I_2 = I_0\), equation (3.2) simplifies to \(I_{12} = 2I_0(1 + \cos\delta)\).

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B-3. Expansion of the absolute value squared of complex amplitudes

Let \(\phi_1 = A e^{i\alpha}\) and \(\phi_2 = B e^{i\beta}\) (where \(A, B\) are positive real numbers). Expand \(|\phi_1 + \phi_2|^2\) and express it in terms of \(A\), \(B\), \(\alpha\), and \(\beta\).

Hint

Use \(|z|^2 = z^* z\) and expand \((\phi_1 + \phi_2)^*(\phi_1 + \phi_2)\) into 4 terms. The cross terms combine into \(\cos(\alpha - \beta)\).

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B-4. Conditions for Interference Maxima and Minima

Let the slit separation be \(d\), the distance from the slits to the screen be \(L\) (\(L \gg d\)), and the de Broglie wavelength of the electron be \(\lambda\). The path difference between the two paths at position \(x\) on the screen (measured from the center) can be approximately written as \(\Delta = dx/L\).

  1. Write the condition for constructive interference (maxima) in terms of \(\Delta\) and \(\lambda\).
  2. Write the condition for destructive interference (minima) in terms of \(\Delta\) and \(\lambda\).
  3. Find the spacing \(\Delta x\) between adjacent maxima.
Hint

The phase difference is \(\delta = 2\pi\Delta/\lambda\). Maxima correspond to \(\delta = 2n\pi\) (\(n\) is an integer), and minima correspond to \(\delta = (2n+1)\pi\).

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B-5. Concrete Calculation of Interference Fringes

An electron is accelerated through a voltage \(V = 150\,\mathrm{V}\). The slit separation is \(d = 1.0\,\mu\mathrm{m}\), and the distance to the screen is \(L = 0.50\,\mathrm{m}\).

  1. Find the de Broglie wavelength \(\lambda\) of the electron. (\(m_e = 9.11 \times 10^{-31}\,\mathrm{kg}\), \(e = 1.60 \times 10^{-19}\,\mathrm{C}\), \(h = 6.63 \times 10^{-34}\,\mathrm{J \cdot s}\))
  2. Find the fringe spacing \(\Delta x\) of the interference pattern on the screen.
Hint

The kinetic energy of an electron accelerated through a voltage \(V\) is \(eV = p^2/(2m_e)\), so \(p = \sqrt{2m_e eV}\). Use \(\lambda = h/p\). Apply the result from D4: \(\Delta x = \lambda L / d\).

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B-6. Normalization of a Probability Distribution

Suppose the arrival probability distribution of electrons is given by \(P(x) = C\cos^2\!\left(\frac{\pi d x}{\lambda L}\right)\) (consider only one period: \(-\frac{\lambda L}{2d} \le x \le \frac{\lambda L}{2d}\)). Find the normalization constant \(C\).

Hint

Use \(\cos^2\theta = \frac{1}{2}(1 + \cos 2\theta)\) to evaluate the integral. Normalize over one period.

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B-7. Visibility of Interference

The visibility of an interference pattern is defined as:

\[\mathcal{V} = \frac{I_{\max} - I_{\min}}{I_{\max} + I_{\min}}\]

For the general case where \(I_1 \neq I_2\) in Eq. (3.2), express \(\mathcal{V}\) in terms of \(I_1\) and \(I_2\).

Hint

\(I_{\max}\) is \(I_{12}\) when \(\cos\delta = +1\), and \(I_{\min}\) is \(I_{12}\) when \(\cos\delta = -1\).

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Medium

M-1. Quantitative Comparison of Adding Probabilities vs Adding Amplitudes

In a double-slit experiment, suppose the path difference from the two slits to a point \(x\) on the screen is \(\Delta(x) = dx/L\). Let the probability amplitudes be

\[\phi_1(x) = \frac{1}{\sqrt{2}} e^{ik r_1(x)}, \quad \phi_2(x) = \frac{1}{\sqrt{2}} e^{ik r_2(x)}\]

where \(k = 2\pi/\lambda\), and \(r_1\), \(r_2\) are the distances from each slit, respectively.

  1. Express the quantum mechanical probability distribution \(P_{12}^{\mathrm{QM}}(x) = |\phi_1 + \phi_2|^2\) in terms of \(k\), \(d\), \(x\), and \(L\).
  2. Find the classical particle probability distribution \(P_{12}^{\mathrm{cl}}(x) = |\phi_1|^2 + |\phi_2|^2\).
  3. Find the interference term \(P_{12}^{\mathrm{QM}} - P_{12}^{\mathrm{cl}}\), and discuss the ranges of \(x\) where it is positive and where it is negative.
Hint

Using the approximation \(r_1 - r_2 \approx dx/L\), the phase difference becomes \(\delta = k \cdot dx/L\). Since \(|\phi_1| = |\phi_2| = 1/\sqrt{2}\), simplify \(P_{12}^{\mathrm{QM}} = \frac{1}{2} + \frac{1}{2} + 2 \cdot \frac{1}{2}\cos\delta\).

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M-2. "Which-Path" Information and the Disappearance of the Interference Term

Based on equation (3.9) in the text, show the following.

When path identification by the light source is complete (\(b = 0\), \(b' = 0\), \(|a|^2 = |a'|^2 = 1\)):

  1. Show that equation (3.9) reduces to \(P_{12}' = |\phi_1|^2 + |\phi_2|^2 = P_1 + P_2\).
  2. Conversely, show that when no path identification is possible (\(a = a' = b = b' = 1/\sqrt{2}\)), equation (3.9) reduces to \(|\phi_1 + \phi_2|^2\).
Hint

For (2), expand the two terms in equation (3.9) separately and check how the cross terms combine. You can use \(|c\phi_1 + c\phi_2|^2 = |c|^2|\phi_1 + \phi_2|^2\).

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M-3. Relationship Between Path Difference and de Broglie Wavelength

An electron (mass \(m_e\), charge \(e\)) accelerated through a potential difference \(V\) passes through a double slit with slit spacing \(d\) and screen distance \(L\).

  1. Express the de Broglie wavelength \(\lambda\) of the electron in terms of \(V\), \(m_e\), \(e\), and \(h\).
  2. Find the position \(x_n\) of the \(n\)-th order interference maximum on the screen.
  3. How does the fringe spacing change when the accelerating voltage is changed from \(V\) to \(4V\)?
Hint

\(\lambda = h/\sqrt{2m_e eV}\). When \(V\) is changed to \(4V\), \(\lambda\) becomes \(1/2\) times its original value.

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M-4. Understanding the Disappearance of Interference through "Conditional Decomposition of Probability"

Let \(A_1\) be the event that an electron passes through slit 1, and \(A_2\) be the event that it passes through slit 2. In classical probability theory, the law of total probability holds:

\[P(x) = P(x|A_1)P(A_1) + P(x|A_2)P(A_2)\]
  1. Show that when interference fringes are observed in the double-slit experiment, this law of total probability does not hold, using \(P_1\), \(P_2\), and \(P_{12}\).
  2. Based on the discussion in the text, explain what property \(A_1\) and \(A_2\) have lost that corresponds to this breakdown.
Hint

If we set \(P(A_1) = P(A_2) = 1/2\) (symmetric slits), the law of total probability predicts \(P_{12} = \frac{1}{2}P_1 + \frac{1}{2}P_2\). Compare this with the experimental result \(P_{12} = |\phi_1 + \phi_2|^2\). The cause of the breakdown lies in the assumption that "\(A_1\) and \(A_2\) exist as definite events."

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Advanced

A-1. Partial Which-Path Information and Interference Visibility

In a double-slit experiment, a "which-path marker" is placed behind each slit. When an electron passes through slit \(j\) (\(j = 1, 2\)), the marker's state becomes \(|m_j\rangle\). The state of the total system is:

\[|\Psi\rangle = \frac{1}{\sqrt{2}}\left(\phi_1(x)|m_1\rangle + \phi_2(x)|m_2\rangle\right)\]

Here \(|m_1\rangle\) and \(|m_2\rangle\) are not necessarily orthonormal in general, and have the inner product \(\langle m_1 | m_2 \rangle = \gamma\) (a real number with \(0 \le \gamma \le 1\)).

  1. Trace out (ignore) the marker states and find the probability distribution \(P(x)\) of the electron on the screen. Express the result explicitly in terms of \(|\phi_1|^2\), \(|\phi_2|^2\), and \(\mathrm{Re}(\phi_1^*\phi_2)\).
  2. In the case \(|\phi_1| = |\phi_2|\), express the interference fringe visibility \(\mathcal{V}\) in terms of \(\gamma\).
  3. Verify that the results are consistent with the discussion in the main text in the limits \(\gamma = 1\) (the marker carries no path information at all) and \(\gamma = 0\) (completely distinguishable).
  4. Defining the path distinguishability as \(\mathcal{D} = \sqrt{1 - \gamma^2}\), show that \(\mathcal{V}^2 + \mathcal{D}^2 = 1\) holds (the equality condition of Englert's complementarity inequality).
Hint

Tracing out the marker means not computing \(P(x) = \langle m_1|\rho_m|m_1\rangle |\phi_1|^2 + \cdots\), but rather computing \(P(x) = \sum_k |\langle e_k|\Psi\rangle|^2\) (where \(|e_k\rangle\) is any complete set in the marker space). More simply, show that \(P(x) = \frac{1}{2}|\phi_1|^2 + \frac{1}{2}|\phi_2|^2 + \mathrm{Re}(\gamma \cdot \phi_1^* \phi_2)\). The key point is that the coefficient of the interference term becomes \(\gamma\).

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A-2. Analysis of the Delayed Choice Experiment

Consider Wheeler's delayed choice experiment. After an electron passes through the slits, one chooses whether to observe interference or to observe which-path information.

Concrete setup: A beam splitter that can be inserted or removed is placed behind the double slit.

  • When the beam splitter is inserted: The amplitudes from the two paths are superposed, and interference is observed from the intensity at the output ports.
  • When the beam splitter is removed: Each path reaches a separate detector directly, and which-path information is obtained.

Let the transmittance of the beam splitter be \(t\) and the reflectance be \(r\) (\(|t|^2 + |r|^2 = 1\)). For incident amplitudes \(\phi_1\), \(\phi_2\), the amplitudes at output ports \(A\), \(B\) are:

\[\phi_A = t\phi_1 + r\phi_2, \quad \phi_B = r\phi_1 + t\phi_2\]
  1. For a 50:50 beam splitter (\(t = r = 1/\sqrt{2}\), but with a \(\pi/2\) phase shift upon reflection so that \(r = i/\sqrt{2}\), \(t = 1/\sqrt{2}\)), express \(|\phi_A|^2\) and \(|\phi_B|^2\) in terms of \(|\phi_1|\), \(|\phi_2|\), and the phase difference \(\delta = \arg(\phi_1) - \arg(\phi_2)\).
  2. When \(|\phi_1| = |\phi_2| = 1/\sqrt{2}\), find \(|\phi_A|^2\) and \(|\phi_B|^2\) as functions of \(\delta\), and show that interference is observed.
  3. Explain that when the beam splitter is removed (\(\phi_A = \phi_1\), \(\phi_B = \phi_2\)), which-path information is obtained at the cost of interference disappearing.
  4. Discuss how the fact that "the choice is made after the electron has passed through the slits" contradicts classical realism (the view that the electron has a definite path determined at the moment of passage).
Hint

(1) Expand \(|\phi_A|^2\) with \(\phi_A = \frac{1}{\sqrt{2}}\phi_1 + \frac{i}{\sqrt{2}}\phi_2\). (4) Use the argument that if the electron's path were determined at the moment of passage, then whether or not the beam splitter is inserted afterward should not change the result (since the path was already determined in the past).

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