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Chapter 5: Problem 36

Consider a freely moving quantum particle with mass \(m\) and speed \(v\). Its energy is \(E=K+U=\frac{1}{2} m v^{2}+0 .\) Determine the phase speed of the quantum wave representing the particle and show that it is different from the speed at which the particle transports mass and energy.

Short Answer

Expert verified
The phase speed of the quantum wave representing the particle is \(\sqrt {\frac{E}{2m}}\), which is different from the speed \(\sqrt {\frac{2E}{m}}\) at which the particle transports mass and energy.

Step by step solution

01

Express momentum in terms of energy

From the energy equation \(E = \frac{1}{2} m v^{2} + 0\), and knowing that the momentum \(p\) of a particle is related to its speed and mass by the equation \(p = m v\), one can express momentum in terms of energy as follows: \(E = \frac{p^{2}}{2m}\). Solving for \(p\) we get \(p = \sqrt {2mE}\).
02

Apply de Broglie formula

The de Broglie formula, which relates the wavelength \(位\) of a particle to its momentum is given by \(\lambda = \frac{h}{p}\) where \(h\) is the Planck constant. Using the expression for momentum obtained in step 1, we can write the wavelength as \(\lambda = \frac{h}{\sqrt{2mE}}\).
03

Determine the phase speed

The phase speed \(v_p\) of a wave is given by the dispersion relationship, which in this case is the De Broglie relation \(v_p = \frac{E}{p}\). Substituting \(p = \sqrt{2mE}\) in \(v_p = \frac{E}{p}\) we get the phase speed as \(v_p = \frac{E}{\sqrt {2mE}} = \sqrt {\frac{E}{2m}}\). This speed depends on the energy of the particle.
04

Comparing phase speed and particle speed

The speed \(v\) of a particle is related to the energy \(E\) by the relation \(v = \sqrt {\frac{2E}{m}}\). Comparing this relation with the phase speed \(v_p = \sqrt {\frac{E}{2m}}\) obtained in step 3, it is evident that the phase speed of the quantum wave differs from the speed at which the particle transports mass and energy.

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Key Concepts

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

Phase Speed
In quantum mechanics, the concept of phase speed is essential for understanding how waves move. Phase speed, denoted as \(v_p\), refers to the speed at which the wavefronts of a wave propagate through space. This is pivotal in distinguishing between the wave nature of particles and their classical behavior.
To calculate the phase speed of a quantum wave, we use the relation \(v_p = \frac{E}{p}\), where \(E\) represents the energy of the wave and \(p\) is the momentum. As explained in the solution, substituting \(p = \sqrt{2mE}\) gives us the phase speed as \(v_p = \sqrt{\frac{E}{2m}}\). This outcome demonstrates that the phase speed is highly dependent on energy.
  • It's important to note that phase speed is not the speed at which the physical particle moves.
  • It captures how fast the wave shape is moving, not where the particle is going.
  • The phase speed can indeed be greater than the speed of light, but this does not imply that energy or matter are moving faster than light.
Understanding phase speed helps illustrate the dual nature of particles, acting both as waves and particles, a key concept in quantum mechanics.
Momentum-Energy Relation
The Momentum-Energy Relation is a fundamental concept in quantum mechanics that helps link the kinematic properties of particles to their wave characteristics. The momentum \(p\) and energy \(E\) of a particle are connected through the equation \(E = \frac{p^2}{2m}\), derived from kinetic energy principles.
This relationship is crucial as it allows us to express the wave properties of a particle in terms of its classical properties:
  • Momentum \(p\) is expressed in terms of energy \(E\) and mass \(m\), \(p = \sqrt{2mE}\).
  • This expression is pivotal in deriving the de Broglie wavelength, which is vital for understanding quantum behavior.
  • Identifying and computing these properties helps us predict the behavior and interaction of particles at small scales, where classical physics fails.
Using the momentum-energy relation, quantum mechanics provides a deeper insight into the actions and interactions that occur at the smallest scales of nature.
De Broglie Wavelength
One of the cornerstones of quantum mechanics is the concept of the De Broglie Wavelength. It proposes that every moving particle or object has an associated wave, a breakthrough that helped develop quantum theories. The wavelength \(\lambda\) is related to the particle's momentum \(p\) by the formula \(\lambda = \frac{h}{p}\), where \(h\) is Planck's constant.
From the momentum-energy relation, we can substitute the expression for \(p\) to find that \(\lambda = \frac{h}{\sqrt{2mE}}\). This relation shows that:
  • The wavelength decreases as the momentum increases, meaning faster particles have shorter wavelengths.
  • It highlights the dual wave-particle nature, where objects exhibit properties of both waves and particles.
  • De Broglie's hypothesis was crucial in the development of the Schr枚dinger equation, underpinning modern quantum mechanics.
Understanding the De Broglie Wavelength allows us to comprehend phenomena like diffraction and interference in a way that is compatible with quantum theories. It connects classical intuitions about particles with the enigmatic, wave-like behaviors observed at the quantum level.

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Most popular questions from this chapter

An electron of momentum \(p\) is at a distance \(r\) from a stationary proton. The system has a kinetic energy \(K=p^{2} / 2 m_{\mathrm{e}}\) and potential energy \(U=-k e^{2} / r .\) Its total energy is \(E=K+U .\) If the electron is bound to the proton to form a hydrogen atom, its average position is at the proton but the uncertainty in its position is approximately equal to the radius, \(r\), of its orbit. The electron's average momentum will be zero, but the uncertainty in its momentum will be given by the uncertainty principle. Treat the atom as a one-dimensional system in the following: (a) Estimate the uncertainty in the electron's momentum in terms of \(r .\) (b) Estimate the electron's kinetic, potential, and total energies in terms of \(r\). (c) The actual value of \(r\) is the one that minimizes the total energy, resulting in a stable atom. Find that value of \(r\) and the resulting total energy. Compare your answer with the predictions of the Bohr theory.

A matter wave packet. (a) Find and sketch the real part of the matter wave pulse shape \(f(x)\) for a Gaussian amplitude distribution \(a(k)\), where $$ a(k)=A e^{-\alpha^{2}\left(k-k_{0}\right)^{2}} $$ Note that \(a(k)\) is peaked at \(k_{0}\) and has a width that decreases with increasing \(\alpha .\) (Hint: In order to put \(f(x)=(2 \pi)^{-1 / 2} \int_{-\infty}^{+\infty} a(k) e^{i k x} d k \quad\) into the standard form \(\int_{-\infty}^{+\infty} e^{-a z^{2}} d z\), complete the square in \(\left.k_{.}\right)\) (b) \(\mathrm{By}\) comparing the result for the real part of \(f(x)\) to the standard form of a Gaussian function with width \(\Delta x\), \(f(x) \propto A e^{-(x / 2 \Delta x)^{2}}\), show that the width of the matter wave pulse is \(\Delta x=\alpha \cdot(c)\) Find the width \(\Delta k\) of \(a(k)\) by writing \(a(k)\) in standard Gaussian form and show that \(\Delta x \Delta k=\frac{1}{2}\), independent of \(\alpha .\)

A ball of mass \(50 \mathrm{~g}\) moves with a speed of \(30 \mathrm{~m} / \mathrm{s}\). If its speed is measured to an accuracy of \(0.1 \%\), what is the minimum uncertainty in its position?

Calculate the de Broglie wavelength of a proton that is accelerated through a potential difference of \(10 \mathrm{MV}\).

The dispersion relation for free relativistic electron waves is $$ \omega(k)=\sqrt{c^{2} k^{2}+\left(m_{\mathrm{e}} c^{2} / \hbar\right)^{2}} $$ Obtain expressions for the phase velocity \(v_{\mathrm{p}}\) and group velocity \(v_{\mathrm{g}}\) of these waves and show that their product is a constant, independent of \(k\). From your result, what can you conclude about \(v_{\mathrm{g}}\) if \(v_{\mathrm{p}}>\ell\) ?
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