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Use appropriate connection formulas to analyze the problem of scattering from a barrier with sloping walls (Figurea).

Hint: Begin by writing the WKB wave function in the form

(x)={1pxAeihxx1px'dx'+Be-ihxr1px'dx',x<x11pxCeihx1'px'dx'+De-1hx1Xpx'dx',X1<X<X21pxFeihx2xpx'dx.x>x2

Do not assume C=0 . Calculate the tunneling probability, T=|F|2/|A|2, and show that your result reduces to Equation 8.22 in the case of a broad, high barrier.

Short Answer

Expert verified

The Tunneling probability isTe-2.

Step by step solution

01

To find patching wave function.

We use a patching wave function pto match the WKB wave functions on either side of the turning points.

Consider the turning point x1(with upward slope).

Let's shift the axes over so that the left hand turning point x1occurs at x=0 .

The WKB wave function takes the form

role="math" localid="1658387748450" x1p(x)[Ae-ihx0p(x')dx'+Beihx0p(x')dx'],x<0,1|px|[Ceih0x|px'|dx'+De-1h0X|px'|dx'],x>0..(1)

and the patching wave function is

px=aAi伪虫+bBi伪虫,..(2)

For appropriate constants a and b .

Also, the linear approximation of the potential in the overlapping regions yields

px2mE-E-V,0x=丑伪3/2-x

In particular, in overlap region 2,

0xpx'dx'丑伪3/20xx'dx'=23h伪齿3/2

Therefore the WKB wave function (Equation (1)) can be written as

x1h3/41/4Ce23伪虫3/2+De-23伪虫3/2

Meanwhile, using the large- z asymptotic forms of the Airy functions, the patching wave function (Equation(2)) in overlap region 2 becomes,

pxa2伪虫1/4e-23伪虫3/2+b伪虫1/4e23伪虫3/2

Comparing the two solutions, we see that

a=4丑伪D,Andb=丑伪C 鈥︹赌︹赌︹赌︹赌︹赌︹赌︹赌︹赌︹赌︹赌(3)

02

To derive the overlap region.

Now we go back and repeat the procedure for overlap region 1. Once again, p(x) is given by Equation9.39, but this time is x negative, so

x0px'dx'h3/2x0-x'dx'=23h-x3/2

Therefore the WKB wave function (Equation (1)) can be written as

x1h3/4-x1/4Ae-i23-伪虫3/2+Bei23-伪虫3/2

Meanwhile, using the large- asymptotic form of the Airy function for large negative z, the patching wave function (Equation (2)) in overlap region 1 becomes

pxa-伪虫1/4sin23-伪虫3/2+4+b-伪虫1/4cos23-伪虫3/2+4px=12-伪虫1/4-iaecei23-伪虫3/2+-iae-颈蟺//4e-c+be颈蟺/4ei23-伪虫3/2+be-颈蟺//4e-i23-伪虫3/2px=12-伪虫1/4e颈蟺//4b-iaei23-伪虫3/2+e颈蟺//4b+iae-i23-伪虫3/2

Comparing the WKB and patching wave functions in overlap region 1,

We find

b-ia2e颈蟺/4=B丑伪andb+ia2e-颈蟺/4=A丑伪

or, putting in Equation (3) for a and b :

B=C2-iDe颈蟺/4andA=C2+iDe-颈蟺/4 鈥︹赌︹赌︹赌︹赌︹赌︹赌︹赌︹赌(4)

These are the connection formulas, joining the WKB solutions at either side of the turning point x1 .

03

To write upward slope and downward slope.

Now we consider the turning points x2 (with downward slope) and apply the same procedure. First we need to rewrite the WKB solution over the region x1<x<x2:

x1pxCe1hx1x2px'dx'-xx2px'dx'+De1h-x1x2px'dx+xx2px'dx

Letting CDe-,D'Ce,where 1hx1x2px'dxis constant (using the same notation as in Equation 9.23), the WKB wave function takes the form (after shifting the origin to x2).

fx1IpxIC'e1hx0px'dx'+De-1hx0px'dx',x<01pxFeihx0px'dx',x>0 鈥︹赌︹赌︹赌︹赌︹赌︹赌..(5)

In this case, the linear approximation of the potential is

VxE-V'ox

Accordingly, the patching wave functionp is given by

px=aAi-伪虫+b'Bi-伪虫 鈥︹赌︹赌︹赌︹赌︹赌︹赌︹赌︹赌.(6)

(The same solution as the upward-slope case with -as role="math" localid="1658393316907" =2mV'0/h21/3we also have,

px=2mE-E+V'0xpx=丑伪3/2x.

04

Using region 2 equation to find constant values.

In region 2, we have

0xpx'dx'丑伪3/20xx'dx'=23h伪虫3/2

Therefore the WKB wave function (Equation (5)) can be written as,

xFh3/4X1/4ei23伪虫3/2

Meanwhile, using the large- z asymptotic form of the Airy function for large negative zz=-伪虫, the patching wave function (Equation (6)) in overlap region 2 becomes

pxa'伪虫1/4sin23伪虫3/2+4+b'伪虫1/4cos23伪虫3/2+4px=12伪虫1/4-ia'e颈蟺/4ei23伪虫3/2+ia'e=-/4e-i23伪虫3/2_+be颈蟺/4ei23伪虫3/2+be=-/4e-i23伪虫3/2px=12伪虫1/4e颈蟺/4b-iaei23伪虫3/2+e-/4b'+ia-i23伪虫3/2

Comparing the two solutions, we see that

b'ia'2e颈蟺/4=F丑伪Andb''+ia'=0

which leads to,

b'=e颈蟺/4丑伪Fanda'=ie-颈蟺/4丑伪F 鈥︹赌︹赌︹赌︹赌︹赌︹赌(7)

05

Using region 1 equation to find other constant values.

In region 1, we have

x0px'dx'x0-x'dx'=23h-伪虫3/2,

Therefore the WKB wave function (Equation (5)) can be written as

x1h3/4-x1/4Ce23-伪虫3/2D'e-23-伪虫3/2

Meanwhile, using the large- z asymptotic form of the Airy function for large positive ZZ=-伪虫, the patching wave function (Equation (6)) in overlap region 1 becomes,

pxa'2-x1/4e-23-伪虫3/2+b'-x1/4e23-伪虫3/2

Comparing the WKB and patching wave functions in overlap region l, we find

D'=丑伪4a'AndC'=丑伪b'

or, putting in Equation (7) for a and b :

D'=i2e-颈蟺/4FAndC'=e-颈蟺/4F

But remember that D'=Ceand data-custom-editor="chemistry" C'=De, then

C=i2e-颈蟺/4e-FAndD=e-颈蟺/4eF 鈥︹赌︹赌︹赌︹赌︹赌(8)

These are the connection formulas, joining the WKB solutions at either side of the turning point x2.

06

To find the tunneling probability.

Putting Equation (8) forand back into Equation (4) for A , we find

A=ie-颈蟺/2e-4+eF

It follows that the tunneling probability is

T=F2A2T=1e-/4+别纬2.

In case of a broad, higher barrier, is very large.

Hence, the result for the tunneling probability reduces toTe-2 .

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Most popular questions from this chapter

Use equation 8.22 calculate the approximate transmission probability for a particle of energy E that encounters a finite square barrier of height V0 > E and width 2a. Compare your answer with the exact result to which it should reduce in the WKB regime T << 1.

Derive the connection formulas at a downward-sloping turning point, and confirm equation8.50.

(x)={D'|p(x)|exp-1hxx1|p(x')|dx'.ifx<x12D'p(x)sin-1hx1xp(x')dx'+4.ifx<x1

Use the WKB approximation to find the allowed energies (En)of an infinite square well with a 鈥渟helf,鈥 of heightV0, extending half-way across

role="math" localid="1658403794484" V(x)={v0,(0<x<a/2)0,(a/2<x<a),(otherwise)

Express your answer in terms ofrole="math" localid="1658403507865" V0andEn0(苍蟺魔)2/2ma2(the nth allowed energy for the infinite square well with no shelf). Assume that, but do not assume that E10>V0. Compare your result with what we got in Section 7.1.2, using first-order perturbation theory. Note that they are in agreement if eitherV0is very small (the perturbation theory regime) or n is very large (the WKB鈥攕emi-classical鈥攔egime).

Consider a particle of massm in the n th stationary state of the harmonic oscillator (angular frequency ).

(a) Find the turning point, x2 .
(b) How far (d) could you go above the turning point before the error in the linearized potential reaches 1%? That is, if V(x2+d)-VIin(x2+d)V(x2)=0.01 what is ?
(c) The asymptotic form of Ai(z) is accurate to 1% as long as localid="1656047781997" z5. For the din part (b), determine the smallest nsuch thatd5 . (For any n larger than this there exists an overlap region in which the liberalized potential is good to 1% and the large-z form of the Airy function is good to 1% .)

Consider the case of a symmetrical double well, such as the one pictured in Figure. We are interested in Figure 8.13bound states with E<V(0) .
(a) Write down the WKB wave functions in regions (i) x>x2 , (ii) x1<x<x2 , and (iii) 0<x<x1. Impose the appropriate connection formulas at and (this has already been done, in Equation 8.46, for x2 ; you will have to work out x1 for yourself), to show that


(x)={DPxexp-1hx2xpx'dx'2DPxsin1hx2xpx'dx'+4DPx2cos1hvx1pxdx+sine1hvx1pxdx

FIGURE 8.13: Symmetrical double well; Problem 8.15.
Where

1hx1x2p(x)dx .

(b) Because v(x) is symmetric, we need only consider even (+)and odd (-)wave functions. In the former case (0)=0 , and in the latter case =(0)=0 . Show that this leads to the following quantization condition:

tan=2ef.

Where

f1h-x1x1p(x)dx

Equation 8.59determines the (approximate) allowed energies (note that Ecomes into 1 and x2, so and are both functions of E ).

(c) We are particularly interested in a high and/or broad central barrier, in which case is large, and eis huge. Equation 8.59 then tells us thatA must be very close to a half-integer multiple of . With this in mind, write localid="1658823154085" (=n+1/2)+o . where localid="1658823172105" ||<<1, and show that the quantization condition becomes

localid="1658823190448" =(n+12)m12e-f.

(d) Suppose each well is a parabola: 16

localid="1658823244259" v(x)={12m2x+a2,ifx<012m2x-a2,ifx<0 .

Sketch this potential, find (Equation 8.58 ), and show that

localid="1658823297094" En=(n+12)hmh2e-f

Comment: If the central barrier were impenetrable localid="1658823318311" (f) , we would simply have two detached harmonic oscillators, and the energies, localid="1658823339386" En=(n+1/2)h , would be doubly degenerate, since the particle could be in the left well or in the right one. When the barrier becomes finite (putting the two wells into "communication"), the degeneracy is lifted. The even (localid="1658823360600" n+) states have slightly lower energy, and the odd ones n. have slightly higher energy.
(e) Suppose the particle starts out in the right well-or, more precisely, in a state of the form

localid="1658823391675" (x,0)=12(n++n)

which, assuming the phases are picked in the "natural" way, will be concentrated in the right well. Show that it oscillates back and forth between the wells, with a period

localid="1658823416179" =22e.

(f) Calculate , for the specific potential in part (d), and show that for , localid="1658823448873" V(0).>>E,~ma2/h.

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