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Chapter 2: Problem 44

If two (or more) distinct \(^{56}\) solutions to the (time-independent) Schr枚dinger 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

Short Answer

Expert verified
There are no degenerate bound states in one dimension; any two solutions are not distinct but multiples of each other.

Step by step solution

01

Identifying the Schr枚dinger equation

The time-independent Schr枚dinger equation in one dimension is given by \[ -\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2} + V(x)\psi = E\psi \]where \( \psi(x) \) is the wave function, \( V(x) \) is the potential energy, \( E \) is the total energy, and \( \hbar \) is the reduced Planck's constant.
02

Multiply the equations

Consider two solutions \( \psi_1 \) and \( \psi_2 \) with the same energy \( E \). Multiply the Schr枚dinger equation for \( \psi_1 \) by \( \psi_2 \) and for \( \psi_2 \) by \( \psi_1 \). We get:1. \( \psi_2 \left( -\frac{\hbar^2}{2m}\frac{d^2\psi_1}{dx^2} + V(x)\psi_1 \right) = E \psi_2 \psi_1 \)2. \( \psi_1 \left( -\frac{\hbar^2}{2m}\frac{d^2\psi_2}{dx^2} + V(x)\psi_2 \right) = E \psi_1 \psi_2 \)
03

Subtract the equations

Subtract the second equation from the first:\[ \psi_2 \left( -\frac{\hbar^2}{2m}\frac{d^2\psi_1}{dx^2} \right) - \psi_1 \left( -\frac{\hbar^2}{2m}\frac{d^2\psi_2}{dx^2} \right) = 0 \]Simplifying, we obtain:\[ -\frac{\hbar^2}{2m} \left( \psi_2 \frac{d^2\psi_1}{dx^2} - \psi_1 \frac{d^2\psi_2}{dx^2} \right) = 0 \]
04

Express as a derivative

The expression can be rewritten using the product rule for differentiation:\[ \frac{d}{dx}\left( \psi_2 \frac{d\psi_1}{dx} - \psi_1 \frac{d\psi_2}{dx} \right) = 0 \]This indicates that the term \( \psi_2 \frac{d\psi_1}{dx} - \psi_1 \frac{d\psi_2}{dx} \) is a constant.
05

Apply boundary conditions

For bound states, the wave functions \( \psi_1 \) and \( \psi_2 \) must go to zero as \( x \rightarrow \pm \infty \). This means that the constant, \( \psi_2 \frac{d\psi_1}{dx} - \psi_1 \frac{d\psi_2}{dx} \), must be zero because if it were not, it would imply non-zero probability density at infinity, which contradicts normalization.
06

Conclude linear dependence

Since \( \psi_2 \frac{d\psi_1}{dx} - \psi_1 \frac{d\psi_2}{dx} = 0 \), it follows from the properties of the wave functions that\[ \frac{d\psi_1}{dx} \] is proportional to \( \frac{d\psi_2}{dx} \). Thus, \( \psi_2 = c \cdot \psi_1 \) for some constant \( c \), implying \( \psi_1 \) and \( \psi_2 \) are linearly dependent and not distinct.

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Key Concepts

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

Time-independent Schr枚dinger Equation
The Time-independent Schr枚dinger Equation is a fundamental part of quantum mechanics that describes how the quantum state of a physical system changes. It鈥檚 a key tool used to determine the allowed energy levels of a quantum system and is represented as:\[ -\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2} + V(x)\psi = E\psi \]In this equation:
  • \( \psi(x) \) is the wave function, which describes the probability amplitude of finding a particle at position \( x \).
  • \( V(x) \) is the potential energy, which varies depending on the position \( x \).
  • \( E \) is the energy eigenvalue, representing the system鈥檚 total energy.
  • \( \hbar \) is the reduced Planck's constant.
This equation outlines how energy is conserved in quantum mechanical systems. Importantly, finding solutions to this equation is what allows physicists to predict how quantum states evolve over time. These solutions must satisfy the boundary conditions specific to the physical problem.
Wave Functions and Normalization
Wave functions (\( \psi(x) \)) play a crucial role in the description of quantum systems. Each wave function corresponds to a state of the quantum system and is used to describe probabilities within that system. An essential feature of wave functions is that they need to be normalized.**Normalization** is the process of adjusting the wave function so that the total probability of finding the particle across all space is equal to one. Mathematically, this is expressed as:\[ \int_{-\infty}^{\infty} |\psi(x)|^2 \, dx = 1 \]Key aspects of normalization:
  • It ensures that the probability of all possible outcomes (positions) sums up to 1.
  • Normalization leads to meaningful physical integers when calculating observables like position or momentum.
  • Wave functions that tend to zero as \( x \to \pm \infty \) are typically normalizable.
Normalizable solutions help distinguish physical, realistic quantum states from mathematical artifacts and ensure the probabilities computed are physically significant.
Bound States in Quantum Mechanics
Bound states refer to quantum states where a particle is confined to a finite region of space by a potential, such as an electron in an atom. In these states, particles have energy levels that are discrete and quantized.In one-dimensional quantum systems, it's particularly relevant to study bound states to understand the motion and energy levels of particles under various potentials. The exercise talks about the degeneracy of these bound states and proves that in one dimension, degenerate bound states cannot exist.**Key features of bound states:**
  • **Discreteness:** The energy levels of a bound system are discrete, not continuous, due to the particle's confinement within potential wells.
  • **Normalization:** The wave functions of particles in bound states must vanish at infinity, ensuring normalizable solutions.
  • **Non-degeneracy in 1D:** In a one-dimensional system, bound states do not exhibit degeneracy under typical potentials, which means that for bound states, \( \psi_1 \) and \( \psi_2 \) with the same energy are not truly distinct solutions; rather, they are multiples of one another, as outlined in the exercise solution.
Bound states provide insight into stable configurations within quantum systems, and the exercise demonstrates why degeneracy is not observed in these conditions for one-dimensional cases.

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Most popular questions from this chapter

The gaussian wave packet. A free particle has the initial wave function \(\Psi(x, 0)=A e^{-a x^{2}}\) where \(A\) and \(a\) are (real and positive) constants. (a) Normalize \(\Psi(x, 0)\) (b) Find \(\Psi(x, t)\). Hint: Integrals of the form $$\int_{-\infty}^{+\infty} e^{-\left(a x^{2}+b x\right)} d x$$ can be handled by "completing the square": Let \(y \equiv \sqrt{a}[x+(b / 2 a)]\) and note that \(\left(a x^{2}+b x\right)=y^{2}-\left(b^{2} / 4 a\right) .\) Answer (c) Find \(|\Psi(x, t)|^{2}\). Express your answer in terms of the quantity \(w \equiv \sqrt{a /\left[1+(2 \hbar a t / m)^{2}\right]}\) Sketch \(|\Psi|^{2}(\text { as a function of } x)\) at \(t=0,\) and again for some very large \(t\) Qualitatively, what happens to \(|\Psi|^{2},\) as time goes on? (d) \(\quad\) Find \(\langle x\rangle,\langle p\rangle,\left\langle x^{2}\right\rangle,\left\langle p^{2}\right\rangle, \sigma_{x},\) and \(\sigma_{p} .\) Partial answer: \(\left\langle p^{2}\right\rangle=a \hbar^{2},\) but it may take some algebra to reduce it to this simple form. (e) Does the uncertainty principle hold? At what time \(t\) does the system come closest to the uncertainty limit?

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.

Consider the potential $$V(x)=-\frac{\hbar^{2} a^{2}}{m} \operatorname{sech}^{2}(a x),$$ where \(a\) is a positive constant, and "sech" stands for the hyperbolic secant. (a) Graph this potential. (b) Check that this potential has the ground state $$\psi_{0}(x)=A \operatorname{sech}(a x),$$ and find its energy. Normalize \(\psi_{0},\) and sketch its graph. (c) Show that the function $$\psi_{k}(x)=A\left(\frac{i k-a \tanh (a x)}{i k+a}\right) e^{i k x},$$ (where \(k \equiv \sqrt{2 m E} / \hbar,\) as usual) solves the Schr枚dinger equation for any (positive) energy \(E .\) since \(\tanh z \rightarrow-1\) as \(z \rightarrow-\infty,\) \(\psi_{k}(x) \approx A e^{i k x}, \quad\) for large negative \(x.\) This represents, then, a wave coming in from the left with no accompanying reflected wave (i.e. no term \(\exp (-i k x)\) ). What is the asymptotic form of \(\psi_{k}(x)\) at large positive \(x\) ? What are \(R\) and \(T\), for this potential? Comment: This is a famous example of a reflectionless potential-every incident particle, regardless its energy, passes right through. 62.

$$\left(1-x^{2}\right) \frac{d^{2} f}{d x^{2}}-2 x \frac{d f}{d x}+\ell(\ell+1) f=0$$ where \(\ell\) is some (non-negative) real number. (a) Assume a power series solution, $$f(x)=\sum_{n=0}^{\infty} a_{n} x^{n}$$ and obtain a recursion relation for the constants \(a_{n}\) (b) Argue that unless the series truncates (which can only happen if \(\ell\) is an integer \(),\) the solution will diverge at \(x=1\) (c) When \(\ell\) is an integer, the series for one of the two linearly independent solutions (either \(f\) even or \(f\) odd depending on whether \(\ell\) is even or odd) will truncate, and those solutions are called Legendre polynomials \(P_{\ell}(x)\) Find \(P_{0}(x), P_{1}(x), P_{2}(x),\) and \(P_{3}(x)\) from the recursion relation. Leave your answer in terms of either \(a_{0}\) or \(a_{1} .^{69}\)

Suppose $$V(x)=\left\\{\begin{array}{ll} m g x, & x>0 \\ \infty, & x \leq 0 \end{array}\right.$$ (a) Solve the (time-independent) Schr枚dinger equation for this potential. Hint: First convert it to dimensionless form: $$-y^{\prime \prime}(z)+z y(z)=\epsilon y(z)$$ by letting \(z \equiv a x\) and \(y(z) \equiv(1 / \sqrt{a}) \psi(x)\) (the \(\sqrt{a}\) is just so \(y(z)\) is normalized with respect to \(z\) when \(\psi(x)\) is normalized with respect to \(x\) ) What are the constants \(a\) and \(\varepsilon\) ? Actually, we might as well set \(a \rightarrow 1-\) this amounts to a convenient choice for the unit of length. Find the general solution to this equation (in Mathematica DSolve will do the job). The result is (of course) a linear combination of two (probably unfamiliar) functions. Plot each of them, for \((-15
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