Quasi-classical states: does the quantum harmonic oscillator ACTUALLY oscillate??

[@yoda7104]

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[@ProfessorMdoesScience]

Really glad to hear this, and good luck with your PhD!

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[@ProfessorMdoesScience]

Thanks for your support, and glad you like them! :)

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[@ProfessorMdoesScience]

Really glad you like the videos! And indeed, we use this videos in our university level teaching :)

[@urty4395]

I was dying to get to know quantum harmonic oscillator since few weeks and finally.... Thanks a lot!

[@ProfessorMdoesScience]

Great timing! :)

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omg, your content is GOLD! So glad to have found!

[@ProfessorMdoesScience]

Glad you like it! :)

[@band5680]

Please keep doing the wrap-up. It's very helpful!

[@ProfessorMdoesScience]

Thanks for the suggestion, we will! :)

[@band5680]

​@@ProfessorMdoesScience On that note, it would be great if you could generally do a wrap-up slide like you do in some other videos. Really love the width and style of your content.

[@ProfessorMdoesScience]

@@band5680 Thanks for the suggestion!

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[@ProfessorMdoesScience]

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[@quantumradio]

Thank you for making this content. Your explanations are very pedagogical and intuitive. I appreciate that they don't sidetrack into material that is interesting but not necessarily relevant for a first go at the explanation of what's physically taking place.

[@ProfessorMdoesScience]

Glad you like the videos!

[@richardthomas3577]

This was outstanding, even better than usual! Thank you! BTW I find that it really helps to work some problems; it is usually pretty easy to find good problems (and sometimes solutions) in the qualifying exam problem banks made available online at many universities.

[@richardthomas3577]

P.s. also I just noticed that, apparently, every coherent state is a minimum uncertainty state. Bizarre!

[@ProfessorMdoesScience]

Glad you really liked this one! It is great to hear about problems, we are considering creating problems+solutions associated with the videos :)

[@mehmetalivat]

7:02 The quantum world is a different world that we describe as strange because we cannot enter that massive beings like us. We just have to accept this fact. Thank you so much this lecture series.

[@ProfessorMdoesScience]

Thanks for watching! :)

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Amazing videos. I wish my classes were this fun! They just made me remember that I actually like studying physics

[@ProfessorMdoesScience]

Glad you like them! May we ask where you study?

[@faizfarhan8231]

Thanks for sharing such lecture

[@ProfessorMdoesScience]

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[@brainfitness6066]

Hi! I´m from Mexico. I've learned a lot of new things in your videos... Congratulations!

[@ProfessorMdoesScience]

Thanks for watching and glad you find the videos useful!

[@jozsefkele7858]

Hi, great video as usual. In the bra-ket expression n= am I mistaken to think the the root 2 belongs in the numerator rather than the denominator. This follows from the fact that in their definition a and a^\dagger have the root 2 in their denominator, thus x and p should have this factor in their numerator.

[@ProfessorMdoesScience]

I think the expression we have is correct. The definitions of a and a^dagger do indeed have the root 2 in the denominator. But when we invert them, we use: a^dagger + a = 2*(1/sqrt(2))*sqrt(m*omega/hbar)*x The "x" term contributes twice, so we get 2/sqrt(2), which gives sqrt(2) in the numerator. Then isolating x moves this root 2 to the other side in the denominator, as in our expression. I hope this helps!

[@narfwhals7843]

Is there an analogous way to construct quasi classical EM waves via the second quantization formalism?

[@ProfessorMdoesScience]

Coherent states were in fact first explored in some detail in quantum optics by Glauber. Wikipedia has a good overview of these: en.wikipedia.org/wiki/Coherent_state I hope this helps!

[@shiqiangec1304]

Great video, incredibly didactic. Regarding the temporal dependence of the solutions, everything makes sense. However, how do we reconcile the frequency dependence in the classical vs. the quantum? For example, if you set everything constant and move along the frequency, you will have a quadratic dependence in the classical energy and a linear one for the quantum energy of the QHO. Of course, if you replace x_0=\sqrt(2hbar/m\omega)l\alpha_0l you get the same dependence. But this feels like I am cheating right? There must be a physical argument relating the amplitude of the CHO x_0 with the coherent eigenvalue \alpha_0. Hope that I was clear enough haha

[@ProfessorMdoesScience]

Glad you like it! I am not sure I follow what you mean by "move along the frequency", could you please clarify this? Regarding the relation between alpha_0 and x_0, there are various ways to think about it, and I believe it will become fully clearer in the video coming up in two week's time on the conceptual understanding of coherent state wave functions. However, this understanding builds from the other videos on coherent states, and roughly speaking, the "displacement operator" (see video: kzbin.info/www/bejne/r6G1hX9_pJJ8f8U ) by an amount alpha_0 shifts the ground state wave function by an amount _alpha, and you can then think of the center of the Gaussian wave function as related to the "classical" position, so the larger the magnitude of alpha_0, the larger the classical amplitude x_0. Overall, I would recommend following the full playlist on coherent states, which we should complete in the next few weeks. I hope this helps!

[@adityaprasad465]

I think I see what you're asking. Here's the way I look at it: we know that the CHO starts at (and has a maximum value of) x_0. Where does the QHO start? Well, we've just proven that it _must_ start at [the nasty sqrt term you wrote out]. In this way, the identification is _not_ arbitrary.

[@omerunlusoy]

you saved my life, thanks from my heart :)

[@ProfessorMdoesScience]

Glad you found the video useful!

[@ManojKumar-cj7oj]

Thanks buddy 😊

[@ProfessorMdoesScience]

Thanks for watching!

[@rodrigoappendino]

18:22 But what alpha? In that time dependent part, there is no scalar alpha. Only a ket alpha. 5:14 The average of the position is 0, but where is 0? I mean, where is the origin of the coordinate system? It's confusing, because it's always said that X|n> = x|n>, right? But there is not an operator X. We have to write it differently according to each problem? This position operator thing confuses me.

[@rodrigoappendino]

Oh. And if a quantum system is in a state n, how can it change to a state alpha? They are just different coordinate systems for the same space? So you can represent all QHO in the coherent state basis?

[@ProfessorMdoesScience]

18:22 Coherent states are eigenstates of the lowering operator, and the symbol in the ket is the eigenvalue associated with the eigenstate, and it is simply a scalar label that allows us to distinguish the different states. In this notation, |alpha> for alpha=alpha0*e^(-i*omega*t) means that the state is an eigenstate of the lowering operator that at time t is the eigenstate that corresponds to the eigenvalue alpha0*e^(-i*omega*t). Another more explicit way to write this would be to directly write the label of the eigenstate in the "expanded" form: |alpha0*e^(-i*omega*t)>. 5:14 The harmonic potential is V(x)=0.5*m*omega^2*x^2, so in our coordinate system the zero is at the minimum of the potential. It is actually very easy to understand why _n=0 in an energy eigenstate by just inspecting the eigenstate wave functions (they are all even or odd functions, so their 'average' is at x=0). We plot them in this video: kzbin.info/www/bejne/ZqCQfaKAh9mresU Regarding how to go from a state |n> to a state |alpha>, this is a nontrivial question that can be phrased more generally: how does one prepare a general quantum state (whether it is |n>, |alpha>, or something else)? This is in general a challenge, and very often ongoing research revolves around the preparation of specific quantum states. In the case of coherent states |alpha>, they are most commonly constructed in the context of quantum optics. I hope this helps!

[@quantum4everyone]

The ground state is often fairly simple to make--just cool the system until it can be cooled no more. One needs a good way for cooling, but if you do, the ground state can be fairly easily made. The higher energy eigenstates are very difficult to make. One has to try some quite nontrivial things to make these energy eigenstates when n is nonzero. But coherent states are not so hard. Once you have made the ground state by cooling, then you just need to give the system a push by impulsing it with a force over a short period of time. Then it will oscillate. Large pushes may be difficult, but moderate ones, not so hard at all. These things have been done for ions in a trap for quite some time now. Also laser light is a coherent state of the oscillations of the electromagnetic fields.

[@joeaverage8329]

Prof M, is there a reason why writing expectation values in coherent basis corresponds to classical case in this harmonic oscillator case? To me, coherent basis having to do with classical description is non-intuitive (To me, it would make more sense if this correspondence was possible in energy or position basis). I think there should be a reason / intuition behind this. Is there Prof M?

[@ProfessorMdoesScience]

One way to see this is taking the opposite view. If we ask the question: which states follow the classical description more closely? Then the answer to this is that it is coherent states. I hope this helps!

[@tomkerruish2982]

Coherent states are also those states that minimize the uncertainty in both position and momentum. Those properties make them somewhat classical.

[@DrMarcoArmenta]

So the eigenstates of the Hamiltonian do not give an oscillation as in the classical case and we have to use the eigenstates of the lowering operator to see the oscillations... what is the intuition behind this? or should I just trust the math? I mean, how are the eigenstates of the lowering operator linked to the Hamiltonian of the system? or is it ALL about the relation H = h^bar w (a^dag a +1/2)?

[@ProfessorMdoesScience]

I am not sure I can provide a very deep intuition. The best I can think of is that, if you ask yourself what states of the quantum harmonic oscillator most closely resemble the classical solutions, then you end up with coherent states, and this derivation does not depend on the fact that they are eigenstates of the lowering operator. In this context, then they just happen to be eigenstates of a. Put anothe way: you can define coherent states equivalently as the eigenstates of the lowering operator, or as the quantum states that most closely resemble the classical motion. The answer to these two (apparently unrelated) questions happens to be the same, and none of these approaches is deeper than the other. I hope this helps!

[@quantum4everyone]

A better reason to understand this is that the coherent state is a displacement of the ground state. This does not seem to be discussed in the video, but is often done when discussing coherent states. The displacement, is like taking the ground state, centered at x=zero, pulling it to x_0, and letting go. Of course, it should oscillate! Just like a mass on a spring does. It will oscillate back and forth to x_0 and -x_0 and back forever, because there is no friction in this problem.

[@herpederpe4320]

this is good stuff! very edible little nuggets of knowledge, like sushi for the mind

[@ProfessorMdoesScience]

Great way to put it! :)

[@geraldpellegrini2782]

What happens if you "measure" the energy of a quantum harmonic oscillator? Does it "collapse" to an energy eigenstate and all motion stops?

[@tomkerruish2982]

Yes, but also, no. Yes, it collapses to an energy eigenstate. However, all motion does not stop. The average momentum will be zero; there's no reason to prefer motion in one direction over the other. However, the average kinetic energy will be half the total energy. Similar remarks apply to average position and average potential energy.

[@zray2937]

I have nothing against the mathematics of the presentation. But calling the coherent states the most classical states is strange to me. In the classical case, the energy is fixed, zero chance for it to be another number. In the quantum case that only happens when the expectation value is measured in an energy eigenfunction. When the root mean deviation is calculated in a coherent state, it is not only non-zero, but it also increases with the modulo of alpha.

[@ProfessorMdoesScience]

You are of course correct, and analogies between quantum and classical systems will always be rough. Having said this, it is common to call a state in which the expectation values coincide with the classical values a "quasi-classical" state or similar, and this is why we use this language. Moving forward, we'll try to make these statements more precise, thanks for the feedback!

[@quantum4everyone]

The expectation value of the energy in the coherent state does not change in time. This is because the Hamiltonian is independent of time. The Hamiltonian commutes with itself, so the expectation value of a coherent state is also independent of time. Recall, time evolution for a time invariant Hamiltonian depends only on the Hamiltonian.

[@zray2937]

@@quantum4everyone You are indeed correct. I was a typo, I indented to write "chance". So, while the energy expectation is constant, the spread in the energy is non-zero.

[@quantum4everyone]

@@zray2937 Yes, that is correct. There are fluctuations in the energy, because it is no longer in an eigenstate, except when alpha=0, where the fluctuations go to zero, because it is an energy eigenstate.

[@garvitmakkar]

Snapshot or screenshot😊

[@ProfessorMdoesScience]

Oh dear :)

[@DanielFBest]

What is your nationality, Professor M? I can't place your accent for my life.

[@ProfessorMdoesScience]

Ah, have definitely moved around over the years, but originally from Spain :)

[@DanielFBest]

@@ProfessorMdoesScience Your accent is very lovely, and expressive!