1.1 The Radiation Laws and the Birth of Quantum Mechanics

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QUANTUM MECHANICS I (Dr. T. Brandes): Example/Solution Sheets 4 1.4.2 ** Math: Gauß (20 min) Look up who Gauß was, where he lived etc. Write down the definition of the Gauss function. Look up examples for areas of mathematics and physics where the Gauss function is used. SOLUTION: The Gauss function is 1 x2 g(x) := √ e− 2σ 2 . (1.12) 2πσ 2 1.4.3 * Math: Gauß Integral 1 (10 min) Use polar coordinates to calculate ∫ ∞ −y 2 √ −∞ dxdye−x2 πṠOLUTION: in order to prove the above ∫ ∞ −∞ dxe−x2 = (∫ ∞ ) 2 dxe −x2 = −∞ 1.4.4 Math: Gauß Integral 2 (10 min) ∫ ∞ −∞ dxdye −x2 −y 2 = 2π = [x = r 2 , dx = 2rdr] = π ∫ ∞ 0 ∫ ∞ Use ∫ ∞ −∞ dxe−x2 = √ π to prove the formula for the Gauß integral SOLUTION: ∫ ∞ −∞ ∫ ∞ 0 drre −r2 dxxe −x = π. (1.13) dxe −ax2 +bx = √ π a eb2 /4a , a > 0. (1.14) −∞ dxe −ax2 +bx = = [y = √ a(x − b/2a)] = 1 √ a ∫ ∞ −∞ ∫ ∞ −∞ 1.4.5 Math: Fourier Transform of Gauss Function (20 min) dxe −a(x−b/2a)2 +b 2 /4a = dye −y2 +b 2 /4a = √ π a eb2 /4a . (1.15) The Gauss function f(x) := 1 x2 √ e− 2σ 2 (1.16) 2πσ 2 is a convenient example to discuss properties of the Fourier transform. Show that it can be decomposed into plane waves by ˜f(k) = ∫ ∞ −∞ dxf(x)e −ikx = e − 1 2 σ2 k 2 , f(x) = 1 2π ∫ ∞ −∞ dke − 1 2 σ2 k 2 e ikx . (1.17) Draw f(x) and ˜f(k) for different values of σ and discuss their relation. SOLUTION: In principle, one first has to show that the formula 1.14 for the Gauss integral also holds for complex b. This can be proven by complex integration, but we will not do it here. Then, simple application of 1.14 with b = −ik yields the result for ˜f(k). The equation for f(x) then is simply the Fourier back–transformation (definition), but you can explicitely verify it again by calculating the Gauss integral. Small σ: corresponds to a narrow Gauss function f(x) in x–space and a broad distribution ˜f(k) of Fourier components in k–space. Large σ: corresponds to a broad Gauss function f(x) in x–space and a narrow distribution ˜f(k) of Fourier components in k–space.
QUANTUM MECHANICS I (Dr. T. Brandes): Example/Solution Sheets 5 1.4.6 * Wave packet (20 min) We assume that a particle with energy E = p 2 /2m can be described by a function that is a superposition of plane waves, Ψ(x, t) = ∫ ∞ −∞ dka(k)e i(kx−ω(k)t) , ω(k) = E = 2 k 2 /(2m). (1.18) Use a(k) = C √ σ 2 /(2π)e −k2 σ 2 /2 to calculate the wave packet Ψ(x, t). Here, C is a constant. Show that ) C Ψ(x, t) = √ (− 1 + i(t/mσ2 ) exp x 2 . 2σ 2 [1 + i(t/mσ 2 )] To simplify your calculation, you can set = 2m = 1 during your calculation and re-install it in the result. Why does this ‘trick’ work Discuss Ψ(x, t) as a function of time. SOLUTION: The solution of this problem is discussed in many text books. Basically, one has to calculate a Gauss integral of the type 1.14. As a function of time, Ψ(x, t) becomes broader. Actually, the physical interesting quantity is the square |Ψ(x, t)| 2 which is the probability density to find the particle in the interval [x, x + dx] at time t. We see from this calculation that this probability density becomes broader with increasing time: If the particle was initially localized near the origin x = 0, it ‘spreads’ out. However, this does not mean that the particle disintegrates into smaller pieces or even into a continuous mass distribution. |Ψ(x, t)| 2 is not a mass density but a probability density: if we ‘look’ at time t if the particle is in the interval [x, x + dx], it is either there (with probability |Ψ(x, t)| 2 dx) or not. 1.5 Position and Momentum in Quantum Mechanics 1.5.1 Normalization (2min) Write down the normalization condition for the wave function Ψ(x, t) of a particle that is necessary to interpret |Ψ(x, t)| 2 as a probability density. SOLUTION: The probability to find the particle somewhere in space must be one and hence ∫ R 3 d 3 x|Ψ(x, t)| 2 = 1. (1.19) 1.5.2 Expectation values in quantum mechanics (5min) Write down the expectation value of the position x and the momentum p of a particle with a normalized wave function Ψ(x, t). SOLUTION: ∫ 〈x〉 t = ∫ dxΨ ∗ (x, t)xΨ(x, t), 〈p〉 t = dxΨ ∗ (x, t) ∂ x Ψ(x, t) (1.20) i We recognize that the position x corresponds to the operator ‘multiplication with x’. On the other hand, the momentum corresponds to the operator −i∂ x .
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1.1 The Radiation Laws and the Birth of Quantum Mechanics

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