Appendix A Solutions¶
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Table of Contents
Basic
- B-1. Multiplication of Complex Numbers
- B-2. Division of Complex Numbers
- B-3. Absolute Value and Complex Conjugate
- B-4. Conversion to Polar Form
- B-5. Powers of \(i\)
- B-6. Rules for Complex Conjugation
- B-7. Calculating Terms of a Maclaurin Expansion
- B-8. Calculating \(e^{i\theta}\)
- B-9. Polar Form and Rewriting with Euler's Formula
- B-10. Calculating \(|e^{i\theta}|\)
Medium
- M-1. General Proof of the Product Rule for Complex Conjugates
- M-2. Derivation of de Moivre's Theorem
- M-3. Derivation of the Division Formula in Polar Form
- M-4. Exponential Representations of \(\cos\theta\) and \(\sin\theta\)
- M-5. Conditions for a Complex Number to Be Real
Advanced
Basic¶
B-1. Multiplication of Complex Numbers¶
Problem:
Express the following products in the form \(a + bi\).
(a) \((2 + 3i)(4 - i)\)
(b) \((1 + i)^3\)
(c) \((-2 + i)(3 + 2i)\)
(a) \((2 + 3i)(4 - i)\):
Strategy: Expand normally and substitute \(i^2 = -1\).
Verification: Check using the product of absolute values. \(|2+3i|^2 = 4+9 = 13\), \(|4-i|^2 = 16+1 = 17\). The squared absolute value of the product is \(13 \times 17 = 221\). For the result, \(|11+10i|^2 = 121 + 100 = 221\). ✓
(b) \((1 + i)^3\):
First compute \((1+i)^2\).
Then multiply by \((1+i)\).
Verification: \(|1+i|^2 = 2\) so \(|1+i|^6 = 8\). \(|-2+2i|^2 = 4+4 = 8\). ✓
(c) \((-2 + i)(3 + 2i)\):
Verification: \(|-2+i|^2 = 4+1=5\), \(|3+2i|^2 = 9+4=13\). Product \(= 65\). \(|-8-i|^2 = 64+1=65\). ✓
B-2. Division of Complex Numbers¶
Problem:
Simplify the following quotients into the form \(a + bi\).
(a) \(\dfrac{2 + i}{1 + 3i}\)
(b) \(\dfrac{5}{2 - i}\)
(c) \(\dfrac{1 + 2i}{3 - 4i}\)
(a) \(\dfrac{2 + i}{1 + 3i}\):
Strategy: Multiply the numerator and denominator by the complex conjugate of the denominator, \(1 - 3i\), to make the denominator real.
Denominator: \((1+3i)(1-3i) = 1 + 9 = 10\)
Numerator: \((2+i)(1-3i) = 2 - 6i + i - 3i^2 = 2 - 5i + 3 = 5 - 5i\)
Verification: \(\left(\frac{1}{2} - \frac{1}{2}i\right)(1+3i) = \frac{1}{2} + \frac{3}{2}i - \frac{1}{2}i - \frac{3}{2}i^2 = \frac{1}{2} + i + \frac{3}{2} = 2 + i\). ✓
(b) \(\dfrac{5}{2 - i}\):
Verification: \((2+i)(2-i) = 4+1 = 5\). ✓
(c) \(\dfrac{1 + 2i}{3 - 4i}\):
Numerator: \((1+2i)(3+4i) = 3 + 4i + 6i + 8i^2 = 3 + 10i - 8 = -5 + 10i\)
Verification: \(\left(-\frac{1}{5}+\frac{2}{5}i\right)(3-4i) = -\frac{3}{5}+\frac{4}{5}i+\frac{6}{5}i-\frac{8}{5}i^2 = -\frac{3}{5}+\frac{10}{5}i+\frac{8}{5} = \frac{5}{5}+2i = 1+2i\). ✓
B-3. Absolute Value and Complex Conjugate¶
Problem:
For each of the following complex numbers, find the complex conjugate \(z^*\) and the absolute value \(|z|\).
(a) \(z = 5 - 12i\)
(b) \(z = -3i\)
(c) \(z = -2 + 2i\)
(d) \(z = 7\) (real number)
(a) \(z = 5 - 12i\):
Check: \(zz^* = (5-12i)(5+12i) = 25 + 144 = 169 = 13^2 = |z|^2\). ✓
(b) \(z = -3i\):
Check: \(zz^* = (-3i)(3i) = -9i^2 = 9 = 3^2\). ✓
(c) \(z = -2 + 2i\):
Check: \(zz^* = (-2+2i)(-2-2i) = 4 + 4 = 8 = (2\sqrt{2})^2\). ✓
(d) \(z = 7\):
The complex conjugate of a real number is itself. \(zz^* = 49 = |z|^2\). ✓
B-4. Conversion to Polar Form¶
Problem:
Express the following complex numbers in polar form \(r(\cos\theta + i\sin\theta)\). Give the argument \(\theta\) in the range \(-\pi < \theta \leq \pi\).
(a) \(z = 1 + \sqrt{3}\,i\)
(b) \(z = -2\)
(c) \(z = -1 - i\)
(d) \(z = 3i\)
(a) \(z = 1 + \sqrt{3}\,i\):
\(\tan\theta = \sqrt{3}/1 = \sqrt{3}\), and since it is in the first quadrant, \(\theta = \pi/3\).
Check: \(2(\cos 60° + i\sin 60°) = 2\left(\frac{1}{2} + i\frac{\sqrt{3}}{2}\right) = 1 + \sqrt{3}\,i\). ✓
(b) \(z = -2\):
Since it points in the negative real axis direction, \(\theta = \pi\).
Check: \(2(-1 + 0i) = -2\). ✓
(c) \(z = -1 - i\):
With \(a = -1, b = -1\), this is in the third quadrant. \(\tan\theta = (-1)/(-1) = 1\), but since it is in the third quadrant, \(\theta = -\frac{3\pi}{4}\) (in the range \(-\pi < \theta \leq \pi\)).
Check: \(\cos(-3\pi/4) = -\frac{\sqrt{2}}{2}\), \(\sin(-3\pi/4) = -\frac{\sqrt{2}}{2}\). \(\sqrt{2}\left(-\frac{\sqrt{2}}{2} - i\frac{\sqrt{2}}{2}\right) = -1 - i\). ✓
(d) \(z = 3i\):
Since it points in the positive imaginary axis direction, \(\theta = \pi/2\).
Check: \(3(0 + i) = 3i\). ✓
B-5. Powers of \(i\)¶
Problem:
Find the following values.
(a) \(i^5\)
(b) \(i^{13}\)
(c) \(i^{-1}\)
(d) \(i^{100}\)
Powers of \(i\) cycle with period 4: \(i^0 = 1,\; i^1 = i,\; i^2 = -1,\; i^3 = -i\).
(a) \(i^5\):
Since \(5 = 4 \times 1 + 1\), we have \(i^5 = i^1 = \boxed{i}\)
(b) \(i^{13}\):
Since \(13 = 4 \times 3 + 1\), we have \(i^{13} = i^1 = \boxed{i}\)
(c) \(i^{-1}\):
\(i^{-1} = \dfrac{1}{i} = \dfrac{i}{i^2} = \dfrac{i}{-1} = -i\). Alternatively, \(i^{-1} = i^3 = \boxed{-i}\)
Check: \((-i) \cdot i = -i^2 = 1\). ✓
(d) \(i^{100}\):
Since \(100 = 4 \times 25 + 0\), we have \(i^{100} = i^0 = \boxed{1}\)
B-6. Rules for Complex Conjugation¶
Problem:
Let \(z_1 = 2 + i\) and \(z_2 = 1 - 3i\). Verify the following by direct calculation.
(a) Confirm that \((z_1 z_2)^* = z_1^* z_2^*\) holds by computing the left-hand side and right-hand side separately.
(b) Confirm that \((z_1 + z_2)^* = z_1^* + z_2^*\) holds.
\(z_1 = 2 + i\), \(z_2 = 1 - 3i\).
(a) Verification of \((z_1 z_2)^* = z_1^* z_2^*\):
Left-hand side:
Right-hand side:
LHS \(= 5 + 5i =\) RHS. ✓ \(\quad \blacksquare\)
(b) Verification of \((z_1 + z_2)^* = z_1^* + z_2^*\):
Left-hand side:
Right-hand side:
LHS \(= 3 + 2i =\) RHS. ✓ \(\quad \blacksquare\)
B-7. Calculating Terms of a Maclaurin Expansion¶
Problem:
Write out the Maclaurin expansion of \(e^x\) (Equation (A.25)) up to the 5th-order term, substitute \(x = 2\), and find the approximate value of \(e^2\) to two decimal places. (Reference: \(e^2 \approx 7.389\))
Writing out the Maclaurin expansion of \(e^x\) up to the 5th-order term:
Substituting \(x = 2\):
Verification: The true value is \(e^2 \approx 7.389\). Adding the 6th-order term \(\frac{2^6}{6!} = \frac{64}{720} \approx 0.089\) gives \(7.36\), which is even closer. This is a reasonable approximation for truncation at 5th order. ✓
B-8. Calculating \(e^{i\theta}\)¶
Problem:
Using Euler's formula \(e^{i\theta} = \cos\theta + i\sin\theta\), express the following values in the form \(a + bi\).
(a) \(e^{i\pi/4}\)
(b) \(e^{i\pi/2}\)
(c) \(e^{i\pi}\)
(d) \(e^{-i\pi/3}\)
We use Euler's formula \(e^{i\theta} = \cos\theta + i\sin\theta\).
(a) \(e^{i\pi/4}\):
(b) \(e^{i\pi/2}\):
(c) \(e^{i\pi}\):
This corresponds to the famous Euler's identity \(e^{i\pi} + 1 = 0\).
(d) \(e^{-i\pi/3}\):
Verification: The absolute value of (a) \(= \sqrt{1/2 + 1/2} = 1\). This is consistent with \(|e^{i\theta}| = 1\). The others similarly have absolute value 1. ✓
B-9. Polar Form and Rewriting with Euler's Formula¶
Problem:
Express the following complex numbers in the form \(re^{i\theta}\) (polar form using Euler's formula).
(a) \(z = 1 + i\)
(b) \(z = -\sqrt{3} + i\)
(c) \(z = -5i\)
(a) \(z = 1 + i\):
(b) \(z = -\sqrt{3} + i\):
\(a = -\sqrt{3},\; b = 1\) (second quadrant). \(\tan\theta = \frac{1}{-\sqrt{3}}\), and since it is in the second quadrant, \(\theta = \frac{5\pi}{6}\).
Verification: \(2(\cos\frac{5\pi}{6} + i\sin\frac{5\pi}{6}) = 2(-\frac{\sqrt{3}}{2} + i\frac{1}{2}) = -\sqrt{3} + i\). ✓
(c) \(z = -5i\):
Verification: \(5(\cos(-\pi/2) + i\sin(-\pi/2)) = 5(0 - i) = -5i\). ✓
B-10. Calculating \(|e^{i\theta}|\)¶
Problem:
Show that \(|e^{i\theta}| = 1\) for any real number \(\theta\), using the relation \(|z|^2 = zz^*\) from equation (A.15).
Strategy: Use the relation \(|z|^2 = zz^*\) from Eq. (A.15).
Let \(z = e^{i\theta}\), then \(z^* = e^{-i\theta}\) (from Euler's formula, the complex conjugate of \(e^{i\theta} = \cos\theta + i\sin\theta\) is \(\cos\theta - i\sin\theta = e^{-i\theta}\)).
Since \(|e^{i\theta}|\) is a non-negative real number,
This holds for any real number \(\theta\). Geometrically, this means that \(e^{i\theta}\) always lies on the unit circle. \(\blacksquare\)
Medium¶
M-1. General Proof of the Product Rule for Complex Conjugates¶
Problem:
For arbitrary complex numbers \(z_1 = a + bi\), \(z_2 = c + di\) (where \(a, b, c, d\) are real), prove that
using the definition of multiplication in Eq. (A.5) and the definition of complex conjugation (Eq. (A.12)).
Proof:
Let \(z_1 = a + bi\) and \(z_2 = c + di\) (where \(a, b, c, d\) are real numbers).
Calculation of the left-hand side:
From equation (A.5),
From the definition of the complex conjugate (equation (A.12)),
Calculation of the right-hand side:
Comparing \((*)\) and \((**)\),
Therefore \((z_1 z_2)^* = z_1^* z_2^*\) holds. \(\blacksquare\)
Verification: This was confirmed concretely in D6(a) for the case \(z_1 = 2+i\), \(z_2 = 1-3i\). It is consistent with the general proof. ✓
M-2. Derivation of de Moivre's Theorem¶
Problem:
Using Euler's formula, show that for any integer \(n\),
holds. Furthermore, by comparing the real and imaginary parts for the case \(n = 2\), derive the double-angle formulas for \(\cos 2\theta\) and \(\sin 2\theta\).
Proof of de Moivre's Theorem:
Strategy: Using Euler's formula \(e^{i\theta} = \cos\theta + i\sin\theta\), rewrite the left-hand side in exponential form.
By the exponent rule \((e^a)^n = e^{na}\),
Applying Euler's formula again,
Combining the above,
holds for any integer \(n\). \(\blacksquare\)
The Case \(n = 2\): Derivation of the Double-Angle Formulas:
Expanding the left-hand side:
Right-hand side (de Moivre's theorem):
Comparing real parts:
Comparing imaginary parts:
These are precisely the double-angle formulas for trigonometric functions.
Verification: Check with \(\theta = \pi/4\). \(\cos(\pi/2) = 0\), \(\cos^2(\pi/4) - \sin^2(\pi/4) = 1/2 - 1/2 = 0\). ✓ \(\sin(\pi/2) = 1\), \(2\cos(\pi/4)\sin(\pi/4) = 2 \cdot \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{2}}{2} = 1\). ✓
M-3. Derivation of the Division Formula in Polar Form¶
Problem:
For two complex numbers \(z_1 = r_1 e^{i\theta_1}\) and \(z_2 = r_2 e^{i\theta_2}\) (with \(r_2 \neq 0\)), show that
holds. Using this result, explain that "division of complex numbers is division of absolute values and subtraction of arguments."
Strategy: Use the laws of exponents to simplify the fraction.
First, compute \(\frac{e^{i\theta_1}}{e^{i\theta_2}}\). Multiplying the numerator and denominator by \(e^{-i\theta_2}\),
Here we used \(e^{i\theta_2} \cdot e^{-i\theta_2} = e^0 = 1\).
Therefore,
\(\blacksquare\)
Interpretation: From this result, for the quotient \(z_1/z_2\):
- The modulus is \(\dfrac{r_1}{r_2} = \dfrac{|z_1|}{|z_2|}\) (division of moduli)
- The argument is \(\theta_1 - \theta_2 = \arg(z_1) - \arg(z_2)\) (subtraction of arguments)
In other words, division of complex numbers decomposes into division of moduli and subtraction of arguments. This is the natural inverse operation of multiplication being "the product of moduli and the sum of arguments" (Eq. (A.11)).
Verification: When \(z_1 = z_2\), we have \(z_1/z_2 = 1\). From the formula: \(\frac{r_1}{r_1} e^{i \cdot 0} = 1 \cdot 1 = 1\). ✓
M-4. Exponential Representations of \(\cos\theta\) and \(\sin\theta\)¶
Problem:
By solving Euler's formula \(e^{i\theta} = \cos\theta + i\sin\theta\) and its complex conjugate \(e^{-i\theta} = \cos\theta - i\sin\theta\) as a system of equations, derive
Furthermore, compare this result with the structure of Equations (A.13) and (A.14), and confirm that the relationship between \(e^{i\theta}\) and \(e^{-i\theta}\) is that of complex conjugates.
Strategy: Solve Euler's formula and its complex conjugate as a system of simultaneous equations.
Euler's formula:
Replace \(\theta\) with \(-\theta\) (or take the complex conjugate of equation (1)):
Derivation of \(\cos\theta\):
Equation (1) + Equation (2):
Derivation of \(\sin\theta\):
Equation (1) − Equation (2):
Comparison with Equations (A.13) and (A.14):
Equation (A.13) states that \(\operatorname{Re}(z) = \frac{z + z^*}{2}\). Setting \(z = e^{i\theta}\) gives \(z^* = e^{-i\theta}\), so
This is consistent with the fact that the real part of \(e^{i\theta} = \cos\theta + i\sin\theta\) is \(\cos\theta\).
Similarly, Equation (A.14) states that \(\operatorname{Im}(z) = \frac{z - z^*}{2i}\), and
This is also consistent with the fact that the imaginary part of \(e^{i\theta}\) is \(\sin\theta\).
From the above, we have confirmed that the exponential representations of \(\cos\theta\) and \(\sin\theta\) are nothing other than special cases (\(z = e^{i\theta}\), \(z^* = e^{-i\theta}\)) of the formulas for extracting real and imaginary parts using complex conjugates (Equations (A.13) and (A.14)). \(\blacksquare\)
Verification: Substitute \(\theta = 0\). \(\cos 0 = \frac{e^0 + e^0}{2} = \frac{2}{2} = 1\). ✓ \(\sin 0 = \frac{e^0 - e^0}{2i} = 0\). ✓
M-5. Conditions for a Complex Number to Be Real¶
Problem:
Prove that a necessary and sufficient condition for a complex number \(z\) to be real is \(z = z^*\). Similarly, prove that a necessary and sufficient condition for \(z\) to be purely imaginary is \(z = -z^*\) (and \(z \neq 0\)).
\(z\) is real \(\iff\) \(z = z^*\):
Proof:
Let \(z = a + bi\) (where \(a, b\) are real numbers).
(\(\Rightarrow\)) If \(z\) is real, then \(b = 0\), so \(z = a\) and \(z^* = a\). Therefore \(z = z^*\).
(\(\Leftarrow\)) If \(z = z^*\), then \(a + bi = a - bi\). Comparing the imaginary parts of both sides gives \(b = -b\), i.e., \(2b = 0\), so \(b = 0\). Therefore \(z = a\) is real. \(\blacksquare\)
\(z\) is purely imaginary (\(z \neq 0\)) \(\iff\) \(z = -z^*\) (and \(z \neq 0\)):
Proof:
(\(\Rightarrow\)) If \(z\) is purely imaginary, then \(a = 0\) and \(b \neq 0\), so \(z = bi\) and \(z^* = -bi\). Therefore \(-z^* = bi = z\).
(\(\Leftarrow\)) If \(z = -z^*\), then \(a + bi = -(a - bi) = -a + bi\). Comparing the real parts of both sides gives \(a = -a\), i.e., \(2a = 0\), so \(a = 0\). From the condition \(z \neq 0\), we have \(b \neq 0\). Therefore \(z = bi\) is purely imaginary. \(\blacksquare\)
Verification (concrete examples): - \(z = 5\) (real): \(z^* = 5 = z\). ✓ - \(z = 3i\) (purely imaginary): \(z^* = -3i\), \(-z^* = 3i = z\). ✓ - \(z = 1 + i\) (neither): \(z^* = 1 - i \neq z\) and \(-z^* = -1 + i \neq z\). ✓
Advanced¶
A-1. The \(n\)-th Roots of a Complex Number and the Regular \(n\)-gon¶
Problem:
Find all solutions to the equation \(z^n = 1\) (where \(n\) is a positive integer).
(a) By setting \(z = re^{i\theta}\) and using the conditions \(|z| = 1\) and the argument constraint, show that the \(n\) solutions are
(b) Explain why plotting these \(n\) solutions on the complex plane gives the vertices of a regular \(n\)-gon inscribed in the unit circle.
(c) Prove, using the geometric series formula, that the sum of the \(n\) "\(n\)-th roots of unity" satisfies $$
\sum_{k=0}^{n-1} z_k = 0$$
(d) In quantum mechanics, the same structure appears in the discrete Fourier transform. For the case \(n = 4\), find the four solutions explicitly and verify that they are \(\{1, i, -1, -i\}\).
(a) The \(n\) solutions of \(z^n = 1\):
Strategy: Let \(z = re^{i\theta}\) and separate the conditions on the modulus and argument.
Comparing moduli: \(r^n = 1\). Since \(r > 0\), we have \(r = 1\).
Comparing arguments: \(n\theta = 0 + 2\pi k\) (\(k\) is an integer). Here the argument has an ambiguity of integer multiples of \(2\pi\).
For \(k = 0, 1, 2, \ldots, n-1\), \(\theta\) takes distinct values \(0, \frac{2\pi}{n}, \frac{4\pi}{n}, \ldots, \frac{2\pi(n-1)}{n}\). When \(k = n\), \(\theta = 2\pi\) represents the same point as \(\theta = 0\), so no new solutions are obtained.
Therefore, the \(n\) solutions of \(z^n = 1\) are
\(\blacksquare\)
(b) Vertices of a regular \(n\)-gon:
All \(n\) solutions \(z_k = e^{2\pi i k/n}\) satisfy \(|z_k| = 1\), so they are points on the unit circle.
The difference in argument between adjacent solutions \(z_k\) and \(z_{k+1}\) is
which is constant. That is, the \(n\) points are arranged at equal intervals on the unit circle.
A polygon inscribed in the unit circle with \(n\) vertices equally spaced is nothing other than a regular \(n\)-gon. The first vertex \(z_0 = 1\) is at the point \((1, 0)\) on the real axis, and the remaining vertices are located at positions rotated counterclockwise by \(2\pi/n\) from there. \(\blacksquare\)
(c) The sum of the \(n\)-th roots of unity is 0:
Strategy: Use the geometric series formula.
Let \(\omega = e^{2\pi i/n}\), so that \(z_k = \omega^k\).
This is a geometric series with first term \(1\), common ratio \(\omega\), and \(n\) terms, so (when \(\omega \neq 1\), i.e., \(n \geq 2\)),
Since \(\omega^n = (e^{2\pi i/n})^n = e^{2\pi i} = \cos 2\pi + i\sin 2\pi = 1\), we have
When \(n = 1\), there is only \(z_0 = 1\) and the sum is \(1 \neq 0\), but we normally consider \(n \geq 2\).
\(\blacksquare\)
Verification (geometric interpretation): By symmetry, the sum of position vectors to the vertices of a regular \(n\)-gon points to the origin. This corresponds to the centroid being at the origin. ✓
(d) The case \(n = 4\):
- \(k = 0\): \(e^{0} = 1\)
- \(k = 1\): \(e^{i\pi/2} = \cos(\pi/2) + i\sin(\pi/2) = i\)
- \(k = 2\): \(e^{i\pi} = \cos\pi + i\sin\pi = -1\)
- \(k = 3\): \(e^{i3\pi/2} = \cos(3\pi/2) + i\sin(3\pi/2) = -i\)
Verification: Sum \(= 1 + i + (-1) + (-i) = 0\). ✓ Also confirming \(z_k^4 = 1\) for each \(z_k\): \(i^4 = (i^2)^2 = (-1)^2 = 1\), \((-1)^4 = 1\), \((-i)^4 = ((-i)^2)^2 = (-1)^2 = 1\). ✓
A-2. Bridge to Quantum Mechanics: Interference of Complex Amplitudes¶
Problem:
In quantum mechanics, the "probability amplitude" for a particle to reach a certain point is given as a complex number. When there are two paths (path 1 and path 2), letting the amplitude for each path be \(\phi_1 = r_1 e^{i\alpha}\) and \(\phi_2 = r_2 e^{i\beta}\), the total amplitude is \(\phi = \phi_1 + \phi_2\), and the detection probability is given by \(P = |\phi|^2\).
(a) Expand \(P = |\phi_1 + \phi_2|^2\) and derive
(b) If probability amplitudes could only take real values (i.e., \(\alpha, \beta\) are restricted to \(0\) or \(\pi\) only), show that the interference term \(2r_1 r_2 \cos(\alpha - \beta)\) can only take the values \(\pm 2r_1 r_2\). On the other hand, explain that when \(\alpha - \beta\) can vary continuously, the interference term changes continuously from \(-2r_1 r_2\) to \(+2r_1 r_2\), and discuss one aspect of why "complex numbers are essentially necessary."
(c) In particular, when \(r_1 = r_2 = r\), show that \(P = 0\) for \(\alpha - \beta = \pi\) (phase difference \(\pi\)). This corresponds to two waves completely canceling each other (destructive interference). Describe in 2–3 sentences how this result differs from the classical intuition that "probabilities are always positive."
(a) Expansion of \(P = |\phi_1 + \phi_2|^2\):
Approach: Expand using \(|\phi|^2 = \phi\phi^*\).
Expanding,
Now substitute \(\phi_1 = r_1 e^{i\alpha}\), \(\phi_2 = r_2 e^{i\beta}\).
The sum of the cross terms is,
Using the exponential representation of \(\cos\) derived in S4, \(\cos\theta = \frac{e^{i\theta} + e^{-i\theta}}{2}\),
Therefore,
\(\blacksquare\)
Verification: When \(\phi_2 = 0\) (no path 2), \(r_2 = 0\) so \(P = r_1^2 = |\phi_1|^2\). ✓ Also when \(\alpha = \beta\) (in phase), \(P = r_1^2 + r_2^2 + 2r_1 r_2 = (r_1 + r_2)^2\). This corresponds to the amplitudes adding together. ✓
(b) Comparison between real and complex amplitudes:
Real amplitude case (where \(\alpha, \beta\) are only \(0\) or \(\pi\)):
The possible values of \(\alpha - \beta\) are \(0, \pi, -\pi\).
- When \(\alpha - \beta = 0\): \(\cos(\alpha - \beta) = \cos 0 = 1\)
- When \(\alpha - \beta = \pm\pi\): \(\cos(\alpha - \beta) = \cos\pi = -1\)
Therefore the interference term \(2r_1 r_2 \cos(\alpha - \beta)\) takes only two values: \(+2r_1 r_2\) or \(-2r_1 r_2\).
Complex amplitude case (where \(\alpha - \beta\) varies continuously):
The phase difference \(\delta = \alpha - \beta\) can vary continuously from \(0\) to \(2\pi\), and \(\cos\delta\) takes all values continuously from \(-1\) to \(+1\). Therefore the interference term varies continuously over the range
Why complex numbers are essentially necessary: With real amplitudes, interference is limited to two choices—"completely constructive" or "completely destructive"—and intermediate interference cannot be expressed. On the other hand, with complex amplitudes, the phase difference becomes a continuous parameter, enabling the description of all degrees of interference, including partial constructive and destructive interference. To correctly describe the interference patterns observed in nature (such as the continuous variation of bright and dark fringes in the double-slit experiment), it is essential that amplitudes be complex numbers.
(c) Complete destructive interference:
When \(r_1 = r_2 = r\), the result from (a) becomes
Substituting \(\alpha - \beta = \pi\),
When the amplitudes of two paths are equal and the phase difference is exactly \(\pi\), the detection probability becomes exactly zero. This is destructive interference.
Contrast with classical intuition:
In classical probability theory, one simply adds the probabilities of passing through each path as \(P_1 = r^2\) and \(P_2 = r^2\), giving
so the probability can never be zero. However, in quantum mechanics, what we add are not probabilities but probability amplitudes (complex numbers), and cancellation can occur at the amplitude level, resulting in zero probability. This is a phenomenon that contradicts the classical intuition that "the particle must pass through one path or the other, yet is never detected," and it is an essential feature of quantum mechanics.
Verification: When \(\alpha - \beta = 0\) (constructive interference), \(P = 2r^2(1+1) = 4r^2 = (2r)^2\). The amplitude doubles to \(2r\), and the probability quadruples. This is twice the classical value of \(2r^2\), clearly showing the effect of interference. ✓
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