Special Problem One

Peter D Alison

May 11, 2009

0.1 Exercise 2.51

P113AID 25732824

Consider the potential

V (x) = − h2a2

msech(ax),

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 possibleon 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 Schrodinger Equation and solve for the en-ergy E.

− h2

2m

∂2ψ

∂x2+ V (x)ψ = Eψ

∂2

∂x2(Asech(ax))− a2h2sech2(ax)

m(Asech(ax)) = E0(Asech(ax))

1

A(−a2sech3(ax)+a2sech(ax) tanh2(ax))−a2h2sech2(ax)

m(Asech(ax)) = E0(Asech(ax))

Solving for E0 we find that

E0 = −a2h2

2m

To normalize ψ0, we use the normalization condition∫ +∞

−∞|ψ0(x)|2 dx = 1

A2∫ +∞

−∞sech2(ax) dx = 1

2A2

a= 1

A =

√a

2

The maximum of ψ0 occurs at√

a2.

(c) Show that the function

ψk(x) = A(ik − a tanh(ax)

ik + aeikx)

(where k =√

2mE/h, as usual) solves the Schrodinger Equation for any (pos-itive) 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 accompany-ing 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: Thisis 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 Schrodinger Equation.

− h2

2m

∂2ψk

∂x2+ V (x)ψk = Ekψk

2

− h2

2m

∂2ψk

∂x2+ V (x)ψk

= −2ia2Ae1kxksech2(ax)

ik + a+

2a3Aeikxsech2(ax) tanh(ax)

ik + a−Ae

ikxk2(ik − a tanh(ax))

ik + a

= − h2a2

msech2(ax)A(

ik − a tanh(ax)

ik + a)eikx

Simplifying and replace k for√

2mE we obtain

−√aei√

2mEkxEkh2(−2i

√Ekm− a tanh(ax))√

2(a+ i√

2mEk)

AlsoEkψk

=

√aEke

i√

2mEkx(i√

2mEk − a tanh(ax))√2(a+ i

√2mEk)

As we can see ψk satisfies the Schrodinger Equation.

We havelim

x→+∞tanh(ax) = 1

For large positive x, tanh(ax)→ 1 as x→ +∞

so ψk → A( ik−aik+a

)eikx. For the transmission coefficient T , we take the ra-tio of the squares of the amplitudes of ψk.

T = (A( ik−a

ik+a)

A)2

T = (ik − aik + a

)(−ik − a−ik + a

) = 1

The transmission coefficient is 1, which implies that R = 0.

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