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Coherent state wave function || Maths 

Professor M does Science
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2 окт 2024

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Комментарии : 24   
@materiasacra
@materiasacra 8 месяцев назад
Very nice exposition, easy to follow. I would do the last step, starting at 14:07, differently. Instead of introducing the phase angle \theta_\alpha, I would show that Re(\alpha)Im(\alpha) is a factor times _\alpha _\alpha, and then absorb this into the x-dependent phase factor. This then becomes exp[ -i _\alpha (x - _\alpha)/\hbar], which is nicely relative to the spatial expectation value, just like the argument of \psi_0.
@ProfessorMdoesScience
@ProfessorMdoesScience 8 месяцев назад
Thanks for the suggested alternative!!
@sudeshregmi330
@sudeshregmi330 Год назад
Wow! I Love you for this
@ProfessorMdoesScience
@ProfessorMdoesScience Год назад
Glad you like this!
@richardthomas3577
@richardthomas3577 2 года назад
This was excellent! Thank you! Interesting -- a coherent state is a displaced ground state, and a coherent state wave function (position rep) is a translated ground state wave function. So at least here, a displacement of a Hilbert space state vector (ground state ket) is in the position representation a translation of the corresponding ground state wave function, with some phase factors thrown in. Which seems to make sense. Since the displacement operator contains both position and momentum operators (pretty symmetrically), I wonder if a similar thing happens in the momentum space representation? I will think about that. Again many thanks!
@ProfessorMdoesScience
@ProfessorMdoesScience 2 года назад
Love to hear that the videos are motivating you to explore new ideas like what happens in the momentum representation, let us know what you find! :)
@richardthomas3577
@richardthomas3577 2 года назад
@@ProfessorMdoesScience This turned out very nicely! I basically just duplicated your work with obvious twists (it occurred to me afterward that maybe I could have just Fourier-transformed the position representation state!). The result is basically symmetrical with the position representation wave function, as I expected, with a couple of sign changes that hopefully are not the result of calculation errors: Psi_alpha(p) = exp[-i(theta_alpha)] x exp[-i(_alpha)p/h-bar]x Psi_0 (p-_alpha) If anyone is interested, here are some details: To find the coherent state wave function in the momentum representation, I needed . Since for operators, A+B = B+A, I was able to switch the order of the exponential factors in the displacement operator (expressed in terms of operators X and P) with just a switch in the sign of the commutator term. This allowed me to apply the P-operator term to = |p+alpha>. The rest was pretty routine, following your lecture. For a moment I was worried when I had a factor i where I expected to have _alpha, but of course that is correct - - there is also a hidden factor of i due to the fact that (alpha - alpha*) is imaginary, so we still have a real expectation value. Thanks again for your excellent lectures!
@ProfessorMdoesScience
@ProfessorMdoesScience 2 года назад
@@richardthomas3577 Nice work! :)
@narfwhals7843
@narfwhals7843 Год назад
From the final result it seems like Psi0 is a coherent state with alpha=0. Both exponentials become 1 and the displacement is 0. Should we exclude the eigenvalue 0 explicitly? This would make some sense, since we must also say "acting on an Eigenstate of the number operator produces another Eigenstate of the number operator, unless alpha=0". Or do we accept Psi0 as a coherent state that corresponds to the classical particle sitting at the origin? From the construction of the Eigenstates, it is also an Eigenstate of the lowering operator, again with alpha=0. Though that seems like a copout, since that is true for all n.
@ProfessorMdoesScience
@ProfessorMdoesScience Год назад
You are correct that Psi0 is also a coherent state. And to put it another way: the ground state of the harmonic oscillator is a coherent state. There is no reason to exclude it. On the contrary, you can build any other coherent state from the coherent ground state by applying the appropriate displacement operator, as discussed in this video: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE--MaF_TzD4Q8.html I hope this helps!
@narfwhals7843
@narfwhals7843 Год назад
@@ProfessorMdoesScience So it is a coherent state and as such an Eigenstate of the lowering operator with Eigenvalue 0. I don't know why i thought the same would be true for all n when I wrote the comment... The other Eigenstates of the Number operator are not coherent states and we can't construct them from the infinite sum. Thank you, as always!
@YossiSirote
@YossiSirote 9 месяцев назад
Excellent
@ProfessorMdoesScience
@ProfessorMdoesScience 9 месяцев назад
Glad you like it!
@mehdisi9194
@mehdisi9194 3 года назад
Thank you very much. Do you have an article or book that talks about the free photon wave function and how to extract electric field quantities such as E and B? I had seen an article that attempted to obtain a photon wave function through coherent states, but it was not very satisfactory.
@ProfessorMdoesScience
@ProfessorMdoesScience 3 года назад
I don't have one in mind, but I suspect that most quantum optics or quantum electrodynamics textbooks should cover this. Have you tried those?
@mehdisi9194
@mehdisi9194 3 года назад
@@ProfessorMdoesScience Thank you. Yes, I'm trying to learn qed and I'll take a look at quantum optics. Thank you for your response
@danielborrero9480
@danielborrero9480 3 года назад
As far as I know: the first to cover the photon wave function were Sipe in (1995) and Birula (1994-1996) they both got the same wave function and wave equation for the photon, even though they did it independently. It must be pointed out that before their work, there wasn't something like a Photon Wave Equation that governs its evolution (this is also known as first quantization). I mean, we know Dirac equation works for fermions (first quantization), but there wasn't such equation for Bosons of spin-1 (photons) until their work(Sipe-Birula). However, I haven't found that much implications on papers using the wave function of photons (first quantization). Most of the papers in quantum optics or photonics use the second quantization of the electromagnetic field to study the phenomena. The second quantization is the one where physical magnitudes became operators (used in this video), for light that is also known as Canonical quantization, check for example the book "quantum optics" by Gerry & Knight. Also, Smith & Raymer in 2007 found a direct connection between the first and second quantizations of light, by using an extraction rule they defined. Thus, it is possible from a wavepacket of Fock states to get the Photon Wave Function. Another important remark, the first quantization and the Photon Wave Equation leads to Maxwell equations if the wave function is the known Riemann-Silberstein vector (combination of electric and magnetic fields in a complex way R = E + iB). To summarize: The wave equation for photons (first quantization) has been all along there in the Maxwell equations. Maxwell was so smart he came up with wave equations way before quantum mechanics was developed (isn't it crazy?). All of this has blown my mind. I studied all of that to do my bachelor thesis in physics. It's crazy how little attention this topic has brought in the Quantum optics/photonics community. But still, there is a lot of papers and books about it. I will left some references: Sipe: journals.aps.org/pra/abstract/10.1103/PhysRevA.52.1875 Birula: www.sciencedirect.com/science/article/abs/pii/S0079663808703160 Smith: www.spiedigitallibrary.org/conference-proceedings-of-spie/5866/0000/The-Maxwell-wave-function-of-the-photon-Invited-Paper/10.1117/12.619359.short?SSO=1 Smith & Raymer: iopscience.iop.org/article/10.1088/1367-2630/9/11/414/meta
@mehdisi9194
@mehdisi9194 3 года назад
@@danielborrero9480 Thank you so much Daniel for your valuable comments and guidance as well as for referencing helpful articles and content.
@danielborrero9480
@danielborrero9480 3 года назад
@@mehdisi9194 You're welcome!
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