Wave function

Wave function#

What you need to know

The Probabilistic Nature of the Quantum World
The quantum realm operates on fundamentally probabilistic principles, where certainty is replaced by probabilities.
Wave Function and Probability Distribution
The absolute square of the wave function, \(|\psi|^2\), represents the probability distribution of finding a quantum particle in space and time.

In the classical wave equation, the wave function has a clear mechanical interpretation: it represents the degree of disturbance in the wave. For example, it can describe the elevation of a guitar string from its resting position.
In contrast, the quantum wave function is less intuitive. The wave function itself does not have direct physical meaning, as it is generally a complex function. To connect it to measurable quantities, we need to extract real values from it that correspond to physical observables.
The key insight is that the absolute square of the wave function gives the probability distribution:

Probabilistic meaning of quantum wave function (square)

\[p(x) = \psi^{*}(x) \cdot \psi(x) = |\psi(x)|^2\]

\(p(x)\) is a probability distribution function. It is describing the likelihood of finding a quantum object at a positions x.
\(p(x)dx\) gives the probability to find particle in a tiny part of space inside \(x, x+dx\) interval.
In three-dimensional space, the analogous expression is:

\[p(x, y, z) = \psi(x, y, z)^{*} \cdot \psi(x, y, z)\]

For the wave function to represent a proper probability distribution, it must be normalizable. If it is not normalizable, the wave function is only proportional to a probability distribution and not equal to it.
Normalization of \(\psi^2\) ensures that there is absolute certainty that the quantum object exists somewhere in space. In an experiment, when searching for a quantum particle across the entire space, normalization guarantees that you will find it somewhere.
Normalization in 1D:

\[\int^{+\infty}_{-\infty} |\psi(x)|^2 dx = \int^{+\infty}_{-\infty} p(x) dx = 1\]
To normalize a wave function \(\psi'\), multiply it by a constant: \(\psi = N\psi'\). The constant \(N\) is determined by plugging this expression into the normalization condition. In other words, normalization helps determine the multiplicative factor in front of wave functions.
Normalization in 3D:

\[\int^{+\infty}_{-\infty} \int^{+\infty}_{-\infty} \int^{+\infty}_{-\infty} |\psi(x, y, z)|^2 dx \, dy \, dz = 1\]

By definition probability distribution function \(p(x) \) allows quantifying various probabilities that a quantum “particle” is located in an infinitesimal slice \([x, x+dx]\) around point \(x\). This then enables us to find probabilities in any finite region \([a,b]\) simply by integrating:

\[p(a<x<b)=\int_a^b |\psi(x)|^2dx\]
In higher dimensions, e.g. 3D, we can locate particle around volume \(dxdydz\) or any finite volume via a similar integration:

\[p(a_x<x<b_x,a_y<y<b_y, a_z<z<b_z )=\int^{b_x}_{a_x} \int^{b_y}_{a_y} \int^{b_z}_{a_z} |\psi(x,y,z)|^2dx dy dz\]

Recall that mean value of x is computed by weighting its values by probabilities, e.g think of average mass of box of candies, we multiply probability or fraction of each candy type by its mass and sum.

\[\langle x \rangle = x_1 p_2+x_2 p_2 + ...\]

\[\langle x \rangle = \int x\cdot p(x) dx\]

Likewise you can compute averge of any function of x, say \(x^2\) or \(sin(x)\).

\[\langle f \rangle = \int f(x) \cdot p(x) dx\]

For quantities like momentum or total energy which are no longer simple functions as in classical mechanics but operators, \(\hat{p}\) and \(\hat{H}\), we simply have to use operators in the defintion of moments!

\[\langle A \rangle = \int \hat{A} p(x) \cdot dx = \int \psi^{*}(x) \cdot \hat{A} \psi(x) \cdot dx \]

Average quantity	Corresponding operator
\(\langle E \rangle=\int \psi^{*}(x) \hat{H} \psi(x) dx\)	\(\hat{H}=-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}+V(x)\)
\(\langle K \rangle=\int \psi^{*}(x) \hat{K}\psi(x) dx \)	\(\hat{K}=-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}\)
\(\langle p \rangle=\int \psi^{*}(x) \hat{p} \psi(x) dx\)	\(\hat{p}=-i\hbar\frac{d}{dx}\)
\(\langle p^2 \rangle=\int \psi^{*}(x) \hat{p}^2 \psi(x) dx\)	\(\hat{p}^2=-\hbar^2\frac{d^2}{dx^2}\)