Problem Set 8

I.

a.

• The potential energy for the finite square well is as follows:

V(x) = -V₀ for –a < x < a

V(x) = 0 otherwise

• In this problem, we are investigating bound states (E < 0) that are also odd: y(-x) = - y(x).
• We need to solve the Schrodinger equation in three regions: x < -a, -a < x < a, and x > a.
• In the region x < -a, V = 0, so the Schrodinger equation is –(ћ²/2m)d²y/dx² = Ey.
• Upon rearrangement: d²y/dx² = -(2mE/ћ²)y.
• This becomes d²y/dx² = a²y by defining a = (-2mE)^1/2/ћ. (Observe that a is real because E < 0.)
• A general solution to this differential equation is y(x) = Ae^ax + Be^-ax.
• Since e^-ax ® ¥ as x ® -¥, B must be 0 for y to be normalizable.
• In the region –a < x < a, V = -V₀, so the Schrodinger equation is –(ћ²/2m)d²y/dx² –V₀y = Ey.
• Upon rearrangement: d²y/dx² = -[2m(E+V₀)/ћ²]y.
• This becomes d²y/dx² = -k²y by defining k = [2m(E+V₀)]^1/2/ћ.
• A general solution to this differential equation is y(x) = Csin(kx) + Dcos(kx), but we're only looking at odd solutions now, so we'll set D = 0.
• In the region x > a, V = 0, so the Schrodinger equation is –(ћ²/2m)d²y/dx² = Ey, just as it was in the region x < -a.
• We rewrite the Schrodinger equation as d²y/dx² = -a²y, where once again a = (-2mE)^1/2/ћ.
• A general solution to this differential equation is y(x) = Fe^ax + Ge^-ax.
• Since e^ax ® ¥ as x ® ¥, F must be 0 for y to be normalizable.
• So our wavefunction is

y(x) = Ae^ax for x < -a

y(x) = Csin(kx) for –a < x < a

y(x) = Ge^-ax for x > a

• To find a formula for allowed energies, we apply the continuity conditions at x = a.
• The continuity of y at x = a gives Csin(ka) = Ge^-aa.
• The continuity of dy/dx at x = a gives kCcos(ka) = -aG^-aa.
• If we divide the second equation by the first, we obtain kcot(ka) = -a, or –cot(ka) = a/k.
• We define z = ka and z₀ = (a/ћ)(2mV₀)^1/2.
• Then our equation becomes –cot(z) = a/k = (-2mE)^1/2/(ћk) through the definition of a.
• Next, -cot(z) = [-2mE/(ћ²k²)]^1/2 = [-(k²ћ² – 2mV₀)/(ћ²k²)]^1/2 by writing E in terms of k (from the definition of k).
• Next, -cot(z) = [(2mV₀)/(ћ²k²) – 1]^1/2 = [(2mV₀)/(ћ²k²) – 1]^1/2 = (z₀²/z² – 1)^1/2. Check it!

b.

• To determine the number of bound states, we must determine z₀.
• According to our definition, z₀ = (a/ћ)(2mV₀)^1/2.
• We arrive at more convenient units if we multiply the equation by c/c: z₀ = [a/(ћc)](2mc²V₀)^1/2 = (1 nm/197 eV^.nm)(2 ´ 511 keV ´ 9 eV)^1/2 = 15.4.
• It is useful to write z₀ as a multiple of p/2: z₀ = 15.4 = 9.8p/2.
• We can see from the figure below that whenever z₀ increases by p/2 (1.57), we pick up a new intersection with tan(z) or –cot(z).
• So in our case, we have nine intersections after the initial intersection, giving us ten intersections.
• Each intersection represents an allowed bound-state energy level and an allowed wavefunction.

6-42.

a.

• In the region x > 0, V = -V₀, so the Schrodinger equation is –(ћ²/2m)d²y/dx² –V₀y = Ey.
• Upon rearrangement: d²y/dx² = -[2m(E+V₀)/ћ²]y.
• This becomes d²y/dx² = -k₂²y by defining k₂ = [2m(E+V₀)]^1/2/ћ.
• A general solution to this differential equation is y₂(x) = Cexp(ik₂x) + Dexp(-ik₂x), so k₂ = [2m(E+V₀)]^1/2/ћ is the wave number in the region of positive x.
• We're given E = 2V₀, so k₂ = (6mV₀)^1/2/ћ.
• We're also given (or could derive) ћ²k₁²/(2m) = 2V₀, so m = ћ²k₁²/(4V₀).
• So k₂ = (3ћ²k₁²/2)^1/2/ћ = (3/2)^1/2k₁.

b.

• In the region x < 0, V = 0, the Schrodinger equation is –(ћ²/2m)d²y/dx² = Ey.
• Upon rearrangement: d²y/dx² = -(2mE/ћ²)y = -k₁²y, where k₁ = (2mE)^1/2/ћ.
• A general solution to this differential equation is y₁(x) = Aexp(ik₁x) + Bexp(-ik₁x).
• The first term above represents a beam moving to the right (the incident wave). The second term represents a beam moving to the left (the reflected wave).
• In the region x > 0, y₂(x) = Cexp(ik₂x). This is the transmitted wave. D = 0 because there is no wave approaching from the right.
• We apply the continuity of y at x = 0: A + B = C.
• We apply the continuity of dy/dx at x = 0: ik₁A – ik₁B = ik₂C.
• We use the first equation to eliminate C from the second equation: k₁A – k₁B = k₂A + k₂B.
• Solving for B: B = (k₁ – k₂)A/(k₁ + k₂).
• The reflection coefficient R = |C|²/|A|² = (k₁ - k₂)²/(k₁ + k₂)².
• Observe that this expression for a downward step is identical to the expression for an upward step. (For a downward step, k₂ > k₁, and for an upward step, k₂ < k₁.)

c.

Since all of the incident beam is either reflected or transmitted, T = 1 – R = 4k₁k₂/(k₁ + k₂)².

d.

• We found in (a) that k₂ = (3/2)^1/2k₁.
• So T = 4(3/2)^1/2k₁²/[k₁ + (3/2)^1/2k₁]² = 4(3/2)^1/2/[1 + (3/2)^1/2]² = 0.99.
• So if a million particles arrive each second at the potential step, about 990,000 are transmitted each second.
• Classically, all the particles should continue unimpeded since the potential energy everywhere is less than the particle energy; in fact, the particles gain energy as they cross the potential step.

6-53.

a.

• Since |y(x)|² is symmetric (even), |y(-x)|² = |y(x)|².
• Since y(-x) and y(x) have the same magnitude, they can only differ by some factor C whose magnitude is 1: y(-x) = Cy(x). (You might want to say immediately that C must be 1 or -1. But we need to prove that it's not complex.)
• Replacing x with –x yields y(x) = Cy(-x). Substituting this into the equation above yields y(-x) = C²y(-x).
• So C² = 1, and C = ±1.

b.

• Inside the well, V = 0, and the Schrodinger equation is –(ћ²/2m)d²y/dx² = Ey.
• Upon rearrangement: d²y/dx² = -(2mE/ћ²)y = -k²y, with k = (2mE)^1/2/ћ.
• A general solution to this differential equation is y(x) = Asin(kx) + Bcos(kx).
• (A less useful general solution is Ae^ikx + Be^-ikx. While we prefer the exponential solutions for reflection/transmission problems, we prefer the sin/cos solutions for bound states.)
• We showed in (a) that y(x) must be even or odd. So y will be either a sine or a cosine.
• Let's work with sine first: y(x) = Asin(kx). y(x) is 0 outside the infinite well, so continuity requires y(x) = 0 at the edge of the well. So 0 = y(L/2) = Asin(kL/2). Sine is 0 when its argument is (integer)p, so kL/2 = (integer)p, or k = 2p(integer)/L. Thus y(x) = Asin[2p(integer)x/L] with (integer) = 1, 2, 3…, or Asin(pnx/L) with n = 2, 4, 6….
• Now we can normalize: òy*ydx = A²òsin²(pnx/L)dx = A²(L/2). (I used the trick that the integral of a sinusoid over one or more complete periods is half the distance between the limits of integration.) This equals 1, so A = (2/L)^1/2.
• Now let's look at cosine: y(x) = Bcos(kx). This must go to 0 at the edge of the well (L/2), so 0 = y(L/2) = Bcos(kL/2). Cosine is 0 when its argument is p/2 + (integer)p, so kL/2 = p/2 + (integer)p, or k = p/L + 2p(integer)/L. Thus y(x) = Bcos[px/L + 2p(integer)x/L] with (integer) = 1, 2, 3…, or Bcos(npx/L) with n = 1, 3, 5….
• Normalization gives B = (2/L)^1/2.

c.

• We'd defined k = (2mE)^1/2/ћ, so E = ћ²k²/(2m).
• We can see from our wavefunctions, sin(kx) and cos(kx), that k = np/L, where n is even for sines and odd for cosines, so all positive integers are allowed.
• So E = ћ²n²p²/(2mL²), which is just what we found when we had a square well between 0 and L.

6-61.

a.

• As always, we must solve the Schrodinger equation in every region.
• In the region x < 0, V = 0, so the Schrodinger equation is –(ћ²/2m)d²y/dx² = Ey.
• Upon rearrangement: d²y/dx² = -(2mE/ћ²)y = -k²y, where k = (2mE)^1/2/ћ.
• A general solution to this differential equation is Ae^ikx + Be^-ikx.
• In the region 0< x < a, V = V₀, so the Schrodinger equation is –(ћ²/2m)d²y/dx² + V₀y = Ey.
• Upon rearrangement: d²y/dx² = [2m(V₀ – E)/ћ²]y = a²y, where a = [2m(V₀ – E)]^1/2/ћ.
• A general solution to this differential equation is Ce^-ax + De^ax.
• In the region x > a, V = 0, so the Schrodinger equation becomes d²y/dx² = -(2mE/ћ²)y = -k²y, where again k = (2mE)^1/2/ћ.
• A general solution to this differential equation is Fe^ikx + Ge^-ikx, but we set G = 0 because there is no wave arriving from the right.
• Now we apply the continuity conditions at x = 0 and x = a.
• The continuity of y at x = 0 yields A + B = C + D. (1)
• The continuity of dy/dx at x = 0 yields ikA – ikB = -aC + aD. (2)
• The continuity of y at x = a yields Ce^-aa + De^aa = Fe^ika. (3)
• The continuity of dy/dx at x = a yields -aCe^-aa + aDe^aa = ikFe^ika. (4)
• The physics ends here. The rest is tedious algebra. Make sure you master all the physics before moving on to the algebra!
• We want to eliminate B, C, and D.
• First, let's use Equations (3) and (4) to solve for C and D in terms of F.
• If we multiply Equation (3) by a and add it to Equation (4), we find 2aDe^aa = (a + ik)Fe^ika, or D = (a + ik)Fe^a(ik-a)/(2a).
• If we multiply Equation (3) by -a and add it to Equation (4), we find -2aCe^-aa = (ik - a)Fe^ika, or C = (a - ik)Fe^a(ik+a)/(2a).
• Now let's plug our equations for C and D into Equation (1):

A + B = (a - ik)Fe^a(ik+a)/(2a) + (a + ik)Fe^a(ik-a)/(2a). (5)

• And let's plug our equations for C and D into Equation (2):

ikA – ikB = -a(a - ik)Fe^a(ik+a)/(2a) + a(a + ik)Fe^a(ik-a)/(2a). (6)

• We can eliminate B by multiplying Equation (5) by ik and adding it to Equation (6): 2ikA = (2aik + k² - a²)Fe^a(ik+a)/(2a) + (2aik – k² + a²)Fe^a(ik-a)/(2a).
• So F = 4aikA/[(2aik + k² - a²)e^a(ik+a) + (2aik – k² + a²)e^a(ik-a)].
• The complex conjugate of the above equation is

F* = -4aikA*/[(-2aik + k² - a²)e^a(-ik+a) + (-2aik – k² + a²)e^a(-ik-a)].

• Notice that I simply replaced every i with –i.
• So F*F =

16a²k²A*A/{[4a²k² + (k² - a²)²]e^2aa - (2aik + k² - a²)² - (2aik – k² + a²)² + [4a²k² + (a² - k²)²]e^-2aa}=

16a²k²A*A/{[4a²k² + (k² - a²)²]e^2aa – [-4a²k² + 4aik(k² - a²) + (k² - a²)²] - [-4a²k² + 4aik(a² - k²) + (a² - k²)²] + [4a²k² + (a² - k²)²]e^-2aa}=

16a²k²A*A/{[4a²k² + (k² - a²)²]e^2aa + 8a²k² - 2(k² - a²)² + [4a²k² + (a² - k²)²]e^-2aa}=

16a²k²A*A/{[4a²k² + (k² - a²)²](e^2aa - 2 + e^-2aa) + 16a²k²} Check it!

= A*A/{sinh²(aa)[4a²k² + (k² - a²)²]/(4a²k²) + 1} Check it!

= A*A/{sinh²(aa)[16E(V₀ – E) + (4E – 2V₀)²]/[16E(V₀ – E)] + 1}

= A*A/{1 + sinh²(aa)[16E(V₀ – E) + 16E² – 16EV₀ + 4V₀²]/[16E(V₀ – E)]}

= A*A/{1 + sinh²(aa)(4V₀²)/[16E(V₀ – E)]}

= A*A/{1 + sinh²(aa)/[(4E/V₀)(1 – E/V₀)]}

• So T = |F|²/|A|² = 1/{1 + sinh²(aa)/[(4E/V₀)(1 – E/V₀)] }.

b.

• When aa >> 1, sinh(aa) = (e^aa + e^-aa)/2 = e^aa/2, and sinh²(aa) = e^2aa/4.
• So T = 1/{1 + e^2aa/[(16E/V₀)(1 – E/V₀)]}.
• The first 1 in the denominator is negligible compared with the very large e^2aa, so T = (16E/V₀)(1 – E/V₀)e^-2aa.