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Introduction to Quantum Mechanics

David J. Griffiths

Physics

2nd Edition 路 469 pages

ISBN: 9780131118928

Chapter 2: Q18P (page 66)

Show that $[A e^{i k x} + B e^{- i k x}]$ and $[C c o s (k x) + D s i n (k x)]$ are equivalent ways of writing the same function of $x$ , and determine the constants $C$ and $D$ in terms of $A$ and $B$ , and vice versa.

Short Answer

Expert verified

The constants $C$ and $D$ in terms of $A$ and $B$ , and vice versa are,

(i) $C = A + B$ and $D = i (A - B)$

(ii) $A = \frac{1}{2} (C - i D); B = \frac{1}{2} (C + i D)$

Step by step solution

01

Step 1:Formulae used

The required formula are

$e^{ikx} = (c o s (k x) + i s i n (k x))$

$e^{- i k x} = (c o s (k x) - i s i n (k x))$

$c o s (k x) = (\frac{e^{i k x} + e^{- i k x}}{2})$

$s i n (k x) = D (\frac{e^{i k x} - e^{- i k x}}{2})$

02

Compute C and D in terms of A and B and vice versa

Apply Euler's formula,

$A e^{i k x} + B e^{- i k x} = A (\cos (k x) + i \sin (k x)) + B (\cos (k x) - i \sin (k x))$

$= (A + B) \cos (k x) + i (A - B) \sin (k x)$

$= C \cos (k x) + D \sin (k x)$

Thus, $C$ and $D$ expressed in terms of arbitrary constants $A$ and $B$ , $C = A + B$

$D = i (A - B)$

Now, Solve both the equations for $A$ and B and get,

$C \cos (k x) + D \sin (k x) = C (\frac{e^{i k x} + e^{i k x}}{2}) + D (\frac{e^{i k x} - e^{k x}}{2 i})$

$= \frac{1}{2} (C - i D) e^{i k x} + \frac{1}{2} (C + i D) e^{i k x}$

$= A e^{A x} + B e^{i x x}$

Therefore, the values of $A$ and $B$ are,

A = \frac{1}{2} (C - i D)

B = \frac{1}{2} (C + i D)

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Most popular questions from this chapter

A particle in the harmonic oscillator potential starts out in the state $唯 (x, 0) = A [3 蠄_{0} (x) + 4 蠄_{1} (x)]$

a) Find $A$ .

b) Construct $唯 (x, t)$ and $| 唯 (x, t)^{2} |$

c) Find $鉄� x 鉄�$ and $鉄� p 鉄�$ . Don't get too excited if they oscillate at the classical frequency; what would it have been had I specified $蠄_{2} (x)$ , instead of $蠄_{1} (x)$ ?Check that Ehrenfest's theorem holds for this wave function.

d) If you measured the energy of this particle, what values might you get, and with what probabilities?

Delta functions live under integral signs, and two expressions $(D_{1} (x) a n d D_{2} (x))$ involving delta functions are said to be equal if

${鈭�}_{- 鈭�}^{+ 鈭�} f (x) D_{1} (x) d x = {鈭�}_{- 鈭�}^{+ 鈭�} f (x) D_{2} (x) d x$ for every (ordinary) function f(x).

(a) Show that

$未 (c x) = \frac{1}{| c |} 未 (x)$ (2.145)

where c is a real constant. (Be sure to check the case where c is negative.)

(b) Let $胃 (x)$ be the step function:

$胃 (x) 鈮� {\begin{cases} 1, x > 0 \\ 0, x > 0 \end{cases}$ (2.146).

(In the rare case where it actually matters, we define $胃 (0)$ to be 1/2.) Show that $d 胃 l d x = 未$

Analyze the odd bound state wave functions for the finite square well. Derive the transcendental equation for the allowed energies and solve it graphically. Examine the two limiting cases. Is there always an odd bound state?

A particle of mass m is in the potential

$V (x) = {\begin{cases} 鈭�, (x < 0) \\ \frac{- 32 h^{2}}{{ma}^{2}}, (0 鈮� x 鈮� a) \\ 0, (x > a) \end{cases}$

How many bound states are there?

In the highest-energy bound state, what is the probability that the particle would be found outside the well (x>a)? Answer: 0.542, so even though it is 鈥渂ound鈥� by the well, it is more likely to be found outside than inside!

A particle of mass m is in the ground state of the infinite square well (Evaluation 2.19). Suddenly the well expands to twice its original size 鈥� the right wall moving from a to 2a 鈥� leaving the wave function (momentarily) undisturbed. The energy of the particle is now measured.

What is the most probable result? What is the probability of getting that result?
What is the next most probable result, and what is its probability?
What is the expectation value of the energy? (Hint: if you find yourself confronted with an infinite series, try another method)

See all solutions

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