Chapter 6: Q13E (page 224)

Show that $ψ (x) = A' e^{i k x} + B' e^{- i k x}$ is equivalent to $ψ (x) = A s i n k x + B c o s k x$ , provided that $A' = \frac{1}{2} (B - i A)$ $B' = \frac{1}{2} (B + i A)$ .

Short Answer

Expert verified

Hence, the proof for the equation is obtained.

Step by step solution

Concept involved

According to theEuler’s formulain complex numbers, it can be written that:

$e^{i φ} = c o s φ + i s i n φ \dots \dots (1)$

Given/known parameters

Consider the given function:

$ψ (x) = A' e^{i k x} + B' e^{- i k x}$

Consider the equations:

$A' = \frac{1}{2} (B - i A) \dots \dots (2)$

$B' = \frac{1}{2} (B + i A) \dots \dots (3)$

Solution

Apply Euler’s formula from equation (1) and solve:

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

Rewrite the above equation as,

$ψ (x) = (A' + B') \cos k x + (A' - B') \sin (k x) \dots . . (4)$

Now, by using equation (2) and (3) in equation (4) solve as:

_{$ψ (x) = (\frac{1}{2} [B - i A] + \frac{1}{2} [B + i A]) \cos (k x) - (\frac{1}{2} [B - i A] - \frac{1}{2} [B + i A]) \sin (k x) ψ (x) = B \cos (k x) + A \sin (k x)$}

Thus, you can say that: $ψ (x) = A' e^{i k x} + B' e^{- i k x}$ is equivalent to

$ψ (x) = B \cos (k x) + A \sin (k x)$ provided that $A' = \frac{1}{2} (B - i A)$ and $B' = \frac{1}{2} (B + i A)$

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Show that $ψ (x) = A' e^{i k x} + B' e^{- i k x}$ is equivalent to $ψ (x) = A s i n k x + B c o s k x$ , provided that $A' = \frac{1}{2} (B - i A)$ $B' = \frac{1}{2} (B + i A)$ .

Short Answer

Step by step solution

Concept involved

Given/known parameters

Solution

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Show that ψ(x)=A'eikx+B'e-ikxis equivalent to ψ(x)=Asinkx+Bcoskx, provided that A'=12(B-iA)B'=12(B+iA).

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Concept involved

Given/known parameters

Solution

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Most popular questions from this chapter

Recommended explanations on Physics Textbooks

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Show that $ψ (x) = A' e^{i k x} + B' e^{- i k x}$ is equivalent to $ψ (x) = A s i n k x + B c o s k x$ , provided that $A' = \frac{1}{2} (B - i A)$ $B' = \frac{1}{2} (B + i A)$ .