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# HW5-sol - PHYSICS 4455 QUANTUM MECHANICS Problem Set 5...

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PHYSICS 4455 — QUANTUM MECHANICS Problem Set 5 — Solutions. 1. One-dimensional problems. Liboff Problem 3.23 Since the particle is free, V ( x ) = 0 for x 0. So, the stationary Schrödinger equation for this problem is 2 2 m d 2 dx 2 ψ( x ) = E ψ( x ) ψ  ( x ) + 2 m 2 E ψ( x ) = 0 The “rigid wall” at x = 0 translates into the boundary condition ψ( 0 ) = 0. Note that there is no information here about a boundary condition at the other end. Following the discussion of “wiggly” eigenfunctions in class, we should demand E > 0 in order to ensure that ψ and ψ  have opposite signs. As a result, the solution will have the form ψ( x ) = Ae ikx + Be ikx with two (real) integration constants A and B . The boundary condition at x = 0 implies A + B = 0 so that ψ( x ) = A ( e ikx e ikx ) = Ā sin kx Inserting ψ( x ) into the Schrödinger equation results in E = 2 k 2 2 m with no further conditions on E . In particular, E in this case is not quantized, i.e., the allowed eigenvalues (the “spectrum”) are continuous. Also, the wave function is somewhat pathological in that it cannot be normalized. The time-dependent solution ψ( x , t ) is given by ψ( x , t ) = Ae i ω t ( e ikx e ikx ) with ω = E = 2 k 2 2 m Liboff Problem 4.1 You can do this in two ways: First, you can of course write down the Schrödinger equation for the one-dimensional box on the interval (− a / 2, + a / 2 ) and solve it. Alternatively, you can start from the eigenfunctions for the box on the interval ( 0, a ) , namely, ϕ n ( x ) = 2 a sin ( k n x ) with k n = n π/ a , n = 1,2,... and replace x by x + a / 2. Note that the new function, called ϕ n ( x ) and defined by ϕ n ( x ) ≡ ϕ n ( x + a 2 ) = 2 a sin k n x + k n a 2 vanishes at x = − a / 2 and at x = + a / 2, i.e., it satisfies the boundary conditions. Obviously, it also satisfies the Schrödinger equation inside the box, namely,

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2 2 m d 2 dx 2 ϕ n ( x ) = 2 k n 2 2 m ϕ n ( x ) = E n ϕ n ( x ) Using basic trigonometry, we can write ϕ n ( x ) = 2 a sin ( k n x ) cos k n a 2 + sin k n a 2 cos ( k n x ) = 2 a sin ( k n x ) cos n π 2 + sin n π 2 cos ( k n x ) Note that cos n π 2 = 0 (− 1 ) n / 2 for n odd n even and sin n π 2 = (−
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HW5-sol - PHYSICS 4455 QUANTUM MECHANICS Problem Set 5...

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