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| de_broglie_matter_waves [2021/02/03 21:02] – [1.v.2 Experimental Confirmation of de Broglie's Hypothesis] admin | de_broglie_matter_waves [2022/10/13 18:00] (current) – [1.v.1 de Broglie Wavelength and Wave Vector] admin |
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| ====== 1.v.1 de Broglie Wavelength and Wave Vector ====== | ====== 1.v.1 de Broglie Wavelength and Wave Vector ====== |
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| The photoelectric effect and Compton scattering show that electromagnetic waves sometimes exhibit particle-like properties. In 1923, de Broglie proposed that matter, which we normally think of as made up of particles like electrons, protons and neutrons, should have wave-like properties. In other words, //everything// has both wave- and particle-like properties. | The photoelectric effect and Compton scattering show that electromagnetic waves sometimes exhibit particle-like properties. In 1923, Louis de Broglie proposed that matter, which we normally think of as made up of particles like electrons, protons and neutrons, should have wave-like properties. In other words, //everything// has both wave- and particle-like properties. |
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| Obviously, electromagnetic radiation does behave like a wave in many circumstances and matter behaves as if it were made up like particles in many circumstances, i.e., all circumstances where classical physics provides an adequate account. Therefore, whether a system exhibits wave-like or particle-like properties depends on the experiment that we are doing. This is known as //**wave-particle duality**//. | Obviously, electromagnetic radiation does behave like a wave in many circumstances and matter behaves as if it were made up like particles in many circumstances, i.e., all circumstances where classical physics provides an adequate account. Therefore, whether a system exhibits wave-like or particle-like properties depends on the experiment that we are doing. This is known as //**wave-particle duality**//. |
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| For a photon, we have $E = pc$ and the quantum postulate says that $E = h\nu$. Combining these gives $p = h\nu /c = h/\lambda$, or $\lambda = h/p$. | For a photon, we have \(E = pc\) and the quantum postulate says that $E = h\nu$. Combining these gives $p = h\nu /c = h/\lambda$, or $\lambda = h/p$. |
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| de Broglie proposed that the same relation should hold for matter particles, so a matter particle with momentum $p$ is associated with a wave of wavelength | de Broglie proposed that the same relation should hold for matter particles, so a matter particle with momentum $p$ is associated with a wave of wavelength |
| ====== 1.v.2 de Broglie Wavelength of the Electron ====== | ====== 1.v.2 de Broglie Wavelength of the Electron ====== |
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| Before computing the de Broglie wavelength of the electron, a word on units. Since Planck's constant and the masses of elementary particles are so small, the typical masses and energy scales involved in quantum mechanics are very small. Therefore, it is common to measure masses and energies in //**electron volts ($\text{eV}$)**// rather than the SI unit Joules ($\text{J}$). | Before computing the de Broglie wavelength of the electron, a word on units. Since Planck's constant and the masses of elementary particles are so small, the typical masses and energy scales involved in quantum mechanics are very small. Therefore, it is common to measure masses and energies in //**electron volts**// ($\text{eV}$) rather than the SI unit Joules ($\text{J}$). |
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| The electron volt is a unit of energy, defined to be the change in electric potential energy of an electron as it moves across a potential difference of $1\,\text{V}$. Thus, the conversion factor is | The electron volt is a unit of energy, defined to be the change in electric potential energy of an electron as it moves across a potential difference of $1\,\text{V}$. Thus, the conversion factor is |
| \[1\,\text{eV} = 1.062\times 10^{-19}\,\text{J}\] | \[1\,\text{eV} = 1.062\times 10^{-19}\,\text{J}\] |
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| The mass of the electron is $9.109 \times 10^{-31}\,\text{kg}$. From special relativity, we know that mass and energy are equivalent. The rest energy of the electron is $m_e c^2$. In SI units this is $(9.109 \times 10^{-31}\,\text{kg})(2.998\times 10^{8}\,\text{ms}^{-1}) = 8.187\times 10^{-14}\,\text{J}$. Because these numbers are so small, it is common to covert the rest energy to electron volts, which gives | The mass of the electron is $9.109 \times 10^{-31}\,\text{kg}$. From special relativity, we know that mass and energy are equivalent. The rest energy of the electron is $m_e c^2$. In SI units this is $(9.109 \times 10^{-31}\,\text{kg})(2.998\times 10^{8}\,\text{ms}^{-1})^2 = 8.187\times 10^{-14}\,\text{J}$. Because these numbers are so small, it is common to covert the rest energy to electron volts, which gives |
| $(8.187\times 10^{-14}\,\text{J})/(1.062\times 10^{-19}\,\text{JeV}^{-1}) = 5.11\times 10^{5}\,\text{eV} = 0.511\,\text{MeV}$. Somewhat confusingly, physicists will often say that the //mass// of the electron is $0.511\,\text{MeV}$, but it is important to remember that this really means the //rest energy// of the electron, so $m_e c^2 = 0.511\,\text{MeV}$. | $(8.187\times 10^{-14}\,\text{J})/(1.062\times 10^{-19}\,\text{JeV}^{-1}) = 5.11\times 10^{5}\,\text{eV} = 0.511\,\text{MeV}$. Somewhat confusingly, physicists will often say that the //mass// of the electron is $0.511\,\text{MeV}$, but it is important to remember that this really means the //rest energy// of the electron, so $m_e c^2 = 0.511\,\text{MeV}$. |
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| When asked to do a calculation where masses and energies are given to you in electron volts, it is always possible to convert everything into SI units first and then do the calculation. However, it is also possible to do the calculation directly in electron volts using the fact that $hc = 1.240 \times 10^{-6} \,\text{eVm}$. The advantage of this is that the exponents in the calculation have smaller magnitude, so it is less likely that you will get a number that is too small/large for your calculator to handle during the calculation. To illustrate this, let's calculate the do Broglie wavelength of an electron with kinetic energy $54\,\text{eV}$ both ways: first by converting to SI units and then keeping everything in electron volts. | When asked to do a calculation where masses and energies are given to you in electron volts, it is always possible to convert everything into SI units first and then do the calculation. However, it is also possible to do the calculation directly in electron volts using the fact that $hc = 1.240 \times 10^{-6} \,\text{eVm}$. The advantage of this is that the exponents in the calculation have smaller magnitude, so it is less likely that you will get a number that is too small/large for your calculator to handle during the calculation. To illustrate this, let's calculate the de Broglie wavelength of an electron with kinetic energy $54\,\text{eV}$ both ways: first by converting to SI units and then keeping everything in electron volts. |
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| Since $54\,\text{eV} \ll 5.11\times 10^{5}\,\text{eV}$, the kinetic energy is a lot smaller than the rest energy of the electron, so using non-relativistic physics should be fine. The kinetic energy of the electron is | Since $54\,\text{eV} \ll 5.11\times 10^{5}\,\text{eV}$, the kinetic energy is a lot smaller than the rest energy of the electron, so using non-relativistic physics should be fine. The kinetic energy of the electron is |
| If matter particles sometimes behave like waves then it ought to be possible to detect wave interference effects like diffraction and double-slit interference. At this point, you should do the in-class activity to determine some more typical de Broglie wavelengths of matter particles in order to understand why it is difficult to detect these interference effects. | If matter particles sometimes behave like waves then it ought to be possible to detect wave interference effects like diffraction and double-slit interference. At this point, you should do the in-class activity to determine some more typical de Broglie wavelengths of matter particles in order to understand why it is difficult to detect these interference effects. |
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| In 1927, Davisson and Germer scattered a monoenergetic beam of electron with kinetic energy $54\,\text{eV}$ off a slab of Nickel crystal. They observed the distribution of the intensity of the scattered electrons as they varied the scattering angle. The following figure illustrates their setup. | In 1927, Davisson and Germer scattered a monoenergetic beam of electrons with kinetic energy $54\,\text{eV}$ off a slab of Nickel crystal. They observed the distribution of the intensity of the scattered electrons as they varied the scattering angle. The following figure illustrates their setup. |
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| {{ :davissongermer.png?direct&400 |}} | {{ :davissongermer.png?direct&400 |}} |
| This is a pretty good agreement with the theoretical de Broglie wavelength of $0.167\,\text{nm}.$ | This is a pretty good agreement with the theoretical de Broglie wavelength of $0.167\,\text{nm}.$ |
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| Many other matter interference experiments have been conducted since the Davisson-Germer experiments. It is now possible to perform the double-slit experiment with electron beams and the size of particles we can detect interference with has been gradually increasing. For example, in 1999 the double-slit experiment was performed with carbon-60 molecules (buckyballs). | Many other matter interference experiments have been conducted since the Davisson-Germer experiment. It is now possible to perform the double-slit experiment with electron beams and the size of particles we can detect interference with has been gradually increasing. For example, in 1999 the double-slit experiment was performed with carbon-60 molecules (buckyballs). |
| ====== 1.v.3 de Broglie Matter Waves ====== | ====== 1.v.4 de Broglie Matter Waves ====== |
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| | de Broglie's hypothesis $\vec{k} = \vec{p}/\hbar$, where $\hbar = h/2\pi$, and the experimental confirmation of matter-interference, suggests that a particle of momentum $\vec{p}$ should be associated with a plane wave |
| | \[\psi(\vec{r},t) = Ae^{i(\vec{k}\cdot\vec{r} - \omega t)}.\] |
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| | Since $E = h\nu$ and $\omega = 2\pi\nu$, we have $\omega = E/\hbar$. Therefore, we can rewrite this plane wave as |
| | \[\psi(\vec{r},t) = Ae^{i(\vec{p}\cdot\vec{r} - E t)/\hbar}.\] |
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| | This is how we would typically write it in quantum mechanics, and it is our first example of a quantum mechanical //**wave function**//. |
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| {{:question-mark.png?direct&50|}} | {{:question-mark.png?direct&50|}} |