Kartazion Posted February 8, 2020 Posted February 8, 2020 (edited) Split reference: When we find a particle with the Schrodinger equation, is it either stationary (fixed position), or in motion with a given velocity with the uncertainty principle? Edited February 8, 2020 by Kartazion
swansont Posted February 8, 2020 Posted February 8, 2020 No. The Schrödinger equation for a particle in a box, for example, gives you energy eigenstates. They don’t vary in time, which is why they are called stationary. The particle has no defined motion, and it’s possible to find the particle outside of a well of finite depth. The solution gives probability of where you might find the particle. 1
Kartazion Posted February 8, 2020 Author Posted February 8, 2020 1 hour ago, swansont said: No. The Schrödinger equation for a particle in a box, for example, gives you energy eigenstates. They don’t vary in time, which is why they are called stationary. The particle has no defined motion ... Then the quantum oscillator is only a of probability of presence of the particle in the form of wave, no? It's a wave probability (for to find the particle) into the well? 1 hour ago, swansont said: And it’s possible to find the particle outside of a well of finite depth. The solution gives probability of where you might find the particle. Ok.
Kartazion Posted February 8, 2020 Author Posted February 8, 2020 2 hours ago, swansont said: and it’s possible to find the particle outside of a well of finite depth. The solution gives probability of where you might find the particle. I found that: The ground state energy for the quantum harmonic oscillator can be shown to be the minimum energy allowed by the uncertainty principle. ... This is a very significant physical result because it tells us that the energy of a system described by a harmonic oscillator potential cannot have zero energy.
swansont Posted February 8, 2020 Posted February 8, 2020 1 hour ago, Kartazion said: Then the quantum oscillator is only a of probability of presence of the particle in the form of wave, no? I recognize the individual words but this makes no sense. A quantum oscillator is one example of a quantum system described by the Schrödinger equation, but not everything described by it is an oscillator. 1 hour ago, Kartazion said: It's a wave probability (for to find the particle) into the well? Huh?
Kartazion Posted February 8, 2020 Author Posted February 8, 2020 40 minutes ago, swansont said: Huh? Sorry, a probability amplitude. It's a probability amplitude into the well?
swansont Posted February 8, 2020 Posted February 8, 2020 1 hour ago, Kartazion said: Sorry, a probability amplitude. It's a probability amplitude into the well? That’s what a wave function is.
Kartazion Posted February 8, 2020 Author Posted February 8, 2020 3 hours ago, swansont said: That’s what a wave function is. But what does the oscillator oscillate?
swansont Posted February 8, 2020 Posted February 8, 2020 31 minutes ago, Kartazion said: But what does the oscillator oscillate? Nothing literally oscillates. One is solving the problem at the quantum level, and we use the same terminology as the macroscopic problem.
Kartazion Posted February 8, 2020 Author Posted February 8, 2020 11 minutes ago, swansont said: Nothing literally oscillates. Ok. 11 minutes ago, swansont said: One is solving the problem at the quantum level, and we use the same terminology as the macroscopic problem. I did'nt understand everything. Could you detail what this means? Which problem precisely?
Mordred Posted February 9, 2020 Posted February 9, 2020 With the uncertainty principle one cannot accurately measure the position and momentum of a particle at the same time. Measuring one observable P or Q will interfere with the other. Also the more accurately you measure one the less accurate you can determine the other. Both observable's will have a probability amplitude.
Kartazion Posted February 9, 2020 Author Posted February 9, 2020 1 hour ago, Mordred said: With the uncertainty principle one cannot accurately measure the position and momentum of a particle at the same time. Measuring one observable P or Q will interfere with the other. Also the more accurately you measure one the less accurate you can determine the other. Both observable's will have a probability amplitude. "One is solving the problem at the quantum level" Which one is solve? "and we use the same terminology as the macroscopic problem" But there is no uncertainty principle at the macroscopic level?
swansont Posted February 9, 2020 Posted February 9, 2020 2 hours ago, Kartazion said: and we use the same terminology as the macroscopic problem" But there is no uncertainty principle at the macroscopic level? It’s small and can safely be ignored on the macroscopic level.
Kartazion Posted February 10, 2020 Author Posted February 10, 2020 (edited) So what is the role and the advantage of involving the quantum harmonic oscillator, if there is no oscillating particle? We oscillate the Schrodinger equation as a wave function? Simpler, we oscillate a corpuscle, and not a particle? Edited February 10, 2020 by Kartazion
Mordred Posted February 10, 2020 Posted February 10, 2020 (edited) It describes the probability wavefunctions. An oscillator isn't restricted to object motion but can also be used to describe any repeating varying value. A sinusoidal waveform in electronics is a good example. Edited February 10, 2020 by Mordred 1
swansont Posted February 10, 2020 Posted February 10, 2020 1 hour ago, Kartazion said: So what is the role and the advantage of involving the quantum harmonic oscillator, if there is no oscillating particle? There are quantum systems behave like that, so we study it. Such as diatomic molecules. The analysis tells us about their energy states, and that informs us on e.g. how photons might interact with them. Quote We oscillate the Schrodinger equation as a wave function? Simpler, we oscillate a corpuscle, and not a particle? We study behavior we can measure of a system in a potential well. Position is not an eigenstate. Energy is. (Do you understand what I mean by eigenstate?) 1
Kartazion Posted February 12, 2020 Author Posted February 12, 2020 On 2/10/2020 at 2:14 PM, swansont said: We study behavior we can measure of a system in a potential well. Position is not an eigenstate. Energy is. (Do you understand what I mean by eigenstate?) The eigenstate interprets the observable states represented by an eingenfunction, which is characterized by a linear operator and its eigenvalue. In the case of the oscillator these eigenstates are measured in a potential well. Position is not an eigenstate. Back to the HUP. The particle identified and located by the uncertainty principle asserts the undeniable presence of particle, right? If yes the particle takes what physical form if it implies a position?
swansont Posted February 12, 2020 Posted February 12, 2020 4 hours ago, Kartazion said: Back to the HUP. The particle identified and located by the uncertainty principle asserts the undeniable presence of particle, right? If yes the particle takes what physical form if it implies a position? The momentum-position version of the HUP tells you that the particle has no definite location.
Kartazion Posted February 12, 2020 Author Posted February 12, 2020 1 hour ago, swansont said: The momentum-position version of the HUP tells you that the particle has no definite location. I understand. The particle is localized by a probability distribution.
swansont Posted February 12, 2020 Posted February 12, 2020 1 hour ago, Kartazion said: I understand. The particle is localized by a probability distribution. Right. It's in a region, but you don't have a specific location, just limits on where it might be found.
Kartazion Posted February 12, 2020 Author Posted February 12, 2020 2 hours ago, swansont said: Right. It's in a region, but you don't have a specific location, just limits on where it might be found. So it starts from the same principle, namely a not specific location, when we trap a photon?
swansont Posted February 12, 2020 Posted February 12, 2020 20 minutes ago, Kartazion said: So it starts from the same principle, namely a not specific location, when we trap a photon? Yes. If by "trap" you mean in an optical cavity — put the radiation in a cavity so you have a standing wave, it just sits there, but you can't localize where the photon would be. That's true for anything with a wave nature.
Kartazion Posted February 13, 2020 Author Posted February 13, 2020 7 hours ago, swansont said: If by "trap" you mean in an optical cavity Yes. 7 hours ago, swansont said: That's true for anything with a wave nature. All leptons and bosons included?
swansont Posted February 13, 2020 Posted February 13, 2020 10 hours ago, Kartazion said: Yes. All leptons and bosons included? Yes, although they will behave differently. The wavelength of a particle is h/p
Kartazion Posted July 1, 2021 Author Posted July 1, 2021 What is the role and the implication of the second quantization in all that we just said in this thread? It another method of to be able to determine the localization of the particle? Or a standardization of the field in a classical and quantum entity? Thanks. On 2/12/2020 at 5:56 PM, swansont said: — put the radiation in a cavity so you have a standing wave, it just sits there, but you can't localize where the photon would be. That's true for anything with a wave nature. Isn't this standing wave the entire location of the photon?
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