1.1 The Radiation Laws and the Birth of Quantum Mechanics
1.1 The Radiation Laws and the Birth of Quantum Mechanics
1.1 The Radiation Laws and the Birth of Quantum Mechanics
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QUANTUM MECHANICS I (Dr. T. Br<strong>and</strong>es): Example/Solution Sheets 17<br />
3.11.3 * Expansion into eigenmodes (40 min)<br />
Consider <strong>the</strong> vector f ∈ H, f(x) = cx(L − x).<br />
a) Calculate <strong>the</strong> constant c such that f is normalized, i.e. ‖f‖ = 1. Show that c = √ 30/L/L 2 .<br />
∫ L<br />
∫ L<br />
1 = ‖f‖ 2 = dxf ∗ (x)f(x) = c 2 x 2 (L − x) 2 =<br />
0<br />
0<br />
∫ L<br />
[ 1<br />
= c 2 dx[x 4 − 2Lx 3 + L 2 x 2 ] = c 2 L 5 5 − 2 4 + 1 ] √<br />
30<br />
c =<br />
3<br />
L<br />
0<br />
b) Show that f can be exp<strong>and</strong>ed in <strong>the</strong> basis ψ n as<br />
1<br />
L 2 .<br />
∞∑<br />
f = c n ψ n ,<br />
n=1<br />
c n = 2 √ 60 1 − (−1)n<br />
n 3 π 3 (3.90)<br />
We have<br />
√<br />
2<br />
c n = 〈ψ n |f〉 = c<br />
L<br />
= √ 60<br />
∫ 1<br />
0<br />
∫ 1<br />
∫ 1<br />
c) Use b) to prove <strong>the</strong> formula<br />
We have<br />
√<br />
30<br />
L<br />
0<br />
0<br />
∫ L<br />
0<br />
dxx(L − x) sin<br />
dy(y − y 2 ) sin(nπy).<br />
dyy sin(nπy) = − cos(nπ)<br />
nπ<br />
dyy 2 sin(nπy) = − 2<br />
n 3 π 3 + 2 − n2 π 2<br />
n 3 π 3<br />
1<br />
L 2 x(L − x) = √ 60<br />
(1/2)(1/4) =<br />
π3<br />
32<br />
=<br />
∞∑<br />
n=1<br />
∞∑<br />
k=0<br />
3.11.4 * Scalar product (20 min)<br />
<br />
∞∑<br />
n=1<br />
∫ 1<br />
0<br />
( nπx<br />
)<br />
= [y = x/L) =<br />
L<br />
cos(nπ)<br />
dy(y − y 2 ) sin(nπy) = 2 1 − (−1)n<br />
n 3 π 3<br />
π 3 ∞<br />
32 = ∑ (−1) k<br />
(2k + 1) . 3<br />
k=0<br />
2 1 − (−1)n<br />
n 3 π 3<br />
2 1 − (−1)n<br />
n 3 π 3<br />
(−1) k<br />
(2k + 1) 3 .<br />
sin<br />
√<br />
2<br />
( nπx<br />
)<br />
L sin , setx = L/2<br />
L<br />
∞∑<br />
( nπ<br />
)<br />
= [n = 2k + 1] =<br />
2<br />
k=0<br />
4(−1) k<br />
π 3 (2k + 1) 3<br />
a) Use <strong>the</strong> bra <strong>and</strong> ket notation to show that for an orthonormal basis {|ψ n 〉} <strong>and</strong> two Hilbert<br />
space vectors |ψ〉 <strong>and</strong> |χ〉, one has<br />
〈ψ|χ〉 =<br />
∞∑<br />
〈ψ|ψ n 〉〈ψ n |χ〉. (3.91)<br />
n=0