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Ch. 3 Solutions

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Basic

Medium

Advanced

Basic

B-1. Calculation of Interference Terms

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Problem:

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 \(2\mathrm{Re}(\phi_1^* \phi_2)\)

Solution strategy: Given the probability amplitudes \(\phi_1\) and \(\phi_2\), we sequentially calculate the squared moduli, the squared modulus of their sum, and the interference term.

1. \(P_1 = |\phi_1|^2\):

\[P_1 = \left|\frac{1}{\sqrt{2}} e^{i\pi/3}\right|^2 = \frac{1}{2} |e^{i\pi/3}|^2 = \frac{1}{2}\]

2. \(P_2 = |\phi_2|^2\):

\[P_2 = \left|\frac{1}{\sqrt{2}} e^{-i\pi/6}\right|^2 = \frac{1}{2} |e^{-i\pi/6}|^2 = \frac{1}{2}\]

3. \(P_{12} = |\phi_1 + \phi_2|^2\):

First, we find the phase difference:

\[\delta = \frac{\pi}{3} - \left(-\frac{\pi}{6}\right) = \frac{\pi}{3} + \frac{\pi}{6} = \frac{2\pi}{6} + \frac{\pi}{6} = \frac{\pi}{2}\]

Using the formula \(|\phi_1 + \phi_2|^2 = |\phi_1|^2 + |\phi_2|^2 + 2|\phi_1||\phi_2|\cos\delta\):

\[P_{12} = \frac{1}{2} + \frac{1}{2} + 2 \cdot \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} \cos\frac{\pi}{2} = 1 + 1 \cdot 0 = 1\]
\[\boxed{P_{12} = 1}\]

4. Interference term \(2\mathrm{Re}(\phi_1^* \phi_2)\):

\[\phi_1^* \phi_2 = \frac{1}{\sqrt{2}} e^{-i\pi/3} \cdot \frac{1}{\sqrt{2}} e^{-i\pi/6} = \frac{1}{2} e^{-i(\pi/3 + \pi/6)} = \frac{1}{2} e^{-i\pi/2}\]
\[2\mathrm{Re}(\phi_1^* \phi_2) = 2 \cdot \mathrm{Re}\left(\frac{1}{2} e^{-i\pi/2}\right) = 2 \cdot \frac{1}{2} \cos\left(-\frac{\pi}{2}\right) = \cos\left(-\frac{\pi}{2}\right) = 0\]
\[\boxed{2\mathrm{Re}(\phi_1^* \phi_2) = 0}\]

Verification: \(P_{12} = P_1 + P_2 + 2\mathrm{Re}(\phi_1^*\phi_2) = \frac{1}{2} + \frac{1}{2} + 0 = 1\) ✓. Since the phase difference is \(\pi/2\), the interference term vanishing is consistent with \(\cos(\pi/2) = 0\).

B-2. Phase Difference and Interference Intensity

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Problem:

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. \(\delta = 2\pi/3\)

Solution strategy: When \(I_1 = I_2 = I_0\), equation (3.2) becomes

\[I_{12} = I_0 + I_0 + 2\sqrt{I_0 \cdot I_0}\cos\delta = 2I_0(1 + \cos\delta)\]

1. \(\delta = 0\):

\[I_{12} = 2I_0(1 + \cos 0) = 2I_0(1 + 1) = \boxed{4I_0}\]

2. \(\delta = \pi/2\):

\[I_{12} = 2I_0(1 + \cos(\pi/2)) = 2I_0(1 + 0) = \boxed{2I_0}\]

3. \(\delta = \pi\):

\[I_{12} = 2I_0(1 + \cos\pi) = 2I_0(1 - 1) = \boxed{0}\]

4. \(\delta = 2\pi/3\):

\[I_{12} = 2I_0\left(1 + \cos\frac{2\pi}{3}\right) = 2I_0\left(1 - \frac{1}{2}\right) = 2I_0 \cdot \frac{1}{2} = \boxed{I_0}\]

Verification: At \(\delta = 0\) the maximum is \(4I_0\) (complete constructive interference), and at \(\delta = \pi\) the minimum is \(0\) (complete destructive interference). This is the typical result for equal-amplitude interference. At \(\delta = \pi/2\), \(2I_0 = I_1 + I_2\) (zero interference term) is also correct.

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

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Problem:

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 \(\beta\).

Solution strategy: Use \(|z|^2 = z^* z\) to expand \((\phi_1 + \phi_2)^*(\phi_1 + \phi_2)\).

\[|\phi_1 + \phi_2|^2 = (\phi_1 + \phi_2)^*(\phi_1 + \phi_2)\]
\[= \phi_1^*\phi_1 + \phi_1^*\phi_2 + \phi_2^*\phi_1 + \phi_2^*\phi_2\]

Computing each term: - \(\phi_1^*\phi_1 = (Ae^{-i\alpha})(Ae^{i\alpha}) = A^2\) - \(\phi_2^*\phi_2 = (Be^{-i\beta})(Be^{i\beta}) = B^2\) - \(\phi_1^*\phi_2 = (Ae^{-i\alpha})(Be^{i\beta}) = ABe^{i(\beta - \alpha)}\) - \(\phi_2^*\phi_1 = (Be^{-i\beta})(Ae^{i\alpha}) = ABe^{i(\alpha - \beta)}\)

Combining the cross terms:

\[\phi_1^*\phi_2 + \phi_2^*\phi_1 = AB\left(e^{i(\beta-\alpha)} + e^{-i(\beta-\alpha)}\right) = 2AB\cos(\alpha - \beta)\]

Therefore:

\[\boxed{|\phi_1 + \phi_2|^2 = A^2 + B^2 + 2AB\cos(\alpha - \beta)}\]

Verification: When \(A = B\), \(\alpha = \beta\), we have \(|2Ae^{i\alpha}|^2 = 4A^2\), and from the formula \(A^2 + A^2 + 2A^2\cos 0 = 4A^2\) ✓. When \(\alpha - \beta = \pi\), we get \(A^2 + B^2 - 2AB = (A-B)^2\), and \(|Ae^{i\alpha} + Be^{i(\alpha+\pi)}|^2 = |A - B|^2 = (A-B)^2\) ✓.

B-4. Conditions for Interference Maxima and Minima

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Problem:

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.

Solution strategy: The phase difference corresponding to the path difference \(\Delta = dx/L\) is \(\delta = 2\pi\Delta/\lambda = 2\pi dx/(\lambda L)\).

1. Maxima (constructive interference) condition:

Constructive interference occurs when \(\cos\delta = +1\), i.e., when \(\delta = 2n\pi\) (where \(n\) is an integer).

\[\frac{2\pi\Delta}{\lambda} = 2n\pi \implies \boxed{\Delta = n\lambda \quad (n = 0, \pm 1, \pm 2, \ldots)}\]

2. Minima (destructive interference) condition:

Destructive interference occurs when \(\cos\delta = -1\), i.e., when \(\delta = (2n+1)\pi\).

\[\frac{2\pi\Delta}{\lambda} = (2n+1)\pi \implies \boxed{\Delta = \left(n + \frac{1}{2}\right)\lambda \quad (n = 0, \pm 1, \pm 2, \ldots)}\]

3. Spacing between adjacent maxima \(\Delta x\):

The position of the \(n\)-th order maximum is given by \(\Delta = n\lambda\), so \(dx_n/L = n\lambda\), i.e., \(x_n = n\lambda L/d\).

The spacing between adjacent maxima is:

\[\boxed{\Delta x = x_{n+1} - x_n = \frac{\lambda L}{d}}\]

Verification: Dimensional analysis: \([\lambda L/d] = \mathrm{m} \cdot \mathrm{m} / \mathrm{m} = \mathrm{m}\) ✓. The fringe spacing increases with larger \(\lambda\) and larger \(L\), and decreases with larger \(d\). This is physically reasonable.

B-5. Concrete Calculation of Interference Fringes

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Problem:

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.

1. de Broglie wavelength \(\lambda\) of the electron:

Momentum of an electron accelerated through a voltage \(V\):

\[eV = \frac{p^2}{2m_e} \implies p = \sqrt{2m_e eV}\]
\[p = \sqrt{2 \times 9.11 \times 10^{-31} \times 1.60 \times 10^{-19} \times 150}\]
\[= \sqrt{2 \times 9.11 \times 1.60 \times 150 \times 10^{-50}}\]
\[= \sqrt{4373 \times 10^{-50}} = \sqrt{4.373 \times 10^{-47}}\]
\[= 6.61 \times 10^{-24}\,\mathrm{kg \cdot m/s}\]

de Broglie wavelength:

\[\lambda = \frac{h}{p} = \frac{6.63 \times 10^{-34}}{6.61 \times 10^{-24}} = 1.00 \times 10^{-10}\,\mathrm{m}\]
\[\boxed{\lambda \approx 1.0 \times 10^{-10}\,\mathrm{m} = 1.0\,\text{Å}}\]

2. Fringe spacing \(\Delta x\):

Using the result from D4:

\[\Delta x = \frac{\lambda L}{d} = \frac{1.0 \times 10^{-10} \times 0.50}{1.0 \times 10^{-6}}\]
\[= \frac{5.0 \times 10^{-11}}{1.0 \times 10^{-6}} = 5.0 \times 10^{-5}\,\mathrm{m}\]
\[\boxed{\Delta x = 50\,\mu\mathrm{m} = 0.050\,\mathrm{mm}}\]

Verification: The de Broglie wavelength of about 1 Å for electrons accelerated through 150 V is consistent with typical values in electron diffraction (similar values were obtained in the Davisson-Germer experiment). The fringe spacing of 50 μm is on a scale readily observable with an optical microscope, which is reasonable.

B-6. Normalization of a Probability Distribution

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Problem:

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 \(C\).

Solution strategy: Determine \(C\) by requiring \(\int P(x)\,dx = 1\) over one period \(-\frac{\lambda L}{2d} \le x \le \frac{\lambda L}{2d}\).

Substituting \(u = \frac{\pi d x}{\lambda L}\), we get \(du = \frac{\pi d}{\lambda L}dx\), i.e., \(dx = \frac{\lambda L}{\pi d}du\).

The range of \(x\), \(\left[-\frac{\lambda L}{2d}, \frac{\lambda L}{2d}\right]\), corresponds to the range of \(u\), \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\).

\[\int_{-\lambda L/(2d)}^{\lambda L/(2d)} C\cos^2\!\left(\frac{\pi d x}{\lambda L}\right)dx = C \cdot \frac{\lambda L}{\pi d} \int_{-\pi/2}^{\pi/2} \cos^2 u\,du\]

Using \(\cos^2 u = \frac{1}{2}(1 + \cos 2u)\):

\[\int_{-\pi/2}^{\pi/2} \cos^2 u\,du = \int_{-\pi/2}^{\pi/2} \frac{1}{2}(1 + \cos 2u)\,du = \frac{1}{2}\left[u + \frac{\sin 2u}{2}\right]_{-\pi/2}^{\pi/2}\]
\[= \frac{1}{2}\left[\left(\frac{\pi}{2} + \frac{\sin\pi}{2}\right) - \left(-\frac{\pi}{2} + \frac{\sin(-\pi)}{2}\right)\right] = \frac{1}{2}\left[\frac{\pi}{2} + \frac{\pi}{2}\right] = \frac{\pi}{2}\]

Therefore:

\[C \cdot \frac{\lambda L}{\pi d} \cdot \frac{\pi}{2} = 1 \implies C \cdot \frac{\lambda L}{2d} = 1\]
\[\boxed{C = \frac{2d}{\lambda L}}\]

Verification: The dimensions of \(C\) are \([1/\text{length}]\) (the dimensions of probability density), and \(d/(\lambda L) = \mathrm{m}/(\mathrm{m}\cdot\mathrm{m}) = \mathrm{m}^{-1}\) ✓. Also, the width of the integration interval is \(\lambda L/d\), and the average value of \(P(x)\) is \(C/2 = d/(\lambda L)\). Average value × interval width \(= d/(\lambda L) \times \lambda L/d = 1\) ✓.

B-7. Visibility of Interference

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Problem:

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 \(I_2\).

Solution strategy: From Eq. (3.2) \(I_{12} = I_1 + I_2 + 2\sqrt{I_1 I_2}\cos\delta\), find \(I_{\max}\) and \(I_{\min}\).

Maximum when \(\cos\delta = +1\):

\[I_{\max} = I_1 + I_2 + 2\sqrt{I_1 I_2}\]

Minimum when \(\cos\delta = -1\):

\[I_{\min} = I_1 + I_2 - 2\sqrt{I_1 I_2}\]

Calculating the visibility:

\[\mathcal{V} = \frac{I_{\max} - I_{\min}}{I_{\max} + I_{\min}} = \frac{(I_1 + I_2 + 2\sqrt{I_1 I_2}) - (I_1 + I_2 - 2\sqrt{I_1 I_2})}{(I_1 + I_2 + 2\sqrt{I_1 I_2}) + (I_1 + I_2 - 2\sqrt{I_1 I_2})}\]
\[= \frac{4\sqrt{I_1 I_2}}{2(I_1 + I_2)}\]
\[\boxed{\mathcal{V} = \frac{2\sqrt{I_1 I_2}}{I_1 + I_2}}\]

Verification: - When \(I_1 = I_2 = I_0\): \(\mathcal{V} = \frac{2\sqrt{I_0^2}}{2I_0} = \frac{2I_0}{2I_0} = 1\) ✓ (perfect visibility) - When \(I_1 \gg I_2\): \(\mathcal{V} \approx \frac{2\sqrt{I_1 I_2}}{I_1} = 2\sqrt{I_2/I_1} \to 0\) ✓ (fringes become hard to see when one source dominates) - That \(0 \le \mathcal{V} \le 1\) can be confirmed from the AM–GM inequality (\(I_1 + I_2 \ge 2\sqrt{I_1 I_2}\)) ✓

Medium

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

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Problem:

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.

1. Quantum Mechanical Probability Distribution \(P_{12}^{\mathrm{QM}}(x)\):

\[P_{12}^{\mathrm{QM}} = |\phi_1 + \phi_2|^2 = \left|\frac{1}{\sqrt{2}}e^{ikr_1} + \frac{1}{\sqrt{2}}e^{ikr_2}\right|^2\]
\[= \frac{1}{2} + \frac{1}{2} + 2 \cdot \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} \cos(k(r_1 - r_2))\]

Using the path difference approximation \(r_1 - r_2 \approx dx/L\), the phase difference is \(\delta = k \cdot dx/L = \frac{2\pi dx}{\lambda L}\).

\[\boxed{P_{12}^{\mathrm{QM}}(x) = 1 + \cos\left(\frac{kdx}{L}\right) = 1 + \cos\left(\frac{2\pi dx}{\lambda L}\right)}\]

2. Classical Particle Picture Probability Distribution \(P_{12}^{\mathrm{cl}}(x)\):

\[P_{12}^{\mathrm{cl}} = |\phi_1|^2 + |\phi_2|^2 = \frac{1}{2} + \frac{1}{2}\]
\[\boxed{P_{12}^{\mathrm{cl}}(x) = 1}\]

(Uniform regardless of position \(x\))

3. Discussion of the Interference Term:

\[P_{12}^{\mathrm{QM}} - P_{12}^{\mathrm{cl}} = \cos\left(\frac{kdx}{L}\right)\]

Range of \(x\) where it is positive (constructive interference):

\[\cos\left(\frac{kdx}{L}\right) > 0 \iff -\frac{\pi}{2} + 2n\pi < \frac{kdx}{L} < \frac{\pi}{2} + 2n\pi\]
\[\iff \left(n - \frac{1}{4}\right)\frac{\lambda L}{d} < x < \left(n + \frac{1}{4}\right)\frac{\lambda L}{d}\]

That is, it is positive in regions of width \(\lambda L/(2d)\) centered on each maximum position \(x_n = n\lambda L/d\).

Range of \(x\) where it is negative (destructive interference):

\[\cos\left(\frac{kdx}{L}\right) < 0 \iff \frac{\pi}{2} + 2n\pi < \frac{kdx}{L} < \frac{3\pi}{2} + 2n\pi\]
\[\iff \left(n + \frac{1}{4}\right)\frac{\lambda L}{d} < x < \left(n + \frac{3}{4}\right)\frac{\lambda L}{d}\]

That is, it is negative in the vicinity of each minimum position.

Physical meaning: The interference term averages to zero over one full period (the average of \(\cos\) is zero). Compared to the classical prediction, quantum mechanics makes electrons more likely to arrive at certain locations (constructive interference due to probability redistribution) and less likely at other locations (destructive interference). The total probability is conserved.

Verification: Integrating \(P_{12}^{\mathrm{QM}}\) over one period in \(x\) gives \(\int_0^{\lambda L/d}(1 + \cos(\cdots))dx = \lambda L/d\). The integral of \(P_{12}^{\mathrm{cl}}\) over the same interval is also \(\lambda L/d\). The total probabilities are equal ✓.

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

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Problem:

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 \(|\phi_1 + \phi_2|^2\).

1. Complete Path Identification (\(b = 0\), \(b' = 0\), \(|a|^2 = |a'|^2 = 1\)):

Substituting into equation (3.9):

\[P_{12}' = |a\phi_1 + b\phi_2|^2 + |b'\phi_1 + a'\phi_2|^2\]

Substituting \(b = 0\), \(b' = 0\):

\[P_{12}' = |a\phi_1|^2 + |a'\phi_2|^2 = |a|^2|\phi_1|^2 + |a'|^2|\phi_2|^2\]

Substituting \(|a|^2 = |a'|^2 = 1\):

\[P_{12}' = |\phi_1|^2 + |\phi_2|^2 = P_1 + P_2\]
\[\boxed{P_{12}' = P_1 + P_2}\]

The interference term has completely vanished. \(\blacksquare\)

Physical interpretation: \(b = 0\) means "the probability of a photon being detected at \(D_1\) when the electron passed through hole 2 is zero," so the electron's path can be completely identified from the photon's detection position. In this case, the two paths correspond to distinguishable final states, so we add probabilities.

2. No Path Identification Possible (\(a = a' = b = b' = 1/\sqrt{2}\)):

Substituting into equation (3.9):

\[P_{12}' = \left|\frac{1}{\sqrt{2}}\phi_1 + \frac{1}{\sqrt{2}}\phi_2\right|^2 + \left|\frac{1}{\sqrt{2}}\phi_1 + \frac{1}{\sqrt{2}}\phi_2\right|^2\]
\[= \frac{1}{2}|\phi_1 + \phi_2|^2 + \frac{1}{2}|\phi_1 + \phi_2|^2 = |\phi_1 + \phi_2|^2\]
\[\boxed{P_{12}' = |\phi_1 + \phi_2|^2}\]

The interference pattern is completely recovered. \(\blacksquare\)

Physical interpretation: When \(a = b\) and \(a' = b'\), regardless of whether the photon enters \(D_1\) or \(D_2\), no information is obtained about which hole the electron passed through. Since the paths are indistinguishable, the rule of adding amplitudes before squaring applies.

Verification: Confirming that the two terms in (2) are identical. First term: \(|a\phi_1 + b\phi_2|^2 = |\frac{1}{\sqrt{2}}(\phi_1 + \phi_2)|^2 = \frac{1}{2}|\phi_1 + \phi_2|^2\). Second term: \(|b'\phi_1 + a'\phi_2|^2 = |\frac{1}{\sqrt{2}}(\phi_1 + \phi_2)|^2 = \frac{1}{2}|\phi_1 + \phi_2|^2\). The sum is \(|\phi_1 + \phi_2|^2\) ✓.

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

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Problem:

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 \(4V\)?

1. de Broglie Wavelength:

The kinetic energy of an electron accelerated through a potential difference \(V\):

\[\frac{p^2}{2m_e} = eV \implies p = \sqrt{2m_e eV}\]

de Broglie wavelength:

\[\boxed{\lambda = \frac{h}{p} = \frac{h}{\sqrt{2m_e eV}}}\]

2. Position of the \(n\)-th Order Interference Maximum:

The maximum condition is \(\Delta = n\lambda\), i.e., \(dx_n/L = n\lambda\):

\[\boxed{x_n = \frac{n\lambda L}{d} = \frac{nhL}{d\sqrt{2m_e eV}}}\]

3. When the Accelerating Voltage is Changed to \(4V\):

New wavelength:

\[\lambda' = \frac{h}{\sqrt{2m_e e(4V)}} = \frac{h}{2\sqrt{2m_e eV}} = \frac{\lambda}{2}\]

New fringe spacing:

\[\Delta x' = \frac{\lambda' L}{d} = \frac{\lambda L}{2d} = \frac{\Delta x}{2}\]
\[\boxed{\text{The fringe spacing becomes } \frac{1}{2} \text{ of the original (the fringes become finer)}}\]

Verification: Increasing the accelerating voltage increases the electron's momentum, shortening the de Broglie wavelength. A shorter wavelength produces finer interference fringes, which is physically reasonable. Dimensional analysis: the dimensions of \(\lambda = h/\sqrt{2m_e eV}\) are \(\mathrm{J\cdot s}/\sqrt{\mathrm{kg \cdot C \cdot V}} = \mathrm{J\cdot s}/\sqrt{\mathrm{kg \cdot J}} = \mathrm{J\cdot s}/\sqrt{\mathrm{kg \cdot kg \cdot m^2/s^2}} = \mathrm{m}\) ✓.

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

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Problem:

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.

1. Breakdown of the Law of Total Probability:

Assuming symmetric slits, \(P(A_1) = P(A_2) = 1/2\). The classical law of total probability states:

\[P(x) = P(x|A_1)P(A_1) + P(x|A_2)P(A_2)\]

Here \(P(x|A_1)\) is "the probability of arriving at \(x\) given passage through hole 1" = \(P_1(x)\) (the appropriately normalized distribution when only hole 1 is open), and similarly \(P(x|A_2) = P_2(x)\). Therefore the law of total probability predicts:

\[P_{12}^{\mathrm{cl}}(x) = \frac{1}{2}P_1(x) + \frac{1}{2}P_2(x)\]

On the other hand, the quantum mechanical probability distribution observed experimentally is:

\[P_{12}^{\mathrm{QM}}(x) = |\phi_1 + \phi_2|^2 = P_1 + P_2 + 2\mathrm{Re}(\phi_1^*\phi_2)\]

(where the normalization is matched such that \(|\phi_1|^2 = P_1/1 = P_1\), etc.)

Since the interference term \(2\mathrm{Re}(\phi_1^*\phi_2) \neq 0\) exists:

\[\boxed{P_{12}^{\mathrm{QM}}(x) \neq \frac{1}{2}P_1(x) + \frac{1}{2}P_2(x) = P_{12}^{\mathrm{cl}}(x)}\]

The law of total probability fails. \(\blacksquare\)

2. Cause of the Breakdown:

The prerequisites for the law of total probability \(P(x) = \sum_i P(x|A_i)P(A_i)\) to hold are:

  1. Events \(A_1\) and \(A_2\) are mutually exclusive (they cannot occur simultaneously)
  2. Events \(A_1\) and \(A_2\) are exhaustive (one of them necessarily occurs)
  3. Each event \(A_i\) exists as a definite fact

When interference fringes are observed in the double-slit experiment, it is condition 3 that breaks down.

The events "the electron passed through hole 1" and "the electron passed through hole 2" do not exist as definite facts unless they are observed. In a situation where the path is not determined (does not have reality), the decomposition into conditional probabilities premised on "which slit it passed through" itself becomes invalid.

This corresponds to the argument presented in the text: "If we assume proposition A (each electron passes through one slit or the other), then \(P_{12} = P_1 + P_2\) follows, which contradicts experimental results." This means that classical realism—"physical quantities possess definite values even without measurement"—collapses.

Verification: When "which slit it passed through" is actually observed (the experiment with the light source turned on), the path becomes a definite fact, so the law of total probability is restored and \(P_{12}' = P_1 + P_2\). This is consistent with equation (3.6) in the text ✓.

Advanced

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

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Problem:

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).

1. Probability Distribution After Tracing Out the Marker:

State of the total system:

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

The probability distribution of the electron on the screen is obtained by tracing out the marker degrees of freedom. Using an arbitrary complete orthonormal basis \(\{|e_k\rangle\}\) of the marker space:

\[P(x) = \sum_k |\langle e_k|\Psi\rangle|^2\]

However, we compute this more directly using the density matrix approach. The probability of the electron arriving at position \(x\) is:

\[P(x) = \langle\Psi|\Psi\rangle \text{ component at } x\]

More precisely, \(P(x)\) is given as a trace over the marker state space:

\[P(x) = \mathrm{Tr}_m\left[|\Psi(x)\rangle\langle\Psi(x)|\right]\]

where \(|\Psi(x)\rangle\) is the marker-space vector \(\frac{1}{\sqrt{2}}(\phi_1(x)|m_1\rangle + \phi_2(x)|m_2\rangle)\).

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

Substituting \(\langle m_1|m_1\rangle = \langle m_2|m_2\rangle = 1\), \(\langle m_1|m_2\rangle = \gamma\) (real), \(\langle m_2|m_1\rangle = \gamma^*= \gamma\):

\[P(x) = \frac{1}{2}|\phi_1|^2 + \frac{1}{2}|\phi_2|^2 + \frac{1}{2}\gamma(\phi_1^*\phi_2 + \phi_2^*\phi_1)\]
\[\boxed{P(x) = \frac{1}{2}|\phi_1|^2 + \frac{1}{2}|\phi_2|^2 + \gamma\,\mathrm{Re}(\phi_1^*\phi_2)}\]

The coefficient of the interference term is suppressed by \(\gamma\).

2. Expressing the Visibility \(\mathcal{V}\) in Terms of \(\gamma\):

Let \(|\phi_1| = |\phi_2| = A\). Writing \(\phi_1^*\phi_2 = A^2 e^{-i\delta}\):

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

Maximum (\(\cos\delta = 1\)): \(P_{\max} = A^2(1 + \gamma)\)

Minimum (\(\cos\delta = -1\)): \(P_{\min} = A^2(1 - \gamma)\)

\[\mathcal{V} = \frac{P_{\max} - P_{\min}}{P_{\max} + P_{\min}} = \frac{A^2(1+\gamma) - A^2(1-\gamma)}{A^2(1+\gamma) + A^2(1-\gamma)} = \frac{2\gamma}{2}\]
\[\boxed{\mathcal{V} = \gamma}\]

3. Verification of the Limiting Cases:

When \(\gamma = 1\) (\(|m_1\rangle = |m_2\rangle\), the marker carries no path information):

\[P(x) = \frac{1}{2}|\phi_1|^2 + \frac{1}{2}|\phi_2|^2 + \mathrm{Re}(\phi_1^*\phi_2) = \frac{1}{2}|\phi_1 + \phi_2|^2\]

(The last equality can be verified from \(|\phi_1+\phi_2|^2 = |\phi_1|^2 + |\phi_2|^2 + 2\mathrm{Re}(\phi_1^*\phi_2)\).)

The complete interference pattern is recovered. \(\mathcal{V} = 1\). This is consistent with the discussion in the main text that "interference is restored when no path identification is possible." ✓

When \(\gamma = 0\) (\(\langle m_1|m_2\rangle = 0\), perfectly distinguishable):

\[P(x) = \frac{1}{2}|\phi_1|^2 + \frac{1}{2}|\phi_2|^2 = \frac{1}{2}(P_1 + P_2)\]

The interference term completely vanishes. \(\mathcal{V} = 0\). This is consistent with the discussion in the main text that "interference disappears with complete path identification." ✓

4. Englert's Complementarity Inequality \(\mathcal{V}^2 + \mathcal{D}^2 = 1\):

From \(\mathcal{V} = \gamma\) and \(\mathcal{D} = \sqrt{1 - \gamma^2}\):

\[\mathcal{V}^2 + \mathcal{D}^2 = \gamma^2 + (1 - \gamma^2) = 1\]
\[\boxed{\mathcal{V}^2 + \mathcal{D}^2 = 1} \quad \blacksquare\]

Physical meaning: The interference visibility (a measure of wave-like behavior) and the path distinguishability (a measure of particle-like behavior) are in a complementary relationship. Obtaining one completely means the other is entirely lost, and in cases of partial information, the two are in a trade-off relationship. This is a quantitative expression of Bohr's complementarity principle.

Verification: - \(\gamma = 1\): \(\mathcal{V} = 1\), \(\mathcal{D} = 0\), \(1 + 0 = 1\) ✓ - \(\gamma = 0\): \(\mathcal{V} = 0\), \(\mathcal{D} = 1\), \(0 + 1 = 1\) ✓ - \(\gamma = 1/\sqrt{2}\): \(\mathcal{V} = 1/\sqrt{2}\), \(\mathcal{D} = 1/\sqrt{2}\), \(1/2 + 1/2 = 1\) ✓

A-2. Analysis of the Delayed Choice Experiment

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Problem:

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).

1. Output Amplitudes at a 50:50 Beam Splitter:

Setting \(t = 1/\sqrt{2}\), \(r = i/\sqrt{2}\):

\[\phi_A = t\phi_1 + r\phi_2 = \frac{1}{\sqrt{2}}\phi_1 + \frac{i}{\sqrt{2}}\phi_2\]
\[\phi_B = r\phi_1 + t\phi_2 = \frac{i}{\sqrt{2}}\phi_1 + \frac{1}{\sqrt{2}}\phi_2\]

Let \(\phi_1 = |\phi_1|e^{i\alpha}\), \(\phi_2 = |\phi_2|e^{i\beta}\), \(\delta = \alpha - \beta\).

Calculation of \(|\phi_A|^2\):

\[|\phi_A|^2 = \left|\frac{1}{\sqrt{2}}\phi_1 + \frac{i}{\sqrt{2}}\phi_2\right|^2 = \frac{1}{2}|\phi_1|^2 + \frac{1}{2}|\phi_2|^2 + \frac{1}{2}((-i)\phi_1^*\cdot\phi_2 + i\phi_2^*\cdot\phi_1) \cdot \frac{1}{1}\]

Carefully computing the cross terms:

\[\frac{1}{\sqrt{2}}\phi_1^* \cdot \frac{i}{\sqrt{2}}\phi_2 + \frac{-i}{\sqrt{2}}\phi_2^* \cdot \frac{1}{\sqrt{2}}\phi_1\]
\[= \frac{i}{2}\phi_1^*\phi_2 + \frac{-i}{2}\phi_2^*\phi_1 = \frac{i}{2}(\phi_1^*\phi_2 - \phi_2^*\phi_1)\]

Since \(\phi_1^*\phi_2 = |\phi_1||\phi_2|e^{i(\beta-\alpha)} = |\phi_1||\phi_2|e^{-i\delta}\):

\[\phi_1^*\phi_2 - \phi_2^*\phi_1 = |\phi_1||\phi_2|(e^{-i\delta} - e^{i\delta}) = -2i|\phi_1||\phi_2|\sin\delta\]

Therefore:

\[\text{cross terms} = \frac{i}{2} \cdot (-2i|\phi_1||\phi_2|\sin\delta) = |\phi_1||\phi_2|\sin\delta\]
\[\boxed{|\phi_A|^2 = \frac{1}{2}|\phi_1|^2 + \frac{1}{2}|\phi_2|^2 + |\phi_1||\phi_2|\sin\delta}\]

Calculation of \(|\phi_B|^2\):

\[\phi_B = \frac{i}{\sqrt{2}}\phi_1 + \frac{1}{\sqrt{2}}\phi_2\]

Cross terms:

\[\frac{-i}{\sqrt{2}}\phi_1^* \cdot \frac{1}{\sqrt{2}}\phi_2 + \frac{1}{\sqrt{2}}\phi_2^* \cdot \frac{i}{\sqrt{2}}\phi_1 = \frac{-i}{2}\phi_1^*\phi_2 + \frac{i}{2}\phi_2^*\phi_1\]
\[= \frac{-i}{2}(\phi_1^*\phi_2 - \phi_2^*\phi_1) = \frac{-i}{2}(-2i|\phi_1||\phi_2|\sin\delta) = -|\phi_1||\phi_2|\sin\delta\]
\[\boxed{|\phi_B|^2 = \frac{1}{2}|\phi_1|^2 + \frac{1}{2}|\phi_2|^2 - |\phi_1||\phi_2|\sin\delta}\]

2. When \(|\phi_1| = |\phi_2| = 1/\sqrt{2}\):

\[|\phi_A|^2 = \frac{1}{4} + \frac{1}{4} + \frac{1}{2}\sin\delta = \frac{1}{2}(1 + \sin\delta)\]
\[|\phi_B|^2 = \frac{1}{4} + \frac{1}{4} - \frac{1}{2}\sin\delta = \frac{1}{2}(1 - \sin\delta)\]
\[\boxed{|\phi_A|^2 = \frac{1}{2}(1 + \sin\delta), \quad |\phi_B|^2 = \frac{1}{2}(1 - \sin\delta)}\]

Both \(|\phi_A|^2\) and \(|\phi_B|^2\) oscillate as functions of the phase difference \(\delta\). Since \(\delta\) depends on the position on the screen (or equivalently, the optical path difference between the two paths), interference is observed.

Specifically: - When \(\delta = \pi/2\): \(|\phi_A|^2 = 1\), \(|\phi_B|^2 = 0\) (all electrons exit at port A) - When \(\delta = -\pi/2\): \(|\phi_A|^2 = 0\), \(|\phi_B|^2 = 1\) (all electrons exit at port B) - When \(\delta = 0\): \(|\phi_A|^2 = |\phi_B|^2 = 1/2\) (equal splitting)

The fact that the detection probability at the output ports varies depending on the phase difference is clear evidence of interference.

Verification: \(|\phi_A|^2 + |\phi_B|^2 = \frac{1}{2}(1+\sin\delta) + \frac{1}{2}(1-\sin\delta) = 1\) ✓ (conservation of probability).

3. When the Beam Splitter Is Removed:

If the beam splitter is removed, then \(\phi_A = \phi_1\), \(\phi_B = \phi_2\).

\[|\phi_A|^2 = |\phi_1|^2 = \frac{1}{2}, \quad |\phi_B|^2 = |\phi_2|^2 = \frac{1}{2}\]

If an electron is detected at detector A, we can conclude it "traveled path 1"; if detected at detector B, it "traveled path 2." Which-path information is obtained.

However, neither \(|\phi_A|^2\) nor \(|\phi_B|^2\) depends on the phase difference \(\delta\)—both are constants (\(1/2\)). No interference is observed.

This is a concrete example of the rule from the main text: "When processes are distinguishable in principle → add probabilities (no interference)."

4. Contradiction with Classical Realism:

Assumption of classical realism: At the moment the electron passes through the slits, it possesses a definite path (hole 1 or hole 2). This path does not change regardless of what is done afterward (past facts are settled).

Argument for the contradiction:

If the path is determined at the moment the electron passes through the slit:

  • Two subensembles exist: "electrons that passed through hole 1" and "electrons that passed through hole 2"
  • Whether a beam splitter is inserted later cannot change the path already taken by the electron
  • Therefore, the detection pattern should be the same (\(P_1 + P_2\)) regardless of whether the beam splitter is present or not

However, the experimental results show: - When the beam splitter is inserted, an interference pattern (\(\sin\delta\) dependence) appears - When the beam splitter is removed, interference disappears and which-path information is obtained

Even if the decision to insert/remove the beam splitter is made after the electron has passed through the slits (delayed choice), the results remain unchanged. This contradicts the assumption that "the path was determined at the moment of passage."

Conclusion: Until the final configuration of the measuring apparatus is established, "which path the electron took" does not exist as a determined fact. Wheeler's delayed-choice experiment suggests that even past events do not possess definite reality until the measurement is completed. This constitutes a fundamental refutation of classical realism (the view that physical quantities possess definite values independent of measurement).

Verification: This conclusion does not violate causality, because the choice of beam splitter cannot "send a signal" to the past. The detection position of each individual electron is random, and the interference pattern emerges only upon accumulating statistics over many electrons. Therefore, it is impossible to use the results of delayed choice to achieve superluminal communication ✓.