2 added 378 characters in body edited Mar 23 '17 at 10:56 Emilio Pisanty 91.2k2323 gold badges229229 silver badges474474 bronze badges Yes, the radiation in those two situations would be drastically different. For concreteness, I'll consider the following two situations: An electron is dropped in as a wavepacket with mean velocity $$\mathbf v$$ and a small spatial extent $$\sigma$$ a large distance away from a proton at rest, with some nonzero impact parameter, in such a way that the total energy is negative (so the electron is bound and only a negligible fraction of the electron population will be scattered to infinity) but arbitrarily close to zero. An electron is suddenly injected in a wavepacket of spatial extent $$\sigma$$ and central velocity $$\mathbf v$$ into a region with a uniform, static magnetic field $$\mathbf B$$ orthogonal to $$\mathbf v$$, with no additional external fields. Since this is an accelerated charge, it will radiate, and therefore slowly spiral down to rest at some location close to the center of its initial circular orbit. In situation 1, the electron will radiate its energy through all the possible radiative decay pathways, i.e. all the possible dipole-allowed steps down the Bohr-model states, until it reaches the ground state. You will observe a bunch of different photon energies, all of them of the form $$\hbar\omega= \mathrm{Ry}\left(\frac{1}{m^2}-\frac{1}{n^2}\right)$$, starting small (since the spacings in the Rydberg region are small) and growing until you get to the Lyman-alpha line. In situation 2, the electron will radiate exclusively at the cyclotron frequency at that magnetic field, $$\omega_c = \frac{eB}{m_e}.$$ (This frequency gets its name from the fact that cyclotrons operate at that frequency, but it is a much more general concept.) From a classical perspective, the electron will follow circular orbits, with the crucial characteristic that the frequency of the orbit is independent of the orbit's size or the electron's velocity or energy. This means that the emitted radiation will always be at that frequency. From a more quantum perspective, the electron has been prepared in a superposition of excited Landau levels, and through coupling to the electromagnetic vacuum it will experience radiative decay down that ladder. Mirroring the classical constant-frequency property, the Landau ladder is harmonic and equispaced, i.e. all the levels are separated by $$\hbar\omega_c$$, so you can only radiate at multiples of that frequency (and, moreover, dipole radiation is only possible between adjacent levels). As to why you think this is useful, or a meaningful comparison, I'm still completely in the dark. And, just to not let this slide, the statement that From this starting position it has to be possible to calculate the top speed with which the electron approaches the nucleus. is completely meaningless - it completely relies on an underlying trajectory, which is meaningless in quantum mechanics. Starting off with counterfactual statements will not get you very far. Yes, the radiation in those two situations would be drastically different. For concreteness, I'll consider the following two situations: An electron is dropped in as a wavepacket with mean velocity $$\mathbf v$$ and a small spatial extent $$\sigma$$ a large distance away from a proton at rest, with some nonzero impact parameter, in such a way that the total energy is negative (so the electron is bound and only a negligible fraction of the electron population will be scattered to infinity) but arbitrarily close to zero. An electron is suddenly injected in a wavepacket of spatial extent $$\sigma$$ and central velocity $$\mathbf v$$ into a region with a uniform magnetic field $$\mathbf B$$ orthogonal to $$\mathbf v$$. In situation 1, the electron will radiate its energy through all the possible radiative decay pathways, i.e. all the possible dipole-allowed steps down the Bohr-model states, until it reaches the ground state. You will observe a bunch of different photon energies, all of them of the form $$\hbar\omega= \mathrm{Ry}\left(\frac{1}{m^2}-\frac{1}{n^2}\right)$$, starting small (since the spacings in the Rydberg region are small) and growing until you get to the Lyman-alpha line. In situation 2, the electron will radiate exclusively at the cyclotron frequency at that magnetic field, $$\omega_c = \frac{eB}{m_e}.$$ From a classical perspective, the electron will follow circular orbits, with the crucial characteristic that the frequency of the orbit is independent of the orbit's size or the electron's velocity or energy. This means that the emitted radiation will always be at that frequency. From a more quantum perspective, the electron has been prepared in a superposition of excited Landau levels, and through coupling to the electromagnetic vacuum it will experience radiative decay down that ladder. Mirroring the classical constant-frequency property, the Landau ladder is harmonic and equispaced, i.e. all the levels are separated by $$\hbar\omega_c$$, so you can only radiate at multiples of that frequency (and, moreover, dipole radiation is only possible between adjacent levels). As to why you think this is useful, or a meaningful comparison, I'm still completely in the dark. And, just to not let this slide, the statement that From this starting position it has to be possible to calculate the top speed with which the electron approaches the nucleus. is completely meaningless - it completely relies on an underlying trajectory, which is meaningless in quantum mechanics. Starting off with counterfactual statements will not get you very far. Yes, the radiation in those two situations would be drastically different. For concreteness, I'll consider the following two situations: An electron is dropped in as a wavepacket with mean velocity $$\mathbf v$$ and a small spatial extent $$\sigma$$ a large distance away from a proton at rest, with some nonzero impact parameter, in such a way that the total energy is negative (so the electron is bound and only a negligible fraction of the electron population will be scattered to infinity) but arbitrarily close to zero. An electron is suddenly injected in a wavepacket of spatial extent $$\sigma$$ and central velocity $$\mathbf v$$ into a region with a uniform, static magnetic field $$\mathbf B$$ orthogonal to $$\mathbf v$$, with no additional external fields. Since this is an accelerated charge, it will radiate, and therefore slowly spiral down to rest at some location close to the center of its initial circular orbit. In situation 1, the electron will radiate its energy through all the possible radiative decay pathways, i.e. all the possible dipole-allowed steps down the Bohr-model states, until it reaches the ground state. You will observe a bunch of different photon energies, all of them of the form $$\hbar\omega= \mathrm{Ry}\left(\frac{1}{m^2}-\frac{1}{n^2}\right)$$, starting small (since the spacings in the Rydberg region are small) and growing until you get to the Lyman-alpha line. In situation 2, the electron will radiate exclusively at the cyclotron frequency at that magnetic field, $$\omega_c = \frac{eB}{m_e}.$$ (This frequency gets its name from the fact that cyclotrons operate at that frequency, but it is a much more general concept.) From a classical perspective, the electron will follow circular orbits, with the crucial characteristic that the frequency of the orbit is independent of the orbit's size or the electron's velocity or energy. This means that the emitted radiation will always be at that frequency. From a more quantum perspective, the electron has been prepared in a superposition of excited Landau levels, and through coupling to the electromagnetic vacuum it will experience radiative decay down that ladder. Mirroring the classical constant-frequency property, the Landau ladder is harmonic and equispaced, i.e. all the levels are separated by $$\hbar\omega_c$$, so you can only radiate at multiples of that frequency (and, moreover, dipole radiation is only possible between adjacent levels). As to why you think this is useful, or a meaningful comparison, I'm still completely in the dark. And, just to not let this slide, the statement that From this starting position it has to be possible to calculate the top speed with which the electron approaches the nucleus. is completely meaningless - it completely relies on an underlying trajectory, which is meaningless in quantum mechanics. Starting off with counterfactual statements will not get you very far. 1 answered Mar 22 '17 at 13:53 Emilio Pisanty 91.2k2323 gold badges229229 silver badges474474 bronze badges Yes, the radiation in those two situations would be drastically different. For concreteness, I'll consider the following two situations: An electron is dropped in as a wavepacket with mean velocity $$\mathbf v$$ and a small spatial extent $$\sigma$$ a large distance away from a proton at rest, with some nonzero impact parameter, in such a way that the total energy is negative (so the electron is bound and only a negligible fraction of the electron population will be scattered to infinity) but arbitrarily close to zero. An electron is suddenly injected in a wavepacket of spatial extent $$\sigma$$ and central velocity $$\mathbf v$$ into a region with a uniform magnetic field $$\mathbf B$$ orthogonal to $$\mathbf v$$. In situation 1, the electron will radiate its energy through all the possible radiative decay pathways, i.e. all the possible dipole-allowed steps down the Bohr-model states, until it reaches the ground state. You will observe a bunch of different photon energies, all of them of the form $$\hbar\omega= \mathrm{Ry}\left(\frac{1}{m^2}-\frac{1}{n^2}\right)$$, starting small (since the spacings in the Rydberg region are small) and growing until you get to the Lyman-alpha line. In situation 2, the electron will radiate exclusively at the cyclotron frequency at that magnetic field, $$\omega_c = \frac{eB}{m_e}.$$ From a classical perspective, the electron will follow circular orbits, with the crucial characteristic that the frequency of the orbit is independent of the orbit's size or the electron's velocity or energy. This means that the emitted radiation will always be at that frequency. From a more quantum perspective, the electron has been prepared in a superposition of excited Landau levels, and through coupling to the electromagnetic vacuum it will experience radiative decay down that ladder. Mirroring the classical constant-frequency property, the Landau ladder is harmonic and equispaced, i.e. all the levels are separated by $$\hbar\omega_c$$, so you can only radiate at multiples of that frequency (and, moreover, dipole radiation is only possible between adjacent levels). As to why you think this is useful, or a meaningful comparison, I'm still completely in the dark. And, just to not let this slide, the statement that From this starting position it has to be possible to calculate the top speed with which the electron approaches the nucleus. is completely meaningless - it completely relies on an underlying trajectory, which is meaningless in quantum mechanics. Starting off with counterfactual statements will not get you very far.