Question

One classic problem in quantum mechanics is the "harmonic oscillator." In this problem a particle is subjected to a one-dimensional potential (taken to be along \(x\) ) of the form \(V(x) \propto x^{2}\) where \(-\infty \leq x \leq \infty .\) The probability distribution function for the particle in the lowest-energy state is $$P(x)=C e^{-a x^{2} / 2}$$ Determine the expectation value for the particle along \(x\) (that is, \(\langle x\rangle\) ). Can you rationalize your answer by considering the functional form of the potential energy?

Short answer

The expectation value for the particle along $x$ is \(\langle x \rangle = 0\), which can be rationalized by observing that the potential energy function is symmetric around the origin (\(V(x) \propto x^{2}\)). Thus, on average, the particle is most likely to be found at the minimum of the potential energy, which is at the center of the potential well.

Step-by-step solution

Review the formula for the expectation value

The expectation value for a one-dimensional variable x is given by: \(\langle x\rangle=\int_{-\infty}^{\infty} x P(x) dx\) Here, \(P(x)\) represents the probability distribution function, which is provided in the exercise.

Integrate and find the expectation value

To find the expectation value, we will plug in the given expression for the probability distribution function, \(P(x) = Ce^{-ax^2/2}\), into the formula for the expectation value. \(\langle x \rangle = \int_{-\infty}^{\infty} x (Ce^{-ax^2/2}) dx\) Now, let's perform the integration: \[\langle x \rangle = C\int_{-\infty}^{\infty} x e^{-ax^2/2} dx\] Note that the integrand is an odd function, meaning that for every positive value x, there exists a negative value -x with equal and opposite sign, and the integrand is symmetric around x=0. Therefore, the integration over the entire range from -∞ to ∞ will cancel out both positive and negative terms, resulting in a value of 0. So, \(\langle x \rangle = 0\)

Rationalize the expectation value result

The expectation value for the particle along x, \(\langle x \rangle = 0\), indicates that, on average, the particle is most likely to be found at the minimum of the potential energy function, \(V(x) \propto x^{2}\). Since the lowest-energy state of the quantum harmonic oscillator is symmetric around the origin, it is expected that the particle is more probable to be found at the center of the potential well where the potential energy is minimum. Thus, the obtained result \(\langle x \rangle = 0\) is rational and consistent with the functional form of the potential energy.

Key concepts

Expectation Value

In quantum mechanics, the expectation value gives us an average measurement of where a particle is expected to be found in a given state. It's not simply where the particle is located but rather a statistical mean of the possible positions. In the case of the quantum harmonic oscillator and the exercise we are examining, the expectation value along the x-axis is expressed mathematically as:
\(\langle x \rangle = \int_{-\infty}^{\infty} x P(x) \, dx\)
where \( P(x) \) is the probability distribution function.For a symmetric potential like the one described, \( V(x) \propto x^{2} \), the integral involves an odd function (the product of \( x \) and the exponential term in the probability distribution). Due to the symmetry of this odd function around the vertical line at \( x=0 \), the positive values cancel out the negative values during integration over an infinite domain. This results in an expectation value of zero:

• Symmetric potentials lead to zero expectation values when centered around the origin.
• The calculation highlights the mean position's correlation with the potential's minimum.

So, the expectation value tells us the particle is equally likely to be on either side of the origin, confirming symmetry in its probability distribution.

Probability Distribution Function

The probability distribution function (PDF) in quantum mechanics is crucial as it gives us insight into where a particle is most likely to be found. For the quantum harmonic oscillator, the PDF in its lowest energy state is given by the expression:
\( P(x) = C e^{-ax^2/2} \)
This function is Gaussian in nature, characterized by its bell-shaped curve centered at \( x=0 \).

• The constant \( C \) is a normalization constant that ensures the total probability over all space equals one.
• \( a \) is a parameter related to the stiffness of the potential. It controls the width of the distribution; larger \( a \) results in a narrower distribution.

In this scenario, the form of the PDF reveals the probability density is maximum at the origin (\( x = 0 \)). The exponential factor rapidly decreases as one moves away from the center, meaning the particle is less likely to be found farther from \( x=0 \). This distribution matches our expectation value result that pinpoints the center of the distribution as the most probable location.

Potential Energy

In the context of the quantum harmonic oscillator, the potential energy is represented as a quadratic function of position:
\( V(x) \propto x^{2} \)
This functional form describes a potential well, which is analogous to a parabolic trap that confines the particle within it.

• The term \( x^{2} \) signifies that the potential energy is zero at the origin and increases symmetrically as one moves away from it.
• This results in the particle experiencing a restoring force trying to keep it near \( x=0 \) - a defining characteristic of a harmonic oscillator.

Since the potential energy is minimized at the origin, it's natural and expected in the lowest-energy quantum state, that the particle's average position or expectation value would be zero, aligning with wave function symmetry. Understanding potential energy in such systems is vital as it explains the forces at work that determine particle behavior and probabilistic distribution within the system.