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Chapter 3: Problem 7

Prove that if \(\psi(x)\) is a solution to the Schr枚dinger equation, then any constant times \(\psi(x)\) is also a solution.

Short Answer

Expert verified
Any constant times \(\psi(x)\) satisfies the Schr枚dinger equation if \(\psi(x)\) is a solution.

Step by step solution

01

Identify the Schr枚dinger Equation

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

Assume a Constant Multiple of Solution

Assume \(C\psi(x)\) is a new solution to the Schr枚dinger equation, where \(C\) is a constant.
03

Apply the Schr枚dinger Equation to the New Function

Substitute \(C\psi(x)\) into the Schr枚dinger equation:\[-\frac{\hbar^2}{2m}\frac{d^2 (C\psi(x))}{dx^2} + V(x)(C\psi(x)) = E(C\psi(x))\]Because \(C\) is a constant, it can be factored out of the derivatives and terms.
04

Simplify the Equation

When you differentiate \(C\psi(x)\), the constant \(C\) remains unchanged:\[-\frac{\hbar^2}{2m} C \frac{d^2 \psi(x)}{dx^2} + CV(x)\psi(x) = CE\psi(x)\]Factorize the constant \(C\) from each term:\[C\left(-\frac{\hbar^2}{2m} \frac{d^2 \psi(x)}{dx^2} + V(x)\psi(x)\right) = CE\psi(x)\]
05

Verify the Simplification

Since \(\psi(x)\) is a solution to the Schr枚dinger equation, the original equation holds:\[-\frac{\hbar^2}{2m} \frac{d^2 \psi(x)}{dx^2} + V(x)\psi(x) = E\psi(x)\]Thus the equation becomes:\[C(E\psi(x)) = CE\psi(x)\]This confirms that \(C\psi(x)\) satisfies the Schr枚dinger equation, proving that any constant multiple of a solution is also a solution.

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Key Concepts

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

Quantum Mechanics
Quantum mechanics is an area of physics that describes the behaviors of very small particles, like atoms and subatomic particles. Unlike classical physics, which uses straightforward formulas and rules, quantum mechanics involves probabilities and uncertainties. Let's take a closer look at some key ideas within quantum mechanics:
  • Particles as Waves: In quantum mechanics, particles are not just solid points with mass; they also exhibit wave-like properties.
  • Uncertainty Principle: Werner Heisenberg introduced the idea that certain properties, such as position and momentum, cannot both be known to absolute precision simultaneously.
  • Quantization: Many quantities, like the energy levels of an electron in an atom, can only take on certain discrete values.
These concepts revolutionize our understanding of matter and energy, providing insights that are crucial in technologies like semiconductors and MRI machines. Quantum mechanics helps us understand the very fabric of the universe, through a lens where probabilities and waves define the behavior of matter at its smallest scales.
Wave Function
The wave function is a fundamental concept in quantum mechanics that represents the quantum state of a system. It is usually symbolized as \( \psi(x) \). Here's what you need to know about wave functions:
  • Probability Amplitudes: The square of the wave function's magnitude, \( |\psi(x)|^2 \), gives us the probability density of finding a particle at a specific position.
  • Complex Functions: They can be complex numbers, containing both real and imaginary components, capturing various possible states of a system.
  • Normalization: Wave functions must be normalized so that the total probability of finding a particle somewhere in space is one.
Wave functions are essential because they contain all the information we can know about a quantum system at a given time. Understanding how to manipulate and interpret wave functions is key to making predictions in quantum mechanics.
Time-Independent Schr枚dinger Equation
The time-independent Schr枚dinger equation is a pivotal equation in quantum mechanics that determines how the quantum state of a physical system changes spatially, rather than over time. The equation is expressed as:\[-\frac{\hbar^2}{2m}\frac{d^2 \psi(x)}{dx^2} + V(x)\psi(x) = E\psi(x)\]Where:
  • \( \hbar \) is the reduced Planck's constant.
  • \( m \) is the mass of the particle.
  • \( V(x) \) represents the potential energy as a function of position.
  • \( E \) is the total energy of the system.
The equation is termed "time-independent" because it describes systems where energy does not change with time, making it easier to solve for quantum states in stable environments. By solving this equation, physicists can predict how a quantum system behaves in a potential field, which is crucial for understanding phenomena such as tunneling, energy levels in atoms, and bonding in molecules.

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Most popular questions from this chapter

Calculate \(\langle x\rangle\) and \(\left\langle x^{2}\right\rangle\) for the \(n=2\) state of a particle in a one-dimensional box of length \(a\). Show that $$ \sigma_{x}=\frac{a}{4 \pi}\left(\frac{4 \pi^{2}}{3}-2\right)^{1 / 2} $$

Prove that \int_{0}^{a} e^{\pm i 2 \pi n x / a} d x=0 \quad n \neq 0

This problem shows that the intensity of a wave is proportional to the square of its amplitude. Figure \(3.7\) illustrates the geometry of a vibrating string. Because the velocity at any point of the string is \(\partial u / \partial t\), the kinetic energy of the entire string is $$ K=\int_{0}^{l} \frac{1}{2} \rho\left(\frac{\partial u}{\partial t}\right)^{2} d x $$ where \(\rho\) is the linear mass density of the string. The potential energy is found by considering the increase of length of the small arc \(P Q\) of length \(d s\) in Figure 3.7. The segment of the string along that arc has increased its length from \(d x\) to \(d s\). Therefore, the potential energy associated with this increase is $$ V=\int_{0}^{l} T(d s-d x) $$ where \(T\) is the tension in the string. Using the fact that \((d s)^{2}=(d x)^{2}+(d u)^{2}\), show that $$ V=\int_{0}^{l} T\left\\{\left[1+\left(\frac{\partial u}{\partial x}\right)^{2}\right]^{1 / 2}-1\right\\} d x $$ Using the fact that \((1+x)^{1 / 2} \approx 1+(x / 2)\) for small \(x\), show that $$ V=\frac{1}{2} T \int_{0}^{l}\left(\frac{\partial u}{\partial x}\right)^{2} d x $$ for small displacements. The total energy of the vibrating string is the sum of \(K\) and \(V\) and so $$ E=\frac{\rho}{2} \int_{0}^{l}\left(\frac{\partial u}{\partial t}\right)^{2} d x+\frac{T}{2} \int_{0}^{l}\left(\frac{\partial u}{\partial x}\right)^{2} d x $$ We showed in Chapter 2 (Equations \(2.23\) through \(2.25\) ) that the \(n\)th normal mode can be written in the form $$ u_{n}(x, l)=D_{n} \cos \left(\omega_{n} t+\phi_{n}\right) \sin \frac{n \pi x}{l} $$ where \(\omega_{n}=v n \pi / l\). Using this equation, show that $$ K_{n}=\frac{\pi^{2} v^{2} n^{2} \rho}{4 l} D_{n}^{2} \sin ^{2}\left(\omega_{n} t+\phi_{n}\right) $$ and $$ V_{n}=\frac{\pi^{2} n^{2} T}{4 l} D_{n}^{2} \cos ^{2}\left(\omega_{n} t+\phi_{n}\right) $$ Using the fact that \(v=(T / \rho)^{1 / 2}\), show that $$ E_{n}=\frac{\pi^{2} v^{2} n^{2} \rho}{4 l} D_{n}^{2} $$ Note that the total energy, or intensity, is proportional to the square of the amplitude. Although we have shown this proportionality only for the case of a vibrating string, it is a general result and shows that the intensity of a wave is proportional to the square of the amplitude. If we had carried everything through in complex notation instead of sines and cosines, then we would have found that \(E_{n}\) is proportional to \(\left|D_{n}\right|^{2}\) instead of just \(D_{n}^{2}\). Generally, there are many normal modes present at the same time, and the complete solution is (Equation 2.25) $$ u(x, t)=\sum_{n=1}^{\infty} D_{n} \cos \left(\omega_{n} t+\phi_{n}\right) \sin \frac{n \pi x}{l} $$ Using the fact that (see Problem 3-16) $$ \int_{0}^{l} \sin \frac{n \pi x}{l} \sin \frac{m \pi x}{l} d x=0 \quad \text { if } m \neq n $$ show that $$ E_{n}=\frac{\pi^{2} v^{2} \rho}{4 l} \sum_{n=1}^{\infty} n^{2} D_{n}^{2} $$

In each case, show that \(f(x)\) is an eigenfunction of the operator given. Find the eigenvalue. $$\hat{A}$$ (a) \(\frac{d^{2}}{d x^{2}}\) (b) \(\frac{d}{d t}\) (c) \(\frac{d^{2}}{d x^{2}}+2 \frac{d}{d x}+3\) (d) \(\frac{\partial}{\partial y}\) $$f(x)$$ $$\begin{array}{l} \cos \omega x \\ e^{i \omega t} \\ e^{\alpha x} \\ x^{2} e^{6 y} \end{array}$$

Derive the equation for the allowed energies of a particle in a one- dimensional box by assuming that the particle is described by standing de Broglie waves within the box.
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