Question

Consider a fictitious one-dimensional system with one electron. The wave function for the electron, drawn below, is \(\psi(x)=\sin x\) from \(x=0\) to \(x=2 \pi\). (a) Sketch the probability density, \(\psi^{2}(x)\), from \(x=0\) to \(x=2 \pi\). (b) At what value or values of \(x\) will there be the greatest probability of finding the electron? (c) What is the probability that the electron will be found at \(x=\pi\) ? What is such a point in a wave function called? [Section 6.5]

Short answer

(a) The probability density function for the electron is \(\psi^2(x) = (\sin x)^2\). (b) The greatest probability of finding the electron occurs at \(x = 0\) and \(x = 2\pi\). (c) The probability of finding the electron at \(x = \pi\) is zero, and this point is called a node in the wave function.

Step-by-step solution

(a) Sketching the Probability Density Function

To sketch the probability density \(\psi^2(x)\), we need to calculate the square of the given wave function \(\psi(x)\). Recall that \(\psi(x) = \sin x\), so we have: \(\psi^2(x) = (\sin x)^2\) Now, we will plot \(\psi^2(x)\) for \(x\) ranging from \(0\) to \(2\pi\).

(b) Maximum Probability

To find the value of \(x\) where the probability of finding the electron is greatest, we must find the maximum of \(\psi^2(x) = (\sin x)^2\). To do this, we can check when the derivative of \(\psi^2(x)\) is equal to zero: \(\frac{d}{dx} (\sin x)^2 = 0\) The derivative of \((\sin x)^2\) is: \(\frac{d}{dx} (\sin x)^2 = 2 \sin x \cos x\) Setting it to zero, we get: \(2 \sin x \cos x = 0\) This leads to two possible solutions: \(x = n\pi\) where \(n\) is an integer, or \(\sin x = 0\). In our given range \(0 \leq x \leq 2\pi\), this corresponds to the values of \(x = 0\), \(x = \pi\), and \(x = 2\pi\). The maximum of \(\psi^2(x)\) occurs at \(x = 0\) and \(x = 2\pi\).

(c1) Probability at \(x = \pi\)

To find the probability that the electron will be found at \(x = \pi\), we need to evaluate the probability density function \(\psi^2(x)\) at this value of \(x\). Recall that \(\psi^2(x) = (\sin x)^2\): \(\psi^2(\pi) = (\sin \pi)^2 = 0\) So, the probability of finding the electron at \(x = \pi\) is zero.

(c2) Significance of the Probability Point at x = π

A point in the wave function where the probability of finding the particle is zero is called a node. In our case, since the probability of finding the electron at \(x = \pi\) is zero, we can say that \(x = \pi\) is a node in the wave function. Recap: (a) We sketched the probability density function for \(\psi^2(x) = (\sin x)^2\). (b) The maximum probability of finding the electron is at \(x = 0\) and \(x = 2\pi\). (c) The probability of finding the electron at \(x = \pi\) is zero, and this point is called a node in the wave function.

Key concepts

Probability Density

In quantum mechanics, the probability density function provides a way to determine how likely it is to find a particle, such as an electron, at a specific location. If you have a wave function like \(\psi(x)\), the probability density \(\psi^2(x)\) is simply the square of the absolute value of that function. For the wave function \(\psi(x) = \sin x\), used from \(x = 0\) to \(x = 2\pi\), the probability density becomes \((\sin x)^2\).

When you plot \((\sin x)^2\), you can easily see the oscillations between 0 and 1. The squaring ensures that all values are non-negative, as probabilities must be. These probabilities tell us where an electron might be found within a specified region. A high probability density indicates that an electron is more likely to be at that spot, while a lower one suggests less likelihood.

Maximum Probability

Maximum probability in quantum mechanics tells us where a particle is most likely to be found. To identify these points, we locate where the probability density function reaches its highest values. This involves finding the maximum values of \(\psi^2(x)\).

For our given \(\psi(x) = \sin x\), the probability density is \((\sin x)^2\). To find where this function is maximized, one examines the derivative and sets it to zero: \(\frac{d}{dx} (\sin x)^2 = 0\). The derivative is \(2 \sin x \cos x\), which equates to zero when either \(\sin x = 0\) or \(\cos x = 0\).

Within the interval \(0 \leq x \leq 2\pi\), the maximum probability occurs at points where \(\sin x\) itself is at its peak, i.e., where \(x = 0\) and \(x = 2\pi\). At these points, the sine function aligns back to its root condition, providing the maximum probability for finding the particle here.

Node

A node is a fundamental concept in wave physics and refers to a point where the wave function, and consequently the probability density, is exactly zero. It is a location where a quantum particle, like an electron, will not be found.

In our exercise, we observe that \(\psi(x) = \sin x\) has one such point where \(\sin x = 0\) at \(x = \pi\). When the wave function is zero, squaring it leads to a probability density of zero, \(\psi^2(\pi) = 0\).

Nodes are significant because they tell us that, due to the nature of waves, there are exact locations where it is impossible to find a particle. This can be visualized in electron orbitals within atoms, where nodes describe regions of zero probability, influencing the shape and size of electron clouds.