Introduction To Quantum Mechanics
Introduction To Quantum Mechanics
3rd Edition
ISBN: 9781107189638
Author: Griffiths, David J., Schroeter, Darrell F.
Publisher: Cambridge University Press
Question
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Chapter 2, Problem 2.42P

(a)

To determine

The normalized wave function.

(a)

Expert Solution
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Answer to Problem 2.42P

The normalized wave function is Ψ(x,0)=(2aπ)1/4eax2eilx_.

Explanation of Solution

Given that the wave function is;

  Ψ(x,0)=(2aπ)1/4eax2eilx        (I)

Normalize the wave function.

  |Ψ|2dx=1|Aeax2eilx|2dx=1A2e2ax2dx=1A2π2a=1A=(2aπ)1/4        (II)

Use the normalization constant to rewrite the wave function.

  Ψ(x,0)=(2aπ)1/4eax2eilx        (III)

Conclusion:

Therefore, the normalized wave function is Ψ(x,0)=(2aπ)1/4eax2eilx_.

(b)

To determine

The wave function at time t, the speed of the Gaussian envelop and the speed of the wave.

(b)

Expert Solution
Check Mark

Answer to Problem 2.42P

The wave function at time t is Ψ(x,t)=(2aπ)1/41γea(xltm)2/γ2eil(xlt2m)_. The speed of the Gaussian envelope is l/m_ and the speed of the wave is l/2m_.

Explanation of Solution

Write the solution to the generic quantum problem, for a free particle;

  ϕ(k)=12πΨ(x,0)eikxdx        (IV)

Use equation (III) in (IV).

  ϕ(k)=12π(2aπ)1/4eax2eilxeikxdx        (V)

Thus, the wave function at time t can be represented as;

  Ψ(x,t)=12π1(2πa)1/4e(kl)24aei(kxk2t2m)dk        (VI)

Let ukl, so k=u+l and dk=du (since l is constant). Now equation (VI) can be written as;

  Ψ(x,t)=12π1(2πa)1/4eu24aei[ux+lx(t/2m)(u2+2ul+l2)]du=12π1(2πa)1/4eil(xlt2m)eu2(14a+it2m)+iu(xltm)du        (VII)

The integral in equation (VII) becomes;

  114a+it2me(xltm)2/4(14a+it2m)ey2dy=2aγea(xltm)2/γ2π

Thus, the wave function at time t can be expressed as;

  Ψ(x,t)=(2aπ)1/41γea(xltm)2/γ2eil(xlt2m)        (VIII)

In the above expression, the first exponential term which represent the Gaussian envelop travels at speed l/m, and the second exponential term which represent the sinusoidal wave travel at speed l/2m.

Conclusion:

Therefore, the wave function at time t is Ψ(x,t)=(2aπ)1/41γea(xltm)2/γ2eil(xlt2m)_. The speed of the Gaussian envelope is l/m_ and the speed of the wave is l/2m_.

(c)

To determine

The probability density |Ψ(x,t)|2 and to plot the same.

(c)

Expert Solution
Check Mark

Answer to Problem 2.42P

The probability density is |Ψ(x,t)|2=2πωe2ω2(xltm)2_ where ω=a1+θ2. The plot of |Ψ(x,t)|2 is shown in Figure 1.

Explanation of Solution

The wave function is obtained from part (b) is given in equation (VIII).

  Ψ(x,t)=(2aπ)1/41γea(xltm)2/γ2eil(xlt2m)

The probability density can be expressed as;

  |Ψ(x,t)|2=(2aπ)1/21|γ|2ea(xltm)2[1γ2+1(γ*)2]        (IX)

Simplify the term in the square bracket.

  [1γ2+1(γ*)2]=1|γ|4[(γ*)2+γ2]=1|γ|4(12itm+1+2itm)=2|γ|4        (X)

The term |γ|2 can be expressed as;

  |γ|2=(1+2iatm)(12iatm)=1+θ2        (XI)

Where, θ2atm.

Thus, equation (IX) can be modified as;

  |Ψ(x,t)|2=(2aπ)1/211+θ2e2a(xltm)2/(1+θ2)=2πωe2ω2(xltm)2        (XI)

Where ω=a1+θ2.with θ=2atm.

Thus, the graph of |Ψ(x,t)|2 will be flattening Gaussian as shown in Figure 1.

Introduction To Quantum Mechanics, Chapter 2, Problem 2.42P

Conclusion:

Therefore, the probability density is |Ψ(x,t)|2=2πωe2ω2(xltm)2_ where ω=a1+θ2. The plot of |Ψ(x,t)|2 is shown in Figure 1.

(d)

To determine

The expectation values x, p, x2, p2 and σx and σp

(d)

Expert Solution
Check Mark

Answer to Problem 2.42P

The expectation values and σ values are obtained as; x=lmt, x2=14ω2+(ltm)2, p=l, p2=2(a+l2), σx=12ω, and σp=a.

Explanation of Solution

Write the expression for the expectation value of x.

  x=x|Ψ(x,t)|2dx        (XII)

Use equation (XI) in (IX) and solve the integral.

  x=2πxωe2ω2(xltm)2dx

Let yxvt, so x=y+vt.

  x=2π(y+vt)ωe2ω2y2dyx=vtx=lmt

Here, the first integral is trivially zero; and the second is 1 by normalization.

The expectation value of p can be found as;

  p=mdxdt        (XIII)

Substitute x in (XIII).

  p=mddt(lmt)p=l

The expectation value x2 can be determined as;

  x2=2ω(y+vt)2ωe2ω2y2dyx2=14ω2+0+(vt)2x2=14ω2+(ltm)2

The expectation value p2 can be found out as;

  p2=2Ψ*2Ψx2dx        (XIV)

  Ψx=[2aγ2(xltm)+il]Ψ2Ψx2=2aγ2Ψ+[2aγ2(xltm)+il]2Ψ2Ψx2=(Ax2+Bx+C)Ψ

Where, A, B, and C are defined as;

  A(2aγ2)B(2aγ2)22ltm4ialγ2B4ialγ2

  C2aγ2+(2aγ2)2(ltm)2+(4ialγ2)(ltm)l2C1γ4(2aγ2+l2)

Write p2 in terms of A, B, and C.

  p2=2Ψ*[Ax2+Bx+C]Ψdx=2[Ax2+Bx+C]        (XV)

Use x2 and x in equation (XV).

  p2=2γ4[4a2(14ω2+(ltm)2)4ial(ltm)(2aγ2+l2)]p2=2γ4{[a+a(2atm)22a4ia22tm]+l2[14iatm+4(atm)2]}p2=2γ4(aγ4l2γ4)p2=2(a+l2)

The values of σx and σp can be found as follows.

  σx2=x2x2σx2=14ω2+(ltm)2(ltm)2σx=14ω2σx=12ω

  σp2=p2p2σp2=2a+2l22l2σp=2aσp=a

Conclusion:

Therefore, the expectation values and σ values are obtained as; x=lmt, x2=14ω2+(ltm)2, p=l, p2=2(a+l2), σx=12ω, and σp=a.

(e)

To determine

Whether the uncertainty principle holds or not.

(e)

Expert Solution
Check Mark

Answer to Problem 2.42P

The uncertainty principle holds for the given case.

Explanation of Solution

The values of σx and σp are obtained as;

  σx=12ω

  σp=a

The product of  σx and σp is obtained as;

  σxσp=12ωa

Use the expression for ω in the uncertainty product.

  σxσp=a2(a/1+(2atm)2)σxσp2

Thus, uncertainty principle holds for this case.

Conclusion:

Therefore, the uncertainty principle holds for the given case.

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