1.1 The Radiation Laws and the Birth of Quantum Mechanics

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QUANTUM MECHANICS I (Dr. T. Brandes): Example/Solution Sheets 10 2.7.3 * Orthonormality (10 min) Consider the Hilbert space H of wave functions ψ(x) of the infinite potential well on the interval [0, L] with ψ(0) = ψ(L) = 0. Show that the basis vectors √ 2 ( nπx ) ψ n (x) = L sin L form an orthonormal system. SOLUTION: We have to show that the ψ n (x) form an orthonormal basis: 0 ∫ L 0 dxψ ∗ n(x)ψ m (x) = δ nm . (2.48) We therefore have to calculate the integral ∫ √ L 2 ( nπx ) √ dx L sin 2 ( mπx ) L L sin . (2.49) L For n = m we have already calculated this integral above when we obtained the normalization of the wave functions. For n ≠ m we have to show that 2.49 is zero: Do this by expanding the sin into exponentials and calculating the integrals, or use a theorem for trigonometric functions, or look it up in a table. 2.7.4 Time Evolution (2 min) Consider a wave function ψ(x) of the infinite potential well on the interval [0, L]. Consider the case when the wave function at time t = 0 is one of the eigenstates of energy E n , i.e. Ψ(x, t = 0) = ψ n (x) and check that the time evolution of a wave function that is an energy eigenstate is just given by multiplication with the time–dependent phase factor e −iEnt/ , that is Ψ(x, t = 0) = ψ n (x) Ψ(x, t) = ψ n (x)e −iEnt/ . (2.50) SOLUTION: In principle, this is in general already clear from the definition of the stationary states (see lecture notes): To solve we had made a separation ansatz Inserting into the Schrödinger equation, we have i ∂ Ψ(x, t) = ĤΨ(x, t), (2.51) ∂t Ψ(x, t) = ψ(x)f(t). (2.52) i∂ t f(t) f(t) = Ĥψ(x) ψ(x) = E, (2.53) where we have separated the t– and the x–dependence. Both sides of depend on t resp. x independently and therefore must be constant = E. Solving the equation for f(t) yields f(t) = exp −iEt/ and therefore Ψ(x, t) = ψ(x)e −iEt/ . (2.54) We recognize: the time evolution of the wave function Ψ(x, t) is solely determined by the factor exp −iEt/. Furthermore, the constant E must be an energy (dimension!). We can also check directly that Ψ(x, t) fulfills the time–dependent Schrödinger equation: Ĥψ n (x) = E n ψ n (x) i ∂ ∂t Ψ(x, t) = i ∂ ∂t ψ n(x)e −iEnt/ = E n ψ n (x)e −iEnt/ = Ĥψ n(x)e −iEnt/ = ĤΨ(x, t). (2.55)
QUANTUM MECHANICS I (Dr. T. Brandes): Example/Solution Sheets 11 2.7.5 Expectation values (15 min) Calculate the expectation value of a) the momentum square p 2 and b) the kinetic energy of a particle in the one–dimensional infinite well on the interval [0, L] with wave function Ψ(x, t) = ψ n (x)e −iEnt/ . SOLUTION: Use ∂x 2ψ n(x) = −(n 2 2 /L 2 )ψ n (x): 〈p 2 〉 t = ∫ L 0 dxψ n (x) 2 ∂ 2 x i 2 ψ n (x) = − 2 ∫ L 0 dxψ n (x) (− n2 π 2 ) L 2 ψ n (x) = n2 2 π 2 L 2 〈 p2 2m 〉 t = n2 2 π 2 2L 2 m 2 = E n. (2.56) This is a very obvious result: since the energy of the wave function ψ n (x) is E n , the expectation value of the kinetic energy E = p 2 /2m (= the total energy), i.e. the value one obtains on averaging the results from many measurements on the same system with the same wave function, must be E n . We can obtain this result even easier by calculating the expectation value of the energy 〈Ĥ〉 t = ∫ L 0 dxψ n (x)Hψ n (x) = ∫ L 0 dxψ n (x)E n ψ n (x) = E n , (2.57) where we have used the Schrödinger equation Ĥψ n = E n ψ n and the orthonormality of the ψ n . 2.7.6 * Time evolution of superposition (10 min) a) What is the time evolution of an arbitrary wave function Ψ(x, t = 0), ∞∑ ∫ L Ψ(x, t = 0) = c n ψ n (x), c n = dxψn ∗ (x)Ψ(x) (2.58) b) Consider the wave function n=0 0 Ψ(x, t = 0) = 1 √ 2 (ψ 1 (x) + ψ 2 (x)) . (2.59) What is the probability density to find the particle at x at time t SOLUTION: a) The Schrödinger equation is a linear partial differential equation, so the answer is simple: it is just given by the superposition ∞∑ Ψ(x, t) = c n ψ n (x)e −iEnt/ (2.60) n=0 i∂ t Ψ(x, t) = = ∞∑ c n E n ψ n (x)e −iEnt/ n=0 ∞∑ c n Ĥψ n (x)e −iEnt/ n=0 where we used the fact that Ĥ is a linear operator. b) The time evolution is obtained as = Ĥ ∞∑ c n ψ n (x)e −iEnt/ n=0 = ĤΨ(x, t), Ψ(x, t) = 1 √ 2 (ψ 1 (x)e −iE 1t/ + ψ 2 (x)e −iE 2t/ ) . (2.61) From this we obtain the probability density |Ψ(x, t)| 2 = 1 ( ψ 2 2 1 (x) + ψ2 2 (x) + 2ψ 1(x)ψ 2 (x) cos[(E 1 − E 2 )t/] ) , (2.62) which is no longer constant as a function of time.
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1.1 The Radiation Laws and the Birth of Quantum Mechanics

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