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Chapter 37: Problem 45

The ground state wave function for a harmonic oscillator is given by \(\Psi_{0}(x)=A_{2} e^{-x^{2} / 2 b^{2}}\). a) Determine the normalization constant \(A\). b) Determine the probability that a quantum harmonic oscillator in the \(n=0\) state will be found in the classically forbidden region.

Short Answer

Expert verified
#Answer# The normalization constant is \(A_{2} = \frac{1}{\sqrt{b\sqrt{\pi}}}\). The probability of finding the particle in the classically forbidden region is given by the integral $$ P(x_1<x<x_2)=\frac{1}{b\sqrt{\pi}}\int_{-\sqrt{\frac{\hbar}{m\omega}}}^{\sqrt{\frac{\hbar}{m\omega}}} e^{-x^{2}/b^{2}} dx $$ This integral is difficult to evaluate analytically and typically requires numerical methods.

Step by step solution

01

Normalize the wave function

To normalize the wave function, the integral of the square of the wave function over the entire space should equal 1: $$ \int_{-\infty}^{\infty}\Psi_{0}(x)^2 dx = A_{2}^2 \int_{-\infty}^{\infty}e^{-x^{2} / b^{2}} dx = 1 $$ We can use the Gaussian integral formula, which states that $$ \int_{-\infty}^{\infty}e^{-x^2} dx = \sqrt{\pi} $$ Let's make a substitution: \(u = x / b\). Then \(du = dx / b\) and \(dx = b du\). So, we can rewrite the integral as $$ \int_{-\infty}^{\infty}e^{-x^{2} / b^{2}} dx = b \int_{-\infty}^{\infty}e^{-u^2} du = b\sqrt{\pi} $$ Now we can find \(A_2\): $$ A_{2}^2 b\sqrt{\pi}= 1 $$ Thus, the normalization constant is $$ A_{2} = \frac{1}{\sqrt{b\sqrt{\pi}}}. $$
02

Define the classically forbidden region

For a harmonic oscillator, the classically forbidden region is defined as the region where the particle's total energy is less than its potential energy. The potential energy in a harmonic oscillator is given by $$ V(x) = \frac{1}{2}kx^2 $$ where \(k\) is the spring constant. The ground state energy for a harmonic oscillator is $$ E_0 = \frac{1}{2}\hbar\omega = \frac{1}{2}\hbar\sqrt{\frac{k}{m}} $$ where \(\omega\) is the angular frequency, \(\hbar\) is the reduced Planck's constant, and \(m\) is the mass of the particle. The classically forbidden region is where \(V(x) > E_0\). Thus, $$ \frac{1}{2}kx^2 > \frac{1}{2}\hbar\sqrt{\frac{k}{m}} $$ Solve for the boundaries of the forbidden region: $$ x^2 > \frac{\hbar}{m\omega} $$
03

Calculate the probability of finding the particle in the forbidden region

The probability of finding the particle in the forbidden region is given by the integral of the square of the wave function in that region. Let the boundaries of the forbidden region be \(x_1\) and \(x_2\), then $$ P(x_1<x<x_2)=\int_{x_1}^{x_2}\Psi_{0}(x)^2 dx = \int_{x_1}^{x_2} A_{2}^2 e^{-x^{2}/b^{2}} dx $$ Substitute the normalization constant and the values of \(x_1\) and \(x_2\) we have found: $$ P(x_1<x<x_2)=\frac{1}{b\sqrt{\pi}}\int_{-\sqrt{\frac{\hbar}{m\omega}}}^{\sqrt{\frac{\hbar}{m\omega}}} e^{-x^{2}/b^{2}} dx $$ This integral is difficult to evaluate analytically and usually requires numerical methods. However, we can find the probability of the quantum harmonic oscillator being in the classically forbidden region using the normalized wave function and the given boundaries of the forbidden area.

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Key Concepts

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

Normalization of Wave Function
The normalization of a wave function is a fundamental process in quantum mechanics to ensure that it represents a valid physical state. When dealing with the wave function of a quantum harmonic oscillator, normalization is key to making accurate predictions about the behavior of the system.

For the ground state wave function of a harmonic oscillator, (Psi_{0}(x) = A_{2} e^{-x^{2} / 2 b^{2}}), normalization requires us to adjust the constant (A_{2}) such that the integral of the probability density (the square of the wave function) over all space equals one. This reflects the principle that the total probability of finding the particle somewhere in space must be 100%.

The normalization condition can be written as (int_{-}^{}Psi_{0}(x)^2 dx = 1), which, after solving the corresponding Gaussian integral, gives us the value of (A_{2}). It's crucial to perform this step correctly, as an unnormalized wave function does not provide meaningful physical predictions.
Classically Forbidden Region
In classical physics, a particle's energy limits its motion—it cannot enter regions where the potential energy is greater than its total energy. However, quantum mechanics introduces the concept of the classically forbidden region, an area where the probability of finding a quantum particle is low but not zero, due to the phenomenon known as quantum tunneling.

For the quantum harmonic oscillator, this region is where the potential energy (V(x) = frac{1}{2}kx^2) exceeds the particle's energy. The calculation of this boundary involves comparing the potential energy to the particle's ground state energy (E_0). For the ground state of a harmonic oscillator, the exercise asks us to find out the extent of this forbidden region and then calculate the probability that the oscillator, while in its lowest energy state, is inside this normally off-limits area. This probability calculation is essential for understanding quantum behavior and highlights the stark differences between classical and quantum mechanics.
Probability Amplitude
Probability amplitude is a core concept in quantum mechanics that relates to the wave-like nature of particles. The square of this amplitude correlates to the probability of finding a particle at a given position. In the context of the quantum harmonic oscillator, the ground state wave function provides these probability amplitudes for all positions in space.

When we investigate areas like the classically forbidden region, we must integrate the squared wave function across this range to find the associated probability. The integral of the probability amplitude squared between specified limits gives us the likelihood of locating the particle within that segment of space. The complexity of this integral in the forbidden region often necessitates numerical methods for accurate computation. This integral is not just a mathematical exercise; it embodies one of the most intriguing aspects of quantum mechanics, where particles have a chance to be found where classical particles could not exist.

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Most popular questions from this chapter

Electrons from a scanning tunneling microscope encounter a potential barrier that has a height of \(U=4.0 \mathrm{eV}\) above their total energy. By what factor does the tunneling current change if the tip moves a net distance of \(0.10 \mathrm{nm}\) farther from the surface?

Example 37.1 calculates the energy of the wave function with the lowest quantum number for an electron confined to a box of width \(2.00 \AA\) in the one-dimensional case. However, atoms are three-dimensional entities with a typical diameter of \(1.00 \AA=10^{-10} \mathrm{~m} .\) It would seem then that the next, better approximation would be that of an electron trapped in a three-dimensional infinite potential well (a potential cube with sides of \(1.00 \mathrm{~A}\) ). a) Derive an expression for the electron wave function and the corresponding energies for a particle in a three dimensional rectangular infinite potential well. b) Calculate the lowest energy allowed for the electron in this case.

A surface is examined using a scanning tunneling microscope (STM). For the range of the working gap, \(L\), between the tip and the sample surface, assume that the electron wave function for the atoms under investigation falls off exponentially as \(|\Psi|=e^{-\left(10.0 \mathrm{nm}^{-1}\right) a}\). The tunneling current through the STM tip is proportional to the tunneling probability. In this situation, what is the ratio of the current when the STM tip is \(0.400 \mathrm{nm}\) above a surface feature to the current when the tip is \(0.420 \mathrm{nm}\) above the surface?

Two long, straight wires that lie along the same line have a separation at their tips of \(2.00 \mathrm{nm}\). The potential energy of an electron in the gap is about \(1.00 \mathrm{eV}\) higher than it is in the conduction band of the two wires. Conduction-band electrons have enough energy to contribute to the current flowing in the wire. What is the probability that a conduction electron in one wire will be found in the other wire after arriving at the gap?

Suppose a quantum particle is in a stationary state (energy eigenstate) with a wave function \(\psi(x, t) .\) The calculation of \(\langle x\rangle,\) the expectation value of the particle's position, is shown in the text. Calculate \(d\langle x\rangle / d t(\operatorname{not}\langle d x / d t\rangle)\).
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