Chapter 2: Problem 38
Find the allowed energies of the half-harmonic oscillator $$ V(x)=\left\\{\begin{array}{ll} (1 / 2) m \omega^{2} x^{2}, & \text { for }(x>0) \\ \infty, & \text { for }(x<0) \end{array}\right. $$ (This represents, for example, a spring that can be stretched, but not compressed.) Hint: This requires some careful thought, but very little actual computation.
Short Answer
Step by step solution
Understand the concept of a half-harmonic oscillator
Identify the wave function boundary conditions
Consider the harmonic oscillator solution
Apply the boundary condition \( \psi(0) = 0 \)
Determine the allowed energies
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Quantum Mechanics
Quantum mechanics introduces the concept of the wave function, a mathematical expression that describes the quantum state of a system. This wave function helps predict the probability of a particle’s position or momentum at any given time. It's a crucial tool in identifying potential outcomes, but it doesn't provide certainty — just probabilities.
- Wave-Particle Duality: Particles exhibit both wave-like and particle-like properties.
- Uncertainty Principle: It's impossible to simultaneously know the precise position and velocity of a particle.
- Entanglement: Particles can be interconnected in such a way that the state of one instantly influences the state of another, regardless of distance.
Wave Function Boundary Conditions
The potential is infinite for any point where \( x < 0 \). Thus, the wave function \( \psi(x) \) must be zero at \( x = 0 \) because a particle cannot exist in an area of infinite potential. This zero boundary means the wave function must also be zero at the left-most point of its effective potential region, making \( \psi(x) = 0 \) at \( x = 0 \) a non-negotiable condition.
- Infinite potentials impose hard boundaries on the allowable wave functions.
- The wave function must reflect these boundaries, usually requiring it to be zero at these points.
- This influences which wave functions are physically meaningful in the system.
Harmonic Oscillator Energy Levels
\[ E_n = \left(n + \frac{1}{2}\right)\hbar\omega \]
Here, \( n \) is an integer (0, 1, 2, ...), \( \hbar \) is the reduced Planck's constant, and \( \omega \) is the angular frequency of the oscillator.
For the half-harmonic oscillator, however, the boundary condition \( \psi(0) = 0 \) means only odd \( n \) values (1, 3, 5, etc.) are allowed as they correspond to sine function solutions that match the zero requirement at this boundary.
- In a regular harmonic oscillator, both even and odd solutions are permitted, as there are no imposed restrictions at \( x = 0 \).
- In the half-harmonic scenario, even solutions are excluded as they don’t reach zero at the boundary due to being maximized at that point.
- Thus, only wave functions associated with odd \( n \) harmonics honor the boundary condition of the half-harmonic oscillator.