A particle is in a state described by the wavefunction \(\psi=(\cos \chi) e^{i k x}+\) \((\sin \chi) \mathrm{e}^{-\mathrm{ikx}}\), where \(\chi\) (chi) is a parameter. What is the probability that the particle will be found with a linear momentum \((a)+k \hbar\), (b) \(-k \hbar ?\) What form would the wavefunction have if it were 90 per cent certain that the particle had linear momentum \(+k \hbar\) ?

When diving into the world of the extremely small, classical physics takes a back seat and quantum mechanics gets behind the wheel. Quantum mechanics is a fundamental theory in physics that provides a description of the physical properties of nature at the scale of atoms and subatomic particles. A central part of quantum mechanics is the wave-like nature of particles. Unlike a baseball that can only be at one place at a time, particles like electrons exist in a haze of probability, with mathematical functions—called wavefunctions—describing the likelihood of finding a particle at a certain location.

Every action at the quantum scale, from a particle's position to its momentum, is described by a wavefunction, typically denoted by the symbol \( \psi \). The square of the absolute value of the wavefunction, \( |\psi|^2 \), indicates the probability density of finding the particle at a specific point in space. This is why we often talk about probabilities, rather than certainties, when it comes to quantum behavior.

One counter-intuitive aspect of quantum mechanics is the superposition principle. This principle states that a particle can exist in all possible states simultaneously until it is measured. Think of it like a spinning coin that's both heads and tails at the same time—it's only when the coin lands (or in quantum terms, you make a measurement) that it 'chooses' a state.

In the case of our wavefunction \( \psi \), the particle is in a superposition of states that correspond to having momentum \( +k \hbar \) and \( -k \hbar \). It's the act of measurement—or observation—that 'collapses' the wavefunction into one of these definite states.

• The term \( e^{ikx} \) accounts for the particle's state having momentum \( +k \hbar \) and \( (\cos \chi) \) tells us about the amplitude of this state.
• Similarly, \( e^{-ikx} \) represents the state with momentum \( -k \hbar \) and \( (\sin \chi) \) provides the amplitude for this state.
• The probabilities for these momenta can then be determined by squaring these respective amplitudes, as the square of the amplitude is directly proportional to a state's probability.

In classical mechanics, momentum is simply mass times velocity. But in quantum mechanics, momentum is a bit more abstract and is associated with the wavelength of a particle's wavefunction. De Broglie showed us that particles could be described as waves, and the momentum of a particle is related to the waveness of the particle, articulated as \( p = \hbar k \), where \( p \) is the momentum, \( k \) is the wave number, and \( \hbar \) (h-bar) is the reduced Planck constant.

This relates to our problem where the wavefunction given represents the particle in states with momentum \( +k \hbar \) and \( -k \hbar \)—or, in wave terms, with wavelengths \( \frac{2\pi}{k} \) and \( -\frac{2\pi}{k} \) (because the wave number \( k \) is the spatial frequency of the wave, inversely proportional to its wavelength).

The probability of finding the particle with a certain momentum is derived by extracting the square of the amplitude of the corresponding term in the wavefunction. As such, this wave-particle duality is at the heart of quantum mechanics, describing entities that behave like both particles and waves.