/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 17 Show that \(\left[A e^{i k x}+B ... [FREE SOLUTION] | 91Ó°ÊÓ

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Show that \(\left[A e^{i k x}+B e^{-i k x}\right]\) and \([C \cos k x+D \sin k x]\) 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. Comment: In quantum mechanics, when \(V=0,\) the exponentials represent traveling waves, and are most convenient in discussing the free particle, whereas sines and cosines correspond to standing waves, which arise naturally in the case of the infinite square well.

Short Answer

Expert verified
Exponential form is equivalent to \(C = A + B\), \(D = i(A - B)\); vice versa \(A = \frac{C - iD}{2}\), \(B = \frac{C + iD}{2}\).

Step by step solution

01

Express Exponential Functions in Terms of Trigonometric Functions

Using Euler’s formulas, we have the identities: \[ e^{i k x} = \cos(kx) + i \sin(kx) \ e^{-i k x} = \cos(kx) - i \sin(kx)\]Substitute these into the expression \(Ae^{i k x} + Be^{-i k x}\).
02

Expand Using Euler's Formulas

Substitute the expressions for \(e^{i k x}\) and \(e^{-i k x}\):\[\begin{align*}Ae^{i k x} + Be^{-i k x} &= A(\cos k x + i \sin k x) + B(\cos k x - i \sin k x) \&= (A + B) \cos k x + i(A - B) \sin k x\end{align*}\]
03

Match Terms with Trigonometric Form

Compare the expanded form with the trigonometric form \(C \cos k x + D \sin k x\). From the comparison, we obtain:\[ C = A + B \quad \text{and} \quad D = i(A - B) \]
04

Solve for A and B in Terms of C and D

Using the equations obtained, express \(A\) and \(B\) in terms of \(C\) and \(D\):\[\begin{align*}A &= \frac{C - iD}{2} \B &= \frac{C + iD}{2}\end{align*}\]
05

Verify Equivalent Forms

Substitute the values of \(C\) and \(D\) back into the expression to verify it gives the original exponential function. This confirms that both forms represent the same function. Thus, \(A e^{i k x} + B e^{-i k x}\) is equivalent to \(C \cos k x + D \sin k x\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Euler's formulas
Euler's formulas are essential mathematical expressions that link the exponential function with trigonometric functions. These formulas are key in converting expressions between exponential and trigonometric forms. Specifically, they are represented as:
  • \( e^{i k x} = \cos(kx) + i \sin(kx) \)
  • \( e^{-i k x} = \cos(kx) - i \sin(kx) \)
These identities essentially show how complex exponentials can be expressed as combinations of sine and cosine functions.
In the context of quantum mechanics, Euler's formulas help us simplify and transform wave functions from one form to another, making calculations more flexible and intuitive.
Understanding these transformations allows us to shift between different representations of wave functions, as seen in the problem where we express a wave function involving exponentials in terms of sine and cosine.
Wave functions
Wave functions are fundamental concepts in quantum mechanics that describe the quantum state of a particle. They provide essential information about a system with properties like position, momentum, and energy. In mathematical terms, a wave function is often represented by the symbol \(\psi(x)\), which is a complex-valued function.The exercise provided presents the wave function as a combination of exponential forms:\( A e^{i k x} + B e^{-i k x}\). These express the wave nature of particles, linking them to their wave vector \(k\), which itself is related to momentum.The exponential form of wave functions is particularly useful when dealing with traveling waves, as it succinctly captures both magnitude and phase information. This is essential when particles move freely, experiencing no forces (\(V = 0\)) in quantum terms.
Transitioning to the trigonometric form through Euler's formulas allows us to represent these functions using sine and cosine, suitable for describing standing waves. Each form - exponential or trigonometric - has its own merits and is adapted to specific situations in quantum mechanics.
Free particle
A free particle in quantum mechanics refers to a particle that moves without external influence, meaning there is no potential energy acting on it (\(V=0\)). In such cases, the Schrödinger equation simplifies, and solutions often involve wave functions that describe traveling waves.The expression \(A e^{i k x} + B e^{-i k x}\) is associated with such a free particle, as the exponentials indicate traveling waves moving in opposite directions. Matching free particle characteristics, these waves reflect the particle's movement through space without constraints.Conversely, the use of sine and cosine functions in the trigonometric form \(C \cos k x + D \sin k x\) corresponds to standing waves. Standing waves are typically encountered in bounded systems, like particles in a box, where the movement is restricted, and the reflected waves create a stationary pattern.
In summary, free particles and wave functions underscore the profound interconnection of quantum states, expressed in both exponential and trigonometric terms, where each representation serves distinct physical scenarios.

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Most popular questions from this chapter

Show that there is no acceptable solution to the (time-independent) Schrödinger equation for the infinite square well with \(E=0\) or \(E<0\). (This is a special case of the general theorem in Problem \(2.2,\) but this time do it by explicitly solving the Schrödinger equation, and showing that you cannot satisfy the boundary conditions.)

Find the allowed energies of the balf harmonic oscillator $$V(x)=\left\\{\begin{array}{ll} (1 / 2) m \omega^{2} x^{2}, & x >0 \\ \infty, & x < 0 \end{array}\right.$$ (This represents, for example, a spring that can be stretched, but not compressed.) Hint: This requires some careful thought, but very little actual calculation.

Evaluate the following integrals: (a) \(\int_{-3}^{+1}\left(x^{3}-3 x^{2}+2 x-1\right) \delta(x+2) d x.\) (b) \(\int_{0}^{\infty}[\cos (3 x)+2] \delta(x-\pi) d x.\) (c) \(\int_{-1}^{+1} \exp (|x|+3) \delta(x-2) d x.\)

In this problem you will show that the number of nodes of the stationary states of a one-dimensional potential always increases with energy. 58 Consider two (real, normalized) solutions \(\left(\psi_{n} \text { and } \psi_{m}\right)\) to the timeindependent Schrödinger equation (for a given potential \(V(x)\) ), with energies \(E_{n}>E_{m}.\) (a) Show that $$\frac{d}{d x}\left(\frac{d \psi_{m}}{d x} \psi_{n}-\psi_{m} \frac{d \psi_{n}}{d x}\right)=\frac{2 m}{\hbar^{2}}\left(E_{n}-E_{m}\right) \psi_{m} \psi_{n}.$$ (b) Let \(x_{1}\) and \(x_{2}\) be two adjacent nodes of the function \(\psi_{m}(x) .\) Show that $$\psi_{m}^{\prime}\left(x_{2}\right) \psi_{n}\left(x_{2}\right)-\psi_{m}^{\prime}\left(x_{1}\right) \psi_{n}\left(x_{1}\right)=\frac{2 m}{\hbar^{2}}\left(E_{n}-E_{m}\right) \int_{x_{1}}^{x_{2}} \psi_{m} \psi_{n} d x.$$ (c) If \(\psi_{n}(x)\) has no nodes between \(x_{1}\) and \(x_{2},\) then it must have the same sign everywhere in the interval. Show that (b) then leads to a contradiction. Therefore, between every pair of nodes of \(\psi_{m}(x), \psi_{n}(x)\) must have at least one node, and in particular the number of nodes increases with energy.

If two (or more) distinct \(^{56}\) solutions to the (time-independent) Schrödinger equation have the same energy \(E,\) these states are said to be degenerate. For example, the free particle states are doubly degenerate- one solution representing motion to the right, and the other motion to the left. But we have never encountered normalizable degenerate solutions, and this is no accident. Prove the following theorem: In one dimension \(^{57}\) \((-\infty

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