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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

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