Griffith's Quantum Mechanics Problem 2.51

Special Problem One Peter D Alison May 11, 2009 0.1

Exercise 2.51

P113A ID 25732824 Consider the potential V (x) = −

h ¯ 2 a2 sech(ax), m

where a is a positive constant, and sech stands for hyperbolic secant. (a) Graph this potential. Let us make all constants one so as to make graphing easier and possible on Mathematica or GnuPlot. (b) Check that this potential has the state ψ0 = Asech(ax), and find its energy. Normalize ψ0 , and sketch its graph. We check that ψ0 satisfies the Schr¨odinger Equation and solve for the energy E. h ¯ 2 ∂ 2ψ − + V (x)ψ = Eψ 2m ∂x2 ∂2 a2 h ¯ 2 sech2 (ax) (Asech(ax)) − (Asech(ax)) = E0 (Asech(ax)) ∂x2 m 1

a2 h ¯ 2 sech2 (ax) (Asech(ax)) = E0 (Asech(ax)) A(−a sech (ax)+a sech(ax) tanh (ax))− m Solving for E0 we find that a2 h ¯2 E0 = − 2m To normalize ψ0 , we use the normalization condition 2

3

2

2

Z +∞ −∞

A2

|ψ0 (x)|2 dx = 1

Z +∞

sech2 (ax) dx = 1

−∞

2A2 =1 a r a A= 2 The maximum of ψ0 occurs at

q

a . 2

(c) Show that the function ψk (x) = A(

ik − a tanh(ax) ikx e ) ik + a

√ (where k = 2mE/¯h, as usual) solves the Schr¨odinger Equation for any (positive) energy E. Since tanh(z) → −1 as z → −∞, ψk (x) ≈ Aeikx , for large negative x. This represents, then, a wave coming in from the left with no no accompanying reflected wave (i.e., no exp(−ikx)). What is the asymptotic form of ψk (x) at large positive x? What are R and T , for this potential? Comment: This is a famous example of a reflectionless potential - every incident particle, regardless of it energy, passes right through. We start by checking that ψk satisfies the Schr¨odinger Equation. −

h ¯ 2 ∂ 2 ψk + V (x)ψk = Ek ψk 2m ∂x2 2

h ¯ 2 ∂ 2 ψk − + V (x)ψk 2m ∂x2 2ia2 Ae1kx ksech2 (ax) 2a3 Aeikx sech2 (ax) tanh(ax) Aeikx k 2 (ik − a tanh(ax)) + − =− ik + a ik + a ik + a h ¯ 2 a2 ik − a tanh(ax) ikx =− sech2 (ax)A( )e m ik + a √ Simplifying and replace k for 2mE we obtain √ √ i√2mEk x ae Ek h ¯ 2 (−2i Ek m − a tanh(ax)) √ − √ 2(a + i 2mEk ) Also √ =

E k ψk √ aEk ei 2mEk x (i 2mEk − a tanh(ax)) √ √ 2(a + i 2mEk ) √

As we can see ψk satisfies the Schr¨odinger Equation. We have lim tanh(ax) = 1

x→+∞

For large positive x, tanh(ax) → 1 as x → +∞ )eikx . For the transmission coefficient T , we take the raso ψk → A( ik−a ik+a tio of the squares of the amplitudes of ψk . ik−a A( ik+a ) 2 T =( ) A

ik − a −ik − a )( )=1 ik + a −ik + a The transmission coefficient is 1, which implies that R = 0. T =(

3