I decided to pause and derive the degeneracy of an nth level. I tried to find a connection to Pascal's triangle since each time you add a quantum of energy, you can think of it as a ball falling into 3 bins(x,y,z) with equal probability. And the question is basically how many ways are there to arrange n-1 balls in 3 bins. After a few minutes, I discovered a pattern. If you start with a triangle with vertices labelled (n-1,0,0), (0,n-1,0) and (0,0,n-1), you can subdivide each side of the triangle with n-2 points(counting the vertices, each side now has n points). Moving along a side is like taking balls from one bin and putting them into the second one. Let's say n=4, one of the sides would look like (3,0,0), (2,0,1), (1,0,2), (0,0,3). If you now use this "scaffolding" and draw a Pascal's triangle, it's easy to see that every two points which are connected "differ by 1 ball". The total amount of points is just 1+2+3...+n+(n+1). Which is the same as obtained in the video. Bonus joke: if you add an infinite amount of energy quanta, degeneracy is -1/12
Awesome video!!! can we maybe get a video about the continuum limit, and how quantum mechanical equations turn into classical equations when taking the continuum limit? i would really love to see that! anyhow amazing video!!!
Glad you like the video! And thanks for the suggestion, we'll add it to our list. In the meantime, we touch on the connection between classical and quantum mechanics in our video on the Ehrenfest theorem: th-cam.com/video/55RQCCtdlko/w-d-xo.html
These equations are beautiful, and as usual the graphs are very helpful to me when trying to visualize the math. No doubt the topic is covered elsewhere, but one thought that keeps distracting me is "why assume three?". Obviously we observe xyz around us, but would the observations be simplified if we chose to analyze the math with time included as a 4th dimension? Would the math become even simpler and/or more consistent if we assumed infinite dimensions, but with diminishing effects so that only the first few are easily observed? I'm only a casual amateur at this, so I assume that the professionals have delved into those questions in depth. I'm only passing on what my thoughts were as I struggled to see the patterns. It's fun, the symmetries make it easier to follow, and I'm encouraged to keep going by the fact that the techniques have proven themselves to be useful when analyzing experimental data. But why three? Why no more, why no less? Anyway, thanks for the video. I'm mostly an experimentalist (making medical instruments), but it's good to dabble in the theoretical basics now and then. Maybe it will help me make better instruments. Thanks again.
You are absolutely correct that nothing limits us to mathematically consider more than three dimensions (or indeed fewer). Three dimensions is what is relevant for our 3D world, but the 2D case would also be relevant (it is possible to synthesize quasi-2D compounds), and a good exercise would be to check how the story changes in that case. And then one can also explore, out of interest, what happens at higher dimensions. You should find that the degree of degeneracy keeps increasing with dimension :)
Nice! I am looking forward to your video treating the isotropic 3-D quantum harmonic oscillator potential as a central potential. Will the eigenstates for a given energy have a different form then? Maybe because degeneracy allows us to make linear combinations of eigenstates? But then what about the ground state, with no degeneracy? Will that look the same? Or will we just arrive at the same eigenstates from a different approach using central potential ideas? So many questions -- please don't answer them here; I can wait for the video! (Although I will probably do some Googling . . .)
This is one of the few sources I've found to which despite having to pause and study it a bit, I can make some logical sense from! 👍 I'm a little perplexed though at a certain point that tells me something is missing to which I'm not certain what exactly. I'm looking at your Gaussian parts to which I can see how you could perceive it as a normal distribution. I don't see from the graphs, why are you throwing in Gaussian derivatives for a given state? Are you coming from the perspective that says, this particular direction for x,y, or z is a Gaussian. So to express that as a vector, mth derivative of your Gaussian is your Eigen value/scalar to correspond to the specific direction? Then that coincides with some other things like LaPlace or related? Im trying to visualize things in 3D and make sense of it and upon viewing this, it gives me another logical explanation on EFE and why its normalized as it is. 👍 it helps-trust me. 🙂👍
Hi! Great vid as always, how do you compute the degeneracy for a 2D harmonic oscillator? I have tried to express with a sum but the outcome is always the same as the 3D. " ny= n-nx" but then sum (n-nx+1). Thx!
Wonderful video. I'm just confused that you said, you would need 4 dimensions for drawing the "3D S'later". The wave-eqn is an imaginary funktion, thus I thought about 3 times 2 dimensions. Ok, hard to draw as well ;-) - but I'm just asking for my understanding. Many greetings to Cambridge and thanks a lot!
In the formulation of the quantum harmonic oscillator we discuss, which is by far the most commonly used, the ground state corresponds to n=0. You can find a justification of this in our video on eigenvalues: th-cam.com/video/GkUXscdLQQ0/w-d-xo.html In other standard problems, such as the infinite square well potential, the corresponding formulation does lead to the ground state being labelled by n=1. I hope this helps!
This is an interesting point! We can solve the isotropic harmonic oscillator using an alternative strategy where we view it as a central potential instead of the approach we take in this video, and we'll actually do this in our next few videos. This provides a direct analogy with the hydrogen atom, which can also be described using a central potential (and we'll also publish the hydrogen videos over the next few months).
Elektron için n kuantum sayısının karesi dejenere durumların sayısına eşittir.örneğin n=3 için 3s 3p 3d orbitalleri:1+3+5= 3exp2=9 fakat harmonik osilatörde de dejenere durumların sayısı örneğin N = 3 için 1f 2p = 7+3= 10 . 10 üçün karesi değildir.her N satırı için dejenere durumların sayısı bir üçgen sayıdır.1,3,6,,10,15.....soru: öyleyse hangi mantıkla dalga fonksiyonunun karesi dejenere durumlarının sayısını verebilir.açıkçası bu durumda dalga fonksiyonunun karesini almak yaklaşımı paradoksal değilmi?eğer mantıklı bir cevabı varsa bana elektron için yaptığımız gibi ( mesela üçün karesini almak gibi) harmonik osilatör potansiyelinde hangi kuantum sayısının karesini aldığımızı gösterin
We discuss the approach using spherical coordinates in these videos: th-cam.com/video/jOQThICjLlw/w-d-xo.html th-cam.com/video/3Ipcr9tMlxU/w-d-xo.html I hope this helps!
@Professor M does Science thanks, I did some simulations on this, also when deformed found the duffing type scenario showed up, and I think it's relevant on many areas (ads/cft) nondeterministic behaviour of deterministic systems, topological degeneracy, topological orders, etc
@@ProfessorMdoesScience check this out Quantum behavior of a superconducting Duffing oscillator at the dissipative phase transition Qi-Ming Chen, Michael Fischer, Yuki Nojiri, Michael Renger, Edwar Xie, Matti Partanen, Stefan Pogorzalek, Kirill G. Fedorov, Achim Marx, Frank Deppe, Rudolf Gross Understanding the non-deterministic behavior of deterministic nonlinear systems has been an implicit dream since Lorenz named it the "butterfly effect". A prominent example is the hysteresis and bistability of the Duffing oscillator, which in the classical description is attributed to the coexistence of two steady states in a double-well potential. However, this interpretation fails in the quantum-mechanical perspective, where a single unique steady state is allowed in the whole parameter space. Here, we measure the non-equilibrium dynamics of a superconducting Duffing oscillator and reconcile the classical and quantum descriptions in a unified picture of quantum metastability. We demonstrate that the two classically regarded steady states are in fact metastable states. They have a remarkably long lifetime in the classical hysteresis regime but must eventually relax into a single unique steady state allowed by quantum mechanics. By engineering the lifetime of the metastable states sufficiently large, we observe a first-order dissipative phase transition, which mimics a sudden change of the mean field in a 11-site Bose-Hubbard lattice. We also reveal the two distinct phases of the transition by quantum state tomography, namely a coherent-state phase and a squeezed-state phase separated by a critical point. Our results reveal a smooth quantum state evolution behind a sudden dissipative phase transition, and they form an essential step towards understanding hysteresis and instability in non-equilibrium systems. I also made a video pointing some interesting correlations between black holes, ads/cft and duffing harmonic oscillator
Hi, thank you so much for the awesome work! I have a doubt: at 21:20 I just cannot understand how it's possible for the sum of 1 over n_x, from 0 to n, to be equal to (n+1): what are we summing up?
Good question! This is mostly down to notation. Imagine we had this different sum: sum_{nx=0}^n nx In each term we replace the argument "nx" by the corresponding value. Spelled out this means: 0 + 1 + 2 + 3 + 4 + ... + n Let's imagine we had another different sum: sum_{nx=0}^n nx^2 Using the same procedure, we replace every "nx" by the corresponding value, but now the sum is over the squares, so we get: 0 + 1^2 + 2^2 + 3^2 + 4^2 + ... + n^2 In our case, we have: sum_{nx=0}^n 1 In principle we should also replace every "nx" by the corresponding value, but the argument this time does not contain nx, so we simply get the same term every time, which is the argument (in this case 1): 1 + 1 + 1 + 1 + ... + 1 where there are as many "1s" as terms in the sum. As there are n+1 terms in the sum (we start at 0 and end at n), then we end up with (n+1). I hope this helps!
@@ProfessorMdoesScience Thank you so much! The overall it's very clear now. I just want to say that all of this is very helpful, as i'm preparing for the QM exam on monday! Thank you once again
thanks, we need more high level educational content like that
Glad you like it! :)
I decided to pause and derive the degeneracy of an nth level. I tried to find a connection to Pascal's triangle since each time you add a quantum of energy, you can think of it as a ball falling into 3 bins(x,y,z) with equal probability. And the question is basically how many ways are there to arrange n-1 balls in 3 bins. After a few minutes, I discovered a pattern. If you start with a triangle with vertices labelled (n-1,0,0), (0,n-1,0) and (0,0,n-1), you can subdivide each side of the triangle with n-2 points(counting the vertices, each side now has n points). Moving along a side is like taking balls from one bin and putting them into the second one. Let's say n=4, one of the sides would look like (3,0,0), (2,0,1), (1,0,2), (0,0,3). If you now use this "scaffolding" and draw a Pascal's triangle, it's easy to see that every two points which are connected "differ by 1 ball". The total amount of points is just 1+2+3...+n+(n+1). Which is the same as obtained in the video.
Bonus joke: if you add an infinite amount of energy quanta, degeneracy is -1/12
Thanks for sharing :)
U guys r doing awesome work. Very neat, to the point and conceptually clear videos. U deserve way more views and subscribers.
Thanks for your kind words, you can help telling your friends! :)
To be in an excited state in two directions at the same time - indeed.
:)
Many thanks, Professor. Your lectures are very illuminating and fascinating! Good luck and have a good time.
Glad that you like the videos!
Awesome video!!! can we maybe get a video about the continuum limit, and how quantum mechanical equations turn into classical equations when taking the continuum limit? i would really love to see that! anyhow amazing video!!!
Glad you like the video! And thanks for the suggestion, we'll add it to our list. In the meantime, we touch on the connection between classical and quantum mechanics in our video on the Ehrenfest theorem: th-cam.com/video/55RQCCtdlko/w-d-xo.html
Thankyou, for this video.
It helps me a lot.
Glad to hear that!
Very explainatory, thank you so much
Glad you like it!
Muchísimas gracias, me has ayudado con mi tarea.
Un saludo desde México
Un saludo!
love the content provided by this channel🙂🙂
Glad you like it! :)
Stumbled across a masterpiece
Want to know about some of the applications.
Glad you like it!
These equations are beautiful, and as usual the graphs are very helpful to me when trying to visualize the math. No doubt the topic is covered elsewhere, but one thought that keeps distracting me is "why assume three?".
Obviously we observe xyz around us, but would the observations be simplified if we chose to analyze the math with time included as a 4th dimension? Would the math become even simpler and/or more consistent if we assumed infinite dimensions, but with diminishing effects so that only the first few are easily observed?
I'm only a casual amateur at this, so I assume that the professionals have delved into those questions in depth. I'm only passing on what my thoughts were as I struggled to see the patterns.
It's fun, the symmetries make it easier to follow, and I'm encouraged to keep going by the fact that the techniques have proven themselves to be useful when analyzing experimental data. But why three? Why no more, why no less?
Anyway, thanks for the video. I'm mostly an experimentalist (making medical instruments), but it's good to dabble in the theoretical basics now and then. Maybe it will help me make better instruments. Thanks again.
You are absolutely correct that nothing limits us to mathematically consider more than three dimensions (or indeed fewer). Three dimensions is what is relevant for our 3D world, but the 2D case would also be relevant (it is possible to synthesize quasi-2D compounds), and a good exercise would be to check how the story changes in that case. And then one can also explore, out of interest, what happens at higher dimensions. You should find that the degree of degeneracy keeps increasing with dimension :)
Thank you, it helped me a lot!
Glad it helped!
Can you cover weak measurements that don't use eigenvalues? thanks
Thanks for the suggestion, we'll add it to our (growing) list of topics!
Nice! I am looking forward to your video treating the isotropic 3-D quantum harmonic oscillator potential as a central potential. Will the eigenstates for a given energy have a different form then? Maybe because degeneracy allows us to make linear combinations of eigenstates? But then what about the ground state, with no degeneracy? Will that look the same? Or will we just arrive at the same eigenstates from a different approach using central potential ideas? So many questions -- please don't answer them here; I can wait for the video! (Although I will probably do some Googling . . .)
This is one of the few sources I've found to which despite having to pause and study it a bit, I can make some logical sense from! 👍
I'm a little perplexed though at a certain point that tells me something is missing to which I'm not certain what exactly.
I'm looking at your Gaussian parts to which I can see how you could perceive it as a normal distribution. I don't see from the graphs, why are you throwing in Gaussian derivatives for a given state?
Are you coming from the perspective that says, this particular direction for x,y, or z is a Gaussian. So to express that as a vector, mth derivative of your Gaussian is your Eigen value/scalar to correspond to the specific direction? Then that coincides with some other things like LaPlace or related?
Im trying to visualize things in 3D and make sense of it and upon viewing this, it gives me another logical explanation on EFE and why its normalized as it is. 👍 it helps-trust me. 🙂👍
Hi! Great vid as always, how do you compute the degeneracy for a 2D harmonic oscillator? I have tried to express with a sum but the outcome is always the same as the 3D. " ny= n-nx" but then sum (n-nx+1). Thx!
The procedure should be analogous to the 3D case. As a hint, for 2D you should get that g_n = n+1. I hope this helps!
@@ProfessorMdoesScience Hi, I know it's n+1 but like how do you go from ny=n-nx to (n+1) ?
Wonderful video. I'm just confused that you said, you would need 4 dimensions for drawing the "3D S'later". The wave-eqn is an imaginary funktion, thus I thought about 3 times 2 dimensions. Ok, hard to draw as well ;-) - but I'm just asking for my understanding. Many greetings to Cambridge and thanks a lot!
You are absolutely correct, the wave function is, in general, complex, so we would need even more dimensions to fully plot it :)
@@ProfessorMdoesScience Ah, great! Thanks for your fast response. I highly appreciate your excellent videos. Please keep on going.
We don't take ground state as n=0. It is n=1
In the formulation of the quantum harmonic oscillator we discuss, which is by far the most commonly used, the ground state corresponds to n=0. You can find a justification of this in our video on eigenvalues:
th-cam.com/video/GkUXscdLQQ0/w-d-xo.html
In other standard problems, such as the infinite square well potential, the corresponding formulation does lead to the ground state being labelled by n=1. I hope this helps!
Good stuff. New subscriber.
Glad you like it, and thanks for subscribing!
Great video ! The degeneracy scales n^2 like the hydrogen atom. It is maybe related to the rotationnal invariance... Interesting...
This is an interesting point! We can solve the isotropic harmonic oscillator using an alternative strategy where we view it as a central potential instead of the approach we take in this video, and we'll actually do this in our next few videos. This provides a direct analogy with the hydrogen atom, which can also be described using a central potential (and we'll also publish the hydrogen videos over the next few months).
The symmetry for the 3d isotropic oscillator is SU(3). It has a different degeneracy pattern than hydrogen, whose symmetry is SO(4).
Elektron için n kuantum sayısının karesi dejenere durumların sayısına eşittir.örneğin n=3 için 3s 3p 3d orbitalleri:1+3+5= 3exp2=9 fakat harmonik osilatörde de dejenere durumların sayısı örneğin N = 3 için 1f 2p = 7+3= 10 . 10 üçün karesi değildir.her N satırı için dejenere durumların sayısı bir üçgen sayıdır.1,3,6,,10,15.....soru: öyleyse hangi mantıkla dalga fonksiyonunun karesi dejenere durumlarının sayısını verebilir.açıkçası bu durumda dalga fonksiyonunun karesini almak yaklaşımı paradoksal değilmi?eğer mantıklı bir cevabı varsa bana elektron için yaptığımız gibi ( mesela üçün karesini almak gibi) harmonik osilatör potansiyelinde hangi kuantum sayısının karesini aldığımızı gösterin
How do you do this in spherical coordinates?
We discuss the approach using spherical coordinates in these videos:
th-cam.com/video/jOQThICjLlw/w-d-xo.html
th-cam.com/video/3Ipcr9tMlxU/w-d-xo.html
I hope this helps!
Can you do the same for supersymmetric quantum harmonic oscillator? Would he interested. Thanks
Must confess we are not familiar with this, but can look into it!
@Professor M does Science thanks, I did some simulations on this, also when deformed found the duffing type scenario showed up, and I think it's relevant on many areas (ads/cft) nondeterministic behaviour of deterministic systems, topological degeneracy, topological orders, etc
@@ProfessorMdoesScience you get multiple ground states of the system (double well) vacuum states
@@JAYMOAP Thanks for the info, it looks interesting!
@@ProfessorMdoesScience check this out
Quantum behavior of a superconducting Duffing oscillator at the dissipative phase transition
Qi-Ming Chen, Michael Fischer, Yuki Nojiri, Michael Renger, Edwar Xie, Matti Partanen, Stefan Pogorzalek, Kirill G. Fedorov, Achim Marx, Frank Deppe, Rudolf Gross
Understanding the non-deterministic behavior of deterministic nonlinear systems has been an implicit dream since Lorenz named it the "butterfly effect". A prominent example is the hysteresis and bistability of the Duffing oscillator, which in the classical description is attributed to the coexistence of two steady states in a double-well potential. However, this interpretation fails in the quantum-mechanical perspective, where a single unique steady state is allowed in the whole parameter space. Here, we measure the non-equilibrium dynamics of a superconducting Duffing oscillator and reconcile the classical and quantum descriptions in a unified picture of quantum metastability. We demonstrate that the two classically regarded steady states are in fact metastable states. They have a remarkably long lifetime in the classical hysteresis regime but must eventually relax into a single unique steady state allowed by quantum mechanics. By engineering the lifetime of the metastable states sufficiently large, we observe a first-order dissipative phase transition, which mimics a sudden change of the mean field in a 11-site Bose-Hubbard lattice. We also reveal the two distinct phases of the transition by quantum state tomography, namely a coherent-state phase and a squeezed-state phase separated by a critical point. Our results reveal a smooth quantum state evolution behind a sudden dissipative phase transition, and they form an essential step towards understanding hysteresis and instability in non-equilibrium systems.
I also made a video pointing some interesting correlations between black holes, ads/cft and duffing harmonic oscillator
Hi, thank you so much for the awesome work!
I have a doubt: at 21:20 I just cannot understand how it's possible for the sum of 1 over n_x, from 0 to n, to be equal to (n+1): what are we summing up?
Good question! This is mostly down to notation. Imagine we had this different sum:
sum_{nx=0}^n nx
In each term we replace the argument "nx" by the corresponding value. Spelled out this means:
0 + 1 + 2 + 3 + 4 + ... + n
Let's imagine we had another different sum:
sum_{nx=0}^n nx^2
Using the same procedure, we replace every "nx" by the corresponding value, but now the sum is over the squares, so we get:
0 + 1^2 + 2^2 + 3^2 + 4^2 + ... + n^2
In our case, we have:
sum_{nx=0}^n 1
In principle we should also replace every "nx" by the corresponding value, but the argument this time does not contain nx, so we simply get the same term every time, which is the argument (in this case 1):
1 + 1 + 1 + 1 + ... + 1
where there are as many "1s" as terms in the sum. As there are n+1 terms in the sum (we start at 0 and end at n), then we end up with (n+1). I hope this helps!
@@ProfessorMdoesScience Thank you so much! The overall it's very clear now.
I just want to say that all of this is very helpful, as i'm preparing for the QM exam on monday! Thank you once again
@@kevindelcuoco Glad to be helpful! May I ask what university you are studying at? And good luck with the exam!
@@ProfessorMdoesScience Of course! actually i'm studying physics at the University of studies of Naples "Federico II"