Let ψ1ψ_1 and ψ2ψ_2 be two solutions of Eq. (40.2340.23) [−h22md2ψ(x)dx2+U(x)ψ(x)=Eψ(x)-\frac{$$h^2$$}{2m}\frac{$$d^2$$\psi(x)}{$$dx^2$$}+U\left(x\right)\psi\left(x\right)=E\psi\left(x\right)] with energies E1E_1 and E2E_2 respectively, where E1≠E2E_1≠$$E_2. Is$$ ψ=Aψ1+Bψ2ψ = Aψ_1 + Bψ_2, where AA and BB are nonzero constants, a solution to Eq. (40.2340.23)? Explain your answer.

Ch 40: Quantum Mechanics I: Wave Functions

Young & Freedman Calc - University Physics 14th Edition

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Young & Freedman Calc 14th Edition

Ch 40: Quantum Mechanics I: Wave Functions

Problem 7

Chapter 40, Problem 7

Let $psi sub 1$ and $psi sub 2$ be two solutions of Eq. ( $40.23$ ) [ $- \frac{h^{2}}{2 m} \frac{d^{2} ψ (x)}{d x^{2}} + U (x) ψ (x) = E ψ (x)$ ] with energies $E sub 1$ and $E sub 2$ respectively, where $E sub 1 is not equal to E sub 2$ . Is $ψ = A ψ_{1} + B ψ_{2}$ , where $A$ and $B$ are nonzero constants, a solution to Eq. ( $40.23$ )? Explain your answer.

Verified step by step guidance

Step 1: Start by understanding the given equation, which is the time-independent Schrödinger equation:

-\(\frac{\hbar^2}{2m}\)\(\frac{d^2\psi(x)}{dx^2}\) + U(x)\(\psi\)(x) = E\(\psi\)(x)

. Here,

\(\psi\)(x)

is the wavefunction,

U(x)

is the potential energy,

E

is the energy,

\(\hbar\)

is the reduced Planck's constant, and

m

is the mass of the particle.

Step 2: Recognize that

\(\psi\)_1

and

\(\psi\)_2

are solutions to the Schrödinger equation with energies

E_1

and

E_2

, respectively. This means they satisfy the equations:

-\(\frac{\hbar^2}{2m}\[\frac{d^2\psi_1}{dx^2}\) + U(x)\(\psi\)_1 = E_1\(\psi\)_1

and

-\(\frac{\hbar^2}{2m}\]\frac{d^2\psi_2}{dx^2}\) + U(x)\(\psi\)_2 = E_2\(\psi\)_2

Step 3: Substitute the proposed solution

\(\psi\) = A\(\psi\)_1 + B\(\psi\)_2

into the Schrödinger equation. This gives:

-\(\frac{\hbar^2}{2m}\)\(\frac{d^2(A\psi_1 + B\psi_2)}{dx^2}\) + U(x)(A\(\psi\)_1 + B\(\psi\)_2) = E(A\(\psi\)_1 + B\(\psi\)_2)

Step 4: Use the linearity of differentiation and the potential term to expand the equation:

A\(\left\)(-\(\frac{\hbar^2}{2m}\[\frac{d^2\psi_1}{dx^2}\) + U(x)\(\psi\)_1\(\right\)) + B\(\left\)(-\(\frac{\hbar^2}{2m}\]\frac{d^2\psi_2}{dx^2}\) + U(x)\(\psi\)_2\(\right\)) = E(A\(\psi\)_1 + B\(\psi\)_2)

. Substitute the known equations for

\(\psi\)_1

and

\(\psi\)_2

, which are

E_1\(\psi\)_1

and

E_2\(\psi\)_2

, respectively.

Step 5: After substitution, the equation becomes:

AE_1\(\psi\)_1 + BE_2\(\psi\)_2 = E(A\(\psi\)_1 + B\(\psi\)_2)

. Since

E_1 \(\neq\) E_2

, this equation cannot hold true for all values of

x

unless

A

B

is zero, which contradicts the problem's condition that both are nonzero. Therefore,

\(\psi\) = A\(\psi\)_1 + B\(\psi\)_2

is not a solution to the Schrödinger equation.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Linear Superposition Principle

The linear superposition principle states that if two or more solutions to a linear differential equation exist, any linear combination of these solutions is also a solution. In quantum mechanics, this principle is fundamental as it allows for the construction of new wave functions from known solutions, enabling the analysis of complex systems.

Time-Independent Schrödinger Equation

The time-independent Schrödinger equation describes how the quantum state of a physical system changes in space, given by the equation -h^2/2m(d^2ψ(x))/(dx^2) + U(x)ψ(x) = Eψ(x). Here, ψ(x) represents the wave function, U(x) is the potential energy, and E is the total energy. Solutions to this equation provide the allowed energy levels and corresponding wave functions of a quantum system.

Orthogonality of Quantum States

In quantum mechanics, eigenstates corresponding to different energy levels are orthogonal, meaning their inner product is zero. This property implies that if E_1 and E_2 are distinct energies, the wave functions ψ_1 and ψ_2 are orthogonal. Consequently, a linear combination of these states, such as ψ = Aψ_1 + Bψ_2, is still a valid solution, but it must be treated with care regarding normalization and physical interpretation.

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An electron is moving as a free particle in the $- x$ -direction with momentum that has magnitude $4.50 \times 10^{- 24}$ kg*m/s. Let $k_{2} = 3 k_{1} = 3 k$ . At $t equals 0$ , the probability distribution function $∣ Ψ (x, t) ∣^{2}$ has a maximum at $x equals 0$ .

(a) What is the smallest positive value of $x$ for which the probability distribution function has a maximum at time $t = \frac{2 π}{ω}$ , where $ω = h k^{2} / 2 m$ ?

(b) From your result in part (a), what is the average speed with which the probability distribution is moving in the $+ x$ -direction?

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A particle is described by a wave function $ψ (x) = A e^{- α x^{2}}$ , where $A$ and $α$ are real, positive constants. If the value of $α$ is increased, what effect does this have on (a) the particle’s uncertainty in position and (b) the particle’s uncertainty in momentum? Explain your answers.

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Textbook Question

(a) Find the lowest energy level for a particle in a box if the particle is a billiard ball ( $m equals 0.20$ kg) and the box has a width of $1.3$ m, the size of a billiard table. (Assume that the billiard ball slides without friction rather than rolls; that is, ignore the rotational kinetic energy.)

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Consider a wave function given by $ψ (x) = A s i n k x$ , where $k = 2 π / λ$ and $A$ is a real constant.

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(b) For which values of $x$ is the probability zero? Explain.

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